STAT_333_Assignment_2_Solutions

STAT_333_Assignment_2_Solutions - STAT 333 Assignment 2...

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STAT 333 Assignment 2 SOLUTIONS Type I Problems 1. Suppose X has probability density function f(x) = 2x 3 for x 1. (This is called a power-law or Pareto distribution.) Suppose Y is independent of X and has a gamma distribution with parameters α = 4 and λ = 2. Obtain P(X > Y ). First obtain the cdf of X, F X (x) = for x 1 [] 2 1 2 1 3 1 2 = = x u du u x x 8 y + = = > 1 2 3 2 1 0 2 3 0 2 3 0 2 3 4 0 3 ) ( 3 8 ) 1 ( 3 ) ( ! 2 ) ( ) ( ) | ( dy e y y dy e y e X e y X dy y f y Y Y X P y y y Y = Then, to find P(X > Y), condition on Y P(X > Y) = > = 3 | dy y Y y P substitution > = 8 dy y y P independence since if y < 1, X is guaranteed to be bigger You can then evaluate the integrals either by integrating by parts, using Maple/other software, or with some help from the Gamma function technique. Obviously I can’t ask you something this hard on a test – the point of this question was to use the conditioning to set up the integral and recognize that you need different things if y is below or above 1. 2. # 37 Let T represent the typist a) We know that X|T = A ~ Poi(2.6), X|T = B ~ Poi(3), and X|T = C ~ Poi(3.4) So E[X] = E[E[X|T]] = E[X|T = A]P(T = A) + E[X|T = B]P(T = B) + E[X|T = C]P(T = C) = 2.6 * 1/3 + 3 * 1/3 + 3.4 * 1/3 = 3 b) For Poisson random variables, the variance and mean are both equal to the parameter λ . So the second moment of a Poisson is λ + λ 2 E[X 2 |T = A] = 2.6 + 2.6 2 etc So E[X 2 ] = E[E[X 2 |T]] = E[X 2 |T = A]P(T = A) + E[X 2 |T = B]P(T = B) + E[X 2 |T = C]P(T = C) = (2.6 + 2.6 2 ) * 1/3 + (3 + 3 2 ) * 1/3 + (3.4 + 3.4 2 ) * 1/3 = 12.1067 And thus Var(X) = 12.1067 - 3 2 = 3.1067 3. # 44 Let N represent the number of customers in the store on a given day. N ~ Poi(10) Let X i represent the amount spent by the i th customer of the day. X i ’s ~ X ~ U(0, 100) Let S represent the total amount of money on the day.
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We must assume that the number of customers and the amounts that any one customers spends are all independent random variables. (This may or may not actually be a good assumption…) Then S = X 1 + X 2 + … + X N is a compound random variable since it satisfies the definition. We know E[N] = 10 and Var(N) = 10 since N is Poisson We know (or can derive) E[X] = 50 and Var(X) = 100 2 /12 since X is Uniform So using the properties for compound random variables, E[S] = E[N]E[X] = 10 * 50 = 500 Var(S) = E[N]Var(X) + (E[X]) 2 Var(N) = 10 * 100 2 /12 + 50 2 * 10 = 33,333.33 4. # 62 Let p A , p B , and p C be the probabilities that A, B, and C win overall, respectively. We must have one winner, so p A + p B + p C = 1. Also, A and B are equivalent (we could just rename them) so p A = p B . We need a third equation in order to solve. Look at the probability A wins overall, conditional on the outcome of the first game. p A = P(A wins overall|wins first)*P(wins first) + P(A wins overall|loses first)*P(loses first) Let’s look at P(A wins overall|loses first) ½ the time, B will win against C, and the probability of A winning overall is 0. The other ½ the time, A will be in the exact same starting position as C – about to play the winner of the current game.
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This note was uploaded on 01/16/2011 for the course STAT 333 taught by Professor Chisholm during the Spring '08 term at Waterloo.

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STAT_333_Assignment_2_Solutions - STAT 333 Assignment 2...

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