Midterm Solutions - Stochastic Process | PSTAT 160A, Exams of Statistics

Material Type: Exam; Class: STOCHASTIC PROCESS; Subject: Statistics & Applied Probability; University: University of California - Santa Barbara;

Typology: Exams

2011/2012
On special offer
30 Points
Discount

Limited-time offer


Uploaded on 03/17/2012

rbaseball4
rbaseball4 🇺🇸

5

(2)

12 documents

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
PSTAT160A - Introduction to Stochastic Processes Brunick
PSTAT160A - Midterm - Solutions
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.
1. (20 pts) Suppose that we place two white balls in an empty urn. We then place a number of black
balls in the urn where the number black balls is determined by the roll on a fair die. After this
procedure, there are between three and eight balls total in the urn. Finally, we select a ball from
the urn. Assume that all balls are equally likely to be selected in this last step. Compute the
probability that we select a white ball conditional on the fact that the roll of the die is even. Please
fully simplify your answer.
Solution: Let Wdenote the event that the ball selected is white, and let Xdenote the number
rolled on the die. The problem tells us that P(W|X=i) = 2/(2 + i), so
P(W|Xis even) = P(Wand Xis even)
P(Xis even)
=P3
i=1 P(W|X= 2i)P(X= 2i)
1/2(by law of total probability)
=2/4·1/6+2/6·1/6+2/8·1/6
1/2= 13/36.
2. (20 pts) Let Ube uniformly distributed on (0,1) and let Xbe a geometric random variables with
parameter p, so P(X=i) = pqi1for iNwhere q= 1 p. Now assume that Uand Xare
independent and compute P(U1/X!), where as usual, the “!” denotes the factorial. Please
simplify your answer. It may help to recall that ex=PnN0xn/n!.
Solution: We have
P(U1/X!) = X
iN
P(U1/i!|X=i)P(X=i) (by law of total probability)
=X
iN
P(U1/i!) P(X=i) (because Uand Xare independent)
=X
iN
pqi1
i!
=p
qX
iN0
qi
i!1=p
q(eq1).
3. Suppose that T, the total number of people who visit a store on a given day, is Poisson with
parameter λ(i.e., P(T=i) = eλλi/i! for iN0). Further suppose that the sex of an individual
who visits the store is male with probability p, female with probability q= 1 p, and independent
of the total number of visitors and sex of all other visitors. Finally, let Mdenote the number of
1/3
pf3
Discount

On special offer

Partial preview of the text

Download Midterm Solutions - Stochastic Process | PSTAT 160A and more Exams Statistics in PDF only on Docsity!

PSTAT160A - Midterm - Solutions

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.

  1. (20 pts) Suppose that we place two white balls in an empty urn. We then place a number of black balls in the urn where the number black balls is determined by the roll on a fair die. After this procedure, there are between three and eight balls total in the urn. Finally, we select a ball from the urn. Assume that all balls are equally likely to be selected in this last step. Compute the probability that we select a white ball conditional on the fact that the roll of the die is even. Please fully simplify your answer. Solution: Let W denote the event that the ball selected is white, and let X denote the number rolled on the die. The problem tells us that P(W | X = i) = 2/(2 + i), so

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)

  1. (20 pts) Let U be uniformly distributed on (0, 1) and let X be a geometric random variables with parameter p, so P(X = i) = pqi−^1 for i ∈ N where q = 1 − p. Now assume that U and X are independent and compute P(U ≤ 1 /X!), where as usual, the “!” denotes the factorial. Please simplify your answer. It may help to recall that ex^ =

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).

  1. Suppose that T , the total number of people who visit a store on a given day, is Poisson with parameter λ (i.e., P(T = i) = e−λλi/i! for i ∈ N 0 ). Further suppose that the sex of an individual who visits the store is male with probability p, female with probability q = 1 − p, and independent of the total number of visitors and sex of all other visitors. Finally, let M denote the number of

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.

  1. (20 pts) Let (Xn)n∈N 0 be a simple random walk which steps up with probability p. Compute P(X 1 = X 4 + 1 = X 8 + 1). Solution: Set ξn = Xn − Xn− 1. Then P(X 1 = X 4 + 1 = X 8 + 1) = P

X 4 − X 1 = −1 and X 8 − X 4 = 0

= P

( ∑^4

i=

ξi = −1 and

∑^8

i=

ξi = 0

= P

( ∑^4

i=

ξi = − 1

P

( ∑^8

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.