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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