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                            62                  Conditional Distributions and Expectation

                                                        2.6 Martingales

                            Consider a sequence of real-valued random variables {X 1 , X 2 ,...}, such that E(|X n |) <
                            ∞ for all n = 1, 2,.... The sequence {X 1 , X 2 ,...} is said to be a martingale if for any
                            n = 1, 2,...,

                                                     E(X n+1 |X 1 ,..., X n ) = X n
                            with probability 1.
                              A martingale may be viewed as the fortunes of a gambler playing a sequence of fair
                            games with X n denoting the fortune of the gambler after the nth game, n = 2, 3,..., and
                            X 1 representing the initial fortune. Since the games are fair, the conditional expected value
                            of X n+1 , the fortune at stage n + 1, given the current fortune X n ,is X n .

                            Example 2.23 (Sums of independent random variables). Let Y 1 , Y 2 ,... denote a sequence
                            of independent random variables each with mean 0. Define
                                                 X n = Y 1 +· · · + Y n ,  n = 1, 2,...

                            Then
                                     E(X n+1 |X 1 ,..., X n ) = E(X n |X 1 ,..., X n ) + E(Y n+1 |X 1 ,..., X n ).

                            Clearly, E(X n |X 1 ,..., X n ) = X n and, since (X 1 ,..., X n )isa function of (Y 1 ,..., Y n ),
                            E(Y n+1 |X 1 ,..., X n ) = 0. It follows that {X 1 , X 2 ,...} is a martingale.


                            Example 2.24 (Polya’s urn scheme). Consider an urn containing b black and r red balls.
                            A ball is randomly drawn from the urn and c balls of the color drawn are added to the urn.
                            Let X n denote the proportion of black balls after the nth draw. Hence, X 0 = b/(r + b),
                                                                                         b
                                   Pr[X 1 = (b + c)/(r + b + c)] = 1 − Pr[X 1 = b/(r + b + c)] =  ,
                                                                                       r + b
                            and so on.
                              Let Y n = 1ifthe nth draw is a black ball and 0 otherwise. Clearly,

                                                   Pr(Y n+1 = 1|X 1 ,..., X n ) = X n .
                            After n draws, there are r + b + nc balls in the urn and the number of black balls is given
                            by (r + b + nc)X n . Hence,
                                                         (r + b + nc)X n + Y n+1 c
                                                  X n+1 =
                                                            r + b + (n + 1)c
                            so that
                                                             (r + b + nc)X n + X n c
                                          E[X n+1 |X 1 ,..., X n ] =            = X n ;
                                                                r + b + (n + 1)c
                            it follows that {X 1 , X 2 ,...} is a martingale.

                              Using the interpretation of a martingale in terms of fair games, it is clear that if the
                            gambler has fortune c after n games, then the gamblers expected fortune after a number of
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