stochastic processes assignment help

Stochastic Processes Assignment Help
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Problem 1. The following identity was used in the proof of Theorem 1 in Lec
1
t > 0
ture 4: sup{θ > 0: M (θ) < exp(Cθ)} = inf tI(C + ) (see the proof for
t
details). Establish this identity.
Hint: Establish the convexity of log M (θ) in the region where M (θ) is finite.
∗
dlogM (θ) dθ
Letting θ = sup{θ> 0: M (θ) < exp(Cθ)} use the convexity property above
∗
to argue that logM (θ) ≥θ C +
C
(θ−θ∗). Use this property to finish
C θ∗
the proof of the identity.
Problem 2. This problem concerns the rate of convergence to the limits for the large deviations bounds. Namely, how quickly does n−1 log P(n−1Sn ∈A)
converge to − infx∈A I(x), where Sn is the sum of n i.i.d. random variables? Of course the question is relevant only to the cases when this convergence takes
place.
(a) Let Sn be the sum of n i.i.d. random variables Xi , 1 ≤ i ≤ n taking values in R. Suppose the moment generating function M (θ) = E[exp(θX)] is finite
everywhere. Let a ≥µ = E[X]. Recall that we have established in class that in this case the convergence limn n−1 log P(n−1Sn ≥a) =
−I(a) takes place. Show that in fact there exists a constant C such that
−1 −1
n C
C
n log P(n S ≥ a) + I (a) ≤ ,
C n
for all sufficiently large n. Namely, the rate of convergence is at least as fast as O(1/n).
Hint: Examine the lower bound part of the Crame´r’s Theorem.
(b) Show that the rate O(1/n) cannot beimproved.
Hint: Consider the case a = µ.
Problem 3. Let Xn be an i.i.d. sequence with a Gaussian distribution N (0, 1)
and let Yn be an i.i.d. sequence with a uniform distribution on [−1,1]. Prove
Problems

� �
� �
that the limit
1
lim log P( i
2
X ) + ( 1
n
i
2
Y ) ≥ 1
n → ∞ n
1
n
1≤i≤n 1≤i≤n
exist and compute it numerically. You may use MATLAB (or any other software of your choice) and an approximate numeric answer is acceptable.
Problem 4. Let Zn be the set of all length-n sequences of 0, 1 such that a) 1 is always followed by 0 (so for, for example 001101 · · · is not a legitimate
sequence, but 001010 · · · is); b) the percentage of 0-s in the sequence is atleast 70%. (Namely, if Xn(z) is the number of zeros in a sequence z ∈Zn, then
1
X (z)/n ≥0.7 for every z ∈Z ). Assuming that the limit lim
n n n→ ∞ logZ n
n
exists, compute its limit. You may use MATLAB (or any other software of your choice) to assist you with computations, and an approximate numerical
answer is acceptable. You need to provide details of your reasoning.
Problem 5.
Problem 6.

1 Problem1
Proof. The lecture note 4 has shown that {θ> 0 : M (θ) < exp(Cθ)} is nonempty. Let
θ∗:= sup{θ > 0: M (θ) < exp(Cθ)}
If θ∗= ∞, which implies that for all θ > 0, M (θ) < exp(Cθ) holds, we have
1
inf tI(C + )= inf sup{t(Cθ −logM (θ))+θ}= ∞ = θ ∗
t> 0 t t>0 θ∈R
Consider the case in which θ∗is ﬁnite. According to the deﬁnition of I(C + 1), t
we have
1 1
∗ ∗
I(C + ) ≥θ (C + )−logM (θ )
t t
1 1
∗ ∗
⇒ inftI(C + ) ≥inf t(θ (C + )−logM (θ ))
t> 0 t t> 0 t
=inf t(θ∗C −logM (θ∗))+θ∗
t>0
≥θ∗ (1)
Next, we will establish the convexity of log M (θ) on {θ∈R : M (θ) < ∞}. For two θ1, θ2 ∈{θ∈R : M (θ) < ∞} and 0 < α < 1, Ho¨lder’sinequality gives
1 1
1 2
α
E[exp((αθ +(1−α)θ )X)] ≤E[(exp(αθ1X)) α ] E[(exp((1 1 1−α
−α)θ X )) ] 1−α
Taking the log operations on both sides gives
logM (αθ1 +(1−α)θ2) ≤α logM (θ1)+ (1−α)M (θ2)
By the convexity of log M (θ), we have
1 1 M˙(θ∗)
∗ ∗
(C + )θ −logM (θ) ≤(C + )θ −θ C − (θ −θ )
t t M (θ∗)
˙ ∗
M (θ ) 1 θ∗
∗
=(C −
M (θ∗)
+
t
)(θ −θ )+
t
Solutions

Thus, we have
1 M˙(θ∗) 1
inf t sup (C + )θ −log M (θ) ≤inf t sup (C − ∗
+ )(θ −θ ) + θ ∗
t> 0 t t> 0 ∗
M (θ ) t
θ∈R θ∈R
(2)
˙ ∗
M(θ )
M (θ∗)
Then we will establish the fact that sufficiently small h > 0such that
≥C. If not, then there existsa
logM (θ∗ −h) −logM (θ∗)
< C
−h
which implies that
logM (θ∗ −h) > logM (θ∗)−Ch
⇒ logM (θ∗ −h) > C(θ∗ −h) ⇒ M (θ∗ −h) ≥exp(C(θ∗ −h))
which contradicts the definition of θ∗.By the facts that ˙ ∗
M (θ ) 1
inf t sup (C− −θ∗) ≥0, (when θ = θ∗)
t> 0 M (θ∗)
+
t
)(θ
θ∈R
∗
and
˙
M (θ )
M (θ∗)
≥C, we havethat
˙ ∗
M (θ ) 1
inf t sup (C− −θ∗) =0
t> 0 M (θ∗)
+
t
)(θ
θ∈R
∗ 1 ˙ ∗
and the infimum is obtained at t > 0 such that C + − = 0. From (2),
t∗
M (θ )
M (θ∗)
we have
1
inf t sup (C + )θ −log M (θ) ≤θ ∗
t> 0 t
θ∈R
1
⇒ inf tI(C + ) ≤θ ∗ (3)
t> 0 t
From (1) and (3), we have the result inft>0 tI(C + 1 )= θ∗. t

��
�
2 Problem 2 (Based on Tetsuya Kaji’s Solution)
(a). Let θ0 be the one satisfying I(a) = θ0a − log M (θ0) and δ be a small positive number. Following the proof of the lower bound of Cramer’s theorem, we
have
n−1 logP(n−1Sn ≥a) ≥ n−1 logP(n−1Sn ∈[a, a +δ))
≥−I(a)−θ0δ −n−1 logP(n−1Sñ −a ∈[0, δ))
Namely, this bound can not be improved.
3 Problem 3
For any n ≥0, define a point Mn in R2 by
˜
n 1 n i
where S = Y + ... + Y and Y ( 1 ≤i ≤n) is i.i.d. random variablefollowing
z
the distribution P(Yi ≤z)= M (θ0)−1 exp(θ0x)dP (x). Recallthat −∞
n
(Yi −a) √
n
P(n S −a ∈[0, δ)) =P
−1 ˜ i=1√ ∈[0, nδ)
n
By the CLT, setting δ = O(n−1/2)gives
P(n−1Sñ −a ∈[0, δ)) =O(1)
Thus, we have
n−1 logP(n−1Sn ≥ a)+ I(a) ≥ −θ0δ −n−1 logP(n−1Sñ −a ∈[0,δ))
= −O(n−1/2)
Combining the result from the upper bound n−1 log P(n−1Sn ≥a) ≤ −I(a),
we have
C
−1 −1
|n logP(n Sn ≥a)+ I(a)|≤ √
n
n
−1 1
≥ μ) → as n → ∞ . Recalling that
2
(b). Take a = μ. It is obvious, P(n S I(μ)= 0, wehave
n
|n−1 log P(n−1S ≥μ)+ I(μ)|∼ C
n
1
n
M n
x = X i
i≤n

�
and 1
yM n =
n
Yi
i≤n
Let B0(1) be the open ball of radius one in R2. From these deﬁnitions, we can rewrite
⎛ ⎞
2 2
n n
1
n
1
n
P⎝ Yi ≥1 ⎠ =P(Mn ∈
/B0(1))
Xi +
i=1 i=1
Wewill apply Cramer’s Theorem in R2:
⎛ ⎞
2 2
n n
1 1 1
lim P⎝ Xi +
(x,y)∈B0(1)
Yi ≥1 ⎠ = − inf C
I(x, y)
n
i=1
n
i=1
n → ∞ n
where I(x, y)= sup (θ1x +θ2y −log(M(θ1, θ2)))
(θ1,θ2)∈R2
with
M (θ1, θ2)= E[exp(θ1X +θ2Y )]
Note that since (X, Y )are presumed independent, log(M(θ1,θ2)) = log(MX (θ1))+ log(MY (θ2)), with MX (θ1) = E[exp(θ1,X)] and MY (θ2) = E[exp(θ2Y
)].
Wecan easily compute that
θ2 MX (θ)= exp(
2
)
and
1 1
� 1
θy �
eθy = (eθ −e−θ)
MY (θ)=E[eY θ ]= dy = e
1 1
2 2θ 2θ
−1 −1
Since (x, y) are decoupled in the deﬁnition of (x, y), we obtain I(x, y) =
IX (x)+ IY (y) with
1
X 1 1 1 1
I (x)= sup g (x , θ )= sup(θ x − )=
2
θ2 x2
2
θ1 θ1
Y 2 2 2
1 θ −θ
2 2
I (y) = supg (y, θ ) = sup(θ y −log( (e −e )))
2θ
θ θ
2 2
Since for all y, θ2, g2(y, θ2)= g2(−y, −θ2), for all y, IY (y)= IY (−y).

Since IX (x) is increasing in |x| and IY (y) is increasing in |y|, the maxi- mum is attained on the circle x2 + y2 = 1, which can be reparametrized
as a one-dimensional search over an angle φ. Optimizing over φ, we find that the minimum of I(x, y) is obtained at x = 1,y = 0, and that the
value is equal to
1 . Weobtain
2
⎛ ⎞
n 2 2
n
1 1 1
lim logP ⎝ Xi + Yi ≥1 ⎠ = − 1
2
n → ∞ n n
i=1
n
i=1
4 Problem 4
We denote Yn the set of all length-n sequences which satisfy condition (a). The first step of our method will be to construct a Markov Chain with the
following properties:
• For every n ≥0, and any sequence (X1, ..., Xn) generated by the Markov Chain, (X1, X2, ..., Xn) belongs to Yn.
• For every n ≥0, and every (x1, ..., xn) ∈Yn, (x1, ..., xn) has positive probability, and all sequences of Yn are “almost” equally likely.
Consider a general Markov Chain with two states (0, 1) and general transition probabilities (P00 , P01; P10, P11). We immediately realize that if P11 > 0,
sequences with two consecutive ones don’t have zero probability (in particular, for n = 2, the sequence (1, 1) has probability ν(1)P11. Therefore, we set
P11 = 0 (and thus P10 = 0), and verify this enforces the first condition.
Let now P00 = p, P01 = 1 −p, and let’s find p such that all sequences are almost equiprobable. What is the probability of a sequence (X1, ...,Xn)?
Every 1 in the sequence (X1, ..., Xn) necessarily transited from a 0, with probability (1 −p).
Zeroes in the sequence (X1, ..., Xn) can come either from another 0, in
which case they contribute a p to the joint probability (X1, ..., Xn), or from a 1, in which case they contribute a 1. Denote N0 and N1 the numbers of 0
and 1 in the sequence (X1, ..., Xn). Since each 1 of the sequence transits to a 0 of the sequence, there are N1 zeroes which contribute a probability of 1,
and thus N0 − N1 zeroes contribute a probability of p. This is only ’almost’ correct, though, since we have to account for the initial state X1, and the final
state Xn. By choosing for initial distribution ν(0) = p and ν(1) = (1 −p), the above reasoning applies correctly to X1.

Our last problem is when the last state is 1, in which case that 1 does not give a 1 to 0 transition, and the probabilities of zero-zero transitions is
therefore N0 −N1 + 1. In summary, under the assumptions given above, we have:
(1 −p)N1pN0−N1, when Xn =0
P(X1, ..., Xn )=
n
(1 −p)N1pN0−N1+1, when X =1
Since N0 + N1 = n, we can rewrite (1 −p)N1pN0−N1 as (1 −p)N1pn−2N1, or
equivalently as (1−p
)N1 pn.Weconclude
p2
1−p N n 1
( ) p , when Xn =0
P(X1, ..., Xn )= p2
(1−p
)N1pn+1,
p2 n
when X =1
Weconclude that if 1−p
= 1, sequences will be almost equally likely. This√
5−1 .
p2
equation has positive solution p = = 0.6180, which we take in the rest
2
of the problem (trivia: 1/p = φ, the golden ratio). The steady state distribution of the resulting Markov Chain can be easily computed to be π = (π0, π1) =
1 1−p
( , ) ∼(0.7236, 0.2764). Wealso obtain the “almost” equiprobablecon-
2−p 2−p
dition:
n
p , when Xn =0
P(X1, ..., Xn )=
pn+1, n
when X =1
Wenow relate this Markov Chain at hand. Note the following: log(|Zn|) =
n
|Yn|
|Z |
log( )+ log( n
|Y |), and therefore,
1
n
1
lim log(Z ) = lim log( n
|Y |) + lim
1 |Zn|
log
n → ∞ n n → ∞ n n → ∞ n |Yn|
1
Let us compute ﬁrst lim n→ ∞ n
log(|Y |
). This is easily done using our Markov
n
Chain. Fix n ≥0, and observe that since our Markov Chain only generates sequences which belong to Yn, we have
P(X1, ..., Xn)
1=
(X1,...,Xn)∈Yn
Note that for any (X1, ..., Xn) ∈Yn, we have pn+1 ≤P (X1, ..., Xn) ≤pn, and so we obtain
pn+1|Yn|≤ 1 ≤pn|Yn|

and so
φn ≤|Yn|≤ φn+1, n logφ ≤log|Yn|≤ (n +1)logφ
1
which gives lim n→ ∞ n
log(|Y | )= logφ.
n
Wenow consider the term |Zn|
. The above reasoning shows that intuitively,
|Yn|
Yn
1 is the probability of the equally likely sequences of (X1, ..., Xn), and that
|Zn| is the number of such sequences with more than 70% zeroes. Basic probability reasoning gives that the ratio is therefore the probability that a
random sequence (X1, ..., Xn) has more than 70% zeroes. Let us ﬁrst prove this for- mally, and then compute the said probability. Denote G(X1, ...,
Xn)the percent of zeroes of the sequence (X1, ..., Xn). Then, for any k ∈[0,1]
P(G(X1, ..., Xn) ≥ k)= P (X1, ..., Xn)
(X1,...,Xn)∈Zn
Reusing the same idea as previously,
1 n
P(G(X , ..., X ) ≥k) ≤ p ≤| n n
Z |p ≤|Z |
n n n
n
p ≤|Z |= |Z |
1/p
n
(1/p)n+1
(X1,...,Xn)∈Zn
Similarly,
n
|Z |
P(G(X1, ..., Xn) ≥k) ≤p
|Yn|
Taking logs, we obtain
1
1 n
lim log P(G(X , ..., X ) ≥k) = lim
1 |Zn|
log
n → ∞ n n → ∞ n |Yn|
I(x)= sup(θx −logλ(θ))
θ
where λ(θ) is the largest eigenvalue of the matrix
Wewill use large deviations for Markov Chain to compute that probability. First note that G(X1,..., Xn) is the same as i F (Xi), when F (0) =
1 and F (1) =
0. By Miller’s Theorem, obtain that for any x,
1
lim
n → ∞ n
P( F(Xi) ≥nk)= − inf I(x)
x≥k
i
with
p exp(θ) 1 −p
M (θ)= exp(θ) 0
n
|Z |
≤(1/p)
|Yn|

The characteristic equation is λ2 −√(pexp(θ))λ −(1 −p) exp(θ) = 0, whose
2
p2 exp(2θ)+4(1−p) exp(θ)
. The ratefunction
largest solution is λ(θ) = p exp(θ)+
of the MC is
I(x)= sup(θx −log(λ(θ)))
θ
Since the mean of F under the steady state distribution π is above 0.7, the min-
n
|Z |
imum minx≥0.7 I(x) = I(μ) = 0. Thus, lim n→ ∞ log = 0, andwe
1
n |Yn|
conclude
1
lim log|Zn| = logφ = 0.4812
lim log|Zn(k)| = logφ = 0.4812
and for k > μ,
lim log| n
Z (k)
| = logφ −sup(θk −log(λ(θ)))
θ
5.1 1(i)
Consider a standard Brownian motion B, and let U be a uniform random variable over [1/2, 1]. Let
B (t), when t = U B(U )= 0,
otherwise
W (t)=
With probability 1, B(U )is not zero, and therefore limt→U W (t) = limr→U B(t)= B(U ) = 0 = W (U ), and W is not continuous in U . For any ﬁnite col- lection of
times t = (t1, ..., tn) and real numbers x = (x1, ..., xn); denote
W (t)= (W (t1), ..., W (tn)), x = (x1, ..., xn)
P(W(t) ≤ x) =P(U ∈
/{ti, 1 ≤ i ≤n})P(W(t) ≤x|U ∈
/{ti, 1 ≤ i ≤n})
+ P(U ∈{ti, 1 ≤i ≤n})P(W (t) ≤x|U ∈{ti, 1 ≤i ≤n})
Note that P(U ∈{ti, 1 ≤ i ≤ n}) = 0, and P(W (t) ≤ x|U ∈
/ {ti,i ≤ n})= P(B(t ≤ x)), and thus the Process W has exactly the same distribution properties as B
(gaussian process, independent and stationary increments with zero mean and variance proportional to the size of the interval).
n → ∞ n
In general, for k ≤μ, we will have
1
n → ∞ n
1
n → ∞ n
5 Problem 5

5.2 1(ii)
Let X be a Gaussian random variable (mean 0, standard deviation 1), and denote
QX the set {q+ x, q∈Q}∪ R+ where Q is the set of rationalnumbers.
B(t), when t∈
/QX {0} B(t)+ 1, whent ∈QX{0}
W (t)=
Through the exact same argument as 1(i), W has the same distribution properties as B (this is because QX , just like {ti, 1 ≤ i ≤ n}, has measure zero for a random
variable with density).
n
i(t−x)10 l
However, note that for any t > 0, |t −x − −n
| ≤ 10 , proving
10n
n
i(t−x)10 l n
i(t−x)10 l
n
that lim (x + ) = t. However, for any n, x + X
∈Q , and
10n 10n
n
so lim W (x + i(t−x)10n l
) = B(t)+ 1 = B(t). This proves W (t) is surely
10n
discontinuous everywhere.
5.3 2 1
Let t ≥0
, and consider the event E n
= {|B(t + ) −B(t) |> E}.Then,since
n
1 1
√
B(t + )−B(t) is equal in distribution to N , where N is a standard normal,
n n
by Chebychevs inequality, we have
n
−1/ 2
P(E ) = P(n | −1/2 4 4 −2
N| > E)= P(|N| > En )= P(N > E n )≤
3
E4n2
1
< ∞, by Borel-Cantelli lemma, we have that there
Since n P(En)= n n2
almost surely exists N such that for all n ≥N , |B(t + 1/n) −B(t)| ≤E, proving limn→∞ B(t +1/n)= B(t) almost surely.
6 Problem 6
The event B ∈AR is included in the event B(2) −B(1) = B(1) −B(0), and
thus
P(B ∈AR) ≤ P (B(2) −B(1) = B(1) −B(0)) = 0
Since the probability that two atomless, independent random variables are equal is zero (easy to prove using conditional probabilities).

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