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Unformatted text preview: P ( X n +1 = (1q ) x n  X n = x n ) = 1x n and P ( X n +1 = q + (1q ) x n  X n = x n ) = x n . Therefore, E ( X n +1  X 1 = x 1 , . . . , X n = x n ) = (1q ) x n Â· P ( X n +1 = (1q ) x n  X n = x n ) + ( q + (1q ) x n ) Â· P ( X n +1 = q + (1q ) x n  X n = x n ) = (1q ) x n Â· (1x n ) + ( q + (1q ) x n ) Â· x n = (1q ) x n(1q ) x 2 n + qx n + (1q ) x 2 n = x nqx 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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 Fall '08
 MichaelKozdron
 Conditional Probability, Probability, Xn

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