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Introduction to Probability Solutions of Final Exam
Due:August 8th, 2011 Solve all the problems
- (15 points) You have three coins, showing “Head” with probabilities p 1 , p 2 and p 3. You perform two different experiments: 1. You choose one coin at random and toss it repeatedly. 2. You repeatedly choose a coin at random and toss it.
In both cases calculate the average number of “Heads” among the first n tosses, and the average time you have to wait for the first “Head”.
Solution. 1. (a)
EN =
∑^3
i=
E[N |p = pi]P(p = pi) =
∑^3
i=
npi
(p 1 + p 2 + pi)
(b)
ET =
∑^3
i=
E(T |p = pi)P(p = pi) =
∑^3
i=
∑^ ∞
n=
nP(T = n|p = pi)
∑^3
i=
∑^ ∞
n=
n(1 − pi)n−^1 pi =
∑^3
i=
pi p^2 i
p 1
p 2
p 3
- (a) Let Xi = 1 if the coin shows a Head on the ith turn, and zero otherwise, for i = 1, · · · , n. Then
N = X 1 + · · · + Xn ⇒ EN = E(X 1 + · · · + Xn) = EX 1 + · · · + EXn.
Since EXk =
i=1 E(Xi|p^ =^ pi)P(p^ =^ pi) =^
1 3
i=1 pi, then
EN =
∑^3
i=
pi + · · · +
∑^3
i=
pi =
n 3
(p 1 + p 2 + p 3 ).
(b) Since P(Xi = 1) =
i=1 P(Xi^ = 1|p^ =^ pi)P(p^ =^ pi) =^
p 1 +p 2 +p 3 3 is independent of i, and Xi are independent, then T is a geometric random variable with parameter p = p^1 +p 32 +p^3. Therefore
ET =
p
p 1 + p 2 + p 3
- (15 points) Let X 1 , · · · , X 4 be four independent random variables, and gi : R^2 → R functions for i = 1, 2. Show that Y 1 = g 1 (X 1 , X 2 ) and Y 2 = g 2 (X 3 , X 4 ) are independent.
Proof. We want to show that
P(Y 1 ∈ F, Y 2 ∈ G) = P(Y 1 ∈ F )P(Y 2 ∈ G),
for any event F and G. There are sets A 1 , A 2 such that
{Y 1 ∈ F } := {g 1 (X 1 , X 2 ) ∈ F } = {X 1 ∈ A 1 , X 2 ∈ A 2 }.
Similarly, there are sets A 3 , A 4 , such that
{Y 2 ∈ G} := {g 2 (X 3 , X 4 ) ∈ G} = {X 3 ∈ A 3 , X 4 ∈ A 4 }.
Therefore,
P(Y 1 ∈ F, Y 2 ∈ G) = P(Xi ∈ Ai, for i = 1, · · · , 4) = P(X 1 ∈ A 1 , X 2 ∈ A 2 )P(X 3 ∈ A 3 , X 4 ∈ A 4 ) = P(Y 1 ∈ F )P(Y 2 ∈ G).
This shows that Yn is a geometric random variable, with parameter p. Furthermore, we can write
P(Y 1 = l 1 , Y 2 = l 1 , · · · , Yn = ln) = P(Y 1 = l 1 ) × · · · × P(Yn = ln).
Therefore Y 1 , · · · , Yn are independent.
- (20 points) Suppose that every day can be described by either S (sunny) or R (rainy). The probability for a day to be sunny is 4/8 if the preceding day was rainy, 6/8 if the preceding day was sunny, but 7/8 if both of the preceding days were sunny. 1. Let S = {S, R}, and let Xn = the weather at nth day. Is Xn a Markov chain? 2. Define a Markov chain on the state space S = {SS, SR, RS, RR} that describes the above weather model, and determine the corresponding transition matrix. 3. Given that the weekend(Saturday and Sunday) was sunny calculate the probability for the next five days to be SSRSS.
Solution. 1. No, because knowing the weather status of today, tomor- row’s weather is not independent of weather of yesterday.
SS
SR RS
RR
P =
P(SSRSS|SS) = P(X 1 = SS, X 2 = SS, X 3 = SR, X 4 = RS, X 5 = SS|X 0 = SS)
- (20 points) Consider the following dice game. A pair of dice are rolled. If the sum is 7 then the game ends and you win 0. If the sum is not 7, then you have the option of either stopping the game and receiving an amount equal to that sum or starting over again. For each value of i, i = 1, · · · , 12 find your expected return if you employ the strategy of stopping the first time that a value at least as large as i appears. What value of i leads to the largest expected return? (Hint: Let Xi denote the return when you use the critical value i. To compute EXi, condition on the first sum)
Solution. Assume EiG denote the expected gain, when our strategy is to stop as soon as the sum is at least i, where possible i’s are 2, · · · , 12. Let’s calculate EiG.
EiG = Ei[G|S < i]P(S < i) + Ei[G|S ≥ i]P(S ≥ i)
∑^ i−^1
k=
Ei[G|S = k]P(S = k) +
∑^12
k=i
Ei[G|S = k]P(S = k)
= Ei[G]
∑^ i−^1
k=2,k 6 =
P(S = k) +
∑^12
k=i,k 6 =
kP(S = k),
where we exclude k = 7 because Ei[G|S = k] = 0, when k = 7. We can solve this equation for EiG. If we observe that P(S < i) =
EiG =
k=i,k 6 =7 kP(S^ =^ k) 1 −
∑i− 1 k=2,k 6 =0 P(S^ =^ k)^
The numerator, and the denominator, are both finite sums, and P(S = k) is easy to calculate. Therefore the numerical value of EiG is computable. The result will be Ei^ attains its largest value when i = 7 and i = 8.
- (20 points) The number of accidents that a person has in a given year is a Poisson random variable with mean λ. However, suppose that the value of λ varies in from person to person. Assume the proportion of the population having a value of λ less than x is equal to 1 − e−x. If a person is chosen at random what is the probability that he will have 1. 0 accidents in a year. 2. 0 accidents in a year given he had no accident the preceding year
(Hint: instead of P(λ < x) = 1 − e−x, first solve the problem for an easier case where P(λ = 2) =. and P(λ = 2.5) = .7. )
Solution. 1. Let L denote the number accidents of this person. We don’t know the mean λ associated to this person. But given λ = x, L is a Poisson random variable N with mean x. Therefore, P(L = 0|λ = x) = P(N = 0) = e−x. (0.1) By the total probability formula for the continuous random variable λ,
P(L = 0) =
0
P(L = 0|λ = x)pλ(x)dx,
where pλ(x) is the density of λ. Since P(λ < x) = 1 − e−x, then λ is an exponential random variable with mean 1. Then pλ(x) = e−x. We also know that given λ = x, then N is a Poisson random variable with mean x therefore P(L = 0) =
0
e−xe−xdx =. 5
- Define two events A and B as follows A := {This driver will not have an accident this year}, B := {This driver didn’t have any accident last year}. Notice that P(A|λ = x) = P(B|λ = x) = e−x, as in (0.1). Then
P(A|B) =
P(A ∩ B)
P(B)
Observe that A, and B are not independent, (if somebody had 10 acci- dents last year, he probably has a larger λ, and this means he is likely to have a few accidents this year too), however, conditionally on λ they are independent (Compare this with Example 1.21 on page 37 of your text). Therefore
P(A ∩ B) =
0
P(A ∩ B|λ = x)pλ(x)dx
0
P(A|λ = x)P(B|λ = x)pλ(x)dx =
0
e−^3 xdx =
Similarly, P(B) = 13 , then the final answer is 2/3.
