hw5solutions

hw5solutions - Solutions to Homework Set No. 5 Probability...

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Solutions to Homework Set No. 5 – Probability Theory (235A), Fall 2011 Solution to 1. (a) Z 0 P ( X x ) dx = Z 0 Z y = x dF ( y ) dx = Z 0 Z y 0 dxdF ( y ) = Z 0 ydF ( y ) = E ( X ) . Here, “ dF ( x )” means f ( x ) dx if X has a density. Otherwise, the integral represents a Lebesgue-Stieltjes integral. Note that if the formula holds for all r.v.’s with density (or discrete r.v.’s, for which it is also easy to verify) then by approximation it can easily be seen to hold for general non-negative r.v.’s. (b) Let P ( | X n | ≥ x ) = P ( | X | ≥ x ) = g ( x ) for all n . Since g ( x ) is decreasing, X n =1 P ( | X n | ≥ n ) Z 0 P ( | X n | ≥ x ) dx = E | X n | = E | X 1 | , and so if E | X 1 | < , by Borel-Cantelli lemma, P ( | X n | ≥ n i.o.) = 0. Similarly, 1 + X n =1 P ( | X n | ≥ n ) E | X 1 | , (1) so if E | X 1 | = , then P ( | X n | ≥ n i.o.) = 1. (c) Let
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This note was uploaded on 11/13/2011 for the course STAT 235A taught by Professor Danromik during the Fall '11 term at UC Davis.

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hw5solutions - Solutions to Homework Set No. 5 Probability...

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