University of Chicago Graduate School of Business. Business 41901: Probability Final Exam Solutions

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1 Name: University of Chicago Graduate School of Business Business 490: Probability Final Exam Solutions Special Notes:. This is a closed-book exam. You may use an 8 piece of paper for the formulas.. Throughout this paper, N(µ, σ ) will denote a normal distribution with mean µ and variance σ. 3. This is a hr exam. Honor Code: By signing my name below, I pledge my honor that I have not violated the Booth Honor Code during this examination. Signature:

2 Problem A. True or False: Please Explain your answers in detail. Partial credit will be given (60 points). (Normal) Let X be normal with mean zero, then the distribution of X is a Rayleigh distribution. Solution: False. X χ, with PDF given by whereas Rayleigh distribution has PDF π x / e x/, x x σ e σ.. (Russian Roulette) I have a revolver (six chambers) loaded with one bullet. I pull the trigger once, click, and hand you the revolver. Do you want to to spin it first, or shoot directly? Solution: True. With spin, we have P (survival) = 5/6. Without spin, P (survival) = 4/5. Hence, it s better to spin. 3. (Children and Apples) You have three children, but only one apple. You want to toss a fair coin to decide who gets the apple. You want to be fair. You flip a coin twice. Assign HH, HT, and T T to the children, and ignore T H if you throw it. This is a fair allocation. Solution: True. Let Y, Y Bern(/) be iid over {H, T }. Each element of the space Z = {HH, HT, T T, T H} has probability p(z = Z) = p(y = y Y = y ) = p(y = y )p(y = y ) = /4. If we exclude T H, then p(z) is simply renormalized; thus, each event still has equal probablility. 4. (Raisins per Cookie) A baker put 500 raisins into his dough, and made 00 cookies. You take a random cookie. The probability of finding exactly raisins is 8. Solution: True. Binomial distributed. Imagine the raisins are thrown in one after the other. For every raisin, p = /00 of being in the cookie. But n = 500. Then λ = np = 5 for the corresponding Poisson distribution, and P (X = ) 8%. 5. (Bayes) If (y θ) N(θ, ) and θ has a Jeffreys flat uniform prior p(θ), then the posterior distribution (θ y) N(y, ).

3 Solution: True. p(θ y) p(y θ)p(θ) = e (y θ) / = e (θ y) / Thus, as a function of θ we have θ y N(y, ) 6. (Cauchy) The reciprocal of a Cauchy distribution has finite mean and variance Solution: False. /C D = C, since C = X/Y where X, Y N(0, ), so /C = Y/X. 7. (Order Statistics) Let X,..., X n be a random sample from an exponential distribution with mean one. Then the probability density of any order statistic is also exponential. Solution: False. The distribution for an order statistic, k, of a continuous distribution X from n samples is given by f X(k) = so when X Exp() and x 0 n! (k )!(n k)! (F X(x)) k ( F X (x)) n k f X (x) n! ( f X(k) = ) e x k ( ( e x ) ) n k e x (k )!(n k)! n! ( = ) e x k e (n k+)x (k )!(n k)! which is not exponentially distributed. 8. (Transformations) Let Y = e X have a Pareto distribution. Then Y is a log-normal distribution. Solution: False. The Pareto distribution is given by f Y (y; α, β) = { αβ α y α+ y β 0 otherwise The transform y = g (z) = z = e x has derivative dy = dg (z) = z = e x, so dz dz { αβ α e αx x ln(β)/ f Z (z; α, β) = 0 otherwise 3

4 which is proportional to an Exponential distribution on its support. The log-normal distribution is given by x πσ e (ln x µ) σ for x 0, σ > 0. The two distributions do not share the same form. 9. (Brownian Motion) Let B t be a standard Brownian motion. Then we can calculate the mean E ( B t B s ) = t s for t > s. Solution: False. Z = B t B s N(0, t s) and Y = Z f Y (y; t s) = πσ x e x4 (t s) 4 x > 0 0 otherwise () and has E[Y ] = /4 π Γ(3/4)(t s) t s. 0. (Moments) The function /( t) cannot be a moment generating function, M X (t), of any random variable, X. Solution: False. It is the moment generating function of the Gamma distribution. M X (t) = E[e tx ] = βα Γ(α) = βα Γ(α) = βα Γ(α) e tx x α e βx dx x α e (β t)x dx x α e (β t)x dx, where the integrand is another gamma PDF with β β t, so Now, let α =, β = = βα Γ(α) Γ(α) (β t) α = β α (β t). α ( t). (Martingales) A submartingale has the property that E(X t ) > X 0 for any t. 4

5 Solution: True. By definition E[X t X s ] > X s, hence by iterated expectation E[X t ] = E[E[X t X s ]] > E[X s ], s < t, including s = 0, which proves the statement.. (Odds) Consider betting x against Secretariat and y against Zenyatta. Crazy Eddie takes in x + y in bets and given his beliefs in the following payout table Horse Eddie s Beliefs Fair odds If horse loses Eddie pays Secretariat : x + x = 3x 3 Zenyatta : y + y = y Table : Eddies Payout Then Eddie will always lose money if we bet x =, y = 3 against him. Solution: True. 5

6 Problem B. (0 points) Let X and Y be independent with a joint distribution given by f X,Y (x, y) = ( π xy exp x y ) where x, y > 0. Identify the following distributions The marginal distribution of X Solution: e (x+y) dx = y e y π 0 xy π Compute the joint distribution of U = X and V = X + Y Solution: ( We ) perform the substitutions y v x and x u and obtain the parameter u vector, which has the following Jacobian: v u Thus, the transformed distribution is Compute the marginal distribution of V. Solution: ( ) 0 = f U,V (u, v) = e v/ π u(v u) v e v/ du = π 0 u(v u) e v/ 6

7 Problem C. (0 points) (Hit and Run Taxi) A certain town has two taxi companies: Blue Birds, whose cabs are blue, and Night Owls, whose cabs are black. Blue Birds has 5 taxis in its fleet, and Night Owls has 75. Late one night, there is a hit-and-run accident involving a taxi. The town s taxis were all on the streets at the time of the accident. A witness saw the accident and claims that a blue taxi was involved. At the request of the police, the witness undergoes a vision test under conditions similar to those on the night in question. Presented repeatedly with a blue taxi and a black taxi, in random order, he shows he can successfully identify the colour of the taxi 4 times out of 5. Which company is more likely to have been involved in the accident? Solution: Let W = witness sees blue, B = blue birds. We have the following probabilities P (W B) P (W B) =. P (B) We have P (B W ) = = %. (Children) I tell you that I have two children, and that one of them is a girl (I say nothing about the other). What is the probability that I have two girls? Solution: Let X = two girls, Y = told one girl, Z = girl opens. We are looking for P (X Y ) = / /3 = 3. You come to visit, and a girl (who is my daughter) opens your door. What is the probability that I have two girls? 7

8 Solution: Now we are looking for P (X Y Z) = /4 /4 + /4 =. 8