This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Denker SPRING 2010 416 Stochastic Modeling  Assignment 3 Solutions Problem 1: (Problem 46, p.173) Show that Cov ( X,Y ) = Cov ( X,E [ Y  X ]) . We first show that E [ XE [ Y  X ]] = E [ XY ]. Indeed if X and Y are discrete E [ XE [ Y  X ]] = summationdisplay x xE [ Y  X = x ] P ( X = x ) = summationdisplay x,y xy P ( X = x,Y = y ) P ( X = x ) P ( X = x ) = E [ XY ] . The case of continuous random variables is similar replacing sums by integrals. Since E [ E ( Y  X )] = E [ Y ] and Cov ( U,V ) = E [ UV ] − E [ U ] E [ V ], we arrive at Cov ( X,E [ Y  X ]) = E [ XE [ Y  X ]] − E [ E ( Y  X )] E [ X ] = E [ XY ] − E [ Y ] E [ X ] = Cov ( X,Y ) . Problem 2: (Problem 49, p. 174) A and B play a series of games with A winning each game with probability p . The overall winner is the first player to have won two more games than the other. (a) Find the probability that A is the overall winner. (b) Find the expected number of games played. Let X i denote the sequence of games played, where X i = 1 if player A wins in the ith game and X i = − 1 if B wins in the ith game. Then S n = X 1 + ... + X n describes when A has won two more than B. This happens when S n = 2....
View
Full
Document
 Spring '08
 DENKER
 Stochastic Modeling, Probability theory

Click to edit the document details