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Problems:
1. (a) Imagine that you first roll a fair 6-sided die and then you flip a fair coin the number of
times shown by the die. Letting X denote the number of these flips that come up heads, find
E[X] and var(X).
2. (b) Repeat part (1a) only now assume you first roll two dice. 2. Sam and Pat are dating. All of
their dates are scheduled to start at 9pm at the Kendall/MIT T stop. Sam always arrives
promptly at 9pm. Pat is highly unorganized, and arrives at a time that is uniformly
distributed between 8pm and 10pm. Sam gets irritated when Pat is late. Let X be the time in
hours between 8pm and the time when Pat arrives. When Pat arrives after 9pm, their date
will last for a time that is uniformly distributed between 0 and (3 − X) hours. When Pat
arrives before 9pm, their date will last exactly 3 hours. The date starts at the time that they
meet. Sam will end the relationship after the second date on which Pat is late by more than
45 minutes. All dates are independent of any other dates.
(a) What is the expected number of hours Pat waits for Sam to arrive?
(b) What is the expected duration of any particular date?
(c) What is the expected number of dates they will have before breaking up?
1. 3. Imagine that the number of people that enter a bar in a period of 15 minutes has a
Poisson distribution with rate λ. Each person who comes in buys a drink. If there are N types
of drinks, and each person is equally likely to choose any type of drink, independently of
what anyone else chooses, find the expected number of different types of drinks the
bartender will have to make.
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4. Let X, Y and Z denote any three discrete random variables. You may find the result of part (a)
useful to show the result of part (b).
(a) Pull-Through Property: For any suitable function g, show that
(b) Tower Property: Show that
5. Consider two random variables X and Y . Assume for simplicity that they both have zero mean.
(a) Show that X and E[X | Y ] are positively correlated.
(b) Show that the correlation coefficient of Y and E[X | Y ] has the same sign as the correlation
coefficient of X and Y .
6. The transform for J, the total number of living groups contacted about the MIT blood drive, is:

(a) Determine the PMF pJ (j).
Let K, the number of people in any particular living group, be an independent random variable
with transform:
(b) Without doing much work, determine the PMF pK(k), and evaluate E[K] and Var(K).
(c) Let L be the total number of people whose living groups are contacted about the blood drive.
Determine either the PMF or the transform for random variable L. Also evaluate E[L] and Var(L).
Suppose the probability that any particular person, whose living group is contacted about the
MIT blood drive, donates blood is 1 4 , and that all such individuals make their decisions
independently.
(d) Let M denote the total number of blood donors from the contacted living groups. Determine
an explicit numerical expression for pM(0). Also evaluate E[M] and Var(M). (Your expressions,
albeit not gracious, should be in closed form.)
7. Widgets are packed into cartons which are packed into crates. The weight (in pounds) of a
widget, X, is a continuous random variable with PDF

The number of widgets in any carton, K, is a random variable with PMF
The number of cartons in a crate, N, is a random variable with PMF
Random variables X, K, and N are mutually independent. Determine
(a) The probability that a randomly selected crate contains exactly one widget.
(b) The expected value and variance of the number of widgets in a crate.
(c) The transform or the PDF for the total weight of the widgets in a crate.
(d) The expected value and variance of the total weight of the widgets in a crate.
Let X1, X2, . . . be independent normal random variables with mean 2 and variance 4. Let N be a
geometric random variable which is independent of the Xi , with parameter p = 2/3. (In particular,
E[N] = 3/2 and E[N2 ] = 3.)

8. (a) If δ is a small positive number, we have P(|X1| ≤ δ) ≈ αδ, for some constant α. Find the
value of α. (Your answer may involve π, no need to evaluate numerically.)
(b) Find E[X1N].
(c) Find the variance of X1N.
(d) Find E[X1 + · · · + XN | N ≥ 2].
(e) Write down the transform associated with N + X1 + · · · + XN .
9. Iwana Passe is taking a quiz with 12 questions. The amount of time she spends answering
question i is Ti , and is exponentially distributed with E[Ti ] = 1 3 hour. The amount of time she
spends on any particular question is independent of the amount of time she spends on any other
question. Once she finishes answering a question, she immediately begins answering the next
question.
Let N be the total number of questions she answers correctly.
Let X be the total amount of time she spends on questions that she answers correctly.
For parts (a) and (b), suppose we know she has probability 2 3 of getting any particular quiz
question correct, independently of her performance on any other quiz question.
(a) Find the expectation and variance of X.

(b) Assuming we know she spent at least 1 6 of an hour on each question, find the transform of
X.
For parts (c) and (d), suppose she has a fixed probability P of getting any particular quiz question
correct, independently of her performance on any other quiz question, and with P uniformly
distributed between 0 and 1. Assume P is the same value for all questions.
(c) Find the expectation and variance of N.
(d) Assuming there is only one question on the quiz, find the transform of N.
10. Consider n independent tosses of a die. Each toss has probability pi of resulting in i. Let Xi be
the number of tosses that result in i. Show that X1 and X2 are negatively correlated (i.e., a large
number of ones suggests a smaller number of twos).
11. When using a multiple access communication channel, a certain number of users N try to
transmit information to a single receiver. If the real-valued random variable Xi represents the
signal transmitted by user i, the received signal Y is

where Z is an additive noise term that is independent of the transmitted signals and is assumed
to be a zero-mean Gaussian random variable with variance σ 2 Z . We assume that the signals
transmitted by different users are mutually independent and, furthermore, we assume that they
are identically distributed, each Gaussian with mean µ and variance σ 2 X.
(a) If N is deterministically equal to 2, find the transform or the PDF of Y .
(b) In most practical schemes, the number of users N is a random variable. Assume now that N
is equally likely to be equal to 0, 1, . . ., 10.
i. Find the transform or the PDF of Y .
ii. Find the mean and variance of Y .
iii. Given that N ≥ 2, find the transform or PDF of Y .
(c) Depending on the time of day, the number of users N has different PMF’s. Between 7 and
10AM, and between 5 and 8PM, N can be assumed to be Poisson distributed with parameter λ =
10. During all other hours, N is equally likely to be 0, 1, . . ., 10. On a random time during the
day, based on the reception of Y , the receiver estimates the transmitted signals. Determine the
linear least-squared estimator of Xk (where k is a positive integer) given reception Y .

1. For both parts (a) and (b) we will make use of the formulas:
Let X be the number of heads, and let Y be the result of the roll. Note that it can be easily verified
that E[Y ] = 7/2 and V ar(Y ) = 35/12.
(a)
and similarly
(b) For this part, let X1 be the number of heads that correspond to the first die roll, and X2 be the
number of heads that correspond to the second die roll. Clearly X = X1 + X2 and X1, X2 are iid.
Thus we have
Solutions:

Similarly
2. (a) Pat only needs to wait for Sam if Pat arrives before 9pm. If Pat arrives after 9pm, waiting time
will simply be zero. Therefore, let T be the waiting time in hours,
(b) Similar to part a), there are two cases. If Pat arrives before 9pm, the expected duration of the
date will be 3 hours. Otherwise, the expected duration will be 3−X 2 , since the duration is uniformly
distributed between 0 and (3 − X) hours. Therefore
, Expected duration

(c) The probability of having Pat late by more than 45 minutes on a date is 1 8 . Let U be the
expected number of dates they will have before breaking up, U = Y1 + Y2, where Y1 and Y2 are
i.i.d. with geometric distribution (p = 1 8 ). We know that E[Y1] = 1 p = 8. Therefore,
3. Let D be the number of types of drinks the bartender makes, and let M be the number of
people to enter the bar. Let X1, . . ., XN be the respective indicator variables of each drink. Thus if
at least one person orders drink i, then Xi = 1 otherwise it equals 0. Note that D = X1 + · · · + XN .
Thus we have:

4. (a) From the definition of expectation: E[Yg(X) | X] = P y yg(X)pY |X(y|x)
(b) Show that
Since

= E[E[Y | X] | X, Z] The Pull-Through and Tower Properties are not limited to discrete random
variables. These properties are also true in the continuous case. We can prove this by using the
same approach we used for the discrete case.
(b) We can actually prove a stronger statement than what is asked for in the problem, namely
that both pairs of random variables have the same covariance (from which it immediately
follows that their correlation coefficients have the same sign. We have
6. (a) The transform MJ (s) given is a transform of a binomial random variable with parameters n
= 10 and p = 2 3 . Thus the PMF for J is:

(b) Again by inspection, K is a geometric random variable shifted to the right by 3 with parameter
p = 1 5 . This is because we can rewrite MK(s) = e 3s 1 5 e s 1− 4 5 e s . Thus,
(c) Note that L = K1 + K2 + ...KJ , thus L is a random sum of random variables. So, determining
the transform of L is easier than determining the PMF for L
The expectation of L is E[L] = E[K]E[J] = 8 ∗ 20 3 = 160 3 The variance of L is

(d) P(person donates) = 1 4 . Let M = total # of donors from all living groups, and define
The PMF for X is just
Then,
Therefore the transform of M is:

The transform of X is (by inspection)
Therefore,
To obtain P(M = 0), we simply evaluate the transform of M at e s = 0.
The expectation of M is E[M] = E[X]E[L] = 40 3 The variance of M is

7. (a) Let the random variable T represent the number of widgets in 1 crate and let Ki represent the
number of widgets in the ith carton.
The transform of T is obtained by substituting the transform of N for every value of e s in the
transform of K.
Since T is a non-negative discrete random variable
Since T is a non-negative discrete random variable, we can solve this problem using a different
method.

We can find pT (0) by taking the limit of the transform as s → −∞.
Substituting pT (0) and MT (s) to find pT (1) we get,
If we take the limit now the numerator and denominator both evaluate to 0. Therefore, we need
to take the derivative of the numerator and denominator before we evaluate the limit.

(b) Since K and N are independent,
and

(c) Let W be the total weight of the widgets in a random crate.
The transform of W is M
(d) Since X and T are independent,
and

9. (a) Let X = T1 + T2 + ... + TN where N is a binomial with parameters p = 2 3 and n = 12. E[Ti ] = 1
λ = 1 3 , thus, Ti is an exponential with rate λ = 3, so fTi (t) = 3e −3t with t ≥ 0.
E[X] = E[Ti ]E[N] = 1 3 ∗ 12 ∗ 2 3 = 8 3
var(X) = var(Ti)E[N] + (E[Ti ])2var(N) = 1 9 ∗ 12 ∗ 2 3 + 1 9 ∗ 12 ∗ 2 3 ∗ 1 3 = 32 27
(b) The fact that Iwana spends AT LEAST 1 6 of an hour on each question shifts the transform in t
by 1 6 , thus fTi (t) = 3e −3(t− 1 6 ) for t ≥ 1 6 . We know that a shift by t in the pdf domain leads
to a multiplation by e ts in the transform domain. Therefore, the new MTi (s) = 3e s 6 3−s . Thus
we have,
(c) By the law of iterated expectations, E[N] = E[E[N|P]]. We can compute E[N|P] from the fact
that N is a binomial with parameter P, where P is a random variable uniformly distributed
between 0 and 1. Thus E[N] = E[12P] = 12E[P] = 12 ∗ 1 2 = 6
We compute the var(N) using the law of conditional variance: var(N) = E[var(N|P)]+ var(E[N|P])
= E[12P(1 − P)] + var(12P)

10. Let At (respectively, Bt) be a Bernoulli random variable which is equal to 1 if and only if the
tth toss resulted in 1 (respectively, 2). We have E[AtBt ] = 0 and E[AtBs] = E[At ]E[Bs] = p1p2 for
s 6= t. We have
11. (a) Here it is easier to find the PDF of Y . Since Y is the sum of independent Gaussian
random variables, Y is Gaussian with mean 2µ and variance 2σ 2 X + σ 2 Z .
(b) i. The transform of N is

Since Y is the sum of
• a random sum of Gaussian random variables
• an independent Gaussian random variable,
In general, this is not the transform of a Gaussian random variable. ii. One can differentiate the
transform to get the moments, but it is easier to use the laws of iterated expectation and
conditional variance:

iii. Now, the new transform for N is
Therefore,
(c) Given Y , the linear least-squared estimator of Xk is given by

To determine the mean and variance of Y we first determine those of N:
Now

Since
then

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