416-3sol - Denker SPRING 2010 416 Stochastic Modeling -...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the 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 i-th game and X i = 1 if B wins in the i-th 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

Page1 / 3

416-3sol - Denker SPRING 2010 416 Stochastic Modeling -...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online