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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 S n = X 1 + . . . + X n . Note that S n n - S n - 1 n - 1 = X n n + S n - 1 n ( n - 1) , so on the event { ( X n ) + n i.o. } ∩ { ( X n ) - n i.o. } (which has probability 1 by the previous section), if S n - 1 n ( n - 1) 0 as n → ∞ (which would have to happen in order for lim n →∞ 1 n S n

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