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Unformatted text preview: Math 280A, Fall 2011 Homework 8 Solutions 1. In this problem, the random variables X 1 , X 2 , . . . and X are defined on a probability space ( , F , P ) with the special property that has only finitely many points, say = { 1 , 2 , . . . , N } , and that each singleton { k } is an element of F with P [ { k } ] > for each k . Show that X n P X if and only if X n X a.s. Solution. Since almost sure convergence implies convergence in probability quite generally, we need only concern ourselves with the reverse implication. Thus assume that X n P X . From class discussion (Dominated Convergence) we know that this means that (1 . 1) lim n E [min(  X n X  , 1)] = 0 . But, defining := min( P [ { 1 } ] , . . . , P [ { N } ]) > 0, we have (1 . 2) E [min(  X n X  , 1)] = N summationdisplay k =1 min(  X n ( k ) X ( k )  , 1) P [ { k } ] N summationdisplay k =1 min(  X n ( k ) X ( k )  , 1) Therefore, (1.1) implies that each term in the sum on the right side of (1.2) converges to 0 as n . That is, lim n min(  X n ( k ) X ( k )  , 1) = 0 , k = 1 , 2 , . . ., N. This in turn implies that lim n  X n ( k ) X ( k )  = 0 , k = 1 , 2 , . . ., N, which means that X n ( ) X ( ) for every . 2. Suppose that X n P X and Y n P Y as n . (a) Let f : R R be a continuous function. Show that f ( X n ) P f ( X ) . (b) Show that X n + Y n P X + Y and that X n Y n P XY . Solution. (a) Suppose X n P X and that f : R R is continuous. Most of the difficulty here stems from the fact that f need not be uniformly continuous. Fix > 0 and > 0. We will produce n = n ( , ) such that n n = P [  f ( X n ) f ( X )  > ] < . First choose N N so large that P [  X  N 1] 1 / 2. The restriction of f to the compact interval [ N, N ] is then uniformly continuous, so there exists > 0 such that (2 . 1)  x  N,  y ...
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This note was uploaded on 12/11/2011 for the course MATH 280a taught by Professor Driver,b during the Fall '08 term at UCSD.
 Fall '08
 Driver,B
 Probability

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