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Material Type: Exam; Class: STOCHASTIC PROCESS; Subject: Statistics & Applied Probability; University: University of California - Santa Barbara;
Typology: Exams
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Please write your name and section time on your solutions and then solve the following problems. Please start each new problem (but not subproblem) on a new sheet of paper. To receive full credit, you must label the relevant events and random variables and justify all the steps in your calculation. Useful justifications include “by independence,” “by the law of total probability,” “by the tower law of conditional expectation,” etc. Err on the side of providing too many details.
P(W | X is even) =
P(W and X is even) P(X is even)
=
i=1 P(W^ |^ X^ = 2i)^ P(X^ = 2i) 1 / 2
(by law of total probability)
n∈N 0 x
n/n!.
Solution: We have
P(U ≤ 1 /X!) =
i∈N
P(U ≤ 1 /i! | X = i) P(X = i) (by law of total probability)
i∈N
P(U ≤ 1 /i!) P(X = i) (because U and X are independent)
i∈N
pqi−^1 i!
p q
i∈N 0
qi i!
p q
(eq^ − 1).
men who visit the store, and let W denote the number of women who visit the store. You may want to recall that the mean and variance of a Poisson random variable with parameter λ are both λ.
(a) (10 pts) Compute the probability that P(M = m, W = n) for m, n ∈ N 0. Do not expect to this expression to simplify completely. Hint: rewrite the desired event in terms of M and T. (b) (10 pts) Compute the expected value of M. (c) Bonus (5pts): Show that M and W are independent Poisson random variables and determine the parameter associated with each random variable.
Solution:
(a) The problem essentially tells us that M ∼ Binomial(p, ) and N ∼ Binomial(q,) conditional on the event {T = `}. (Of course, M , W , and T are not independent as M + W = T .) As a result, P(M = m, W = n) = P(M = m, T = m + n) = P(M = m | T = m + n)P(T = m + n)
=
m + n m
pmqne−λλm+n/(n + m)!.
(b) As M ∼ Binomial(p, ) conditional on the event {T =}, we have E[M | T = ] = p. As a result, we have
E[M ] =
n∈N 0
E[M | T = ] P(T =) (by tower law)
n∈N 0
pP(T =) = pE[T ] = pλ.
(c) To solve the bonus part, we rewrite this expression as:
P(M = m, W = n) =
(m + n)! m!n!
pmqne−λλm+n/(n + m)! =
e−pλ(pλ)m/m!
e−qλ(qλ)n/n!
for m, n ∈ N 0. Summing over n shows that M ∼ Poisson(pλ), summing over m shows that W ∼ Poisson(qλ), and the product structure of the joint probability mass function then shows that the random variables are independent.
X 4 − X 1 = −1 and X 8 − X 4 = 0
i=
ξi = −1 and
i=
ξi = 0
i=
ξi = − 1
i=
ξi = 0
(because the steps are independent)
= P(one up and two down) P(two up and two down)
=
p^1 q^2
p^2 q^2
= 9p^3 q^4.