solutions05 - P ( X n +1 = (1-q ) x n | X n = x n ) = 1-x n...

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Stat 351 Fall 2007 Solutions to Assignment #5 1. (a) We begin by calculating E ( Y 1 ). That is, E ( Y 1 ) = 1 · P ( Y = 1) + ( - 1) · P ( Y = - 1) = p - (1 - p ) = 2 p - 1 . We now notice that S n +1 = S n + Y n +1 . Therefore, E ( S n +1 | X 1 , . . . , X n ) = E ( S n + Y n +1 | X 1 , . . . , X n ) = E ( S n | X 1 , . . . , X n ) + E ( Y n +1 | X 1 , . . . , X n ) = S n + E ( Y n +1 ) = S n + 2 p - 1 and so E ( X n +1 | X 1 , . . . , X n ) = E ( S n +1 - ( n + 1)(2 p - 1) | X 1 , . . . , X n ) = E ( S n +1 | X 1 , . . . , X n ) - ( n + 1)(2 p - 1) = S n + 2 p - 1 - ( n + 1)(2 p - 1) = S n - n (2 p - 1) = X n . We can now conclude that { X n , n = 1 , 2 , . . . } is, in fact, a martingale. 1. (b) Notice that Z n +1 = ± 1 - p p ² S n +1 = ± 1 - p p ² S n + Y n +1 = ± 1 - p p ² S n ± 1 - p p ² Y n +1 Therefore, E ( Z n +1 | X 1 , . . . , X n ) = E ³ ± 1 - p p ² S n ± 1 - p p ² Y n +1 | X 1 , . . . , X n ! = ± 1 - p p ² S n E ³ ± 1 - p p ² Y n +1 | X 1 , . . . , X n ! = ± 1 - p p ² S n E ³ ± 1 - p p ² Y n +1 ! We now compute E ³ ± 1 - p p ² Y n +1 ! = p ± 1 - p p ² 1 + (1 - p ) ± 1 - p p ² - 1 = (1 - p ) + p = 1 and so we conclude E ( Z n +1 | X 1 , . . . , X n ) = ± 1 - p p ² S n = Z n . Hence, { Z n , n = 1 , 2 , . . . } is, in fact, a martingale.
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2. We interpret the conditional probabilities given in the problem to mean
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Unformatted text preview: P ( X n +1 = (1-q ) x n | X n = x n ) = 1-x n and P ( X n +1 = q + (1-q ) x n | X n = x n ) = x n . Therefore, E ( X n +1 | X 1 = x 1 , . . . , X n = x n ) = (1-q ) x n P ( X n +1 = (1-q ) x n | X n = x n ) + ( q + (1-q ) x n ) P ( X n +1 = q + (1-q ) x n | X n = x n ) = (1-q ) x n (1-x n ) + ( q + (1-q ) x n ) x n = (1-q ) x n-(1-q ) x 2 n + qx n + (1-q ) x 2 n = x n-qx n + qx n = x n . In other words, E ( X n +1 | X 1 , . . . , X n ) = X n and so we conclude that { X n , n = 1 , 2 , . . . } is, in fact, a martingale....
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This note was uploaded on 03/26/2012 for the course STAT 351 taught by Professor Michaelkozdron during the Fall '08 term at University of Regina.

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solutions05 - P ( X n +1 = (1-q ) x n | X n = x n ) = 1-x n...

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