THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE

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1 THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT131 PROBABILITY AND STATISTICS I EXAMPLE CLASS 8 Review Conditional Distributions and Conditional Expectation For any two events E and F, the conditional probability of E given F is defined by P (E F ) P (E F ) P (F ) provided that P(F) > Let (X, Y ) be a discrete bivariate random vector with joint pmf p(x, y) and marginal pmfs p x (x) and p y (y). The conditional pmf of Y given that X x is the function of y denoted by p Y X (y x), where p X (x) > p Y X (y x) P (Y y X x) If X is independent of Y, then the conditional pmf becomes p Y X (y x) P (Y y, X x) P (X x) p(x, y) p X (x) p X(x)p Y (y) p X (x) p Y (y) p(x, y) p X (x) For continuous random variables, the conditional distributions are defined as: f Y X (y x) f X Y (x y) f(x, y) f X (x) f(x, y) f Y (y) provided that f X (x) > provided that f Y (y) > Definitions and Formula: (a) Conditional Distribution Function of Y given X x { i y F Y X (y x) P (Y y X x) p Y X(i x) y f Y X(t x)dt discrete case continuous case 1

2 (b) Conditional Expectation Function of g(y) given X x { i E(g(Y ) X x) g(i)p Y X(i x) + g(t)f Y X(t x)dt discrete case continuous case (c) Conditional Mean of Y given X x : E(Y X x) (d) Conditional Variance of Y given X x V ar(y X x) E((Y E(Y X x)) 2 X x) E(Y 2 X x) (E(Y X x)) 2 (e) Computing Expectations by Conditioning E(X) E(E(X) Y ) V ar(x) E(V ar(x Y )) + V ar(e(x Y )) In general, E(u(X)) E(E(u(X)) Y ) V ar(u(x)) E(V ar(u(x) Y )) + V ar(e(u(x) Y )) Problems Problem 1 Let the continuous random vector (X, Y ) have joint pdf: f(x, y) e y, < x < y < (a) Compute the conditional pdf of Y given X x. (b) Evaluate E(Y Xx) and Var(Y X x). (c) Calculate E(Y) and Var(Y). (a) f X (x) f(x, y)dy { x x e y dy e x x > Thus X has an exponential distribution with parameter λ 1. For any x > as the value for f X (x) >, then we have the following conditional pdf f Y X (y x) f(x, y) f X (x) { e y e x e (y x) y > x e x y x 2

3 (b) E(Y X x) x ye (y x) dy 1 + x V ar(y X x) E(Y 2 X x) (E(Y X x)) 2 1 x y 2 e (y x) dy (1 + x) 2 (c) { f Y (y) f(x, y)dx y y e y dx ye y y > As a result, Y has a gamma distribution with parameter α 2, λ 1 E(Y ) α λ 2, V ar(y ) α λ 2 2 Alternative method: E(Y ) E(E(Y X)) E(1 + X) 1 + E(X) 2 E(V ar(y X)) E(1) 1 V ar(e(y X)) V ar(1 + X) V ar(x) 1 V ar(y ) E(V ar(y X)) + V ar(e(y X)) 2 Problem 2 A prisoner is trapped in a cell containing 3 doors. The first door leads to a tunnel that returns to his cell after 2 days travel. The second leads to a tunnel that returns him to his cell after 4 days travel. The third door leads to freedom after 1 day of travel. If it is assumed that the prisoner will always select doors 1, 2 and 3 with respective probabilities.5,.3 and.2, what is the expected number of days until the prisoner reaches freedom? Let X be the number of days until the prisoner reaches freedom. Conditional on the prisoner s choice on the first day E(X).5E(X 1st door) +.3E(X 2nd door) +.2E(X 3rd door).5(2 + E(X)) +.3(4 + E(X)) +.2(1) E(X) Then E(X)

4 The expected number of days until the prisoner reaches freedom is 12. Problem 3 The number of customers using the automatic teller machine in a particular day follows a Poisson distribution with λ 18. The amount of money withdrawn by each customer is a random variable with mean $3 and standard deviation $5. (A negative withdrawal means that money was deposited). Find the mean and variance of the total daily withdrawal. Let N be the number of customers used the ATM in one day. Then N P oisson(18). Let X i, i 1, 2,..., N be the amount of each transaction. The daily total transaction, Y, equal to N i1 X i. Therefore, E(Y N) N E(X i ) 3N, V ar(y N) i1 N V ar(x i ) 25N i1 (Note: X i are assumed to be independent given N) E(Y ) E(E(Y N)) E(3N) 3E(N) V ar(y ) E(V ar(y X)) + V ar(e(y X)) E(25N) + V ar(3n) Problem 4 An insurance company supposes that each person has an accident parameter and the yearly number of accidents of someone whose accident parameter is Λ is Poisson distributed. They also suppose that given a newly insured person has n accidents in her first year, the conditional pdf of her accident parameter Λ is a gamma distribution Γ(n + α, β + 1). Determine the expected number of accidents that she will have in the following year. Let N be the number of accidents of an insured person in the first year. Then, Λ N n Γ(n + α, β + 1) 4

5 Let M be the number of accidents of the same insured person in the second year. By the second assumption of the Poisson process, given the value of Λ, M and N are independent. Then E(M Λ, N n) E(M Λ) Λ As a result, E(M N n) E(E(M Λ, N n) N n) E(Λ N n) n + α β + 1 Problem 5 An insect lays a large number of eggs, each surviving with probability p. Assume the number of eggs laid Y has a Poisson distribution with parameter λ and each egg s survival is independent. Let X number of survivors. (a) On the average, how many eggs will survive? (b) What is the distribution of X? (a) Y P oisson(λ), X Y Binomial(Y, p) E(X) E(E(X Y )) E(pY ) pλ 5

6 (b) P (X x) As a result, X P oisson(λp). P (X x, Y y) y P (X x Y y)p (Y y) y ( C y x p x (1 p) y x) ( ) e λ λ y yx e λ (λp) x e λ (λp) x yx ((1 p)λ) y x (y x)! ((1 p)λ) t y e λ (λp) x e (1 p)λ e λp (λp) x t! let t y-x Problem 6 Consider that there are a large number insect mothers and it is no longer clear that the number of eggs laid follows the same Poisson distribution for each mother. Now one mother is chosen at random. Let X number of survivors and Y number of eggs the mother laid. The model is Λ Exp(β), Y Λ P oisson(λ), X Y Binomial(Y, p). (a) On average, how many eggs will survive? (b) What is the pmf of Y? (c) What is the condition pdf of Λ Y? (a) E(X) E(E(X Y )) E(pY ) pe(y ) pe(e(y Λ)) pe(λ) p β 6

7 (b) P (Y y) P (Y y, < Λ < ) β f(y, λ)dλ f Y Λ (y λ)f Λ (λ)dλ ( ) e λ λ y βe βλ dλ λ y e (1+β)λ dλ ) y+1 β ( 1 Γ(y + 1) 1 + β β ( ) y 1, y, 1,... β β (c) The joint pdf is: ( ) e λ λ y f Y,Λ (y, λ) βe βλ, y, 1,..., λ > Then the condition pdf of Λ Y is : f Λ Y (λ y) f Y,Λ(y, λ) f Y (y) ( ) e λ λ y β β+1 ( βe βλ 1 1+β ) y, λ > 7