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Corrected asymptotics for a multi-server queue
                 in the Halfin-Whitt regime
                 A.J.E.M. Janssen∗           J.S.H. van Leeuwaarden⋄             B. Zwart♭

                                    First version: May 12, 2007
                                  Second version: October 8, 2007


                                                 Abstract
            To obtain insight in the quality of heavy-traffic approximations for queues with many
        servers, we consider the steady-state number of waiting customers in an M/D/s queue
        as s → ∞. In the Halfin-Whitt regime, it is well known that this random variable con-
        verges to the supremum of a Gaussian random walk. This paper develops methods that
        yield more accurate results in terms of series expansions and inequalities for the proba-
        bility of an empty queue, and the mean and variance of the queue length distribution.
        This quantifies the relationship between the limiting system and the queue with a small
        or moderate number of servers. The main idea is to view the M/D/s queue through
        the prism of the Gaussian random walk: as for the standard Gaussian random walk, we
        provide scalable series expansions involving terms that include the Riemann zeta function.
        Keywords: M/D/s queue; Halfin-Whitt scaling; Gaussian random walk; all-time maxi-
        mum; Riemann zeta function; Lerch’s transcendent; Spitzer’s identity; queues in heavy
        traffic; Lambert W Function; corrected diffusion approximation.
        AMS 2000 Subject Classification: 11M06, 30B40, 60G50, 60G51, 65B15.


1       Introduction
Heavy-traffic analysis is a popular tool to analyze stochastic networks, since the analysis of
a complicated network often reduces to the analysis of a much simpler (reflected) diffusion,
which may be of lower dimension than the original system. This makes the analysis of complex
systems tractable, and from a mathematical point of view, these results are appealing since
they can be made rigorous.
  A downside of heavy-traffic analysis is that the results are of an asymptotic nature, and
only form an approximation for a finite-sized system. In a pioneering paper, Siegmund [31]
proposed a corrected diffusion approximation for the waiting time in a single-server queue

    ∗
     Philips Research. Digital Signal Processing Group, WO-02, 5656 AA Eindhoven, The Netherlands. Email:
a.j.e.m.janssen@philips.com.
   ⋄
     Eindhoven University of Technology. Math and Computer Science department, 5612 AZ Eindhoven, The
Netherlands. Email: j.s.h.v.leeuwaarden@tue.nl (corresponding author).
   ♭
     Georgia Institute of Technology. H. Milton Stewart School of Industrial and Systems Engineering, 765 Ferst
Drive, 30332 Atlanta, USA. Email: bertzwart@gatech.edu.
   We thank M. Vlasiou for pointing out a connection to the Lambert W function. JvL was supported by the
Netherlands Organisation for Scientific Research (NWO).



                                                      1
(actually, Siegmund formulated his result in terms of a random walk). In heavy traffic,
the workload distribution is approximated by an exponential distribution. Siegmund gave
a precise estimate of the correction term, nowadays a classic result and textbook material,
cf. Asmussen [2], p. 369. Siegmund’s first order correction has been extended recently by
Blanchet & Glynn [5], who give a full series expansion for the tail probability of the GI/GI/1
waiting time distribution in heavy traffic.
   The results in [5, 31] correspond to the conventional heavy-traffic scaling. The present
paper considers corrected diffusion approximations for a heavy-traffic scaling known as the
Halfin-Whitt [12] regime. This regime considers queues where the number of servers grows
large as the system becomes critically loaded. The number of servers s is chosen according
               √
to s = λ + β λ, where β is some positive constant. As the√        scaling parameter λ tends to
infinity, the traffic intensity tends to one according to 1 − O(1/ λ). The Halfin-Whitt regime
is also known as the QED (Quality and Efficiency Driven) regime, due to the fact that, in
the limit, a system can be highly utilized (efficiency) while the√   waiting times stay relatively
small (quality). Also, setting the number of servers as s = λ + β λ is often called square-root
staffing. This terminology is motivated by the emergence of large customer contact centers
which need to be staffed with agents, thus calling for accurate and scalable approximations
of multi-server queues. We refer to Gans et al. [11] and Mandelbaum [26] for overviews.
   The Halfin-Whitt regime was formally introduced in [12] for a system with exponential
service times (G/M/s queue), although in [19] the same type of scaling was already applied
to the Erlang loss model (M/M/s/s queue). The extension of the results on the G/M/s
queue to non-exponential service times turned out to be challenging. The past few years
have witnessed a substantial effort to rise to this challenge, resulting in several papers on the
Halfin-Whitt regime for various types of service time distributions, cf. Puhalskii & Reiman
[29], Jelenkovi´ et al. [20], Whitt [40], Mandelbaum & Momˇilovic [25], and Reed [30].
               c                                               c
   Although these results offer important qualitative insights and are useful to solve concrete
staffing problems, one would like to have a better understanding of the quality of the asymp-
totic approximations. For instance, how fast does convergence to the heavy-traffic limit take
place? It would be helpful to have asymptotic estimates or even inequalities from which we
could judge just how close the scaled queueing model is to its heavy-traffic limit. Borst et
al. [4] consider optimal staffing of an M/M/s queue in the Halfin-Whitt regime, and show
numerically that optimizing the system based on the Halfin-Whitt approximation (with s
infinite instead of finite) of the cost function is rarely off by more than a single agent from
systems with as few as 10 servers. As mentioned in the conclusions of [4], these observations
call for a theoretical foundation−a task we take up in the present paper.

1.1   Goals, results and insights
We now give a general description of the results obtained in this paper. We consider a multi-
server queue with arrival rate λ, s servers and deterministic service times (set to 1). We let
                                                           √
the arrival rate of the system grow large and set s = λ + β λ for some constant β > 0. Our
main performance measure is the probability that the queue is empty. The model at hand
has been considered before by Jelenkovi´ et al. [20] who showed that the scaled number of
                                         c
                     ˆ
waiting customers Qλ converges to the maximum Mβ of a Gaussian random walk with drift
                                                                             √
−β, for which the emptiness probability is known. As λ → ∞, for β < 2 π, there is the




                                               2
result                                                     ∞
                                         √           β           ζ(1/2 − r) −β 2   r
              ˆ
            P(Qλ = 0) → P(Mβ = 0) =          2β exp √                                  ,    (1)
                                                      2π         r!(2r + 1)  2
                                                           r=0

with ζ the Riemann zeta function, see Chang & Peres [7], and Janssen & Van Leeuwaarden [15,
16]. The limiting result for P(Mβ = 0) has the appealing property that the time to compute
it does not depend on the number of servers, which is the case for standard computational
procedures for the M/D/s queue, see e.g. Tijms [37] and Franx [10] and references therein.
   The main aim of this paper is to obtain series expansions refining this asymptotic result.
These series expansions can be used in two ways. First of all, the series expansions quantify
the relationship between the limiting system and the queue with a small or moderate number
of servers. In addition, the first term (or first few terms) of these expansions have the correct
behavior as the number of servers grows large.
   One insight we find particularly interesting is that our approximations are not based on the
parameter β, but on a modification of it, which depends on s and is given by

                                α(s) = (−2s(1 − ρ + ln ρ))1/2 ,                             (2)

with ρ = λ/s. This function converges to β as s → ∞, cf. Lemma 7. Another insight we obtain
                                                                                       ˆ
is that the resulting approximation P(Mα(s) = 0) is, in fact, a lower bound for P(Qλ = 0).
We also obtain an upper bound, again involving the function α(s).
   The model we consider may seem rather specific, but one should realize that obtaining
series expansions and bounds of this type is by no means a trivial task. The state of the
art for traditional corrected diffusion approximations does not go beyond the random walk,
and relies on the availability of the Wiener-Hopf factorization. In the Halfin-Whitt regime,
the limiting no-wait probability has been found in two cases only, namely for exponential
service times and for deterministic service times. We believe that the latter case is the most
challenging one.
   We apply the methods developed in this paper to the M/M/s queue in [17], in which case the
Halfin-Whitt regime results in a non-degenerate limit for the Erlang C formula (probability
that a customer has to wait). There we obtain the same important insight: the Halfin-Whitt
approximation can be substantially improved when β is replaced with α(s); this function is
the same for both models.
   We finally like to point out that the results in this paper are all formulated for the special
case of Poisson arrivals, but the methodology we develop is applicable to more general models
(see Section 6). An additional motivation for considering deterministic service times is that
the number of waiting customers in the queue is related to a discrete bulk-service queue, which
has proven its relevance in a variety of applications (see [24], Chapter 5, for an overview).

1.2      Methodology
We now turn to a discussion and motivation of the techniques we use and develop in this
                      ˆ
paper. The ratio of P(Qλ = 0) and P(Mβ = 0) serves as a primary measure of convergence
and should tend to one as λ grows large. This ratio can be expressed as (using Spitzer’s
identity, cf. (17))

                      ˆ                  ∞                 √         √
                    P(Qλ = 0)                  1    ˆ
                              = exp              (P(Aλl ≤ β l) − P (β l)) ,                 (3)
                    P(Mβ = 0)                  l
                                         l=1


                                                 3
√
      ˆ
where Aλl = (Alλ − lλ)/ lλ and Alλ a Poisson random variable with mean lλ, and
                                                                  x
                                              1                             2 /2
                                     P (x) = √                        e−u          du                   (4)
                                              2π              −∞

the normal distribution function. To estimate (3) one can use Berry-Esseen bounds, but these
do not lead to sharp results (cf. Lemma 1). In order to get more precise estimates one can use
classical approximations for sums of i.i.d. random variables like saddlepoint approximations
                                                                                        √
                                                                                 ˆ
or Edgeworth expansions (see [3, 21]). However, these require each quantity P(Aλl ≤ β l) −
     √
P (β l) to be approximated separately and uniformly in l.
                                                                                         ˆ
  √ get convenient asymptotic expansions, we follow a different approach: we bring P(Aλl ≤
   To
β l) into quasi-Gaussian form, a method that is standard in asymptotic analysis (for an
illuminating discussion see De Bruijn [6], pp. 67-71). The resulting asymptotic expansion for
e.g. the probability of an empty queue then contains terms of the type
                             ∞                  ∞     1       2
                  Gk (a) =         lk+1/2           e− 2 lsx z(x)dx,               a, s ∈ R+ , k ∈ Z,   (5)
                             l=1            a


where z(x) is some function that does not depend on l. This approach seems technical at
first sight but we believe it to be elegant and even intuitively appealing, as there is a clear
interpretation in terms of a change-of-measure argument, see the end of Section 2.
  A large part of this paper deals with obtaining the quasi-Gaussian form, analyzing z(x),
and reformulating and estimating Gk (a) which is done in Section 4. A key result is Theorem
3, which gives a representation of Gk (a) for a large class of functions z(x); the only condition
that is imposed on z(x) is that z : [0, ∞) → C is a continuous function satisfying z(x) =
O(exp(εx2 )) for any ε > 0 and that z(x) has a Taylor series around zero. To illustrate the
generality of our result, we note that Chang & Peres’ result (1) on the Gaussian random walk
can be viewed as a special case by taking z(x) ≡ 1.
  We focus on the case in which Aλ has a Poisson distribution, which ensures a particularly
tractable form of z(x) yielding convenient computational schemes. This form is given in
Subsection 2.2 and studied in detail in Appendix A. The derivative of z(x) is related to
the Lambert W function; our treatment is self-contained, produces some important auxiliary
results, and is based on earlier results obtained by Szeg¨ [33]. We include our analysis in a
                                                            o
separate appendix, since we believe it is interesting in its own right.
  Theorem 3 yields a series expansion which can be truncated at a convenient point to obtain
high precision estimates of performance measures. Using classical methods, we can even
estimate the optimal truncation point of the series expansion. We illustrate these general
ideas by specializing them to the M/D/s queue in Subsection 4.3.

1.3   Organization
This paper is organized as follows. In Section 2 we introduce our model and provide short
proofs of results which can also be found in [20]. In particular we establish convergence of
the number of waiting customers to the maximum of the Gaussian random walk, and give a
rough Berry-Esseen bound. These results form a point of departure for the rest of the paper.
We also explain in Section 2 how our asymptotic analysis will be carried out. In Section 3,
for the emptiness probability, and the mean and variance of the queue length distribution,
we rewrite the Spitzer-type expressions into quasi-Gaussian expressions. The reformulation

                                                          4
and estimation of Gk (a) is carried out in Section 4. Section 5 focuses on lower and upper
bounds which have the correct asymptotic behavior in the Halfin-Whitt regime. We use the
quasi-Gaussian expression for the emptiness probability obtained in Section 3 to derive these
bounds. Conclusions and possible extensions are presented in Section 6.


2     The M/D/s queue and the Halfin-Whitt regime
We consider the M/D/s queue and keep track of the number of customers waiting in the
queue (without those in service) at the end of intervals equal to the constant service time
(which we set to one). Customers arrive according to a Poisson process with rate λ and are
served by at most s servers. Let Qλ,n denote the number of customers waiting in the queue
at the end of interval n. The queueing process is then described by

                          Qλ,n+1 = (Qλ,n + Aλ,n − s)+ ,            n = 0, 1, . . . .      (6)

where x+ = max{0, x}, and Aλ,n denotes the number of customers that arrived at the queue
during interval n. Obviously, the Aλ,n are i.i.d. for all n, and copies of a Poisson random
variable Aλ with mean λ. It should be noted that due to the assumption of constant service
times, the customers which are being serviced at the end of the considered interval should
start within this interval, and for the same reason, the customers whose service is completed
during this interval should start before its beginning.
  Assume that EAλ,n = λ < s and let Qλ denote the random variable that follows the
stationary queue length distribution, i.e., Qλ is the weak limit of Qλ,n . Let
                                               √
                                     s = λ + β λ, β > 0.                                  (7)

Let {Sn : n ≥ 0} be a random walk with S0 = 0, Sn = X1 + . . . + Xn and X, X1 , X2 , . . .
i.i.d. random variables with EX < 0, and let M := max{Sn : n ≥ 0} denote the all-time
maximum. When X is normally distributed with mean −β < 0 and variance 1 we speak
of the Gaussian random walk and denote its all-time maximum by Mβ . We often use the
following notation which is standard in asymptotic analysis:
                                                    ∞
                                         f (x) ∼         fn (x),
                                                   n=0

by which we denote that, for every fixed integer k ≥ 1,
                                        k−1
                              f (x) −         fn (x) = fk (x)(1 + o(1)).
                                        n=0

      d
Let → denote convergence in distribution.

2.1       Basic results
The following theorem can be proved using a similar approach as in Jelenkovi´ et al. [20].
                                                                            c
                              √
                     ˆ
Theorem 1. Define Qλ = Qλ / λ. As λ → ∞,


                                                    5
ˆ d
  (i) Qλ → Mβ ;
        ˆ
 (ii) P(Qλ = 0) → P(Mβ = 0);
        ˆ
(iii) E[Qk ] → E[Mβ ] for any k > 0.
                  k
          λ

Proof. Proof of (i): Note that
                                    ˆ d ˆ      ˆ
                                    Qλ = (Qλ + Aλ − β)+ ,                                  (8)
                      √
      ˆ                           ˆ
with Aλ = (Aλ − λ)/ λ. Since Aλ converges in distribution to the standard normal random
                                                                                      ˆ
variable as λ → ∞, (i) follows from Theorem X.6.1 in Asmussen [2], if the family (Aλ , λ ≥ 0)
                                                                      ˆ
is uniformly integrable. But this follows simply from the fact that E[Aλ2 ] = 1 for all λ.

Proof of (ii): The result lim supλ→∞ P(Q  ˆ λ = 0) ≤ P(Mβ = 0) follows from (i). To show the
lim inf, note that from Spitzer’s identity (see (14))
                                                      ∞
                                  ˆ                       1 ˆ
                             ln P(Qλ = 0) = −               P(Alλ > lβ).                    (9)
                                                          l
                                                    l=1
                                                                              √
                                                             ˆ
Taking the lim inf, applying Fatou’s lemma, and using that P(Alλ > lβ) → P (−β l) yields
                                              ∞            √
                                ˆ                   1
                   lim inf ln P(Qλ = 0) ≥ −           P (−β l) = ln P(Mβ = 0),             (10)
                    λ→∞                             l
                                              l=1

which proves (ii). Statement (iii) follows from (i) if we can prove the additional uniform
                                      ˆ
integrability condition supλ>N E[Qk ] < ∞ for some constant N and any k. To prove this,
                                        λ
                                                                      ˆ
note that the Cram´r-Lundberg-Kingman inequality states that P(Qλ > x) ≤ e−sx , for any
                     e
                       ˆλ −β)
s > 0 such that E[es(A        ] ≤ 1. After some straightforward computation, this inequality can
be rewritten into
                                          √         s
                                   λ es/ λ − 1 − √      − sβ ≤ 0.                           (11)
                                                     λ
Since ex − 1 − x ≤ 1 x2 ex , we see that any s is admissible that satisfies
                   2

                                       s2 s/√λ
                                         e     − sβ ≤ 0.                                   (12)
                                       2
It is easy to see that s = β satisfies this inequality if λ ≥ N := (β/ ln 2)2 . We conclude that
                                         ˆ
                                       P(Qλ > x) ≤ e−βx                                    (13)
                                                                             ˆ
for any x ≥ 0 and any λ > N . The uniform integrability condition supλ>N E[Qk ] < ∞ now
                                                                               λ
                                                 ˆ       ∞        ˆ
follows directly using for example the formula E[Qk ] = 0 kxk−1 P(Qλ > x)dx.
                                                   λ

                                                                           ˆ
  As a consequence of Theorem 1 we know that P(Qλ = 0) (which equals P(Qλ = 0)) tends
to P(Mβ = 0) as λ tends to infinity. We are interested in how fast the M/D/s queue in
the Halfin-Whitt regime approaches the Gaussian random walk, and so we take the ratio of
P(Qλ = 0) and P(Mβ = 0) as our measure of convergence. From Spitzer’s identity for random
walks (see Theorem 3.1 in [32]) we have
                                                      ∞
                                                            1
                               − ln P(M = 0) =                P(Sl > 0),                   (14)
                                                            l
                                                      l=1


                                                  6
which gives for the M/D/s queue
                                     ∞                         ∞          ∞
                                          1                         1                   (lλ)j
                 − ln P(Qλ = 0) =           P(Alλ > ls) =                        e−lλ         ,                  (15)
                                          l                         l                     j!
                                    l=1                       l=1       j=ls+1
                                      √
where we choose λ such that s = λ+β λ is integer-valued, i.e. λ = 1 (2s+β 2 −(4sβ 2 +β 4 )1/2 )
                                                                  2
with s = 1, 2, . . .. For the Gaussian random walk we have ln P(Mβ = 0) as in (10). The
following can be proved using a Berry-Esseen bound.
Lemma 1. For ω := 4 ζ( 3 ) ≈ 2.0899 there are the bounds
                  5    2

                              −ω           P(Qλ = 0)       ω
                          exp √       ≤              ≤ exp √ .                                                   (16)
                                λ          P(Mβ = 0)        λ
Proof. Along the same lines as Theorem 2 in [20]. From (15) and (10) we get
                                             ∞             √
                    P(Qλ = 0)                      1
                                 = exp               (P (−β l) − P(Alλ > ls))
                    P(Mβ = 0)                      l
                                             l=1
                                              ∞            √
                                                   1
                                 ≤ exp               |P (−β l) − P(Alλ > ls)| .                                  (17)
                                                   l
                                             l=1

Rewriting                  √                                √          √
                                                     ˆ
                     |P (−β l) − P(Alλ > ls)| = |P(Aλl ≤ β l) − P (β l)|                   (18)
                        √
      ˆ
with Aλl = (Aλl − λl)/ λl and using the Berry-Esseen bound for the Poisson case (see Michel
[27])
                          √         √              4    30.6        1        4
                  ˆ
              |P(Aλl ≤ β l) − P (β l)| ≤ min         ,            √     ≤ √ .              (19)
                                                   5 1 + β 3 l3/2   lλ     5 lλ
yields, upon substituting (19) into (17), the second inequality in (16). The first inequality in
(16) follows in a similar way.

 We should stress that the occurrence of ζ( 3 ) in Lemma 1 is unrelated to the result (1) of
                                            2
Chang & Peres [7].

2.2    Quasi-Gaussian form: motivation and outline
The bound in (16) does not reveal much information, except that convergence takes place at
          √
rate O(1/ λ). In order to get more precise estimates one can use a saddlepoint approximation
or an Edgeworth expansion. √
                    √         However, these are not very convenient, as they require each
            ˆ
element P(Aλl ≤ β l) − P (β l) to be approximated separately due to its dependence on
l. One example would be the Edgeworth expansion for the Poisson distribution (see [3],
Eq. (4.18) on p. 96)
                          √         √       1      1 2
                   ˆ
                 P(Alλ ≤ β l) = P (β l) − √     e− 2 β l (β 2 l − 1) + O(1/λl),                                  (20)
                                         6 2πlλ
which leads to the approximation
      ∞                √         √                       ∞                        ∞
            1   ˆ                         1                    β 2 − 1 β2l               1        1     2l
              P(Aλl ≤ β l) − P (β l) ≈ − √                         e 2 −                       e− 2 β        .   (21)
      l=1
            l                           6 2πλ           l=1
                                                              l1/2               l=1
                                                                                        l3/2

                                                   7
It may not come as a surprise that (21) is not a good approximation because we neglect
all O(1/λl) terms in (20). Although including more terms in the Edgeworth expansion is
                                                                               √
                                                                        ˆ
an option, we choose to get more convenient asymptotic expansions for P(Aλl ≤ β l) by
bringing it into quasi-Gaussian form.
   Specifically, we prove the following theorem in Section 3.

Theorem 2.
                                       ∞                            ∞                        √
                                             p(ls) 1                           2 /2
                   − ln P(Qλ = 0) =               √                 √    e−x          y ′ (x/ ls)dx,   (22)
                                               l    2π             α l
                                       l=1

in which
                                                          λ      λ             1/2
                               α =      − 2s 1 −            + ln                      ,                (23)
                                                          s      s
α → β as λ → ∞,                                   √
                                     p(n) = nn e−n 2πn/n!,                                             (24)
                                         √
and y ′ is a function analytic in |x| < 2 π (see Appendix A, (138)).

  For p there is Stirling’s formula, see Abramowitz-Stegun [1], 6.1.37 on p. 257,
                                                                           ∞
                                1    1                                          pk
                    p(n) ∼ 1 −    +      + ... =                                   ,       n → ∞,      (25)
                               12n 288n2                                        nk
                                                                         k=0

and for y ′ there is the power series representation
                                                                     ∞
                              2    1                                                             √
                   y (x) = 1 − x + x2 + . . . =
                    ′
                                                                           bi xi ,        |x| < 2 π.   (26)
                              3   12
                                                                     i=0

From an aesthetic viewpoint, expression (22) conveys much understanding about the character
of the convergence, since we have complete agreement with the Gaussian random walk (10)
when we would have λ → ∞. The deviations from the quasi-Gaussian random walk are
embodied by p ≡ 1, y ′ ≡ 1 and α ≡ β. From (22) we see that there is the asymptotic
expansion
                                            ∞
                                        1                            √
                    − ln P(Qλ = 0) ∼ √         pk s−k+1/2 G−(k+1) (α/ s),              (27)
                                        2π k=0
where
                                       ∞                      ∞      1     2
                            Gk (a) =         lk+1/2               e− 2 lsx y ′ (x)dx.                  (28)
                                       l=1                a

  Similar expressions, though somewhat more complicated than the one in (27), exist for
EQλ and VarQλ (see Subsection 3.2) and these involve Gk with k = 0, −1, −2 . . . and
k = 1, 0, −1 . . ., respectively. We shall study Gk thoroughly, leading to series expansions,
asymptotics and bounds.
  We close this section by giving an interpretation of the quasi-Gaussian form (22). Using,
see Appendix A, (132),
                                                ∞
                                   1        n        1   2
                                       =          e− 2 nx y ′ (x)dx                      (29)
                                  p(n)     2π −∞



                                                      8
3

                    2.5

                      2
                                                  y ′ (x)
                    1.5

                      1

                    0.5

                      0
                      −2             −1             0                   1         2                3               4



                               Figure 1: The function y ′ (x) for x ∈ [−2, 4].

and                          ∞
                                       2 /2          √       √                ∞
                                                                                            2 /2
                            √    e−x          y ′ (x/ ls)dx = ls               √
                                                                                    e−lsx          y ′ (x)dx             (30)
                           α l                                               α/ s
we find from (22) that
                                                        ∞                ∞ −lx2 /2 ′   √
                                                                1       α e       y (x/ s)dx
                             − ln P(Qλ = 0) =                           ∞ −lx2 /2 ′    √     .                           (31)
                                                                l       −∞ e      y (x/ s)dx
                                                        l=1

As mentioned in the introduction, the resulting formula reveals that the summands of the
random walk associated with the M/D/s queue, and the summands of the Gaussian ran-
dom walk are absolutely continuous with respect to each other. The connecting measure
                                                                   √
between the two densities has a density as well, and equals y ′ (·/ s). Another interpretation
          ˆ
is that P(Qλ = 0) is obtained by twisting the Gaussian distribution associated with Mβ . The
                                                                               √
associated Radon-Nikodym derivative can again be described in terms of y ′ (·/ s).


3     From Spitzer-type expressions to quasi-Gaussian forms
In this section we show how to obtain the expression (22). In addition, we present similar
results for the mean and variance of the queue length.

3.1    Proof of Theorem 2
For n = 0, 1, . . . we let
                                                            n
                                                                    zk
                                                sn (z) =               ,     z ∈ C.                                      (32)
                                                                    k!
                                                            k=0
With ρ = λ/s and n = ls (so that λl = nρ), and
                                                 q(ξ) = e1−ξ ξ,             ξ ∈ C,                                       (33)
we then have from Szeg¨ [33], p. 50 (also see Abramowitz-Stegun [1], 6.5.13 on p. 262),
                      o
                     ∞                                                                                  ρ
                                    (lλ)j                      nn+1 e−n
                             e−lλ         = 1 − e−λl sn (λl) =                                              q n (ξ)dξ.   (34)
                                      j!                          n!                                0
                   j=n+1


                                                                    9
Using this relation we can rewrite the Spitzer-type expression (15) as
                                                                   ∞                  ρ
                                                             1/2         p(ls)
                             − ln P(Qλ = 0) = s                          √                q ls (ξ)dξ,             (35)
                                                                   l=1
                                                                           2πl    0


with p(n) as defined in (25). We then consider the equation

                                              f (y) := − ln q(1 − y) = 1 x2 ,
                                                                       2                                          (36)

with x ∈ C from which y is to be solved. We note that

                                       f (y) = 1 y 2 + 1 y 3 + 1 y 4 + . . . ,
                                               2       3       4                                                  (37)

whence there is an analytic solution y(x) around x = 0 that satisfies y(x) = x + O(x2 ) as
x → 0. Furthermore, since f increases from 0 to ∞ as y increases from 0 to 1, we have
that y(x) increases from 0 to ∞, and for any x ≥ 0 there is a unique non-negative solution
y(x) = y of (36). Furthermore, we let

                                     γ = −2(1 − ρ + ln ρ),               α = (sγ)1/2 .                            (38)

Then it holds that
                                                             1                1   2
                                              q ls (ρ) = e− 2 lsγ = e− 2 α l ,                                    (39)
and               ρ                    ∞         1    2             1                 ∞      1 2        √
                      q ls (ξ)dξ =    √
                                              e− 2 lsx y ′ (x)dx = √               √      e− 2 x y ′ (x/ ls)dx.   (40)
              0                           γ                          ls           α l

Substituting (40) into (35) yields (22).

Lemma 2. The parameters α and β are related according to
                                              √
                              √          y(α/ s)
                           β/ s =             √       .                                                           (41)
                                    (1 − y(α/ s))1/2
                                            √
Proof. Follows from 1 − ρ = y(γ 1/2 ) = y(α/ s), see Table 3.2, and
                                                      1/2
                        s−λ   s−λ                 λ         1   β        β          √
        1−ρ =               = √                             √ = √ ρ1/2 = √ (1 − y(α/ s))1/2 .                     (42)
                         s      λ                 s          s   s        s



We have that y(x)(1 − y(x))−1/2 = x + 1 x2 +
                                      6
                                                                  5 3
                                                                 72 x    + . . . (see Appendix A), and hence

                                          β =α+         √ α2
                                                        1
                                                                 +    5   3
                                                       6 s           72s α    + ....                              (43)




                                                             10
3.2     Mean and variance of the queue length
Our primary characteristic in this paper is the probability of an empty queue. However, the
techniques that we develop can be applied to other characteristics like the mean and variance
of the queue length. From Spitzer’s identity it follows that the mean and variance of the
maximum M are given by ∞ 1 E((Sl+ )k ) with k = 1 and k = 2, respectively. For the
                              l=1 l
M/D/s queue this yields
                                                           ∞           ∞
                                                                 1                                 (lλ)j
                                    EQλ =                                   (j − ls)e−lλ                 ,                                      (44)
                                                                 l                                   j!
                                                           l=1       j=ls+1
                                                            ∞          ∞
                                                                 1                                  (lλ)j
                                 VarQλ =                                    (j − ls)2 e−lλ                .                                     (45)
                                                                 l                                    j!
                                                           l=1       j=ls+1

This leads after considerable rewriting to
                                             ∞                                                          ρ
                                       1/2         p(ls)
                          EQλ = s                  √     ρq ls (ρ) − ls(1 − ρ)                              q ls (ξ)dξ                          (46)
                                             l=1
                                                     2πl                                            0

and
                    ∞                                                                                                           ρ
              1/2         p(ls)
  VarQλ = s               √     − ρ(ls(1 − ρ) − 1)q ls (ρ) + ((1 − ρ)2 l2 s2 + lsρ)                                                 q ls (ξ)dξ . (47)
                    l=1
                            2πl                                                                                             0

In a similar way as for P(Qλ = 0), (46) and (47) can then be brought into the forms
                                   ∞                   1    2         ∞                                                         √
                          √               e− 2 α l                             p(ls)                ∞
                                                                                                                  2 /2
          EQλ =               s     ρp(ls) √       −                     αR(ρ) √                    √       e−x          y ′ (x/ ls)dx ,        (48)
                                l=1
                                             2πl                     l=1
                                                                                 2π                α l
                               ∞                                            ∞                        √
                                                   p(ls)                               2 /2
        VarQλ = s                 (α2 lR2 (ρ) + ρ) √                        √    e−x          y ′ (x/ ls)dx
                              l=1
                                                     2π                    α l
                                       ∞      √           √           1 2
                                                                   e− 2 α l
                                −            α lρR(ρ) − ρ/ ls p(ls) √       ,                                                                   (49)
                                       l=1
                                                                      2π

where
                                              1−ρ     1
                                        R(ρ) = √ = 1 − (1 − ρ) + . . . .                                                                        (50)
                                                γ     3
For the Gaussian random walk we have that (see [15])
                                       ∞           1   2
                                             e− 2 β l     1                      ∞
                                                                                           2 /2
                    EMβ =                     √       − β√                       √   e−x          dx ,                                          (51)
                                    l=1
                                               2πl         2π                 β l
                                     ∞                                     ∞
                                                          1                            2 /2         β        1 2
               VarMβ =                       (β 2 l + 1) √                  √    e−x          dx − √ l1/2 e− 2 β l .                            (52)
                                    l=1
                                                           2π              β l                      2π
                     √
Ignoring the factors s and s, we again have complete agreement with the Gaussian random
walk when λ → ∞. The deviations from the Gaussian random walk are embodied by p ≡ 1,
y ′ ≡ 1, β ≡ α and the fact that R(ρ) ≡ 1 when ρ < 1. The introduced notation is summarized
in Table 3.2.

                                                                      11
√
               s       =       λ+β λ
              ρ        =       λ/s
              γ        =       −2(1 − ρ + ln ρ)
                               √
              α        =         sγ
                                  √      √
              a        =       α/ s = γ
                                                                   √
             y(x)      =       x − 1 x2 + 36 x3 + . . . ; |x| < 2 π
                                           1
                               √ 3 √                   √ −1/2
              β        =         s y(α/ s)(1 − y(α/ s))
                                                                       √      √
              β        =       α + 1 s−1/2 α2 + 72 s−1 α3 + . . . ; |α/ s| < 2 π
                                    6 √
                                                 5
                                                         1         1
             p(n)      =       nn e−n 2πn/n! ∼ 1 + 12 n−1 + 288 n−2 + . . .


               Table 1: Interrelations between some parameters and functions.

4     Results for Gk
In this section we give a reformulation of the function Gk in terms of a principal series
expansion. The level of generality is higher than needed for the M/D/s queue, as we consider
a large class of functions z(x) of which y ′ (x) is just a special case. In Subsection 4.1 we derive
the Taylor series for the most general case. We also discuss some special cases that lead
to considerable reductions in complexity of the expressions. The principal series expansion
                                           √
comprises terms involving s, z, a = α/ s and elementary functions, as well as a constant
                                 √
Lk , not depending on a = α/ s, which is more complicated. For this Lk we present an
asymptotic series as s → ∞ that can be used conveniently when the radius of convergence
                                                         √
r0 of z(x) = ∞ bj z j is not small (for instance 2 π as in the pilot case z(x) = y ′ (x)). In
                  j=0
Subsection 4.1 we derive the principal series expansion. In Subsection 4.2 we investigate the
numerical evaluation of Lk in terms of the optimal truncation value of the series expansions.
In Subsection 4.3 we use this general result for the specific case of z(x) = y ′ (x) to derive series
expansions and asymptotics for P(Qλ = 0). There is a clear connection with the Gaussian
random walk. In fact, results for the Gaussian random walk involve the function Gk for the
special case z(x) ≡ 1.

4.1   Principal series expansion
We let z : [0, ∞) → C be a continuous function satisfying z(x) = O(exp(εx2 )) for any ε > 0,
and we assume that there is an r0 > 0 such that z(x) is represented by its Taylor series
  ∞       j
  j=0 bj x for 0 ≤ x < r0 . We consider for s > 0 and integer k the function
                                   ∞                   ∞     1        2
                        Gk (a) =         lk+1/2            e− 2 lsx z(x)dx,       a > 0.       (53)
                                   l=1             a

In the case that z(x) = xi we have
                                                       i+1          √
                                    Gk (a) = s−         2    Tk,i (a s),                       (54)
where Tk,i is defined as
                                            ∞                     ∞       1   2
                               Tk,i (b) =         lk+1/2              e− 2 lx xi dx            (55)
                                            l=1               b


                                                       12
with i = 0, 1, . . . and k ∈ Z. The functions Tk,i have been thoroughly investigated in [16],
Section 5, leading to analytic expressions. We now generalize this result to Gk .
Theorem 3. For k ∈ Z and a < 2 π/s we have that
                             k+3/2                        2k+1                                                         a              2k+2    j
                     2                                           bj aj−2k−2                                                z(x) −     j=0 bj x
Gk (a) =                               Γ(k + 3/2)                           − b2k+2 ln a −                                                      dx
                     s                                           2k + 2 − j                                        0               x2k+3
                                                          j=0
                              ∞
                                                            (− 1 s)r             a
         + Lk −                    ζ(−k − r − 1/2)             2
                                                                                     x2r z(x)dx,                                                 (56)
                                                              r!             0
                             r=0
where
          ∞                                                  2k+2                               2k+1                                  j+1
                                   ∞
                    k+1/2                 1
                                        − 2 lsx2                             j1                                    j+1          2      2
Lk =            l                      e            z(x) −          bj x dx +                             bj Γ                              ζ(−k + j/2)
                               0                                              2                                     2           s
          l=1                                                j=0                                j=0
                              k+3/2                          k
                         2                                           1       √
     + b2k+2                               Γ(k + 3/2)                    − ln 2s .                                                                 (57)
                         s                                        2j + 1
                                                            j=0

Proof. We have
                                                                         ∞
                                                                                                    1     2
                                                   G′ (a) = −z(a)
                                                    k                                lk+1/2 e− 2 lsa .                                           (58)
                                                                         l=1
The right-hand side of (58) can be expressed in terms of Lerch’s transcendent Φ, defined as
the analytic continuation of the series
                                                                        ∞
                                                     Φ(z, t, v) =                (v + n)−t z n ,                                                 (59)
                                                                        n=0
which converges for any real number v = 0, −1, −2, . . . if z and t are any complex numbers
with either |z| < 1, or |z| = 1 and Re(t) > 1. Note that ζ(t) = Φ(1, t, 1). Thus,
                                                                    1    2              1   2
                                           G′ (a) = −z(a)e− 2 sa Φ(e− 2 sa , −k − 1 , 1).
                                            k                                     2                                                              (60)
We then use the important result derived by Bateman [9], §1.11(8) (with ζ(t, v) := Φ(1, t, v)
the Hurwitz zeta function)
                                                                                                ∞
                                      Γ(1 − t)                                                                         (ln z)r
                         Φ(z, t, v) =          (ln 1/z)t−1 + z −v                                       ζ(t − r, v)            ,                 (61)
                                        zv                                                                                r!
                                                                                            r=0

which holds for | ln z| < 2π, t = 1, 2, 3, . . ., and v = 0, −1, −2, . . ., as to obtain
                                                            k+3/2                           ∞
                                                      2                                                                      (− 1 sa2 )r
     G′ (a) = −z(a) Γ(k + 3/2)
      k                                                             a−2k−3 +                    ζ(−k − r − 1/2)                 2
                                                                                                                                         .       (62)
                                                      s                                  r=0
                                                                                                                                 r!
Therefore
                                                                   k+3/2 2k+2
                                                             2
                         G′ (a) + Γ(k + 3/2)
                          k                                                             bj aj−2k−3 =
                                                             s
                                                                                 j=0
                                       k+3/2                2k+2                                              ∞
                               2                                                                                                        (− 1 sa2 )r
      −Γ(k + 3/2)                                  z(a) −          bj aj a−2k−3 − z(a)                            ζ(−k − r − 1/2)          2
                                                                                                                                                    .
                               s                                                                                                            r!
                                                            j=0                                            r=0
                                                                                                                                                 (63)

                                                                        13
√
The series on the second line of (63) converges uniformly in a with s1/2 a ∈ [0, c] and c < 2 π,
so upon integrating the identity in (63) we get for 1 sa2 ≤ c < 2π
                                                     2

                                                                     k+3/2       2k+1
                                                             2                             bj aj−2k−2
                              Gk (a) + Γ(k + 3/2)                                                     + b2k+2 ln a =
                                                             s                             j − 2k − 2
                                                                                     j=0
                                                                 k+3/2           a                2k+2
                                                            2
                                Lk − Γ(k + 3/2)                                       z(x) −                bj xj x−2k−3 dx
                                                            s                0                    j=0
                                       ∞
                                                                       (− 1 s)r            a
                                  −          ζ(−k − r − 1/2)              2
                                                                                               x2r z(x)dx,                                      (64)
                                       r=0
                                                                            r!         0

where
                                                                      k+3/2          2k+1
                                                                 2                             bj aj−2k−2
             Lk = lim Gk (a) + Γ(k + 3/2)                                                                 + b2k+2 ln a                  .       (65)
                        a↓0                                      s                             j − 2k − 2
                                                                                      j=0

We shall determine Lk . It holds that, as a ↓ 0,
                   ∞                   ∞                             2k+2                       2k+2        ∞                    ∞
                                              1       2                                                                                 1   2
 Gk (a) =                lk+1/2            e− 2 lsx       z(x) −            bi xi dx +                 bi       lk+1/2               e− 2 lsx xi dx
                   l=1             a                                 i=0                         i=0        l=1              a
                    ∞                  ∞                             2k+2                                   2k+2
                                              1       2                                                                    i+1          √
            =            lk+1/2            e− 2 lsx       z(x) −            bi xi dx + o(1) +                      bi s−    2    Tk,i (a s).
                   l=1             0                                 i=0                                     i=0
                                                                                                                                                (66)
                                                                                                                     2k+2    i
Here (54) has been used and the o(1) comes from the fact that z(x) −                                                 i=0 bi x         = O(x2k+3 )
so that
                         a                 2k+2
                              1    2                            1
                           e− 2 lsx z(x) −      bi xi dx = O         .                                                                          (67)
                       0                                     (ls)k+2
                                                                 i=0

Now from Janssen & van Leeuwaarden [16], Section 5,

                  i+1          √       i+1 Γ(k + 3/2)           √
             s−    2    Tk,i (a s) = s− 2              2k+3/2 (a s)i−2k−2 + Lk,i + O(a)
                                            2k + 2 − i
                                                 2 k+3/2 ai−2k−2          i+1
                                   = Γ(k + 3/2)                       + s− 2 Lk,i + O(a)                                                        (68)
                                                 s         2k + 2 − i

for i = 0, 1, . . . , 2k + 1 and
          2k+3             √       2k+3                      √
     s−     2    Tk,2k+2 (a s) = s− 2 − Γ(k + 3/2)2k+3/2 ln(a s) + Lk,2k+2 + O(a)
                                                                            k+3/2
                                                                      2                   √
                                       = −Γ(k + 3/2)                                  ln(a s) + s−(k+3/2) Lk,2k+2 + O(a).
                                                                      s
                                                                                                                                                (69)

Here
                                1 i + 1 i+1
                          Lk,i = Γ      2 2 ζ(−k + i/2),                                   i = 0, 1, . . . , 2k + 1,                            (70)
                                2   2

                                                                      14
k
                                                                      k+3/2                    1   1
                           Lk,2k+2 = Γ(k + 3/2)2                                                  − ln 2 .                                 (71)
                                                                                            2j + 1 2
                                                                                   j=0

Therefore, as a ↓ 0, we get
               ∞                     ∞                                 2k+2
                         k+1/2           − 1 lsx2
  Gk (a) =           l                   e 2                  z(x) −           bi xi dx + o(1)
               l=1               0                                     i=0
               2k+1                                                                                                        i+1
                                                              k+3/2
                                                         2             ai−2k−2   1 i+1                                 2    2
           +             bi Γ(k + 3/2)                                          + Γ                                              ζ(−k + i/2)
                                                         s            2k + 2 − i 2  2                                  s
               i=0
                                                         k+3/2                                        k
                                                 2                                       √                    1       √
           + b2k+2 Γ(k + 3/2)                                        − ln a − ln                s+                − ln 2 .                 (72)
                                                 s                                                         2j + 1
                                                                                                     j=0

Adding
                                                             k+3/2    2k+1
                                                 2                            bj aj−2k−2
                          Γ(k + 3/2)                                                     + b2k+2 ln a                                      (73)
                                                 s                            j − 2k − 2
                                                                       j=0

at either side of (72) and letting a ↓ 0, we find that Lk has the required value (57). Then (56)
follows from (64).

  Some values of the Riemann zeta function ζ are given in Table 4.1.
                              x                 ζ(x)                          x            ζ(x)
                             -5.5        -0.00267145801990                   0.0    -0.50000000000000
                             -5.0        -0.00396825396825                   0.5    -1.46035450880959
                             -4.5        -0.00309166924722                   1.5    2.61237534868549
                             -4.0                0                           2.0    1.64493406684823
                             -3.5        0.00444101133548                    2.5    1.34148725725092
                             -3.0        0.00833333333333                    3.0    1.20205690315959
                             -2.5        0.00851692877785                    3.5    1.12673386731706
                             -2.0                0                           4.0    1.08232323371114
                             -1.5        -0.02548520188983                   4.5    1.05470751076145
                             -1.0        -0.08333333333333                   5.0    1.03692775514337
                             -0.5        -0.20788622497735                   5.5    1.02520457995469

                         Table 2: Some values of the Riemann zeta function ζ.

  We now give several special cases of Theorem 3. The next two corollaries focus on negative
values of k.
Corollary 1. For 1 sa2 < 2π and k = −2, −3, . . . we have that
                 2

                                                         k+3/2                              a
                                                 2
                 Gk (a) = −                                      Γ(k + 3/2)                     x−2k−3 z(x)dx
                                                 s                                      0
                                                     ∞
                                                                                         (− 1 s)r             a
                                 + Lk −                       ζ(−k − r − 1/2)               2
                                                                                                                  x2r z(x)dx,              (74)
                                                                                           r!             0
                                                     r=0

                                 ∞           1       2
where Lk = ∞ lk+1/2 0 e− 2 lsx z(x)dx (which follows from the definition of Lk in (57) in
               l=1
which all series over j vanish for k = −2, −3, . . .).

                                                                       15
Corollary 2. For 1 sa2 < 2π and k = −1 we have that
                 2

                                                                  1/2                       a
                                                            2π                                  z(x) − b0
                               G−1 (a) = −                                ln a +                          dx
                                                             s                          0           x
                                                                 ∞
                                                                                            (− 1 s)r       a
                                                + L−1 −                ζ(−r + 1/2)             2
                                                                                                               x2r z(x)dx,              (75)
                                                                                              r!       0
                                                                 r=0

                           ∞ −1/2 ∞ − 1 lsx2                                                2π 1/2
                                                                                                     √
where L−1 =                l=1 l  0 e
                                      2      (z(x)                     − b0 )dx −            s     ln 2s.

  Theorem 3 is meant for the case that a and the convergence radius r0 of ∞ bj xj are gen-
                                                                             j=0
eral. In the case that a < r0 the expressions can be simplified considerably, as demonstrated
below. If a < r0 we have
                                            a              2k+2    j                        ∞
                                                z(x) −     j=0 bj x                                bj aj−2k−2
                                                                     dx         =                             ,                         (76)
                                        0               x2k+3                                      j − 2k − 2
                                                                                        j=2k+3

                                                        a                       ∞
                                                                                        bj aj+2r+1
                                                            x2r z(x)dx =                           .                                    (77)
                                                    0                                   j + 2r + 1
                                                                             j=0

As a consequence of (76) we have that the expression on the first line of (56)
                                2k+1                                                    a              2k+2    j
                                       bj aj−2k−2                                           z(x) −     j=0 bj x
                                                  − b2k+2 ln a −                                                 dx                     (78)
                                       2k + 2 − j                                   0               x2k+3
                                j=0

simplifies to
                                                        ∞
                                                                      bj aj−2k−2
                                                                                 − b2k+2 ln a.                                          (79)
                                                                      2k + 2 − j
                                                  j=0,j=2k+2

Together with (77) this gives expressions for Gk (a) that are, apart from the Lk to which we
turn next, convenient for computation when a is small.

Lemma 3. For the first line of (57)
  ∞                                                2k+2                         2k+1                                  j+1
                       ∞
           k+1/2           − 1 lsx2                               j    1                            j+1           2    2
       l                   e 2          z(x) −               bj x dx +                      bj Γ                            ζ(−k + j/2) (80)
                   0                                                   2                             2            s
 l=1                                                j=0                         j=0

there is the asymptotic expression
                                            ∞                                       j+1
                             1                                   j+1        2        2
                           ∼                        bj Γ                                    ζ(−k + j/2),           s → ∞.               (81)
                             2                                    2         s
                                     j=0,j=2k+2

                               j+1
In case that bj Γ               2      = O(B j ) for some B > 0, the asymptotic series in (81) is convergent
when s > 2B 2 , with sum equal to (80).




                                                                           16
Proof. Using
       ∞                                           (j+1)/2        ∞                                           (j+1)/2
           − 1 lsx2 j       1                2                         −u (j−1)/2         1         2                           j+1
           e 2       x dx =                                           e   u          du =                               Γ                  , (82)
   0                        2                ls               0                           2         ls                           2

we find that
               ∞                ∞                            2k+2                       ∞           ∞                       ∞
                                         1     2                                                                                   1   2
                   lk+1/2           e− 2 lsx       z(x) −             bj xj dx ∼               bj         lk+1/2                e− 2 lsx xj dx
           l=1              0                                j=0                      j=2k+3        l=1                 0
                                             ∞                (j+1)/2                    ∞
                                     1                   2                     j+1
                                =                   bj                    Γ                   l−j/2+k .                                      (83)
                                     2                   s                      2
                                         j=2k+3                                         l=1

                                                             ∞ −j/2+k
This yields (81) since ζ(j/2 − k) =                          l=1 l    .


Remark 4. The series expansion (56) for Gk (a) comprises, as a ↓ 0, leading order terms
involving aj−2k−2 , j = 0, 1, . . . , 2k + 1, and ln a when k = −1, 0, 1, . . ., and Gk (a) stays
bounded as a ↓ 0 for k = −2, −3, −4, . . .. In most cases we are interested in, the value of a
is quite small (say ≤ 0.1). The formula in (56) can be used conveniently for computation of
                                              √
Gk (a) for values of a from 0 to as large as π. For larger values of a, we present in Appendix
A formula (163) as an attractive alternative to compute Gk (a). This alternative shows, for
instance, quite clearly an exp(− 1 (s + 1)a2 )-behavior of Gk (a) as a gets large.
                                    2

Remark 5. Chang & Peres [7], Theorem 1.1, proved that
                                                                               ∞
                                                     √           β                   ζ(1/2 − r) −β 2            r
                             P(Mβ = 0) =                 2β exp √                                                   ,                        (84)
                                                                  2π                 r!(2r + 1)  2
                                                                               r=0
             √                                                                           √
for 0 < β < 2 π. This result follows easily from Theorem 3, for the case z(x) ≡ 1, a = β/ s
and k = −1.
                                              √
  For general k, setting z(x) ≡ 1 and a = β/ s in Theorem 3 leads to the following result.
                       √
Lemma 4. For β < 2 π and k ∈ Z we have that
                        ∞                                             ∞
                                    √        β                            ζ(−k − r − 1 ) −β 2             r
                            lk P (−β l) = − √                                          2
                                                                                                              + Rk (β),                      (85)
                                              2π                            r!(2r + 1)    2
                        l=1                                        r=0
                                    √
where R−1 (β) = − ln                    2β and

                                      1 Γ(k + 3 ) k+ 3 −2k−2 1
                                              2
                            Rk (β) = √           2 2β       + ζ(−k),                                      k = −1.                            (86)
                                      2π 2k + 2              2

4.2        Optimal truncation value
Lemma 3 can be deployed in two ways. We can take only the first few terms to get a good
idea of how things behave (see Subsection 4.3), or for the numerical evaluation of Lk , we take
as many terms as needed using optimal truncation. The optimal truncation value J of (81)

                                                                          17
is so large (see developments below) that we can replace ζ(−k + J/2) by 1. The truncation
error made by approximating (80) by
                                 J                                       (j+1)/2
                         1                          j+1          2
                                           bj Γ                                     ζ(−k + j/2)                     (87)
                         2                           2           s
                             j=0,j=2k+2

is of the order
                                                                           (J+2)/2
                                         1            J +2           2
                                           bJ+1 Γ                                     .                             (88)
                                         2              2            s
We replace, furthermore, bJ+1 = (J + 2)aJ+2 by its asymptotic bound, see Appendix A,
Lemma 13,
                                     J + 2 −1/2   1   J+2
                           |bJ+1 | ≤              √       .                     (89)
                                       2         2 π
Thus                                       J +2
                                                                          1/2                         (J+2)/2
           1            J +2         2       2               1                       J +2       1
             bJ+1 Γ                               ≤                             Γ                               .   (90)
           2              2          s                    2(J + 2)                     2       2πs
The factor (1/2(J + 2))1/2 is rather unimportant for determination of the optimal truncation
value J, and we focus on
                                        J +2       1 (J+2)/2
                               DJ = Γ                        .                          (91)
                                          2      2πs
Noting that Γ(J/2 + 3/2)/Γ(J/2 + 1) ≈ (J/2 + 1)1/2 , we see that
                                                                 1/2                1/2
                                     DJ+1             J +2                 1
                                          ≈                                               .                         (92)
                                      DJ                2                 2πs

The right-hand side of (92) decreases in J until J/2 + 1 = 2πs; this J is (near to) the optimal
truncation point. At this point we estimate the right-hand side of (90) by Stirling’s formula
as
         1/2                   2πs                  1/2                     √                       2πs
    1                    1                   1                                                 1              e−2πs
               Γ(2πs)                ≈                    (2πs)2πs−1/2 e−2πs 2π                           =    √ . (93)
   8πs                  2πs                 8πs                                               2πs             s 8π

For instance, for s = 10 this equals 10−29 .

Remark 6. Observe how important it is that we have managed to show the good bound (89)
                                     √
on |bJ+1 |. If, for instance, the 1/2 π in this bound were to be replaced by 1, the e−2πs on the
                                                                                    √
far right of (93) would have to be replaced by e−s/2 and the resulting quantity 2πe−s/2 /s
would be 0.0017 for s = 10.

4.3    Accurate approximations for the M/D/s queue
We can apply Theorem 3 to obtain accurate approximations for the emptiness probability
and the mean and variance of the queue length. By way of illustration, we do this in some
detail for P(Qλ = 0) and briefly indicate at the end of this section how one can proceed for
the other cases.



                                                            18
We have from (25) and (27) that
                                     ∞
                              1                                 √
− ln P(Qλ = 0) ∼             √            pk s−k+1/2 G−(k+1) (α/ s)
                              2π    k=0
                              1              √                                                  √                                    √
                    =        √   s1/2 G−1 (α/ s) −                     1 −1/2
                                                                       12 s   G−2 (α/                s) +    1 −3/2
                                                                                                            288 s   G−3 (α/              s) + . . . .
                              2π
                                                                                                                                                (94)
                                                     √
The G−2 , G−3 , . . . are bounded functions of a = α/ s while G−1 (a) behaves like
                                                   2π       1/2       √
                                           −                      ln a 2s         as   a ↓ 0.                                                (95)
                                                    s
Accurate approximations to − ln P(Qλ = 0) are obtained by including 1, 2, 3, . . . terms of the
second line of (94) in which the G’s must be approximated. For the number of terms of the
asymptotic series in (94) to be included one could follow a truncation strategy (based on
                                        π
(139), (152) and the bound G−k (a) ≤ ( 2s )1/2 ζ(k), k = 2, 3, . . .) pretty much as was done in
Subsection 4.2. We shall not pursue this point here.
  We shall compute accurate approximations to G−k (a) for k = 1, 2, . . .. We have from (74)
                   √
and (75) for α < 2 π
                                                                                                 √
                                                                1/2                            α/ s
                        √                              2π                   √                         y ′ (x) − 1
                 G−1 (α/ s) = −                                        ln α/ s +                                  dx
                                                        s                                  0                x
                                                            ∞                                          √
                                                                            (− 1 s)r                 α/ s
                                      + L−1 −                     ζ(−r + 1 ) 2
                                                                         2                                  x2r y ′ (x)dx,                   (96)
                                                                              r!                 0
                                                            r=0

and for k = 2, 3, . . .,
                                                                                                  √
                                                         −k+3/2                                 α/ s
                      √                            2
               G−k (α/ s) = −                                         Γ(−k + 3/2)                        x2k−3 y ′ (x)dx
                                                   s                                     0
                                                         ∞                                                 √
                                                                                      1 r                α/ s
                                                                                 1 (− 2 s)
                                     + L−k −                    ζ(k − r −        2)                             x2r y ′ (x)dx.               (97)
                                                                                     r!              0
                                                         r=0

Here,
                               ∞                                                                           1/2
                                                   ∞        1     2                               2π               √
                   L−1 =             l−1/2             e− 2 lsx (y ′ (x) − 1)dx −                                ln 2s,                      (98)
                                               0                                                   s
                               l=1
and for k = 2, 3, . . .,
                                                   ∞                        ∞      1   2
                                     L−k =               l−k+1/2                e− 2 lsx y ′ (x)dx.                                          (99)
                                                   l=1                  0

  Below we specify the missing ingredients in (96)-(99).
    • We have
                              √                                         √
                            α/ s                                      α/ s ∞                               ∞                 j
                                   y ′ (x) − 1                                                                   bj   α
                                               dx =                               bj xj−1 dx =                        √          ,         (100)
                        0                x                        0                                              j      s
                                                                            j=1                           j=1
                                                                 √     √
        and the computation of the series is feasible when 0 ≤ α/ s ≤ 2 π, the bj being
                             √ j
        computable and O(1/(2 π) ).

                                                                       19
β = 0.01 (0.0141)                                     β = 0.1 (0.1334)
      s      α         true      (94)-1    (94)-2        (94)-3      α        true    (94)-1    (94)-2        (94)-3
      1   0.0100      0.0268     0.0256 0.0267           0.0267   0.0983     0.2351   0.2265 0.2345           0.2343
      2   0.0100      0.0225     0.0219 0.0225           0.0225   0.0988     0.2022   0.1980 0.2021           0.2021
      5   0.0100      0.0190     0.0188 0.0190           0.0190   0.0993     0.1747   0.1730 0.1747           0.1746
     10   0.0100      0.0174     0.0173 0.0174           0.0174   0.0995     0.1617   0.1609 0.1617           0.1617
     20   0.0100      0.0164     0.0163 0.0164           0.0164   0.0996     0.1529   0.1525 0.1529           0.1529
     50   0.0100      0.0155     0.0155 0.0155           0.0155   0.0998     0.1455   0.1453 0.1455           0.1455
    100   0.0100      0.0151     0.0150 0.0151           0.0151   0.0998     0.1419   0.1418 0.1419           0.1419
    200   0.0100      0.0148     0.0148 0.0148           0.0148   0.0999     0.1393   0.1393 0.1393           0.1393
    500   0.0100      0.0145     0.0145 0.0145           0.0145   0.0999     0.1371   0.1371 0.1371           0.1371
                                β = 0.2 (0.2518)                                      β = 0.5 (0.5293)
      s      α         true      (94)-1    (94)-2        (94)-3      α        true    (94)-1    (94)-2        (94)-3
      1   0.1932      0.4105     0.3979 0.4092           0.4089   0.4573     0.7182   0.7049 0.7137           0.7134
      2   0.1952      0.3613     0.3549 0.3611           0.3610   0.4699     0.6656   0.6586 0.6642           0.6641
      5   0.1970      0.3185     0.3159 0.3185           0.3185   0.4811     0.6156   0.6125 0.6151           0.6151
     10   0.1979      0.2979     0.2966 0.2979           0.2978   0.4867     0.5899   0.5883 0.5897           0.5897
     20   0.1985      0.2838     0.2831 0.2837           0.2837   0.4906     0.5719   0.5710 0.5717           0.5717
     50   0.1991      0.2716     0.2714 0.2716           0.2716   0.4941     0.5560   0.5557 0.5560           0.5560
    100   0.1993      0.2657     0.2655 0.2657           0.2657   0.4958     0.5481   0.5479 0.5481           0.5481
    200   0.1995      0.2615     0.2615 0.2615           0.2615   0.4970     0.5426   0.5425 0.5425           0.5425
    500   0.1997      0.2579     0.2579 0.2579           0.2579   0.4981     0.5377   0.5376 0.5377           0.5377
                                  β = 1 (0.8005)                                       β = 2 (0.9762)
      s      α         true      (94)-1    (94)-2        (94)-3      α        true    (94)-1    (94)-2        (94)-3
      1   0.8299      0.9055     0.8973 0.8948           0.8945   1.3670     0.9835   0.9793 0.9636           0.9633
      2   0.8790      0.8787     0.8746 0.8737           0.8736   1.5296     0.9799   0.9787 0.9674           0.9672
      5   0.9236      0.8511     0.8493 0.8489           0.8489   1.6948     0.9774   0.9770 0.9703           0.9703
     10   0.9462      0.8364     0.8354 0.8352           0.8352   1.7835     0.9766   0.9764 0.9723           0.9723
     20   0.9622      0.8259     0.8253 0.8252           0.8252   1.8473     0.9763   0.9762 0.9738           0.9738
     50   0.9762      0.8165     0.8163 0.8162           0.8162   1.9040     0.9762   0.9761 0.9750           0.9750
    100   0.9832      0.8118     0.8117 0.8117           0.8117   1.9324     0.9762   0.9761 0.9755           0.9755
    200   0.9881      0.8085     0.8084 0.8084           0.8084   1.9524     0.9762   0.9762 0.9759           0.9759
    500   0.9925      0.8056     0.8055 0.8055           0.8055   1.9700     0.9762   0.9762 0.9761           0.9761


Table 3: Series expansions for P(Qλ = 0) based on (94). The values of P(Mβ = 0) are given
between brackets.

   • We have
                √                                √
              α/ s                             α/ s ∞                       ∞                       n+j+1
                      n ′                                                          bj         α
                     x y (x)dx =                         bj xn+j dx =                         √                ,       (101)
          0                                0                                     n+j+1          s
                                                   j=0                     j=0
                                                              √     √
     and the computation of the series is feasible when 0 ≤ α/ s ≤ 2 π. Furthermore
                                      √                                    ∞
                                    α/ s                                                            j+1
                                                                                     bj       α
                     (− 1 s)r
                        2
                                            2r ′
                                           x y (x)dx =        (− 1 α2 )r
                                                                 2
                                                                                              √           .            (102)
                                0                                                2r + j + 1    s
                                                                           j=0


     Since, see [15], Sec. 6, ζ(−r + 1 )/r! = O(1/(2π)r ), the computation of the series over r
                                     2                       √
     at the right-hand side of (96) is feasible when α < 2 π. A similar result holds for the
     series over r at the right-hand side of (97).




                                                            20
• We have by Lemma 3
          ∞                                                                   ∞                                      j+1
                              ∞
                   −1/2             1
                                  − 2 lsx2     ′             1                                  j+1              2    2
               l                  e          (y (x) − 1)dx ∼                        bj Γ                                   ζ(1 + j/2),   (103)
                          0                                  2                                   2               s
         l=1                                                                  j=1

                                         ∞                                        j+1
                                    1                     j+1             2        2
                          L−k     ∼            bj Γ                                     ζ(k + j/2),              k = 2, 3, . . . ,       (104)
                                    2                      2              s
                                        j=0

     for the series expressions at the right-hand sides of (98) and (99). The left-hand sides
     of (103) and (104) can be accurately approximated by using the optimal truncation
     approach of Subsection 4.2. Alternatively, assume that we include all three terms on
     the second line of (94) (so that the truncation error is O(s−7/2 )). We then include in
     the right-hand side of (103) the terms with j = 1, 2, 3, 4, and in the right-hand side of
     (104) the terms with j = 1, 2.

  When we want to compute accurate approximations to EQλ and VarQλ , we can use (48)
and (49), and then it becomes necessary to approximate Gk (a) with k = 0 and k = 1 as
well. This can still be done by using Theorem 3 with its simplifications as pointed out in
                                              √
Corollary 2 since z(x) = y ′ (x) has bj = O((2 π)−j ). Of course, there are a variety of ways to
proceed here, just like in case of − ln P(Qλ = 0) treated above. For the latter case, we have
just worked out one of the more straightforward methods.
  Table 3 displays approximations to P(Qλ = 0) based on the series expansion (94). Results
are given for 1, 2, and 3 terms of the second line of (94), and the G’s are approximated as
described in this subsection. Clearly, the expansions provide sharp approximations, and in
most cases, one term suffices to get accurate results, i.e.,
                                                  √
                                                    s         √
                            P(Qλ = 0) ≈ exp −    √ G−1 (α/ s) .                            (105)
                                                   2π

5    Bounds and approximations for the emptiness probability
The Gaussian form (22) for P(Qλ = 0) is rather complicated due to the presence of p(ls) and
z(x) = y ′ (x), which both can be expressed as infinite series. In this section we obtain bounds
on P(Qλ = 0) by using inequalities for p(ls) and y ′ (x).

Lemma 5.
                                                         ∞                    ∞
                                                   1/2             1                    1   2
       P(Qλ = 0) ≥ exp                   −s                    √          √
                                                                                  e− 2 lsx y ′ (x)dx =: LB,                              (106)
                                                         l=1
                                                                   2πl        γ
                                                          ∞                                          ∞
                                                                   1       1                                 1   2
       P(Qλ = 0) ≤ exp                   − s1/2                √       1−                           √
                                                                                                         e− 2 lsx y ′ (x)dx =: U B. (107)
                                                                   2πl    12ls                       γ
                                                         l=1

Proof. Follows directly from rewriting (22) as
                                                                     ∞                      ∞
                                                               1/2            p(ls)                      1   2
                          P(Qλ = 0) = exp                 −s                  √             √
                                                                                                    e− 2 lsx y ′ (x)dx                   (108)
                                                                     l=1
                                                                                2πl             γ




                                                                         21
and applying
                                  en      1    1             en
                         n−n−1/2 √    1−     ≤    ≤ n−n−1/2 √ .                               (109)
                                   2π    12n   n!             2π



Lemma 6. There is the inequality
                                             ∞               ∞
                          UB        1              1 1                1   2
                             ≤ exp                   √        √    e− 2 x dx .                (110)
                          LB       12s             l2 2π     α l
                                             l=1

Proof. Follows from (106)-(107) and y ′ (x) ≤ 1 (see Lemma 15).

  We next show that the right-hand side of (110) is a decreasing function of λ when β > 0 is
kept fixed. Indeed, this is a direct consequence of the following lemma.

Lemma 7. α = (−2s(1 − ρ + ln ρ))1/2 increases as a function of λ when β > 0 is kept fixed.
In fact, α increases from 0 to β as λ increases from 0 to ∞.

Proof. Letting ν = 1 α2 we have
                   2

                                         √    λ             λ
           ν = −s(1 − ρ + ln ρ) = −(λ + β λ) 1 −√ + ln        √
                                           λ+β λ         λ+β λ
                   √         √      λ                         x
            = −β λ − (λ + β λ) ln     √ = −x β + (x + β) ln                               ,   (111)
                                  λ+β λ                     x+β
                     √
where we have set x = λ. Now
      dν                        x            x                  x+β         β
            = −β − (x + β) ln       − x ln         + (x + β) ·         ·
      dx                       x+β          x+β                   x      (x + β)2
                                  x
            = −2β − (2x + β) ln        = 2x(−t + (1 + 1 t) ln(1 + t)),
                                                      2                                       (112)
                                x+β
where we have set t = β/x > 0. Since it holds that
                                                     t
                                    ln(1 + t) ≥          ,   t ≥ 0,                           (113)
                                                  1 + 1t
                                                       2

(equality at t = 0) and thus
                                     1        1          t   ′
                   (ln(1 + t))′ =       ≥       1 2 =      1   ,                 t ≥ 0,       (114)
                                    1+t   (1 + 2 t)   1 + 2t

the proof is complete.


Lemma 8. The following inequalities hold:

  (i) For α > 0,
                                          UB        π2
                                             ≤ exp      .                                     (115)
                                          LB       144s

                                                   22
√
 (ii) For 0 < α < 2 π,
                                                                                  ∞
                  UB        1 π2 1 2   1                                               ζ(3/2 − r)(−1/2)r 2r+1
                     ≤ exp       + α −√                                                                 α                                 .     (116)
                  LB       12s 12 2     2π                                    r=0
                                                                                           r!(2r + 1)

Proof. (i) Follows from (110) and observing that
                       ∞                 ∞                                        ∞
                             1 1               − 1 x2          1                      1    1      1 2
                               √         √    e  2        dx ≤                           = ζ(2) =    π .                                        (117)
                             l2 2π      α l                    2                      l2   2      12
                       l=1                                                    l=1

(ii) Follows from rewriting the right-hand side of (110) in terms of Lerch’s transcendent and
applying the Bateman series (61).


Lemma 9. There are the inequalities
                                         ∞                        ∞
                                              1 1                             1   2
               LB ≥ exp             −          √                √     e− 2 x dx ,                                                               (118)
                                              l 2π             α l
                                        l=1
                                         ∞                        ∞                                   ∞
                                              1 1                      − 1 x2               2                  1            1    2l
               LB ≤ exp             −          √                √     e  2            dx + √                        e− 2 α                .     (119)
                                              l 2π             α l                        3 2πs            l3/2
                                        l=1                                                          l=1

Proof. Follows from (106) and 1 − 2 x ≤ y ′ (x) ≤ 1 (see Lemma 15).
                                  3

 Note that the right-hand side of (118) equals P(Mα = 0), i.e., the probability that the
maximum of a Gaussian random walk with drift −α (instead of −β) equals zero.

Lemma 10.
                                     ∞                        ∞                               ∞                     ∞
                                           1 1                         1      2          1          1 1                          1 2
              U B ≤ exp         −           √              √      e− 2 x dx +                         √             √       e− 2 x dx
                                           l 2π           α l                            12         l2 2π          α l
                                     l=1                                                      l=1
                                                    ∞                                                ∞
                                    2                         1           1   2           1                1            1       2l
                                 + √                          3/2
                                                                  e− 2 α l −              √                        e− 2 α             .         (120)
                                  3 2πs             l=1
                                                          l                            18s 2πs      l=1
                                                                                                          l3/2

Proof. Follows from (107) and 1 − 2 x ≤ y ′ (x) (see Lemma 15).
                                  3

The right-hand side of (120) can be written as
              ∞              √                ∞              √
                                    1                                 1 2      1 2                                       12s − 1
  exp    −          l−1 P (−α l) +                  l−2 P (−α l) + e− 2 α Φ(e− 2 α , 3 , 1)
                                                                                     2
                                                                                                                           √                  . (121)
                                   12                                                                                   18s 2πs
              l=1                             l=1
                                                                             √
Using Lemma 4 for k = −1, −2, and (61), we find that (121) satisfies, for α < 2 π,

 √             π2            12s − 1     ζ( 1 ) ζ( 3 ) √  12s − 1
     2α exp        + ζ( 3 )
                        2
                               √     + α √ 2 − √ − 2π
                                                   2        √                                                                    + O(α2 ) , (122)
               144          18s 2πs        2π 12 2π      18s 2πs
which yields a sharp approximation for (121) for small values of α.


                                                                      23
Considering the leading component in the exponents of (119) and (120), it makes sense to
use the approximation
                                      ∞              ∞                                 ∞
                                            1 1            − 1 x2         2                   1        1   2l
          P(Qλ = 0) ≈ exp       −            √       √    e  2      dx + √                          e− 2 α
                                      l=1
                                            l 2π    α l                 3 2πs          l=1
                                                                                             l3/2
                                                              ∞
                                                     2               1        1   2l
                      = P(Mα = 0) · exp             √                      e− 2 α
                                                   3 2πs      l=1
                                                                    l3/2
                                          2      1 2      1 2
                      = P(Mα = 0) · exp √     e− 2 α Φ(e− 2 α , 3 , 1) .
                                                                2                                               (123)
                                        3 2πs
We note the approximations (105) and (123) are strongly related. In fact, approximation
(123) follows from (105) by approximating y ′ (x) by 1 − 2 x.
                                                   √     3          √
  From (1), (61) and (123) we then get, for β < 2 π (and hence α < 2 π),

             P(Qλ = 0)        P(Mα = 0)            2      1 2      1 2
                          =              · exp √       e− 2 α Φ(e− 2 α , 3 , 1)
                                                                         2
             P(Mβ = 0)        P(Mβ = 0)        3 2πs
                                             ∞
                              α        1       ζ( 1 − r)(− 1 )r 2r+1
                          ≈     · exp √           2         2
                                                                α       − β 2r+1
                              β         2π r=0    r!(2r + 1)
                                                                     ∞
                                          2    √          ζ( 3 − r)(− 1 )r 2r
                                                             2        2
                              · exp      √    − 2πα +                     α                                .    (124)
                                        3 2πs                    r!
                                                      r=0

For small values of α we then propose the approximation

                  P(Qλ = 0)  α      ζ( 1 )        2 ζ( 3 ) 1
                            ≈ · exp √ 2 (α − β) +   √2 −α √ .                                                   (125)
                  P(Mβ = 0)  β        2π          3   2π    s

5.1    Numerical experiments
We now evaluate the approximations and bounds derived in this section. True values of
P(Qλ = 0) are computed from (15). Some numerical results are displayed in Tables 4 and 5.
In Table 4, the values of P(Mβ = 0) are given in between brackets.
   Both the lower bound P(Mα = 0) and the upper bound (120) may serve as conservative
approximations, but the quality of approximation (123) is striking for small and moderate
values of β. For large values, it seems that the limiting value is reached quite rapidly. In par-
ticular, for the case β = 2, the Halfin-Whitt limit P(Mβ = 0) is reached almost immediately,
which suggests to use P(Mβ = 0) as an approximation for P(Qλ = 0). Indeed, as such, the
Halfin-Whitt limit outperforms all other approximations, including the series expansions in
Table 3. For all other values of β (especially small values), all alternative approximations of
the emptiness probability are closer to the true values than the Halfin-Whitt limit.


6     Conclusions and outlook
The approach in this paper consists of three major steps:

    1. Quasi-Gaussian form. Section 3 is the bridge that permits us to pass from the general
       form of Spitzer’s formula (14) to the more convenient quasi-Gaussian form (22). The

                                                    24
β = 0.01 (0.0141)                                 β = 0.1 (0.1334)
   s      α     (118)     true      (123)     (120)      α       (118)      true     (123)     (120)
   1   0.0100   0.0140   0.0268    0.0279    0.0282   0.0983     0.1313    0.2351 0.2468      0.2488
   2   0.0100   0.0140   0.0225    0.0228    0.0240   0.0988     0.1319    0.2022 0.2061      0.2149
   5   0.0100   0.0140   0.0190 0.01910      0.0203   0.0993     0.1325    0.1747 0.1756      0.1857
  10   0.0100   0.0141   0.0174    0.0175    0.0187   0.0995     0.1328    0.1617 0.1620      0.1718
  20   0.0100   0.0141   0.0164    0.0164    0.0175   0.0996     0.1329    0.1529 0.1531      0.1625
  50   0.0100   0.0141   0.0155    0.0155    0.0166   0.0998     0.1331    0.1455 0.1455      0.1546
 100   0.0100   0.0141   0.0151    0.0151    0.0161   0.0998     0.1332    0.1419 0.1419      0.1507
 200   0.0100   0.0141   0.0148    0.0148    0.0158   0.0999     0.1333    0.1393 0.1393      0.148
 500   0.0100   0.0141   0.0145    0.0145    0.0155   0.0999     0.1333    0.1371 0.1371      0.1457
                         β = 0.2 (0.2518)                                  β = 0.5 (0.5293)
   s      α     (118)     true      (123)     (120)      α       (118)      true     (123)     (120)
   1   0.1932   0.2442   0.4105    0.4332    0.4356   0.4573     0.4961    0.7182 0.7628      0.7635
   2   0.1952   0.2464   0.3613    0.3693    0.3829   0.4699     0.5061    0.6656 0.6831      0.6994
   5   0.1970   0.2484   0.3185    0.3207    0.3367   0.4811     0.5148    0.6156 0.6208      0.6413
  10   0.1979   0.2494   0.2979    0.2987    0.3145   0.4867     0.5192    0.5899 0.5921      0.6127
  20   0.1985   0.2501   0.2838    0.2841    0.2994   0.4906     0.5222    0.5719 0.5728      0.5930
  50   0.1991   0.2507   0.2716    0.2717    0.2865   0.4941     0.5248    0.5560 0.5563      0.5760
 100   0.1993   0.2510   0.2657    0.2657    0.2801   0.4958     0.5262    0.5481 0.5483      0.5676
 200   0.1995   0.2512   0.2615    0.2616    0.2757   0.4970     0.5271    0.5426 0.5426      0.5618
 500   0.1997   0.2514   0.2579    0.2579    0.2719   0.4981     0.5279    0.5377 0.5377      0.5566
                          β = 1 (0.8005)                                    β = 2 (0.9762)
   s      α     (118)     true      (123)     (120)      α       (118)      true     (123)     (120)
   1   0.8299   0.7294   0.9055    0.9574    0.9554   1.3670     0.9019    0.9835 1.0202      1.0177
   2   0.8790   0.7518   0.8787    0.9002    0.9106   1.5296     0.9302    0.9799 0.9938      0.9966
   5   0.9236   0.7709   0.8511    0.8581    0.8718   1.6948     0.9514    0.9774 0.9814      0.9847
  10   0.9462   0.7800   0.8364    0.8395    0.8533   1.7835     0.9603    0.9766 0.9782      0.9812
  20   0.9622   0.7863   0.8259    0.8273    0.8408   1.8473     0.9657    0.9763 0.9770      0.9797
  50   0.9762   0.7917   0.8165    0.8170    0.8302   1.9040     0.9700    0.9762 0.9764      0.9788
 100   0.9832   0.7943   0.8118    0.8120    0.8250   1.9324     0.9720    0.9762 0.9763      0.9785
 200   0.9881   0.7962   0.8085    0.8086    0.8214   1.9524     0.9733    0.9762 0.9762      0.9784
 500   0.9925   0.7978   0.8056    0.8056    0.8183   1.9700     0.9745    0.9762 0.9762      0.9783


           Table 4: Bounds, approximations and true values of P(Qλ = 0).

  key facilitator is, see (40),
                                          √             √
                                                      α/ s
                                      p(s) s                      1   2
                           P(Aλ ≤ s) = √                       e− 2 sx y ′ (x)dx,                      (126)
                                         2π           −∞

  where, as before, Aλ denotes a Poisson random variable with mean λ.
2. Analysis of the function y. Appendix A contains a detailed study of the function y
   (and y ′ ). It is shown that there is a strong connection with the Gamma function,
   the reciprocal Gamma function and the Lambert W function. For this paper, the
   most important result is the power series for y, see (147), which is shown to have
   rapidly decaying coefficients. The latter property makes the quasi-Gaussian form (126)
   extremely effective, because only the first few terms of y ′ have to be taken into account.
3. Principal series expansions. Starting from the quasi-Gaussian form, we derive in Sec-
   tion 4 principal series expansions for P(Qλ = 0). The expansions comprise elementary
   functions and are used to derive asymptotic expressions and bounds for P(Qλ = 0). The
   power series for y leads to even more explicit expansions.

                                               25
β = 0.01                                           β = 0.1
     s    true    (124)    lb(16)    ub(16)        (21)     true    (124)    lb(16)    ub(16)    (21)
     1   1.9047   1.9870   0.1224    8.1692       1.1868   1.7622   1.8500   0.1111    8.9980   1.1608
     2   1.5993   1.6251   0.2270    4.4062       1.1285   1.5154   1.5447   0.2163    4.6226   1.1095
     5   1.3537   1.3595   0.3919    2.5516       1.0794   1.3092   1.3164   0.3845    2.6007   1.0670
    10   1.2406   1.2426   0.5159    1.9385       1.0554   1.2119   1.2145   0.5110    1.9570   1.0466
    20   1.1653   1.1661   0.6264    1.5965       1.0389   1.1463   1.1473   0.6234    1.6041   1.0326
    50   1.1018   1.1021   0.7440    1.3442       1.0244   1.0905   1.0908   0.7426    1.3467   1.0204
   100   1.0710   1.0712   0.8113    1.2326       1.0172   1.0633   1.0634   0.8106    1.2337   1.0144
   200   1.0497   1.0499   0.8626    1.1593       1.0121   1.0444   1.0444   0.8622    1.1599   1.0101
   500   1.0312   1.0313   0.9108    1.0980       1.0077   1.0279   1.0279   0.9106    1.0982   1.0064
                                β = 0.2                                           β = 0.5
     s    true    (124)    lb(16)    ub(16)        (21)     true    (124)    lb(16)    ub(16)    (21)
     1   1.6307   1.7207   0.0993 10.0675         1.1328   1.3568   1.4411   0.0688    14.537   1.0565
     2   1.4351   1.4669   0.2047    4.8840       1.0894   1.2575   1.2905   0.1717    5.8237   1.0368
     5   1.2652   1.2738   0.3763    2.6574       1.0542   1.1629   1.1729   0.3517    2.8433   1.0217
    10   1.1831   1.1865   0.5055    1.9781       1.0375   1.1145   1.1187   0.4891    2.0446   1.0148
    20   1.1271   1.1284   0.6201    1.6127       1.0261   1.0803   1.0822   0.6101    1.6391   1.0102
    50   1.0789   1.0794   0.7410    1.3495       1.0163   1.0504   1.0511   0.7362    1.3582   1.0063
   100   1.0553   1.0555   0.8097    1.2350       1.0115   1.0355   1.0358   0.8071    1.2390   1.0044
   200   1.0388   1.0389   0.8617    1.1605       1.0081   1.0250   1.0251   0.8604    1.1623   1.0031
   500   1.0244   1.0244   0.9104    1.0984       1.0051   1.0158   1.0158   0.9098    1.0991   1.0019
                                 β=1                                                β=2
     s    true    (124)    lb(16)    ub(16)        (21)     true    (124)    lb(16)    ub(16)    (21)
     1   1.1311   1.1960   0.0340 29.4157         0.9645   1.0074   1.0450   0.0064 155.3166    0.9292
     2   1.0976   1.1245   0.1237    8.0841       0.9779   1.0037   1.0179   0.0576   17.3720   0.9593
     5   1.0632   1.0719   0.3114    3.2114       0.9876   1.0011   1.0052   0.2365    4.2284   0.9792
    10   1.0448   1.0487   0.4614    2.1675       0.9918   1.0003   1.0020   0.4057    2.4649   0.9870
    20   1.0316   1.0334   0.5931    1.6862       0.9944   1.0000   1.0007   0.5580    1.7920   0.9915
    50   1.0199   1.0206   0.7282    1.3733       0.9966   1.0000   1.0001   0.7116    1.4054   0.9951
   100   1.0141   1.0144   0.8028    1.2457       0.9977   1.0000   1.0000   0.7938    1.2598   0.9966
   200   1.0099   1.0101   0.8580    1.1654       0.9984   1.0000   1.0000   0.8533    1.1719   0.9977
   500   1.0063   1.0063   0.9088    1.1003       0.9990   1.0000   1.0000   0.9069    1.1027   0.9986


         Table 5: Bounds, approximations and true values of P(Qλ = 0)/P(Mβ = 0).


  Steps 1 and 2 are very case-specific, requiring properties of the Poisson distribution and the
associated function y ′ . Step 3 is far more general. The expansions presented hold for a large
class of functions z, of which z = y ′ (Poisson case) and z ≡ 1 (Gaussian random walk) are
just special cases. We now briefly indicate some ideas that enable the extension of the first
two steps. A more general version of our model is
                                              d
                                      Qλ = (Qλ + A∗ − s)+ .
                                                  λ

If we can write the random variable A∗ as the sum of s i.i.d. random variables, which each
                                        λ
converge weakly to a random variable with unit mean as λ → ∞, then under some addi-
tional assumptions (for example a Lindeberg-type condition) one has still weak convergence
               √
of (A∗ − s)/ s to a Gaussian random variable with negative mean. To write the associated
      λ
performance measure, such as the emptiness probability, in a convenient form, our method
of attack would still work if the cumulative distribution function P(A∗ ≤ s) is of type (126)
                                                                      λ
with y ′ (x) replaced by some function that depends on A∗ . Temme [35] describes a large class
                                                         λ
of distributions that can be brought in this shape.


                                                      26
A     Analysis of the function y
We shall present an analysis to obtain some results on the function y(x) that appeared in
Subsection 2.2, especially in Eq. (22). As before, y(x) is, for |x| sufficiently small, the solution
y of the equation
                               f (y) := −y − ln(1 − y) = 1 x2  2                             (127)
that is real and positive when x is real, positive and sufficiently small. We note that for
|y| < 1,
                              f (y) = 1 y 2 + 1 y 3 + 1 y 4 + . . . ,
                                      2       3       4                             (128)
whence we can write Eq. (127) for small x and y as

                                  y(1 + 2 y + 1 y 2 + . . .)1/2 = x
                                        3     2                                                     (129)

with the principal value of the square root. Hence,

                                       y(x) = x − 1 x2 + O(x3 ).
                                                  3                                                 (130)

From (127) we furthermore see that y(x) increases from 0 to ∞ as x increases from 0 to 1,
and that y(x) increases from −∞ to 0 as x increases from −∞ to 0. The function y(x) occurs
in connection with the asymptotic expansions of Γ(t + 1) and 1/Γ(t) as t → ∞. Thus one
substitutes in the integral representation
                                                              ∞
                                       Γ(t + 1) =                 e−u ut du                         (131)
                                                          0

subsequently u = t(1 + v), v = −y and y = y(x) to obtain
                                                           ∞
                            Γ(t + 1) = e−t tt+1                   et(−v+ln(1+v)) dv =
                                                          −1
                                  1                                       ∞        1   2
                       e−t tt+1        e−tf (y) dy = e−t tt+1                  e− 2 tx y ′ (x)dx.   (132)
                                  −∞                                     −∞

We refer to [6], Section 4.5 on pp. 69-71, and to [28], Ch. 3, §8 on pp. 85-86. By (129), we
have that y(x) has a power series
                                                          ∞
                                           y(x) =              an xn ,                              (133)
                                                       n=1
converging for |x| sufficiently small. Then by Watson’s lemma, see e.g. [28], pp. 112-116, the
asymptotics of Γ(t + 1) follows from (132) as
                                           ∞                  ∞            1   2
                                  −t t+1
                 Γ(t + 1) ∼ e t                  nan               xn−1 e− 2 tx dx
                                           n=1            −∞
                                                  ∞
                                      √                   (2k + 1)(2k − 1) · . . . · 1
                            = e−t tt+1 2π                                              a2k+1 ,      (134)
                                                                     tk
                                                 k=0

as t → ∞. Note that the terms with odd n in the first series in (134) vanish. In a similar
fashion the asymptotics of 1/Γ(t) can be obtained from the Hankel representation
                                       1      i
                                           =              (−u)−t e−u du,                            (135)
                                      Γ(t)   2π       C


                                                      27
where C is a contour that starts at ∞ + i0, encircles the origin in positive sense and ends at
∞ − i0. This is carried out in [36], Subsection 3.6.3 on pp. 69-70. Thus by substitutions as
in (132) we now get
                                1       i    1
                                    =      −t tt−1
                                                       etf (y) dy,                       (136)
                              Γ(t)     2π e        1+C
where it is observed that Ref (y) → −∞ as y → ∞ ± i0. Now the integration path 1 + C
is deformed so as to pass through the saddle point y = 0 where the real axis is crossed
perpendicularly from the upper half plane into the lower half plane. Around the saddle point
we make the substitution f (y) = − 1 x2 = 1 (ix)2 where x ∈ R decreases from +δ to −δ, with
                                   2      2
δ > 0 sufficiently small, and we get, as t → ∞,
                                             δ
                1           1    1                     1   2
                      ∼        −t tt−1
                                                 e− 2 tx y ′ (ix)dx
               Γ(t)        2π e             −δ
                                            ∞                      δ
                            1    1                                             1   2
                      ∼                          nan in−1              xn−1 e− 2 tx dx
                           2π e−t tt−1   n=1                      −δ
                                                 ∞
                                  1                              (2k + 1)(2k − 1) · . . . · 1
                      ∼                 √              (−1)k                                  a2k+1 .   (137)
                           e−t tt−1/2       2π                              tk
                                                 k=0

In particular, we have that the asymptotics of the p(n) of (25) is given by
                       √          ∞
               nn−1 e−n 2πn               (2k + 1)(2k − 1) · . . . · 1
     p(n) =                   ∼     (−1)k                              a2k+1 , n → ∞,                   (138)
                    Γ(n)                             nk
                                      k=0

so that the pk ’s in (25) are given as

                  pk = (−1)k (2k + 1)(2k − 1) · . . . · 1 · a2k+1 ,                k = 0, 1, . . . .    (139)

We shall now present and establish some results for the coefficients an of the power series
 ∞       n
 n=1 an x of y(x). In [38], p. 16, there is given the result
                                                       ∞
                              1        et                      ck Γ(k + 3/2)
                                   ∼ t+1/2                                   ,     t → ∞,               (140)
                          Γ(t + 1)  πt                               tk
                                                     k=0

with ck the coefficients of a function φ(τ ) that is regular inside the circle |τ | = 2π. Apparently,
                                                                         √
since tΓ(t) = Γ(t + 1) and Γ(k + 3/2) = (k + 1/2)(k − 1/2) · . . . · 1 · π, we have that
                                                                      2

                            a2k+1 = (−1)k 2−k−1/2 ck ,                  k = 0, 1, . . . ,               (141)
                             1
so that a2k+1 decays like ( 4π + ǫ)k for any ǫ > 0. However, with the asymmetric integration
ranges (unlike those in (132) and (137)) that occur in the integrals at the right-hand side of
(22), the decay behaviour of the coefficients a2k is equally important. We shall thus conduct
an analysis for y(x) that is somewhat similar to the analysis in [38], Secs. 7-9 on pp. 9-15,
for the function φ(τ ) = ∞ ck τ k . This analysis is greatly facilitated by the study of the
                             k=0
properties of the mapping
                                     z ∈ C → e1−z z = w                                  (142)
as given in Szeg¨ [33]. Note that exp(f (y)) = (1 − y)ey so that many of the observations
                 o
of Szeg¨ directly apply to our function f . The mapping given in (142) is clearly related to
       o

                                                           28
the Lambert W function, which is the inverse of the mapping W ∈ C → W eW , see [8, 18].
Explicitly, our y(x) and Lambert’s W function are related according to
                                                          1 2
                                  y(x) = 1 + W (−e−(1+ 2 x ) ).                             (143)

It appears that Szeg¨’s analysis of the mapping in (142) has largely escaped the attention of
                    o
the Lambert W community. In the proof of the lemma below, we heavily rely on this analysis
and omit some of the details that are contained implicitly or explicitly in [33], §2.
Lemma 11. The functions y(x) extends to an analytic function on all of C with the exception
                         √       1              1
of two branch cuts from 2 π · e± 4 πi to ∞ · e± 4 πi .
Proof. (sketch) Our approach is to take any angle θ with −π ≤ θ ≤ π, and to see how far
y(x) admits analytic continuation when x moves in the direction eiθ from 0. Clearly, from
(130), y(x) starts moving from 0 in the direction eiθ when x does so. We thus plot curves Cθ
in the y-plane that start at the origin and leave from there under an angle θ ∈ [−π, π] with
the positive real axis and on which arg[−y − ln(1 − y)] = 2θ. Writing y = 1 − z and z = ηeiϕ
with η > 0 and real ϕ, there should hold

                    η cos ϕ − 1 − ln η + i(η sin ϕ − ϕ) = r cos 2θ + ir sin 2θ              (144)

with r ≥ 0 for y to lie on Cθ . Consequently, we should have
                                     η sin ϕ − ϕ
                                                     = tan 2θ.                              (145)
                                  η cos ϕ − 1 − ln η
   With reference to Fig. 2, we have that the cases with tan 2θ = 0 give rise to the curves
with θ = 0, ±π/2, ±π. The cases that tan 2θ = ∞ give rise to the four curves with θ =
±π/4, ±3π/4. We note that f ′ (y) = y/(1 − y) so around any y0 = 0, 1, the mapping y → f (y)
is locally invertible as an analytic function. Furthermore, |f (y)| increases along any of the
curves Cθ . Indeed, this is so, by (130), when y is close to 0, and |f (y)| cannot have stationary
points as y = 0, 1 moves along Cθ , for otherwise ln f (y) = ln |f (y)| + 2iθ + 2πin (n some
integer) would have a stationary point, contradicting f ′ (y) = 0. Along the curves we define
ln(1 − y) as the integral of −1/(1 − z) with z moving along Cθ from 0 to y, and thus obtain
a log(1 − y) that is analytic around Cθ . The curves with θ = ±π/4 are particular since they
re-enter the origin (where f ′ = 0), and therefore do not admit an analytic continuation of
                                                                                          √
y(x) beyond there. The values of x at y = 0 in these re-entrance cases are i±1/2 2π for
θ = ±π/4, respectively, since −y − ln(1 − y) has the respective values ±2πi at re-entry to
zero along C±π/4 . The C±π/4 constitute mirrored versions of Szeg¨’s curve, see [33], Fig. 1 on
                                                                      o
p. 51, passed with negative orientation (+π/4) and positive orientation (−π/4), respectively.
In the cases that 0 ≤ |θ| < π/4, we have that Cθ is bounded by the closed curves C±π/4 ,
and that y → 1. The asymptotic behaviour of y is described in these cases in terms of η, ϕ
(see (144),(145)) by the spiral η = exp(−1 + ϕ/ tan 2θ) when ϕ · sgn(θ) → ∞ (observe that
in these cases θ and ϕ have opposite signs). Consequently, when 0 < |θ| < π/4, we have that
|x| = |2f (y)|1/2 → ∞ along the curve Cθ . In the cases that π < |θ| < π, we have that Cθ
                                                                   4
lies completely outside the region enclosed by C±π/4 , and along these curves we have that
|y| → ∞. Therefore | ln(1 − y)/y| → 0 along these curves, and since we have required that
arg[−y − ln(1 − y)] = 2θ, we have that

                       ϕ∞ :=      lim       arg(y) = 2θ − π (mod[−π, π]).                   (146)
                               |y|→∞,y∈Cθ


                                                 29
2
                                      π
                               ϕ∞ =   2                                        ϕ∞ = 0
                         1.5                               π
                                                      θ=   2



                          1

                                                                     π
                                                                θ=   4            π
                         0.5          3π
                                                                             θ=   4
                                                                                      −ε
                                θ=     4



                          0



                               θ = − 3π
                                      4
                        −0.5
                                                               θ = −π
                                                                    4

                                                                                  π
                         −1                                                  θ=   4
                                                                                      +ε

                                                      θ = −π
                                                           2
                        −1.5

                                                                                  ϕ∞ = 0
                               ϕ∞ = − π
                                      2
                         −2
                          −1       −0.5           0        0.5           1    1.5          2




                         Figure 2: Curves Cθ in the y-plane. ε = 0.1.


Also, |x| = |2f (y)|1/2 → ∞ in these cases. We thus conclude that y(x) can be continued
analytically along all rays x = reiθ , r ≥ 0, when θ ∈ [−π, π], except for θ = ±π/4. In the
                                                                          √
latter cases the analytic continuation can only be carried out until r = 2 π.

Lemma 12. We have
                                              ∞
                                                                         √
                                  y(x) =              an xn ,     |x| < 2 π,                              (147)
                                             n=1

with a1 = 1 and the an ’s recursively defined as
                                              k
                         −1
               ak+2 =       ak+1 +    (n + 1)an+1 ak+2−n ,                            k = 0, 1, . . . .   (148)
                        k+3        n=1
                                                                             √
Proof. From Lemma 11 we have that y(x) is analytic in the disk |x| < 2 π, with branch
               √
points at x = 2 π · i±1/2 . This gives (147). From (127) we get by differentiation with respect
to x and some rewriting the equation

                                          y ′ (x)y(x) = x − xy(x).                                        (149)




                                                         30
Using a1 = 1, this can be written in terms of the an ’s as
                         k−1
                               (n + 1)an+1 ak−n = −ak−1 ,          k = 2, 3, . . . ,        (150)
                         n=0

and this gives (148). Also see [28], Ex. 8.3 on p. 88 and [36], p. 70.

  The first five coefficients an are given by
                                     1        1          1           1
                     a1 = 1, a2 = − , a3 = , a4 =           , a5 =      ,                   (151)
                                     3        36        270        4320
and this rapid decay is further substantiated by the following result.
Lemma 13. There is the asymptotic form (rapid decay to 0)
                             sin π (n − 1) −
                                 4
                                                2π     π
                                              6n−9 cos 4 (n     − 1)
                      an =                        √ n                  1 + O(1/n) .         (152)
                                      ( 1 n)3/2 (2 π)
                                        2
Proof. The precise asymptotic behavior of the an can be determined by Darboux’s method,
                                                                                          √
see Szeg¨ [34], Section 8.4 on pp. 206-208. For this we examine y(e±πi/4 v) with 0 ≤ v < 2 π
        o
as the solution y near 0 and on C±π/4 of the equation
                    2πi − y − ln(1 − y) = 2πi + 1 y 2 + 1 y 3 + . . . =
                                                2       3
                                                                                   1 2
                                                                                   2 iv .   (153)
Thus                                                       √                       √
                   y(eπi/4 v) = e−πi/4       4π − v 2 + O(2 π − v),           v ↑ 2 π,      (154)
and similarly,
                                                           √                       √
                   y(e−πi/4 v) = eπi/4       4π − v 2 + O(2 π − v),           v ↑ 2 π.      (155)
When we use Darboux’s method in a first order form, we get the leading asymptotics of an
as
                               sin π (n − 1)
                                   4
                        an = 1 3/2 √ n + lower order.                             (156)
                             ( 2 n) (2 π)
Note that the leading term at the right-hand side of (156) vanishes when n = 4k + 1, k =
0, 1, . . . , and so more precise information is required. This can be obtained when we write
                   −y(eπi/4 v) − ln(1 − y(eπi/4 v)) = 1 (e−πi/4 (4π − v 2 )1/2 )2 ,
                                                      2                                     (157)
                        ∞       n
so that from y(x) =     n=1 an x    we get
                                             ∞
                             y(eπi/4 v) =         an (e−πi/4 (4π − v 2 )1/2 )n ,            (158)
                                            n=1

and a similar expression for    y(e−πi/4 v).
                                           We are now in a position to use the full strength of
Darboux’s theorem, see Thm 8.4 in [34] (noting that it should be −e−iϕk instead of −eiϕk at
the right-hand side of (8.4.8)). For instance, from the fact that a2 = −1/3, we get (152). The
particular form 2π/(6n − 9) of the coefficient in front of cos π (n − 1) arises when the binomials
                                                              4
occurring in Darboux’s results are manipulated; also it has been used that (−1)n 1/2 is well
                                                                                       n
approximated as −1/((2n − 1) π(n + 1/2)). Note that the leading factor in (152) is non-
vanishing for any n = 1, 2, . . . .


  The accuracy of the approximation (152) is demonstrated by Fig. 3.

                                                     31
4.5

                       4

                    3.5

                       3

                    2.5

                       2

                    1.5

                       1

                    0.5

                       0

                   −0.5
                           1       5            10         15           20         25          30       35      40




Figure 3: The ratio of an and the approximation in (152), minus 1, for the first 40 coefficients.

Lemma 14. It holds that
                                                   ∞
                                                         mm−1 − 1 mx2
                           y(x) = 1 −                         e 2     ,                 |arg(x)| ≤ π/4.                (159)
                                                         m!em
                                                m=1

Proof. Letting z = 1 − y and u = exp(−1 − 1 x2 ), we can write −y − ln(1 − y) = 1 x2 as
                                          2                                     2

                                                              ze−z = u,                                                (160)

so that by the B¨rmann-Lagrange inversion formula
                u
                            ∞                           m−1                                         ∞
                                   1           d                  z          m
                                                                                           m                 mm−1 m
               z(u) =                                                                   u      =                 u .   (161)
                                   m!          dz               ze−z             z=0                          m!
                           m=1                                                                      m=1

The right-hand side of (161) is analytic in |u| < 1/e and continuous in |u| ≤ 1/e. This leads
to the formula in (159), with x restricted to |arg(x)| ≤ π/4.

  An interesting consequence of (159) is that
                       ∞                                            ∞                  ∞
                               1   2                                         1     2         mm−1           1   2
                           e− 2 lsx y ′ (x)dx =                         e− 2 lsx                m
                                                                                                  · mx · e− 2 mx dx
                   a                                            a                            m!e
                                                                                       m=1
                   ∞                       ∞                                           ∞            1          2
                           mm                       1           2                           mm e− 2 (m+ls)a
              =                                e− 2 (m+ls)x xdx =                                           .          (162)
                           m!em        a                                                    m!em (m + ls)
                  m=1                                                              m=1

This series expansion can be then be inserted into (53) to yield
                                               ∞                                 ∞
                                                     mm − 1 m(a2 +2)                    lk+1/2 − 1 lsa2
                           Gk (a) =                     e 2                                    e 2      .              (163)
                                                     m!                                 m + ls
                                           m=1                                   l=1


                                                                    32
For somewhat larger values of a, this seems to offer an attractive alternative to evaluate the
Gk (a) numerically. Furthermore, (159) can be used to establish a formal connection between
the function y and Lerch’s transcendent. Indeed, we have

                kk−1    p(k)    1     1       1          1
                     = √     ∼ √          −         +         + ... .                                   (164)
                k!ek   k 2πk     2π k 3/2   12k 5/2   288k7/2
Inserting this asymptotic expansion into (159) and using the definition (59) of Lerch’s tran-
scendent, we find a formal identity between 1 − y(x) and
               1 2
            e− 2 x             1   2            1      1 2             1       1 2
             √           Φ(e− 2 x , 3 , 1) −
                                    2             Φ(e− 2 x , 5 , 1) +
                                                             2            Φ(e− 2 x , 7 , 1) + . . . .
                                                                                     2                  (165)
               2π                              12                     288
                                                                                         √
Observe that each of the Φ-functions in (165) has a convergent power series in x, |x| < 2 π,
per formula (61).
  We conclude our analysis of y with the following result.
Lemma 15. There is the inequality

                                         1 − 2 x ≤ y ′ (x) ≤ 1,
                                             3                      x ≥ 0.                              (166)

Proof. Solve f (y) = −y − ln(1 − y) = 1 x2 for the unique solution y(x) ∈ [0, 1). From the
                                      2
inequalities, to be proved below,
             x                x     1        x            x
      −           − ln 1 −         ≥ x2 ≥ −     − ln 1 −     ,                               x ≥ 0,     (167)
          1 + x/3          1 + x/3  2       1+x          1+x
and monotonicity of f we infer that
                                          x              x
                                               ≥ y(x) ≥     ,           x ≥ 0.                          (168)
                                       1 + x/3          1+x
From (149) we get
                                                       1
                                        y ′ (x) = x        −1 ,       x ≥ 0.                            (169)
                                                      y(x)
Inserting the inequalities (168) into (169), we find
                2    1 + x/3                   1+x
              1− x=x         − 1 ≤ y ′ (x) ≤ x     − 1 = 1,                               x ≥ 0,        (170)
                3       x                       x
as required. We now prove the two inequalities in (167). As to the second one, we need to
show that
                                                    1     1
               g2 (x) := −x + (1 + x) ln(1 + x) ≤ x2 + x3 =: h2 (x), x ≥ 0.         (171)
                                                    2     2
Noting that g2 (0) = h2 (0) = 0, g2 (0) = h′ (0) = 0, and
                                  ′
                                           2

                                ′′          1       3
                               g2 (x) =        ≤ 1 + x = h′′ (x),
                                                          2                  x ≥ 0,                     (172)
                                           1+x      2
we get (171). Note that the second inequality in (167) is the sharpest in its kind: when α > 0,
the inequality y(x) ≥ x/(1 + (1 − α)x) fails to hold for large x ≥ 0, since this would require
                               x              1 − αx     1
                     −                − ln              ≤ x2 ,                   0 ≤ x < ∞,             (173)
                         1 + (1 − α)x      1 + (1 − α)x  2

                                                         33
to hold. We next show the first inequality in (167). To that end we must prove that
                                                       1    1
 g1 (x) := −x − (1 + x/3)(ln(1 − 2x/3) − ln(1 + x/3)) ≥ x2 + x3 =: h1 (x),                    x ≥ 0. (174)
                                                       2    6
It holds that g1 (0) = h1 (0) = 0, g1 (0) = h′ (0) = 0, g1 (0) = h′′ (0) = 1, while
                                    ′
                                             1
                                                         ′′
                                                                  1

          ′′′           4/27         1/27          8/9
         g1 (x) =               −            +             ≥ 1 = h′′′ (x),
                                                                  3                  x ≥ 0,          (175)
                    (1 − 2x/3)2   (1 + x/3)2   (1 − 2x/3)3
again with equality at x = 0. This proves the lemma.

References
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 [2] Asmussen, S. (2003). Applied Probability and Queues (second edition). Springer-Verlag, New York.
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[20] Jelenkovi´, P., A. Mandelbaum, P. Momˇilovic (2004). Heavy traffic limits for queues with many deter-
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                                                     35

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Corrected asymptotics for a multi-server queue in the Halfin-Whitt regime

  • 1. Corrected asymptotics for a multi-server queue in the Halfin-Whitt regime A.J.E.M. Janssen∗ J.S.H. van Leeuwaarden⋄ B. Zwart♭ First version: May 12, 2007 Second version: October 8, 2007 Abstract To obtain insight in the quality of heavy-traffic approximations for queues with many servers, we consider the steady-state number of waiting customers in an M/D/s queue as s → ∞. In the Halfin-Whitt regime, it is well known that this random variable con- verges to the supremum of a Gaussian random walk. This paper develops methods that yield more accurate results in terms of series expansions and inequalities for the proba- bility of an empty queue, and the mean and variance of the queue length distribution. This quantifies the relationship between the limiting system and the queue with a small or moderate number of servers. The main idea is to view the M/D/s queue through the prism of the Gaussian random walk: as for the standard Gaussian random walk, we provide scalable series expansions involving terms that include the Riemann zeta function. Keywords: M/D/s queue; Halfin-Whitt scaling; Gaussian random walk; all-time maxi- mum; Riemann zeta function; Lerch’s transcendent; Spitzer’s identity; queues in heavy traffic; Lambert W Function; corrected diffusion approximation. AMS 2000 Subject Classification: 11M06, 30B40, 60G50, 60G51, 65B15. 1 Introduction Heavy-traffic analysis is a popular tool to analyze stochastic networks, since the analysis of a complicated network often reduces to the analysis of a much simpler (reflected) diffusion, which may be of lower dimension than the original system. This makes the analysis of complex systems tractable, and from a mathematical point of view, these results are appealing since they can be made rigorous. A downside of heavy-traffic analysis is that the results are of an asymptotic nature, and only form an approximation for a finite-sized system. In a pioneering paper, Siegmund [31] proposed a corrected diffusion approximation for the waiting time in a single-server queue ∗ Philips Research. Digital Signal Processing Group, WO-02, 5656 AA Eindhoven, The Netherlands. Email: a.j.e.m.janssen@philips.com. ⋄ Eindhoven University of Technology. Math and Computer Science department, 5612 AZ Eindhoven, The Netherlands. Email: j.s.h.v.leeuwaarden@tue.nl (corresponding author). ♭ Georgia Institute of Technology. H. Milton Stewart School of Industrial and Systems Engineering, 765 Ferst Drive, 30332 Atlanta, USA. Email: bertzwart@gatech.edu. We thank M. Vlasiou for pointing out a connection to the Lambert W function. JvL was supported by the Netherlands Organisation for Scientific Research (NWO). 1
  • 2. (actually, Siegmund formulated his result in terms of a random walk). In heavy traffic, the workload distribution is approximated by an exponential distribution. Siegmund gave a precise estimate of the correction term, nowadays a classic result and textbook material, cf. Asmussen [2], p. 369. Siegmund’s first order correction has been extended recently by Blanchet & Glynn [5], who give a full series expansion for the tail probability of the GI/GI/1 waiting time distribution in heavy traffic. The results in [5, 31] correspond to the conventional heavy-traffic scaling. The present paper considers corrected diffusion approximations for a heavy-traffic scaling known as the Halfin-Whitt [12] regime. This regime considers queues where the number of servers grows large as the system becomes critically loaded. The number of servers s is chosen according √ to s = λ + β λ, where β is some positive constant. As the√ scaling parameter λ tends to infinity, the traffic intensity tends to one according to 1 − O(1/ λ). The Halfin-Whitt regime is also known as the QED (Quality and Efficiency Driven) regime, due to the fact that, in the limit, a system can be highly utilized (efficiency) while the√ waiting times stay relatively small (quality). Also, setting the number of servers as s = λ + β λ is often called square-root staffing. This terminology is motivated by the emergence of large customer contact centers which need to be staffed with agents, thus calling for accurate and scalable approximations of multi-server queues. We refer to Gans et al. [11] and Mandelbaum [26] for overviews. The Halfin-Whitt regime was formally introduced in [12] for a system with exponential service times (G/M/s queue), although in [19] the same type of scaling was already applied to the Erlang loss model (M/M/s/s queue). The extension of the results on the G/M/s queue to non-exponential service times turned out to be challenging. The past few years have witnessed a substantial effort to rise to this challenge, resulting in several papers on the Halfin-Whitt regime for various types of service time distributions, cf. Puhalskii & Reiman [29], Jelenkovi´ et al. [20], Whitt [40], Mandelbaum & Momˇilovic [25], and Reed [30]. c c Although these results offer important qualitative insights and are useful to solve concrete staffing problems, one would like to have a better understanding of the quality of the asymp- totic approximations. For instance, how fast does convergence to the heavy-traffic limit take place? It would be helpful to have asymptotic estimates or even inequalities from which we could judge just how close the scaled queueing model is to its heavy-traffic limit. Borst et al. [4] consider optimal staffing of an M/M/s queue in the Halfin-Whitt regime, and show numerically that optimizing the system based on the Halfin-Whitt approximation (with s infinite instead of finite) of the cost function is rarely off by more than a single agent from systems with as few as 10 servers. As mentioned in the conclusions of [4], these observations call for a theoretical foundation−a task we take up in the present paper. 1.1 Goals, results and insights We now give a general description of the results obtained in this paper. We consider a multi- server queue with arrival rate λ, s servers and deterministic service times (set to 1). We let √ the arrival rate of the system grow large and set s = λ + β λ for some constant β > 0. Our main performance measure is the probability that the queue is empty. The model at hand has been considered before by Jelenkovi´ et al. [20] who showed that the scaled number of c ˆ waiting customers Qλ converges to the maximum Mβ of a Gaussian random walk with drift √ −β, for which the emptiness probability is known. As λ → ∞, for β < 2 π, there is the 2
  • 3. result ∞ √ β ζ(1/2 − r) −β 2 r ˆ P(Qλ = 0) → P(Mβ = 0) = 2β exp √ , (1) 2π r!(2r + 1) 2 r=0 with ζ the Riemann zeta function, see Chang & Peres [7], and Janssen & Van Leeuwaarden [15, 16]. The limiting result for P(Mβ = 0) has the appealing property that the time to compute it does not depend on the number of servers, which is the case for standard computational procedures for the M/D/s queue, see e.g. Tijms [37] and Franx [10] and references therein. The main aim of this paper is to obtain series expansions refining this asymptotic result. These series expansions can be used in two ways. First of all, the series expansions quantify the relationship between the limiting system and the queue with a small or moderate number of servers. In addition, the first term (or first few terms) of these expansions have the correct behavior as the number of servers grows large. One insight we find particularly interesting is that our approximations are not based on the parameter β, but on a modification of it, which depends on s and is given by α(s) = (−2s(1 − ρ + ln ρ))1/2 , (2) with ρ = λ/s. This function converges to β as s → ∞, cf. Lemma 7. Another insight we obtain ˆ is that the resulting approximation P(Mα(s) = 0) is, in fact, a lower bound for P(Qλ = 0). We also obtain an upper bound, again involving the function α(s). The model we consider may seem rather specific, but one should realize that obtaining series expansions and bounds of this type is by no means a trivial task. The state of the art for traditional corrected diffusion approximations does not go beyond the random walk, and relies on the availability of the Wiener-Hopf factorization. In the Halfin-Whitt regime, the limiting no-wait probability has been found in two cases only, namely for exponential service times and for deterministic service times. We believe that the latter case is the most challenging one. We apply the methods developed in this paper to the M/M/s queue in [17], in which case the Halfin-Whitt regime results in a non-degenerate limit for the Erlang C formula (probability that a customer has to wait). There we obtain the same important insight: the Halfin-Whitt approximation can be substantially improved when β is replaced with α(s); this function is the same for both models. We finally like to point out that the results in this paper are all formulated for the special case of Poisson arrivals, but the methodology we develop is applicable to more general models (see Section 6). An additional motivation for considering deterministic service times is that the number of waiting customers in the queue is related to a discrete bulk-service queue, which has proven its relevance in a variety of applications (see [24], Chapter 5, for an overview). 1.2 Methodology We now turn to a discussion and motivation of the techniques we use and develop in this ˆ paper. The ratio of P(Qλ = 0) and P(Mβ = 0) serves as a primary measure of convergence and should tend to one as λ grows large. This ratio can be expressed as (using Spitzer’s identity, cf. (17)) ˆ ∞ √ √ P(Qλ = 0) 1 ˆ = exp (P(Aλl ≤ β l) − P (β l)) , (3) P(Mβ = 0) l l=1 3
  • 4. ˆ where Aλl = (Alλ − lλ)/ lλ and Alλ a Poisson random variable with mean lλ, and x 1 2 /2 P (x) = √ e−u du (4) 2π −∞ the normal distribution function. To estimate (3) one can use Berry-Esseen bounds, but these do not lead to sharp results (cf. Lemma 1). In order to get more precise estimates one can use classical approximations for sums of i.i.d. random variables like saddlepoint approximations √ ˆ or Edgeworth expansions (see [3, 21]). However, these require each quantity P(Aλl ≤ β l) − √ P (β l) to be approximated separately and uniformly in l. ˆ √ get convenient asymptotic expansions, we follow a different approach: we bring P(Aλl ≤ To β l) into quasi-Gaussian form, a method that is standard in asymptotic analysis (for an illuminating discussion see De Bruijn [6], pp. 67-71). The resulting asymptotic expansion for e.g. the probability of an empty queue then contains terms of the type ∞ ∞ 1 2 Gk (a) = lk+1/2 e− 2 lsx z(x)dx, a, s ∈ R+ , k ∈ Z, (5) l=1 a where z(x) is some function that does not depend on l. This approach seems technical at first sight but we believe it to be elegant and even intuitively appealing, as there is a clear interpretation in terms of a change-of-measure argument, see the end of Section 2. A large part of this paper deals with obtaining the quasi-Gaussian form, analyzing z(x), and reformulating and estimating Gk (a) which is done in Section 4. A key result is Theorem 3, which gives a representation of Gk (a) for a large class of functions z(x); the only condition that is imposed on z(x) is that z : [0, ∞) → C is a continuous function satisfying z(x) = O(exp(εx2 )) for any ε > 0 and that z(x) has a Taylor series around zero. To illustrate the generality of our result, we note that Chang & Peres’ result (1) on the Gaussian random walk can be viewed as a special case by taking z(x) ≡ 1. We focus on the case in which Aλ has a Poisson distribution, which ensures a particularly tractable form of z(x) yielding convenient computational schemes. This form is given in Subsection 2.2 and studied in detail in Appendix A. The derivative of z(x) is related to the Lambert W function; our treatment is self-contained, produces some important auxiliary results, and is based on earlier results obtained by Szeg¨ [33]. We include our analysis in a o separate appendix, since we believe it is interesting in its own right. Theorem 3 yields a series expansion which can be truncated at a convenient point to obtain high precision estimates of performance measures. Using classical methods, we can even estimate the optimal truncation point of the series expansion. We illustrate these general ideas by specializing them to the M/D/s queue in Subsection 4.3. 1.3 Organization This paper is organized as follows. In Section 2 we introduce our model and provide short proofs of results which can also be found in [20]. In particular we establish convergence of the number of waiting customers to the maximum of the Gaussian random walk, and give a rough Berry-Esseen bound. These results form a point of departure for the rest of the paper. We also explain in Section 2 how our asymptotic analysis will be carried out. In Section 3, for the emptiness probability, and the mean and variance of the queue length distribution, we rewrite the Spitzer-type expressions into quasi-Gaussian expressions. The reformulation 4
  • 5. and estimation of Gk (a) is carried out in Section 4. Section 5 focuses on lower and upper bounds which have the correct asymptotic behavior in the Halfin-Whitt regime. We use the quasi-Gaussian expression for the emptiness probability obtained in Section 3 to derive these bounds. Conclusions and possible extensions are presented in Section 6. 2 The M/D/s queue and the Halfin-Whitt regime We consider the M/D/s queue and keep track of the number of customers waiting in the queue (without those in service) at the end of intervals equal to the constant service time (which we set to one). Customers arrive according to a Poisson process with rate λ and are served by at most s servers. Let Qλ,n denote the number of customers waiting in the queue at the end of interval n. The queueing process is then described by Qλ,n+1 = (Qλ,n + Aλ,n − s)+ , n = 0, 1, . . . . (6) where x+ = max{0, x}, and Aλ,n denotes the number of customers that arrived at the queue during interval n. Obviously, the Aλ,n are i.i.d. for all n, and copies of a Poisson random variable Aλ with mean λ. It should be noted that due to the assumption of constant service times, the customers which are being serviced at the end of the considered interval should start within this interval, and for the same reason, the customers whose service is completed during this interval should start before its beginning. Assume that EAλ,n = λ < s and let Qλ denote the random variable that follows the stationary queue length distribution, i.e., Qλ is the weak limit of Qλ,n . Let √ s = λ + β λ, β > 0. (7) Let {Sn : n ≥ 0} be a random walk with S0 = 0, Sn = X1 + . . . + Xn and X, X1 , X2 , . . . i.i.d. random variables with EX < 0, and let M := max{Sn : n ≥ 0} denote the all-time maximum. When X is normally distributed with mean −β < 0 and variance 1 we speak of the Gaussian random walk and denote its all-time maximum by Mβ . We often use the following notation which is standard in asymptotic analysis: ∞ f (x) ∼ fn (x), n=0 by which we denote that, for every fixed integer k ≥ 1, k−1 f (x) − fn (x) = fk (x)(1 + o(1)). n=0 d Let → denote convergence in distribution. 2.1 Basic results The following theorem can be proved using a similar approach as in Jelenkovi´ et al. [20]. c √ ˆ Theorem 1. Define Qλ = Qλ / λ. As λ → ∞, 5
  • 6. ˆ d (i) Qλ → Mβ ; ˆ (ii) P(Qλ = 0) → P(Mβ = 0); ˆ (iii) E[Qk ] → E[Mβ ] for any k > 0. k λ Proof. Proof of (i): Note that ˆ d ˆ ˆ Qλ = (Qλ + Aλ − β)+ , (8) √ ˆ ˆ with Aλ = (Aλ − λ)/ λ. Since Aλ converges in distribution to the standard normal random ˆ variable as λ → ∞, (i) follows from Theorem X.6.1 in Asmussen [2], if the family (Aλ , λ ≥ 0) ˆ is uniformly integrable. But this follows simply from the fact that E[Aλ2 ] = 1 for all λ. Proof of (ii): The result lim supλ→∞ P(Q ˆ λ = 0) ≤ P(Mβ = 0) follows from (i). To show the lim inf, note that from Spitzer’s identity (see (14)) ∞ ˆ 1 ˆ ln P(Qλ = 0) = − P(Alλ > lβ). (9) l l=1 √ ˆ Taking the lim inf, applying Fatou’s lemma, and using that P(Alλ > lβ) → P (−β l) yields ∞ √ ˆ 1 lim inf ln P(Qλ = 0) ≥ − P (−β l) = ln P(Mβ = 0), (10) λ→∞ l l=1 which proves (ii). Statement (iii) follows from (i) if we can prove the additional uniform ˆ integrability condition supλ>N E[Qk ] < ∞ for some constant N and any k. To prove this, λ ˆ note that the Cram´r-Lundberg-Kingman inequality states that P(Qλ > x) ≤ e−sx , for any e ˆλ −β) s > 0 such that E[es(A ] ≤ 1. After some straightforward computation, this inequality can be rewritten into √ s λ es/ λ − 1 − √ − sβ ≤ 0. (11) λ Since ex − 1 − x ≤ 1 x2 ex , we see that any s is admissible that satisfies 2 s2 s/√λ e − sβ ≤ 0. (12) 2 It is easy to see that s = β satisfies this inequality if λ ≥ N := (β/ ln 2)2 . We conclude that ˆ P(Qλ > x) ≤ e−βx (13) ˆ for any x ≥ 0 and any λ > N . The uniform integrability condition supλ>N E[Qk ] < ∞ now λ ˆ ∞ ˆ follows directly using for example the formula E[Qk ] = 0 kxk−1 P(Qλ > x)dx. λ ˆ As a consequence of Theorem 1 we know that P(Qλ = 0) (which equals P(Qλ = 0)) tends to P(Mβ = 0) as λ tends to infinity. We are interested in how fast the M/D/s queue in the Halfin-Whitt regime approaches the Gaussian random walk, and so we take the ratio of P(Qλ = 0) and P(Mβ = 0) as our measure of convergence. From Spitzer’s identity for random walks (see Theorem 3.1 in [32]) we have ∞ 1 − ln P(M = 0) = P(Sl > 0), (14) l l=1 6
  • 7. which gives for the M/D/s queue ∞ ∞ ∞ 1 1 (lλ)j − ln P(Qλ = 0) = P(Alλ > ls) = e−lλ , (15) l l j! l=1 l=1 j=ls+1 √ where we choose λ such that s = λ+β λ is integer-valued, i.e. λ = 1 (2s+β 2 −(4sβ 2 +β 4 )1/2 ) 2 with s = 1, 2, . . .. For the Gaussian random walk we have ln P(Mβ = 0) as in (10). The following can be proved using a Berry-Esseen bound. Lemma 1. For ω := 4 ζ( 3 ) ≈ 2.0899 there are the bounds 5 2 −ω P(Qλ = 0) ω exp √ ≤ ≤ exp √ . (16) λ P(Mβ = 0) λ Proof. Along the same lines as Theorem 2 in [20]. From (15) and (10) we get ∞ √ P(Qλ = 0) 1 = exp (P (−β l) − P(Alλ > ls)) P(Mβ = 0) l l=1 ∞ √ 1 ≤ exp |P (−β l) − P(Alλ > ls)| . (17) l l=1 Rewriting √ √ √ ˆ |P (−β l) − P(Alλ > ls)| = |P(Aλl ≤ β l) − P (β l)| (18) √ ˆ with Aλl = (Aλl − λl)/ λl and using the Berry-Esseen bound for the Poisson case (see Michel [27]) √ √ 4 30.6 1 4 ˆ |P(Aλl ≤ β l) − P (β l)| ≤ min , √ ≤ √ . (19) 5 1 + β 3 l3/2 lλ 5 lλ yields, upon substituting (19) into (17), the second inequality in (16). The first inequality in (16) follows in a similar way. We should stress that the occurrence of ζ( 3 ) in Lemma 1 is unrelated to the result (1) of 2 Chang & Peres [7]. 2.2 Quasi-Gaussian form: motivation and outline The bound in (16) does not reveal much information, except that convergence takes place at √ rate O(1/ λ). In order to get more precise estimates one can use a saddlepoint approximation or an Edgeworth expansion. √ √ However, these are not very convenient, as they require each ˆ element P(Aλl ≤ β l) − P (β l) to be approximated separately due to its dependence on l. One example would be the Edgeworth expansion for the Poisson distribution (see [3], Eq. (4.18) on p. 96) √ √ 1 1 2 ˆ P(Alλ ≤ β l) = P (β l) − √ e− 2 β l (β 2 l − 1) + O(1/λl), (20) 6 2πlλ which leads to the approximation ∞ √ √ ∞ ∞ 1 ˆ 1 β 2 − 1 β2l 1 1 2l P(Aλl ≤ β l) − P (β l) ≈ − √ e 2 − e− 2 β . (21) l=1 l 6 2πλ l=1 l1/2 l=1 l3/2 7
  • 8. It may not come as a surprise that (21) is not a good approximation because we neglect all O(1/λl) terms in (20). Although including more terms in the Edgeworth expansion is √ ˆ an option, we choose to get more convenient asymptotic expansions for P(Aλl ≤ β l) by bringing it into quasi-Gaussian form. Specifically, we prove the following theorem in Section 3. Theorem 2. ∞ ∞ √ p(ls) 1 2 /2 − ln P(Qλ = 0) = √ √ e−x y ′ (x/ ls)dx, (22) l 2π α l l=1 in which λ λ 1/2 α = − 2s 1 − + ln , (23) s s α → β as λ → ∞, √ p(n) = nn e−n 2πn/n!, (24) √ and y ′ is a function analytic in |x| < 2 π (see Appendix A, (138)). For p there is Stirling’s formula, see Abramowitz-Stegun [1], 6.1.37 on p. 257, ∞ 1 1 pk p(n) ∼ 1 − + + ... = , n → ∞, (25) 12n 288n2 nk k=0 and for y ′ there is the power series representation ∞ 2 1 √ y (x) = 1 − x + x2 + . . . = ′ bi xi , |x| < 2 π. (26) 3 12 i=0 From an aesthetic viewpoint, expression (22) conveys much understanding about the character of the convergence, since we have complete agreement with the Gaussian random walk (10) when we would have λ → ∞. The deviations from the quasi-Gaussian random walk are embodied by p ≡ 1, y ′ ≡ 1 and α ≡ β. From (22) we see that there is the asymptotic expansion ∞ 1 √ − ln P(Qλ = 0) ∼ √ pk s−k+1/2 G−(k+1) (α/ s), (27) 2π k=0 where ∞ ∞ 1 2 Gk (a) = lk+1/2 e− 2 lsx y ′ (x)dx. (28) l=1 a Similar expressions, though somewhat more complicated than the one in (27), exist for EQλ and VarQλ (see Subsection 3.2) and these involve Gk with k = 0, −1, −2 . . . and k = 1, 0, −1 . . ., respectively. We shall study Gk thoroughly, leading to series expansions, asymptotics and bounds. We close this section by giving an interpretation of the quasi-Gaussian form (22). Using, see Appendix A, (132), ∞ 1 n 1 2 = e− 2 nx y ′ (x)dx (29) p(n) 2π −∞ 8
  • 9. 3 2.5 2 y ′ (x) 1.5 1 0.5 0 −2 −1 0 1 2 3 4 Figure 1: The function y ′ (x) for x ∈ [−2, 4]. and ∞ 2 /2 √ √ ∞ 2 /2 √ e−x y ′ (x/ ls)dx = ls √ e−lsx y ′ (x)dx (30) α l α/ s we find from (22) that ∞ ∞ −lx2 /2 ′ √ 1 α e y (x/ s)dx − ln P(Qλ = 0) = ∞ −lx2 /2 ′ √ . (31) l −∞ e y (x/ s)dx l=1 As mentioned in the introduction, the resulting formula reveals that the summands of the random walk associated with the M/D/s queue, and the summands of the Gaussian ran- dom walk are absolutely continuous with respect to each other. The connecting measure √ between the two densities has a density as well, and equals y ′ (·/ s). Another interpretation ˆ is that P(Qλ = 0) is obtained by twisting the Gaussian distribution associated with Mβ . The √ associated Radon-Nikodym derivative can again be described in terms of y ′ (·/ s). 3 From Spitzer-type expressions to quasi-Gaussian forms In this section we show how to obtain the expression (22). In addition, we present similar results for the mean and variance of the queue length. 3.1 Proof of Theorem 2 For n = 0, 1, . . . we let n zk sn (z) = , z ∈ C. (32) k! k=0 With ρ = λ/s and n = ls (so that λl = nρ), and q(ξ) = e1−ξ ξ, ξ ∈ C, (33) we then have from Szeg¨ [33], p. 50 (also see Abramowitz-Stegun [1], 6.5.13 on p. 262), o ∞ ρ (lλ)j nn+1 e−n e−lλ = 1 − e−λl sn (λl) = q n (ξ)dξ. (34) j! n! 0 j=n+1 9
  • 10. Using this relation we can rewrite the Spitzer-type expression (15) as ∞ ρ 1/2 p(ls) − ln P(Qλ = 0) = s √ q ls (ξ)dξ, (35) l=1 2πl 0 with p(n) as defined in (25). We then consider the equation f (y) := − ln q(1 − y) = 1 x2 , 2 (36) with x ∈ C from which y is to be solved. We note that f (y) = 1 y 2 + 1 y 3 + 1 y 4 + . . . , 2 3 4 (37) whence there is an analytic solution y(x) around x = 0 that satisfies y(x) = x + O(x2 ) as x → 0. Furthermore, since f increases from 0 to ∞ as y increases from 0 to 1, we have that y(x) increases from 0 to ∞, and for any x ≥ 0 there is a unique non-negative solution y(x) = y of (36). Furthermore, we let γ = −2(1 − ρ + ln ρ), α = (sγ)1/2 . (38) Then it holds that 1 1 2 q ls (ρ) = e− 2 lsγ = e− 2 α l , (39) and ρ ∞ 1 2 1 ∞ 1 2 √ q ls (ξ)dξ = √ e− 2 lsx y ′ (x)dx = √ √ e− 2 x y ′ (x/ ls)dx. (40) 0 γ ls α l Substituting (40) into (35) yields (22). Lemma 2. The parameters α and β are related according to √ √ y(α/ s) β/ s = √ . (41) (1 − y(α/ s))1/2 √ Proof. Follows from 1 − ρ = y(γ 1/2 ) = y(α/ s), see Table 3.2, and 1/2 s−λ s−λ λ 1 β β √ 1−ρ = = √ √ = √ ρ1/2 = √ (1 − y(α/ s))1/2 . (42) s λ s s s s We have that y(x)(1 − y(x))−1/2 = x + 1 x2 + 6 5 3 72 x + . . . (see Appendix A), and hence β =α+ √ α2 1 + 5 3 6 s 72s α + .... (43) 10
  • 11. 3.2 Mean and variance of the queue length Our primary characteristic in this paper is the probability of an empty queue. However, the techniques that we develop can be applied to other characteristics like the mean and variance of the queue length. From Spitzer’s identity it follows that the mean and variance of the maximum M are given by ∞ 1 E((Sl+ )k ) with k = 1 and k = 2, respectively. For the l=1 l M/D/s queue this yields ∞ ∞ 1 (lλ)j EQλ = (j − ls)e−lλ , (44) l j! l=1 j=ls+1 ∞ ∞ 1 (lλ)j VarQλ = (j − ls)2 e−lλ . (45) l j! l=1 j=ls+1 This leads after considerable rewriting to ∞ ρ 1/2 p(ls) EQλ = s √ ρq ls (ρ) − ls(1 − ρ) q ls (ξ)dξ (46) l=1 2πl 0 and ∞ ρ 1/2 p(ls) VarQλ = s √ − ρ(ls(1 − ρ) − 1)q ls (ρ) + ((1 − ρ)2 l2 s2 + lsρ) q ls (ξ)dξ . (47) l=1 2πl 0 In a similar way as for P(Qλ = 0), (46) and (47) can then be brought into the forms ∞ 1 2 ∞ √ √ e− 2 α l p(ls) ∞ 2 /2 EQλ = s ρp(ls) √ − αR(ρ) √ √ e−x y ′ (x/ ls)dx , (48) l=1 2πl l=1 2π α l ∞ ∞ √ p(ls) 2 /2 VarQλ = s (α2 lR2 (ρ) + ρ) √ √ e−x y ′ (x/ ls)dx l=1 2π α l ∞ √ √ 1 2 e− 2 α l − α lρR(ρ) − ρ/ ls p(ls) √ , (49) l=1 2π where 1−ρ 1 R(ρ) = √ = 1 − (1 − ρ) + . . . . (50) γ 3 For the Gaussian random walk we have that (see [15]) ∞ 1 2 e− 2 β l 1 ∞ 2 /2 EMβ = √ − β√ √ e−x dx , (51) l=1 2πl 2π β l ∞ ∞ 1 2 /2 β 1 2 VarMβ = (β 2 l + 1) √ √ e−x dx − √ l1/2 e− 2 β l . (52) l=1 2π β l 2π √ Ignoring the factors s and s, we again have complete agreement with the Gaussian random walk when λ → ∞. The deviations from the Gaussian random walk are embodied by p ≡ 1, y ′ ≡ 1, β ≡ α and the fact that R(ρ) ≡ 1 when ρ < 1. The introduced notation is summarized in Table 3.2. 11
  • 12. s = λ+β λ ρ = λ/s γ = −2(1 − ρ + ln ρ) √ α = sγ √ √ a = α/ s = γ √ y(x) = x − 1 x2 + 36 x3 + . . . ; |x| < 2 π 1 √ 3 √ √ −1/2 β = s y(α/ s)(1 − y(α/ s)) √ √ β = α + 1 s−1/2 α2 + 72 s−1 α3 + . . . ; |α/ s| < 2 π 6 √ 5 1 1 p(n) = nn e−n 2πn/n! ∼ 1 + 12 n−1 + 288 n−2 + . . . Table 1: Interrelations between some parameters and functions. 4 Results for Gk In this section we give a reformulation of the function Gk in terms of a principal series expansion. The level of generality is higher than needed for the M/D/s queue, as we consider a large class of functions z(x) of which y ′ (x) is just a special case. In Subsection 4.1 we derive the Taylor series for the most general case. We also discuss some special cases that lead to considerable reductions in complexity of the expressions. The principal series expansion √ comprises terms involving s, z, a = α/ s and elementary functions, as well as a constant √ Lk , not depending on a = α/ s, which is more complicated. For this Lk we present an asymptotic series as s → ∞ that can be used conveniently when the radius of convergence √ r0 of z(x) = ∞ bj z j is not small (for instance 2 π as in the pilot case z(x) = y ′ (x)). In j=0 Subsection 4.1 we derive the principal series expansion. In Subsection 4.2 we investigate the numerical evaluation of Lk in terms of the optimal truncation value of the series expansions. In Subsection 4.3 we use this general result for the specific case of z(x) = y ′ (x) to derive series expansions and asymptotics for P(Qλ = 0). There is a clear connection with the Gaussian random walk. In fact, results for the Gaussian random walk involve the function Gk for the special case z(x) ≡ 1. 4.1 Principal series expansion We let z : [0, ∞) → C be a continuous function satisfying z(x) = O(exp(εx2 )) for any ε > 0, and we assume that there is an r0 > 0 such that z(x) is represented by its Taylor series ∞ j j=0 bj x for 0 ≤ x < r0 . We consider for s > 0 and integer k the function ∞ ∞ 1 2 Gk (a) = lk+1/2 e− 2 lsx z(x)dx, a > 0. (53) l=1 a In the case that z(x) = xi we have i+1 √ Gk (a) = s− 2 Tk,i (a s), (54) where Tk,i is defined as ∞ ∞ 1 2 Tk,i (b) = lk+1/2 e− 2 lx xi dx (55) l=1 b 12
  • 13. with i = 0, 1, . . . and k ∈ Z. The functions Tk,i have been thoroughly investigated in [16], Section 5, leading to analytic expressions. We now generalize this result to Gk . Theorem 3. For k ∈ Z and a < 2 π/s we have that k+3/2 2k+1 a 2k+2 j 2 bj aj−2k−2 z(x) − j=0 bj x Gk (a) = Γ(k + 3/2) − b2k+2 ln a − dx s 2k + 2 − j 0 x2k+3 j=0 ∞ (− 1 s)r a + Lk − ζ(−k − r − 1/2) 2 x2r z(x)dx, (56) r! 0 r=0 where ∞ 2k+2 2k+1 j+1 ∞ k+1/2 1 − 2 lsx2 j1 j+1 2 2 Lk = l e z(x) − bj x dx + bj Γ ζ(−k + j/2) 0 2 2 s l=1 j=0 j=0 k+3/2 k 2 1 √ + b2k+2 Γ(k + 3/2) − ln 2s . (57) s 2j + 1 j=0 Proof. We have ∞ 1 2 G′ (a) = −z(a) k lk+1/2 e− 2 lsa . (58) l=1 The right-hand side of (58) can be expressed in terms of Lerch’s transcendent Φ, defined as the analytic continuation of the series ∞ Φ(z, t, v) = (v + n)−t z n , (59) n=0 which converges for any real number v = 0, −1, −2, . . . if z and t are any complex numbers with either |z| < 1, or |z| = 1 and Re(t) > 1. Note that ζ(t) = Φ(1, t, 1). Thus, 1 2 1 2 G′ (a) = −z(a)e− 2 sa Φ(e− 2 sa , −k − 1 , 1). k 2 (60) We then use the important result derived by Bateman [9], §1.11(8) (with ζ(t, v) := Φ(1, t, v) the Hurwitz zeta function) ∞ Γ(1 − t) (ln z)r Φ(z, t, v) = (ln 1/z)t−1 + z −v ζ(t − r, v) , (61) zv r! r=0 which holds for | ln z| < 2π, t = 1, 2, 3, . . ., and v = 0, −1, −2, . . ., as to obtain k+3/2 ∞ 2 (− 1 sa2 )r G′ (a) = −z(a) Γ(k + 3/2) k a−2k−3 + ζ(−k − r − 1/2) 2 . (62) s r=0 r! Therefore k+3/2 2k+2 2 G′ (a) + Γ(k + 3/2) k bj aj−2k−3 = s j=0 k+3/2 2k+2 ∞ 2 (− 1 sa2 )r −Γ(k + 3/2) z(a) − bj aj a−2k−3 − z(a) ζ(−k − r − 1/2) 2 . s r! j=0 r=0 (63) 13
  • 14. √ The series on the second line of (63) converges uniformly in a with s1/2 a ∈ [0, c] and c < 2 π, so upon integrating the identity in (63) we get for 1 sa2 ≤ c < 2π 2 k+3/2 2k+1 2 bj aj−2k−2 Gk (a) + Γ(k + 3/2) + b2k+2 ln a = s j − 2k − 2 j=0 k+3/2 a 2k+2 2 Lk − Γ(k + 3/2) z(x) − bj xj x−2k−3 dx s 0 j=0 ∞ (− 1 s)r a − ζ(−k − r − 1/2) 2 x2r z(x)dx, (64) r=0 r! 0 where k+3/2 2k+1 2 bj aj−2k−2 Lk = lim Gk (a) + Γ(k + 3/2) + b2k+2 ln a . (65) a↓0 s j − 2k − 2 j=0 We shall determine Lk . It holds that, as a ↓ 0, ∞ ∞ 2k+2 2k+2 ∞ ∞ 1 2 1 2 Gk (a) = lk+1/2 e− 2 lsx z(x) − bi xi dx + bi lk+1/2 e− 2 lsx xi dx l=1 a i=0 i=0 l=1 a ∞ ∞ 2k+2 2k+2 1 2 i+1 √ = lk+1/2 e− 2 lsx z(x) − bi xi dx + o(1) + bi s− 2 Tk,i (a s). l=1 0 i=0 i=0 (66) 2k+2 i Here (54) has been used and the o(1) comes from the fact that z(x) − i=0 bi x = O(x2k+3 ) so that a 2k+2 1 2 1 e− 2 lsx z(x) − bi xi dx = O . (67) 0 (ls)k+2 i=0 Now from Janssen & van Leeuwaarden [16], Section 5, i+1 √ i+1 Γ(k + 3/2) √ s− 2 Tk,i (a s) = s− 2 2k+3/2 (a s)i−2k−2 + Lk,i + O(a) 2k + 2 − i 2 k+3/2 ai−2k−2 i+1 = Γ(k + 3/2) + s− 2 Lk,i + O(a) (68) s 2k + 2 − i for i = 0, 1, . . . , 2k + 1 and 2k+3 √ 2k+3 √ s− 2 Tk,2k+2 (a s) = s− 2 − Γ(k + 3/2)2k+3/2 ln(a s) + Lk,2k+2 + O(a) k+3/2 2 √ = −Γ(k + 3/2) ln(a s) + s−(k+3/2) Lk,2k+2 + O(a). s (69) Here 1 i + 1 i+1 Lk,i = Γ 2 2 ζ(−k + i/2), i = 0, 1, . . . , 2k + 1, (70) 2 2 14
  • 15. k k+3/2 1 1 Lk,2k+2 = Γ(k + 3/2)2 − ln 2 . (71) 2j + 1 2 j=0 Therefore, as a ↓ 0, we get ∞ ∞ 2k+2 k+1/2 − 1 lsx2 Gk (a) = l e 2 z(x) − bi xi dx + o(1) l=1 0 i=0 2k+1 i+1 k+3/2 2 ai−2k−2 1 i+1 2 2 + bi Γ(k + 3/2) + Γ ζ(−k + i/2) s 2k + 2 − i 2 2 s i=0 k+3/2 k 2 √ 1 √ + b2k+2 Γ(k + 3/2) − ln a − ln s+ − ln 2 . (72) s 2j + 1 j=0 Adding k+3/2 2k+1 2 bj aj−2k−2 Γ(k + 3/2) + b2k+2 ln a (73) s j − 2k − 2 j=0 at either side of (72) and letting a ↓ 0, we find that Lk has the required value (57). Then (56) follows from (64). Some values of the Riemann zeta function ζ are given in Table 4.1. x ζ(x) x ζ(x) -5.5 -0.00267145801990 0.0 -0.50000000000000 -5.0 -0.00396825396825 0.5 -1.46035450880959 -4.5 -0.00309166924722 1.5 2.61237534868549 -4.0 0 2.0 1.64493406684823 -3.5 0.00444101133548 2.5 1.34148725725092 -3.0 0.00833333333333 3.0 1.20205690315959 -2.5 0.00851692877785 3.5 1.12673386731706 -2.0 0 4.0 1.08232323371114 -1.5 -0.02548520188983 4.5 1.05470751076145 -1.0 -0.08333333333333 5.0 1.03692775514337 -0.5 -0.20788622497735 5.5 1.02520457995469 Table 2: Some values of the Riemann zeta function ζ. We now give several special cases of Theorem 3. The next two corollaries focus on negative values of k. Corollary 1. For 1 sa2 < 2π and k = −2, −3, . . . we have that 2 k+3/2 a 2 Gk (a) = − Γ(k + 3/2) x−2k−3 z(x)dx s 0 ∞ (− 1 s)r a + Lk − ζ(−k − r − 1/2) 2 x2r z(x)dx, (74) r! 0 r=0 ∞ 1 2 where Lk = ∞ lk+1/2 0 e− 2 lsx z(x)dx (which follows from the definition of Lk in (57) in l=1 which all series over j vanish for k = −2, −3, . . .). 15
  • 16. Corollary 2. For 1 sa2 < 2π and k = −1 we have that 2 1/2 a 2π z(x) − b0 G−1 (a) = − ln a + dx s 0 x ∞ (− 1 s)r a + L−1 − ζ(−r + 1/2) 2 x2r z(x)dx, (75) r! 0 r=0 ∞ −1/2 ∞ − 1 lsx2 2π 1/2 √ where L−1 = l=1 l 0 e 2 (z(x) − b0 )dx − s ln 2s. Theorem 3 is meant for the case that a and the convergence radius r0 of ∞ bj xj are gen- j=0 eral. In the case that a < r0 the expressions can be simplified considerably, as demonstrated below. If a < r0 we have a 2k+2 j ∞ z(x) − j=0 bj x bj aj−2k−2 dx = , (76) 0 x2k+3 j − 2k − 2 j=2k+3 a ∞ bj aj+2r+1 x2r z(x)dx = . (77) 0 j + 2r + 1 j=0 As a consequence of (76) we have that the expression on the first line of (56) 2k+1 a 2k+2 j bj aj−2k−2 z(x) − j=0 bj x − b2k+2 ln a − dx (78) 2k + 2 − j 0 x2k+3 j=0 simplifies to ∞ bj aj−2k−2 − b2k+2 ln a. (79) 2k + 2 − j j=0,j=2k+2 Together with (77) this gives expressions for Gk (a) that are, apart from the Lk to which we turn next, convenient for computation when a is small. Lemma 3. For the first line of (57) ∞ 2k+2 2k+1 j+1 ∞ k+1/2 − 1 lsx2 j 1 j+1 2 2 l e 2 z(x) − bj x dx + bj Γ ζ(−k + j/2) (80) 0 2 2 s l=1 j=0 j=0 there is the asymptotic expression ∞ j+1 1 j+1 2 2 ∼ bj Γ ζ(−k + j/2), s → ∞. (81) 2 2 s j=0,j=2k+2 j+1 In case that bj Γ 2 = O(B j ) for some B > 0, the asymptotic series in (81) is convergent when s > 2B 2 , with sum equal to (80). 16
  • 17. Proof. Using ∞ (j+1)/2 ∞ (j+1)/2 − 1 lsx2 j 1 2 −u (j−1)/2 1 2 j+1 e 2 x dx = e u du = Γ , (82) 0 2 ls 0 2 ls 2 we find that ∞ ∞ 2k+2 ∞ ∞ ∞ 1 2 1 2 lk+1/2 e− 2 lsx z(x) − bj xj dx ∼ bj lk+1/2 e− 2 lsx xj dx l=1 0 j=0 j=2k+3 l=1 0 ∞ (j+1)/2 ∞ 1 2 j+1 = bj Γ l−j/2+k . (83) 2 s 2 j=2k+3 l=1 ∞ −j/2+k This yields (81) since ζ(j/2 − k) = l=1 l . Remark 4. The series expansion (56) for Gk (a) comprises, as a ↓ 0, leading order terms involving aj−2k−2 , j = 0, 1, . . . , 2k + 1, and ln a when k = −1, 0, 1, . . ., and Gk (a) stays bounded as a ↓ 0 for k = −2, −3, −4, . . .. In most cases we are interested in, the value of a is quite small (say ≤ 0.1). The formula in (56) can be used conveniently for computation of √ Gk (a) for values of a from 0 to as large as π. For larger values of a, we present in Appendix A formula (163) as an attractive alternative to compute Gk (a). This alternative shows, for instance, quite clearly an exp(− 1 (s + 1)a2 )-behavior of Gk (a) as a gets large. 2 Remark 5. Chang & Peres [7], Theorem 1.1, proved that ∞ √ β ζ(1/2 − r) −β 2 r P(Mβ = 0) = 2β exp √ , (84) 2π r!(2r + 1) 2 r=0 √ √ for 0 < β < 2 π. This result follows easily from Theorem 3, for the case z(x) ≡ 1, a = β/ s and k = −1. √ For general k, setting z(x) ≡ 1 and a = β/ s in Theorem 3 leads to the following result. √ Lemma 4. For β < 2 π and k ∈ Z we have that ∞ ∞ √ β ζ(−k − r − 1 ) −β 2 r lk P (−β l) = − √ 2 + Rk (β), (85) 2π r!(2r + 1) 2 l=1 r=0 √ where R−1 (β) = − ln 2β and 1 Γ(k + 3 ) k+ 3 −2k−2 1 2 Rk (β) = √ 2 2β + ζ(−k), k = −1. (86) 2π 2k + 2 2 4.2 Optimal truncation value Lemma 3 can be deployed in two ways. We can take only the first few terms to get a good idea of how things behave (see Subsection 4.3), or for the numerical evaluation of Lk , we take as many terms as needed using optimal truncation. The optimal truncation value J of (81) 17
  • 18. is so large (see developments below) that we can replace ζ(−k + J/2) by 1. The truncation error made by approximating (80) by J (j+1)/2 1 j+1 2 bj Γ ζ(−k + j/2) (87) 2 2 s j=0,j=2k+2 is of the order (J+2)/2 1 J +2 2 bJ+1 Γ . (88) 2 2 s We replace, furthermore, bJ+1 = (J + 2)aJ+2 by its asymptotic bound, see Appendix A, Lemma 13, J + 2 −1/2 1 J+2 |bJ+1 | ≤ √ . (89) 2 2 π Thus J +2 1/2 (J+2)/2 1 J +2 2 2 1 J +2 1 bJ+1 Γ ≤ Γ . (90) 2 2 s 2(J + 2) 2 2πs The factor (1/2(J + 2))1/2 is rather unimportant for determination of the optimal truncation value J, and we focus on J +2 1 (J+2)/2 DJ = Γ . (91) 2 2πs Noting that Γ(J/2 + 3/2)/Γ(J/2 + 1) ≈ (J/2 + 1)1/2 , we see that 1/2 1/2 DJ+1 J +2 1 ≈ . (92) DJ 2 2πs The right-hand side of (92) decreases in J until J/2 + 1 = 2πs; this J is (near to) the optimal truncation point. At this point we estimate the right-hand side of (90) by Stirling’s formula as 1/2 2πs 1/2 √ 2πs 1 1 1 1 e−2πs Γ(2πs) ≈ (2πs)2πs−1/2 e−2πs 2π = √ . (93) 8πs 2πs 8πs 2πs s 8π For instance, for s = 10 this equals 10−29 . Remark 6. Observe how important it is that we have managed to show the good bound (89) √ on |bJ+1 |. If, for instance, the 1/2 π in this bound were to be replaced by 1, the e−2πs on the √ far right of (93) would have to be replaced by e−s/2 and the resulting quantity 2πe−s/2 /s would be 0.0017 for s = 10. 4.3 Accurate approximations for the M/D/s queue We can apply Theorem 3 to obtain accurate approximations for the emptiness probability and the mean and variance of the queue length. By way of illustration, we do this in some detail for P(Qλ = 0) and briefly indicate at the end of this section how one can proceed for the other cases. 18
  • 19. We have from (25) and (27) that ∞ 1 √ − ln P(Qλ = 0) ∼ √ pk s−k+1/2 G−(k+1) (α/ s) 2π k=0 1 √ √ √ = √ s1/2 G−1 (α/ s) − 1 −1/2 12 s G−2 (α/ s) + 1 −3/2 288 s G−3 (α/ s) + . . . . 2π (94) √ The G−2 , G−3 , . . . are bounded functions of a = α/ s while G−1 (a) behaves like 2π 1/2 √ − ln a 2s as a ↓ 0. (95) s Accurate approximations to − ln P(Qλ = 0) are obtained by including 1, 2, 3, . . . terms of the second line of (94) in which the G’s must be approximated. For the number of terms of the asymptotic series in (94) to be included one could follow a truncation strategy (based on π (139), (152) and the bound G−k (a) ≤ ( 2s )1/2 ζ(k), k = 2, 3, . . .) pretty much as was done in Subsection 4.2. We shall not pursue this point here. We shall compute accurate approximations to G−k (a) for k = 1, 2, . . .. We have from (74) √ and (75) for α < 2 π √ 1/2 α/ s √ 2π √ y ′ (x) − 1 G−1 (α/ s) = − ln α/ s + dx s 0 x ∞ √ (− 1 s)r α/ s + L−1 − ζ(−r + 1 ) 2 2 x2r y ′ (x)dx, (96) r! 0 r=0 and for k = 2, 3, . . ., √ −k+3/2 α/ s √ 2 G−k (α/ s) = − Γ(−k + 3/2) x2k−3 y ′ (x)dx s 0 ∞ √ 1 r α/ s 1 (− 2 s) + L−k − ζ(k − r − 2) x2r y ′ (x)dx. (97) r! 0 r=0 Here, ∞ 1/2 ∞ 1 2 2π √ L−1 = l−1/2 e− 2 lsx (y ′ (x) − 1)dx − ln 2s, (98) 0 s l=1 and for k = 2, 3, . . ., ∞ ∞ 1 2 L−k = l−k+1/2 e− 2 lsx y ′ (x)dx. (99) l=1 0 Below we specify the missing ingredients in (96)-(99). • We have √ √ α/ s α/ s ∞ ∞ j y ′ (x) − 1 bj α dx = bj xj−1 dx = √ , (100) 0 x 0 j s j=1 j=1 √ √ and the computation of the series is feasible when 0 ≤ α/ s ≤ 2 π, the bj being √ j computable and O(1/(2 π) ). 19
  • 20. β = 0.01 (0.0141) β = 0.1 (0.1334) s α true (94)-1 (94)-2 (94)-3 α true (94)-1 (94)-2 (94)-3 1 0.0100 0.0268 0.0256 0.0267 0.0267 0.0983 0.2351 0.2265 0.2345 0.2343 2 0.0100 0.0225 0.0219 0.0225 0.0225 0.0988 0.2022 0.1980 0.2021 0.2021 5 0.0100 0.0190 0.0188 0.0190 0.0190 0.0993 0.1747 0.1730 0.1747 0.1746 10 0.0100 0.0174 0.0173 0.0174 0.0174 0.0995 0.1617 0.1609 0.1617 0.1617 20 0.0100 0.0164 0.0163 0.0164 0.0164 0.0996 0.1529 0.1525 0.1529 0.1529 50 0.0100 0.0155 0.0155 0.0155 0.0155 0.0998 0.1455 0.1453 0.1455 0.1455 100 0.0100 0.0151 0.0150 0.0151 0.0151 0.0998 0.1419 0.1418 0.1419 0.1419 200 0.0100 0.0148 0.0148 0.0148 0.0148 0.0999 0.1393 0.1393 0.1393 0.1393 500 0.0100 0.0145 0.0145 0.0145 0.0145 0.0999 0.1371 0.1371 0.1371 0.1371 β = 0.2 (0.2518) β = 0.5 (0.5293) s α true (94)-1 (94)-2 (94)-3 α true (94)-1 (94)-2 (94)-3 1 0.1932 0.4105 0.3979 0.4092 0.4089 0.4573 0.7182 0.7049 0.7137 0.7134 2 0.1952 0.3613 0.3549 0.3611 0.3610 0.4699 0.6656 0.6586 0.6642 0.6641 5 0.1970 0.3185 0.3159 0.3185 0.3185 0.4811 0.6156 0.6125 0.6151 0.6151 10 0.1979 0.2979 0.2966 0.2979 0.2978 0.4867 0.5899 0.5883 0.5897 0.5897 20 0.1985 0.2838 0.2831 0.2837 0.2837 0.4906 0.5719 0.5710 0.5717 0.5717 50 0.1991 0.2716 0.2714 0.2716 0.2716 0.4941 0.5560 0.5557 0.5560 0.5560 100 0.1993 0.2657 0.2655 0.2657 0.2657 0.4958 0.5481 0.5479 0.5481 0.5481 200 0.1995 0.2615 0.2615 0.2615 0.2615 0.4970 0.5426 0.5425 0.5425 0.5425 500 0.1997 0.2579 0.2579 0.2579 0.2579 0.4981 0.5377 0.5376 0.5377 0.5377 β = 1 (0.8005) β = 2 (0.9762) s α true (94)-1 (94)-2 (94)-3 α true (94)-1 (94)-2 (94)-3 1 0.8299 0.9055 0.8973 0.8948 0.8945 1.3670 0.9835 0.9793 0.9636 0.9633 2 0.8790 0.8787 0.8746 0.8737 0.8736 1.5296 0.9799 0.9787 0.9674 0.9672 5 0.9236 0.8511 0.8493 0.8489 0.8489 1.6948 0.9774 0.9770 0.9703 0.9703 10 0.9462 0.8364 0.8354 0.8352 0.8352 1.7835 0.9766 0.9764 0.9723 0.9723 20 0.9622 0.8259 0.8253 0.8252 0.8252 1.8473 0.9763 0.9762 0.9738 0.9738 50 0.9762 0.8165 0.8163 0.8162 0.8162 1.9040 0.9762 0.9761 0.9750 0.9750 100 0.9832 0.8118 0.8117 0.8117 0.8117 1.9324 0.9762 0.9761 0.9755 0.9755 200 0.9881 0.8085 0.8084 0.8084 0.8084 1.9524 0.9762 0.9762 0.9759 0.9759 500 0.9925 0.8056 0.8055 0.8055 0.8055 1.9700 0.9762 0.9762 0.9761 0.9761 Table 3: Series expansions for P(Qλ = 0) based on (94). The values of P(Mβ = 0) are given between brackets. • We have √ √ α/ s α/ s ∞ ∞ n+j+1 n ′ bj α x y (x)dx = bj xn+j dx = √ , (101) 0 0 n+j+1 s j=0 j=0 √ √ and the computation of the series is feasible when 0 ≤ α/ s ≤ 2 π. Furthermore √ ∞ α/ s j+1 bj α (− 1 s)r 2 2r ′ x y (x)dx = (− 1 α2 )r 2 √ . (102) 0 2r + j + 1 s j=0 Since, see [15], Sec. 6, ζ(−r + 1 )/r! = O(1/(2π)r ), the computation of the series over r 2 √ at the right-hand side of (96) is feasible when α < 2 π. A similar result holds for the series over r at the right-hand side of (97). 20
  • 21. • We have by Lemma 3 ∞ ∞ j+1 ∞ −1/2 1 − 2 lsx2 ′ 1 j+1 2 2 l e (y (x) − 1)dx ∼ bj Γ ζ(1 + j/2), (103) 0 2 2 s l=1 j=1 ∞ j+1 1 j+1 2 2 L−k ∼ bj Γ ζ(k + j/2), k = 2, 3, . . . , (104) 2 2 s j=0 for the series expressions at the right-hand sides of (98) and (99). The left-hand sides of (103) and (104) can be accurately approximated by using the optimal truncation approach of Subsection 4.2. Alternatively, assume that we include all three terms on the second line of (94) (so that the truncation error is O(s−7/2 )). We then include in the right-hand side of (103) the terms with j = 1, 2, 3, 4, and in the right-hand side of (104) the terms with j = 1, 2. When we want to compute accurate approximations to EQλ and VarQλ , we can use (48) and (49), and then it becomes necessary to approximate Gk (a) with k = 0 and k = 1 as well. This can still be done by using Theorem 3 with its simplifications as pointed out in √ Corollary 2 since z(x) = y ′ (x) has bj = O((2 π)−j ). Of course, there are a variety of ways to proceed here, just like in case of − ln P(Qλ = 0) treated above. For the latter case, we have just worked out one of the more straightforward methods. Table 3 displays approximations to P(Qλ = 0) based on the series expansion (94). Results are given for 1, 2, and 3 terms of the second line of (94), and the G’s are approximated as described in this subsection. Clearly, the expansions provide sharp approximations, and in most cases, one term suffices to get accurate results, i.e., √ s √ P(Qλ = 0) ≈ exp − √ G−1 (α/ s) . (105) 2π 5 Bounds and approximations for the emptiness probability The Gaussian form (22) for P(Qλ = 0) is rather complicated due to the presence of p(ls) and z(x) = y ′ (x), which both can be expressed as infinite series. In this section we obtain bounds on P(Qλ = 0) by using inequalities for p(ls) and y ′ (x). Lemma 5. ∞ ∞ 1/2 1 1 2 P(Qλ = 0) ≥ exp −s √ √ e− 2 lsx y ′ (x)dx =: LB, (106) l=1 2πl γ ∞ ∞ 1 1 1 2 P(Qλ = 0) ≤ exp − s1/2 √ 1− √ e− 2 lsx y ′ (x)dx =: U B. (107) 2πl 12ls γ l=1 Proof. Follows directly from rewriting (22) as ∞ ∞ 1/2 p(ls) 1 2 P(Qλ = 0) = exp −s √ √ e− 2 lsx y ′ (x)dx (108) l=1 2πl γ 21
  • 22. and applying en 1 1 en n−n−1/2 √ 1− ≤ ≤ n−n−1/2 √ . (109) 2π 12n n! 2π Lemma 6. There is the inequality ∞ ∞ UB 1 1 1 1 2 ≤ exp √ √ e− 2 x dx . (110) LB 12s l2 2π α l l=1 Proof. Follows from (106)-(107) and y ′ (x) ≤ 1 (see Lemma 15). We next show that the right-hand side of (110) is a decreasing function of λ when β > 0 is kept fixed. Indeed, this is a direct consequence of the following lemma. Lemma 7. α = (−2s(1 − ρ + ln ρ))1/2 increases as a function of λ when β > 0 is kept fixed. In fact, α increases from 0 to β as λ increases from 0 to ∞. Proof. Letting ν = 1 α2 we have 2 √ λ λ ν = −s(1 − ρ + ln ρ) = −(λ + β λ) 1 −√ + ln √ λ+β λ λ+β λ √ √ λ x = −β λ − (λ + β λ) ln √ = −x β + (x + β) ln , (111) λ+β λ x+β √ where we have set x = λ. Now dν x x x+β β = −β − (x + β) ln − x ln + (x + β) · · dx x+β x+β x (x + β)2 x = −2β − (2x + β) ln = 2x(−t + (1 + 1 t) ln(1 + t)), 2 (112) x+β where we have set t = β/x > 0. Since it holds that t ln(1 + t) ≥ , t ≥ 0, (113) 1 + 1t 2 (equality at t = 0) and thus 1 1 t ′ (ln(1 + t))′ = ≥ 1 2 = 1 , t ≥ 0, (114) 1+t (1 + 2 t) 1 + 2t the proof is complete. Lemma 8. The following inequalities hold: (i) For α > 0, UB π2 ≤ exp . (115) LB 144s 22
  • 23. √ (ii) For 0 < α < 2 π, ∞ UB 1 π2 1 2 1 ζ(3/2 − r)(−1/2)r 2r+1 ≤ exp + α −√ α . (116) LB 12s 12 2 2π r=0 r!(2r + 1) Proof. (i) Follows from (110) and observing that ∞ ∞ ∞ 1 1 − 1 x2 1 1 1 1 2 √ √ e 2 dx ≤ = ζ(2) = π . (117) l2 2π α l 2 l2 2 12 l=1 l=1 (ii) Follows from rewriting the right-hand side of (110) in terms of Lerch’s transcendent and applying the Bateman series (61). Lemma 9. There are the inequalities ∞ ∞ 1 1 1 2 LB ≥ exp − √ √ e− 2 x dx , (118) l 2π α l l=1 ∞ ∞ ∞ 1 1 − 1 x2 2 1 1 2l LB ≤ exp − √ √ e 2 dx + √ e− 2 α . (119) l 2π α l 3 2πs l3/2 l=1 l=1 Proof. Follows from (106) and 1 − 2 x ≤ y ′ (x) ≤ 1 (see Lemma 15). 3 Note that the right-hand side of (118) equals P(Mα = 0), i.e., the probability that the maximum of a Gaussian random walk with drift −α (instead of −β) equals zero. Lemma 10. ∞ ∞ ∞ ∞ 1 1 1 2 1 1 1 1 2 U B ≤ exp − √ √ e− 2 x dx + √ √ e− 2 x dx l 2π α l 12 l2 2π α l l=1 l=1 ∞ ∞ 2 1 1 2 1 1 1 2l + √ 3/2 e− 2 α l − √ e− 2 α . (120) 3 2πs l=1 l 18s 2πs l=1 l3/2 Proof. Follows from (107) and 1 − 2 x ≤ y ′ (x) (see Lemma 15). 3 The right-hand side of (120) can be written as ∞ √ ∞ √ 1 1 2 1 2 12s − 1 exp − l−1 P (−α l) + l−2 P (−α l) + e− 2 α Φ(e− 2 α , 3 , 1) 2 √ . (121) 12 18s 2πs l=1 l=1 √ Using Lemma 4 for k = −1, −2, and (61), we find that (121) satisfies, for α < 2 π, √ π2 12s − 1 ζ( 1 ) ζ( 3 ) √ 12s − 1 2α exp + ζ( 3 ) 2 √ + α √ 2 − √ − 2π 2 √ + O(α2 ) , (122) 144 18s 2πs 2π 12 2π 18s 2πs which yields a sharp approximation for (121) for small values of α. 23
  • 24. Considering the leading component in the exponents of (119) and (120), it makes sense to use the approximation ∞ ∞ ∞ 1 1 − 1 x2 2 1 1 2l P(Qλ = 0) ≈ exp − √ √ e 2 dx + √ e− 2 α l=1 l 2π α l 3 2πs l=1 l3/2 ∞ 2 1 1 2l = P(Mα = 0) · exp √ e− 2 α 3 2πs l=1 l3/2 2 1 2 1 2 = P(Mα = 0) · exp √ e− 2 α Φ(e− 2 α , 3 , 1) . 2 (123) 3 2πs We note the approximations (105) and (123) are strongly related. In fact, approximation (123) follows from (105) by approximating y ′ (x) by 1 − 2 x. √ 3 √ From (1), (61) and (123) we then get, for β < 2 π (and hence α < 2 π), P(Qλ = 0) P(Mα = 0) 2 1 2 1 2 = · exp √ e− 2 α Φ(e− 2 α , 3 , 1) 2 P(Mβ = 0) P(Mβ = 0) 3 2πs ∞ α 1 ζ( 1 − r)(− 1 )r 2r+1 ≈ · exp √ 2 2 α − β 2r+1 β 2π r=0 r!(2r + 1) ∞ 2 √ ζ( 3 − r)(− 1 )r 2r 2 2 · exp √ − 2πα + α . (124) 3 2πs r! r=0 For small values of α we then propose the approximation P(Qλ = 0) α ζ( 1 ) 2 ζ( 3 ) 1 ≈ · exp √ 2 (α − β) + √2 −α √ . (125) P(Mβ = 0) β 2π 3 2π s 5.1 Numerical experiments We now evaluate the approximations and bounds derived in this section. True values of P(Qλ = 0) are computed from (15). Some numerical results are displayed in Tables 4 and 5. In Table 4, the values of P(Mβ = 0) are given in between brackets. Both the lower bound P(Mα = 0) and the upper bound (120) may serve as conservative approximations, but the quality of approximation (123) is striking for small and moderate values of β. For large values, it seems that the limiting value is reached quite rapidly. In par- ticular, for the case β = 2, the Halfin-Whitt limit P(Mβ = 0) is reached almost immediately, which suggests to use P(Mβ = 0) as an approximation for P(Qλ = 0). Indeed, as such, the Halfin-Whitt limit outperforms all other approximations, including the series expansions in Table 3. For all other values of β (especially small values), all alternative approximations of the emptiness probability are closer to the true values than the Halfin-Whitt limit. 6 Conclusions and outlook The approach in this paper consists of three major steps: 1. Quasi-Gaussian form. Section 3 is the bridge that permits us to pass from the general form of Spitzer’s formula (14) to the more convenient quasi-Gaussian form (22). The 24
  • 25. β = 0.01 (0.0141) β = 0.1 (0.1334) s α (118) true (123) (120) α (118) true (123) (120) 1 0.0100 0.0140 0.0268 0.0279 0.0282 0.0983 0.1313 0.2351 0.2468 0.2488 2 0.0100 0.0140 0.0225 0.0228 0.0240 0.0988 0.1319 0.2022 0.2061 0.2149 5 0.0100 0.0140 0.0190 0.01910 0.0203 0.0993 0.1325 0.1747 0.1756 0.1857 10 0.0100 0.0141 0.0174 0.0175 0.0187 0.0995 0.1328 0.1617 0.1620 0.1718 20 0.0100 0.0141 0.0164 0.0164 0.0175 0.0996 0.1329 0.1529 0.1531 0.1625 50 0.0100 0.0141 0.0155 0.0155 0.0166 0.0998 0.1331 0.1455 0.1455 0.1546 100 0.0100 0.0141 0.0151 0.0151 0.0161 0.0998 0.1332 0.1419 0.1419 0.1507 200 0.0100 0.0141 0.0148 0.0148 0.0158 0.0999 0.1333 0.1393 0.1393 0.148 500 0.0100 0.0141 0.0145 0.0145 0.0155 0.0999 0.1333 0.1371 0.1371 0.1457 β = 0.2 (0.2518) β = 0.5 (0.5293) s α (118) true (123) (120) α (118) true (123) (120) 1 0.1932 0.2442 0.4105 0.4332 0.4356 0.4573 0.4961 0.7182 0.7628 0.7635 2 0.1952 0.2464 0.3613 0.3693 0.3829 0.4699 0.5061 0.6656 0.6831 0.6994 5 0.1970 0.2484 0.3185 0.3207 0.3367 0.4811 0.5148 0.6156 0.6208 0.6413 10 0.1979 0.2494 0.2979 0.2987 0.3145 0.4867 0.5192 0.5899 0.5921 0.6127 20 0.1985 0.2501 0.2838 0.2841 0.2994 0.4906 0.5222 0.5719 0.5728 0.5930 50 0.1991 0.2507 0.2716 0.2717 0.2865 0.4941 0.5248 0.5560 0.5563 0.5760 100 0.1993 0.2510 0.2657 0.2657 0.2801 0.4958 0.5262 0.5481 0.5483 0.5676 200 0.1995 0.2512 0.2615 0.2616 0.2757 0.4970 0.5271 0.5426 0.5426 0.5618 500 0.1997 0.2514 0.2579 0.2579 0.2719 0.4981 0.5279 0.5377 0.5377 0.5566 β = 1 (0.8005) β = 2 (0.9762) s α (118) true (123) (120) α (118) true (123) (120) 1 0.8299 0.7294 0.9055 0.9574 0.9554 1.3670 0.9019 0.9835 1.0202 1.0177 2 0.8790 0.7518 0.8787 0.9002 0.9106 1.5296 0.9302 0.9799 0.9938 0.9966 5 0.9236 0.7709 0.8511 0.8581 0.8718 1.6948 0.9514 0.9774 0.9814 0.9847 10 0.9462 0.7800 0.8364 0.8395 0.8533 1.7835 0.9603 0.9766 0.9782 0.9812 20 0.9622 0.7863 0.8259 0.8273 0.8408 1.8473 0.9657 0.9763 0.9770 0.9797 50 0.9762 0.7917 0.8165 0.8170 0.8302 1.9040 0.9700 0.9762 0.9764 0.9788 100 0.9832 0.7943 0.8118 0.8120 0.8250 1.9324 0.9720 0.9762 0.9763 0.9785 200 0.9881 0.7962 0.8085 0.8086 0.8214 1.9524 0.9733 0.9762 0.9762 0.9784 500 0.9925 0.7978 0.8056 0.8056 0.8183 1.9700 0.9745 0.9762 0.9762 0.9783 Table 4: Bounds, approximations and true values of P(Qλ = 0). key facilitator is, see (40), √ √ α/ s p(s) s 1 2 P(Aλ ≤ s) = √ e− 2 sx y ′ (x)dx, (126) 2π −∞ where, as before, Aλ denotes a Poisson random variable with mean λ. 2. Analysis of the function y. Appendix A contains a detailed study of the function y (and y ′ ). It is shown that there is a strong connection with the Gamma function, the reciprocal Gamma function and the Lambert W function. For this paper, the most important result is the power series for y, see (147), which is shown to have rapidly decaying coefficients. The latter property makes the quasi-Gaussian form (126) extremely effective, because only the first few terms of y ′ have to be taken into account. 3. Principal series expansions. Starting from the quasi-Gaussian form, we derive in Sec- tion 4 principal series expansions for P(Qλ = 0). The expansions comprise elementary functions and are used to derive asymptotic expressions and bounds for P(Qλ = 0). The power series for y leads to even more explicit expansions. 25
  • 26. β = 0.01 β = 0.1 s true (124) lb(16) ub(16) (21) true (124) lb(16) ub(16) (21) 1 1.9047 1.9870 0.1224 8.1692 1.1868 1.7622 1.8500 0.1111 8.9980 1.1608 2 1.5993 1.6251 0.2270 4.4062 1.1285 1.5154 1.5447 0.2163 4.6226 1.1095 5 1.3537 1.3595 0.3919 2.5516 1.0794 1.3092 1.3164 0.3845 2.6007 1.0670 10 1.2406 1.2426 0.5159 1.9385 1.0554 1.2119 1.2145 0.5110 1.9570 1.0466 20 1.1653 1.1661 0.6264 1.5965 1.0389 1.1463 1.1473 0.6234 1.6041 1.0326 50 1.1018 1.1021 0.7440 1.3442 1.0244 1.0905 1.0908 0.7426 1.3467 1.0204 100 1.0710 1.0712 0.8113 1.2326 1.0172 1.0633 1.0634 0.8106 1.2337 1.0144 200 1.0497 1.0499 0.8626 1.1593 1.0121 1.0444 1.0444 0.8622 1.1599 1.0101 500 1.0312 1.0313 0.9108 1.0980 1.0077 1.0279 1.0279 0.9106 1.0982 1.0064 β = 0.2 β = 0.5 s true (124) lb(16) ub(16) (21) true (124) lb(16) ub(16) (21) 1 1.6307 1.7207 0.0993 10.0675 1.1328 1.3568 1.4411 0.0688 14.537 1.0565 2 1.4351 1.4669 0.2047 4.8840 1.0894 1.2575 1.2905 0.1717 5.8237 1.0368 5 1.2652 1.2738 0.3763 2.6574 1.0542 1.1629 1.1729 0.3517 2.8433 1.0217 10 1.1831 1.1865 0.5055 1.9781 1.0375 1.1145 1.1187 0.4891 2.0446 1.0148 20 1.1271 1.1284 0.6201 1.6127 1.0261 1.0803 1.0822 0.6101 1.6391 1.0102 50 1.0789 1.0794 0.7410 1.3495 1.0163 1.0504 1.0511 0.7362 1.3582 1.0063 100 1.0553 1.0555 0.8097 1.2350 1.0115 1.0355 1.0358 0.8071 1.2390 1.0044 200 1.0388 1.0389 0.8617 1.1605 1.0081 1.0250 1.0251 0.8604 1.1623 1.0031 500 1.0244 1.0244 0.9104 1.0984 1.0051 1.0158 1.0158 0.9098 1.0991 1.0019 β=1 β=2 s true (124) lb(16) ub(16) (21) true (124) lb(16) ub(16) (21) 1 1.1311 1.1960 0.0340 29.4157 0.9645 1.0074 1.0450 0.0064 155.3166 0.9292 2 1.0976 1.1245 0.1237 8.0841 0.9779 1.0037 1.0179 0.0576 17.3720 0.9593 5 1.0632 1.0719 0.3114 3.2114 0.9876 1.0011 1.0052 0.2365 4.2284 0.9792 10 1.0448 1.0487 0.4614 2.1675 0.9918 1.0003 1.0020 0.4057 2.4649 0.9870 20 1.0316 1.0334 0.5931 1.6862 0.9944 1.0000 1.0007 0.5580 1.7920 0.9915 50 1.0199 1.0206 0.7282 1.3733 0.9966 1.0000 1.0001 0.7116 1.4054 0.9951 100 1.0141 1.0144 0.8028 1.2457 0.9977 1.0000 1.0000 0.7938 1.2598 0.9966 200 1.0099 1.0101 0.8580 1.1654 0.9984 1.0000 1.0000 0.8533 1.1719 0.9977 500 1.0063 1.0063 0.9088 1.1003 0.9990 1.0000 1.0000 0.9069 1.1027 0.9986 Table 5: Bounds, approximations and true values of P(Qλ = 0)/P(Mβ = 0). Steps 1 and 2 are very case-specific, requiring properties of the Poisson distribution and the associated function y ′ . Step 3 is far more general. The expansions presented hold for a large class of functions z, of which z = y ′ (Poisson case) and z ≡ 1 (Gaussian random walk) are just special cases. We now briefly indicate some ideas that enable the extension of the first two steps. A more general version of our model is d Qλ = (Qλ + A∗ − s)+ . λ If we can write the random variable A∗ as the sum of s i.i.d. random variables, which each λ converge weakly to a random variable with unit mean as λ → ∞, then under some addi- tional assumptions (for example a Lindeberg-type condition) one has still weak convergence √ of (A∗ − s)/ s to a Gaussian random variable with negative mean. To write the associated λ performance measure, such as the emptiness probability, in a convenient form, our method of attack would still work if the cumulative distribution function P(A∗ ≤ s) is of type (126) λ with y ′ (x) replaced by some function that depends on A∗ . Temme [35] describes a large class λ of distributions that can be brought in this shape. 26
  • 27. A Analysis of the function y We shall present an analysis to obtain some results on the function y(x) that appeared in Subsection 2.2, especially in Eq. (22). As before, y(x) is, for |x| sufficiently small, the solution y of the equation f (y) := −y − ln(1 − y) = 1 x2 2 (127) that is real and positive when x is real, positive and sufficiently small. We note that for |y| < 1, f (y) = 1 y 2 + 1 y 3 + 1 y 4 + . . . , 2 3 4 (128) whence we can write Eq. (127) for small x and y as y(1 + 2 y + 1 y 2 + . . .)1/2 = x 3 2 (129) with the principal value of the square root. Hence, y(x) = x − 1 x2 + O(x3 ). 3 (130) From (127) we furthermore see that y(x) increases from 0 to ∞ as x increases from 0 to 1, and that y(x) increases from −∞ to 0 as x increases from −∞ to 0. The function y(x) occurs in connection with the asymptotic expansions of Γ(t + 1) and 1/Γ(t) as t → ∞. Thus one substitutes in the integral representation ∞ Γ(t + 1) = e−u ut du (131) 0 subsequently u = t(1 + v), v = −y and y = y(x) to obtain ∞ Γ(t + 1) = e−t tt+1 et(−v+ln(1+v)) dv = −1 1 ∞ 1 2 e−t tt+1 e−tf (y) dy = e−t tt+1 e− 2 tx y ′ (x)dx. (132) −∞ −∞ We refer to [6], Section 4.5 on pp. 69-71, and to [28], Ch. 3, §8 on pp. 85-86. By (129), we have that y(x) has a power series ∞ y(x) = an xn , (133) n=1 converging for |x| sufficiently small. Then by Watson’s lemma, see e.g. [28], pp. 112-116, the asymptotics of Γ(t + 1) follows from (132) as ∞ ∞ 1 2 −t t+1 Γ(t + 1) ∼ e t nan xn−1 e− 2 tx dx n=1 −∞ ∞ √ (2k + 1)(2k − 1) · . . . · 1 = e−t tt+1 2π a2k+1 , (134) tk k=0 as t → ∞. Note that the terms with odd n in the first series in (134) vanish. In a similar fashion the asymptotics of 1/Γ(t) can be obtained from the Hankel representation 1 i = (−u)−t e−u du, (135) Γ(t) 2π C 27
  • 28. where C is a contour that starts at ∞ + i0, encircles the origin in positive sense and ends at ∞ − i0. This is carried out in [36], Subsection 3.6.3 on pp. 69-70. Thus by substitutions as in (132) we now get 1 i 1 = −t tt−1 etf (y) dy, (136) Γ(t) 2π e 1+C where it is observed that Ref (y) → −∞ as y → ∞ ± i0. Now the integration path 1 + C is deformed so as to pass through the saddle point y = 0 where the real axis is crossed perpendicularly from the upper half plane into the lower half plane. Around the saddle point we make the substitution f (y) = − 1 x2 = 1 (ix)2 where x ∈ R decreases from +δ to −δ, with 2 2 δ > 0 sufficiently small, and we get, as t → ∞, δ 1 1 1 1 2 ∼ −t tt−1 e− 2 tx y ′ (ix)dx Γ(t) 2π e −δ ∞ δ 1 1 1 2 ∼ nan in−1 xn−1 e− 2 tx dx 2π e−t tt−1 n=1 −δ ∞ 1 (2k + 1)(2k − 1) · . . . · 1 ∼ √ (−1)k a2k+1 . (137) e−t tt−1/2 2π tk k=0 In particular, we have that the asymptotics of the p(n) of (25) is given by √ ∞ nn−1 e−n 2πn (2k + 1)(2k − 1) · . . . · 1 p(n) = ∼ (−1)k a2k+1 , n → ∞, (138) Γ(n) nk k=0 so that the pk ’s in (25) are given as pk = (−1)k (2k + 1)(2k − 1) · . . . · 1 · a2k+1 , k = 0, 1, . . . . (139) We shall now present and establish some results for the coefficients an of the power series ∞ n n=1 an x of y(x). In [38], p. 16, there is given the result ∞ 1 et ck Γ(k + 3/2) ∼ t+1/2 , t → ∞, (140) Γ(t + 1) πt tk k=0 with ck the coefficients of a function φ(τ ) that is regular inside the circle |τ | = 2π. Apparently, √ since tΓ(t) = Γ(t + 1) and Γ(k + 3/2) = (k + 1/2)(k − 1/2) · . . . · 1 · π, we have that 2 a2k+1 = (−1)k 2−k−1/2 ck , k = 0, 1, . . . , (141) 1 so that a2k+1 decays like ( 4π + ǫ)k for any ǫ > 0. However, with the asymmetric integration ranges (unlike those in (132) and (137)) that occur in the integrals at the right-hand side of (22), the decay behaviour of the coefficients a2k is equally important. We shall thus conduct an analysis for y(x) that is somewhat similar to the analysis in [38], Secs. 7-9 on pp. 9-15, for the function φ(τ ) = ∞ ck τ k . This analysis is greatly facilitated by the study of the k=0 properties of the mapping z ∈ C → e1−z z = w (142) as given in Szeg¨ [33]. Note that exp(f (y)) = (1 − y)ey so that many of the observations o of Szeg¨ directly apply to our function f . The mapping given in (142) is clearly related to o 28
  • 29. the Lambert W function, which is the inverse of the mapping W ∈ C → W eW , see [8, 18]. Explicitly, our y(x) and Lambert’s W function are related according to 1 2 y(x) = 1 + W (−e−(1+ 2 x ) ). (143) It appears that Szeg¨’s analysis of the mapping in (142) has largely escaped the attention of o the Lambert W community. In the proof of the lemma below, we heavily rely on this analysis and omit some of the details that are contained implicitly or explicitly in [33], §2. Lemma 11. The functions y(x) extends to an analytic function on all of C with the exception √ 1 1 of two branch cuts from 2 π · e± 4 πi to ∞ · e± 4 πi . Proof. (sketch) Our approach is to take any angle θ with −π ≤ θ ≤ π, and to see how far y(x) admits analytic continuation when x moves in the direction eiθ from 0. Clearly, from (130), y(x) starts moving from 0 in the direction eiθ when x does so. We thus plot curves Cθ in the y-plane that start at the origin and leave from there under an angle θ ∈ [−π, π] with the positive real axis and on which arg[−y − ln(1 − y)] = 2θ. Writing y = 1 − z and z = ηeiϕ with η > 0 and real ϕ, there should hold η cos ϕ − 1 − ln η + i(η sin ϕ − ϕ) = r cos 2θ + ir sin 2θ (144) with r ≥ 0 for y to lie on Cθ . Consequently, we should have η sin ϕ − ϕ = tan 2θ. (145) η cos ϕ − 1 − ln η With reference to Fig. 2, we have that the cases with tan 2θ = 0 give rise to the curves with θ = 0, ±π/2, ±π. The cases that tan 2θ = ∞ give rise to the four curves with θ = ±π/4, ±3π/4. We note that f ′ (y) = y/(1 − y) so around any y0 = 0, 1, the mapping y → f (y) is locally invertible as an analytic function. Furthermore, |f (y)| increases along any of the curves Cθ . Indeed, this is so, by (130), when y is close to 0, and |f (y)| cannot have stationary points as y = 0, 1 moves along Cθ , for otherwise ln f (y) = ln |f (y)| + 2iθ + 2πin (n some integer) would have a stationary point, contradicting f ′ (y) = 0. Along the curves we define ln(1 − y) as the integral of −1/(1 − z) with z moving along Cθ from 0 to y, and thus obtain a log(1 − y) that is analytic around Cθ . The curves with θ = ±π/4 are particular since they re-enter the origin (where f ′ = 0), and therefore do not admit an analytic continuation of √ y(x) beyond there. The values of x at y = 0 in these re-entrance cases are i±1/2 2π for θ = ±π/4, respectively, since −y − ln(1 − y) has the respective values ±2πi at re-entry to zero along C±π/4 . The C±π/4 constitute mirrored versions of Szeg¨’s curve, see [33], Fig. 1 on o p. 51, passed with negative orientation (+π/4) and positive orientation (−π/4), respectively. In the cases that 0 ≤ |θ| < π/4, we have that Cθ is bounded by the closed curves C±π/4 , and that y → 1. The asymptotic behaviour of y is described in these cases in terms of η, ϕ (see (144),(145)) by the spiral η = exp(−1 + ϕ/ tan 2θ) when ϕ · sgn(θ) → ∞ (observe that in these cases θ and ϕ have opposite signs). Consequently, when 0 < |θ| < π/4, we have that |x| = |2f (y)|1/2 → ∞ along the curve Cθ . In the cases that π < |θ| < π, we have that Cθ 4 lies completely outside the region enclosed by C±π/4 , and along these curves we have that |y| → ∞. Therefore | ln(1 − y)/y| → 0 along these curves, and since we have required that arg[−y − ln(1 − y)] = 2θ, we have that ϕ∞ := lim arg(y) = 2θ − π (mod[−π, π]). (146) |y|→∞,y∈Cθ 29
  • 30. 2 π ϕ∞ = 2 ϕ∞ = 0 1.5 π θ= 2 1 π θ= 4 π 0.5 3π θ= 4 −ε θ= 4 0 θ = − 3π 4 −0.5 θ = −π 4 π −1 θ= 4 +ε θ = −π 2 −1.5 ϕ∞ = 0 ϕ∞ = − π 2 −2 −1 −0.5 0 0.5 1 1.5 2 Figure 2: Curves Cθ in the y-plane. ε = 0.1. Also, |x| = |2f (y)|1/2 → ∞ in these cases. We thus conclude that y(x) can be continued analytically along all rays x = reiθ , r ≥ 0, when θ ∈ [−π, π], except for θ = ±π/4. In the √ latter cases the analytic continuation can only be carried out until r = 2 π. Lemma 12. We have ∞ √ y(x) = an xn , |x| < 2 π, (147) n=1 with a1 = 1 and the an ’s recursively defined as k −1 ak+2 = ak+1 + (n + 1)an+1 ak+2−n , k = 0, 1, . . . . (148) k+3 n=1 √ Proof. From Lemma 11 we have that y(x) is analytic in the disk |x| < 2 π, with branch √ points at x = 2 π · i±1/2 . This gives (147). From (127) we get by differentiation with respect to x and some rewriting the equation y ′ (x)y(x) = x − xy(x). (149) 30
  • 31. Using a1 = 1, this can be written in terms of the an ’s as k−1 (n + 1)an+1 ak−n = −ak−1 , k = 2, 3, . . . , (150) n=0 and this gives (148). Also see [28], Ex. 8.3 on p. 88 and [36], p. 70. The first five coefficients an are given by 1 1 1 1 a1 = 1, a2 = − , a3 = , a4 = , a5 = , (151) 3 36 270 4320 and this rapid decay is further substantiated by the following result. Lemma 13. There is the asymptotic form (rapid decay to 0) sin π (n − 1) − 4 2π π 6n−9 cos 4 (n − 1) an = √ n 1 + O(1/n) . (152) ( 1 n)3/2 (2 π) 2 Proof. The precise asymptotic behavior of the an can be determined by Darboux’s method, √ see Szeg¨ [34], Section 8.4 on pp. 206-208. For this we examine y(e±πi/4 v) with 0 ≤ v < 2 π o as the solution y near 0 and on C±π/4 of the equation 2πi − y − ln(1 − y) = 2πi + 1 y 2 + 1 y 3 + . . . = 2 3 1 2 2 iv . (153) Thus √ √ y(eπi/4 v) = e−πi/4 4π − v 2 + O(2 π − v), v ↑ 2 π, (154) and similarly, √ √ y(e−πi/4 v) = eπi/4 4π − v 2 + O(2 π − v), v ↑ 2 π. (155) When we use Darboux’s method in a first order form, we get the leading asymptotics of an as sin π (n − 1) 4 an = 1 3/2 √ n + lower order. (156) ( 2 n) (2 π) Note that the leading term at the right-hand side of (156) vanishes when n = 4k + 1, k = 0, 1, . . . , and so more precise information is required. This can be obtained when we write −y(eπi/4 v) − ln(1 − y(eπi/4 v)) = 1 (e−πi/4 (4π − v 2 )1/2 )2 , 2 (157) ∞ n so that from y(x) = n=1 an x we get ∞ y(eπi/4 v) = an (e−πi/4 (4π − v 2 )1/2 )n , (158) n=1 and a similar expression for y(e−πi/4 v). We are now in a position to use the full strength of Darboux’s theorem, see Thm 8.4 in [34] (noting that it should be −e−iϕk instead of −eiϕk at the right-hand side of (8.4.8)). For instance, from the fact that a2 = −1/3, we get (152). The particular form 2π/(6n − 9) of the coefficient in front of cos π (n − 1) arises when the binomials 4 occurring in Darboux’s results are manipulated; also it has been used that (−1)n 1/2 is well n approximated as −1/((2n − 1) π(n + 1/2)). Note that the leading factor in (152) is non- vanishing for any n = 1, 2, . . . . The accuracy of the approximation (152) is demonstrated by Fig. 3. 31
  • 32. 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 −0.5 1 5 10 15 20 25 30 35 40 Figure 3: The ratio of an and the approximation in (152), minus 1, for the first 40 coefficients. Lemma 14. It holds that ∞ mm−1 − 1 mx2 y(x) = 1 − e 2 , |arg(x)| ≤ π/4. (159) m!em m=1 Proof. Letting z = 1 − y and u = exp(−1 − 1 x2 ), we can write −y − ln(1 − y) = 1 x2 as 2 2 ze−z = u, (160) so that by the B¨rmann-Lagrange inversion formula u ∞ m−1 ∞ 1 d z m m mm−1 m z(u) = u = u . (161) m! dz ze−z z=0 m! m=1 m=1 The right-hand side of (161) is analytic in |u| < 1/e and continuous in |u| ≤ 1/e. This leads to the formula in (159), with x restricted to |arg(x)| ≤ π/4. An interesting consequence of (159) is that ∞ ∞ ∞ 1 2 1 2 mm−1 1 2 e− 2 lsx y ′ (x)dx = e− 2 lsx m · mx · e− 2 mx dx a a m!e m=1 ∞ ∞ ∞ 1 2 mm 1 2 mm e− 2 (m+ls)a = e− 2 (m+ls)x xdx = . (162) m!em a m!em (m + ls) m=1 m=1 This series expansion can be then be inserted into (53) to yield ∞ ∞ mm − 1 m(a2 +2) lk+1/2 − 1 lsa2 Gk (a) = e 2 e 2 . (163) m! m + ls m=1 l=1 32
  • 33. For somewhat larger values of a, this seems to offer an attractive alternative to evaluate the Gk (a) numerically. Furthermore, (159) can be used to establish a formal connection between the function y and Lerch’s transcendent. Indeed, we have kk−1 p(k) 1 1 1 1 = √ ∼ √ − + + ... . (164) k!ek k 2πk 2π k 3/2 12k 5/2 288k7/2 Inserting this asymptotic expansion into (159) and using the definition (59) of Lerch’s tran- scendent, we find a formal identity between 1 − y(x) and 1 2 e− 2 x 1 2 1 1 2 1 1 2 √ Φ(e− 2 x , 3 , 1) − 2 Φ(e− 2 x , 5 , 1) + 2 Φ(e− 2 x , 7 , 1) + . . . . 2 (165) 2π 12 288 √ Observe that each of the Φ-functions in (165) has a convergent power series in x, |x| < 2 π, per formula (61). We conclude our analysis of y with the following result. Lemma 15. There is the inequality 1 − 2 x ≤ y ′ (x) ≤ 1, 3 x ≥ 0. (166) Proof. Solve f (y) = −y − ln(1 − y) = 1 x2 for the unique solution y(x) ∈ [0, 1). From the 2 inequalities, to be proved below, x x 1 x x − − ln 1 − ≥ x2 ≥ − − ln 1 − , x ≥ 0, (167) 1 + x/3 1 + x/3 2 1+x 1+x and monotonicity of f we infer that x x ≥ y(x) ≥ , x ≥ 0. (168) 1 + x/3 1+x From (149) we get 1 y ′ (x) = x −1 , x ≥ 0. (169) y(x) Inserting the inequalities (168) into (169), we find 2 1 + x/3 1+x 1− x=x − 1 ≤ y ′ (x) ≤ x − 1 = 1, x ≥ 0, (170) 3 x x as required. We now prove the two inequalities in (167). As to the second one, we need to show that 1 1 g2 (x) := −x + (1 + x) ln(1 + x) ≤ x2 + x3 =: h2 (x), x ≥ 0. (171) 2 2 Noting that g2 (0) = h2 (0) = 0, g2 (0) = h′ (0) = 0, and ′ 2 ′′ 1 3 g2 (x) = ≤ 1 + x = h′′ (x), 2 x ≥ 0, (172) 1+x 2 we get (171). Note that the second inequality in (167) is the sharpest in its kind: when α > 0, the inequality y(x) ≥ x/(1 + (1 − α)x) fails to hold for large x ≥ 0, since this would require x 1 − αx 1 − − ln ≤ x2 , 0 ≤ x < ∞, (173) 1 + (1 − α)x 1 + (1 − α)x 2 33
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