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**Unformatted text preview: **Stat 310A/Math 230A Theory of Probability Homework 6 Solutions Andrea Montanari Due on November 11, 2010 Exercises on the law of large numbers and central limit theorem Exercise [2.3.13] Clearly, | X n | = | X n- 1 || U n | , resulting with log | X n | = n X k =1 log | U k | + log | X | . As P ( | U 1 | r ) = r 2 for 0 r 1, it follows from Corollary ?? and integration by parts that E log | U 1 | = R 1 2 r log rdr =- 1 / 2. Further, log | U k | are i.i.d. so by the strong law of large numbers we have that n- 1 log | X n | a.s. - 1 / 2. Exercise [2.3.21] Let Y k = c k X k for the truncated random variables X k = X k I {| X k | k } . We are given that | c k | M for some M < and all k . Consequently, Var( Y k ) = c 2 k Var( X k ) M 2 Var( X k ) and it follows from ( ?? ) that X k =1 k- 2 Var( Y k ) M 2 X k =1 k- 2 Var( X k ) 2 M 2 E | X 1 | < . Therefore, by Theorem ?? , the random series k k- 1 ( Y k- E Y k ) of independent zero-mean random variables converges w.p.1. Using Kroneckers lemma for b n = n and x n = Y n ( )- E Y n , it follows that almost surely, 1 n n X k =1 Y k ( )- 1 n n X k =1 E Y k . Recall that since X k are identically distributed, by dominated convergence we have that E X k = E X 1 I {| X 1 | k } E X 1 = 0 as k , implying by boundedness of c k that E Y k = c k E X k 0 as well. Therefore, the Ces aro averages n- 1 n k =1 E Y k 0, implying that almost surely n- 1 n k =1 Y k ( ) 0 as n . Finally, recall that k P ( c k X k 6 = Y k ) = k P ( | X 1 | > k ) is finite by integrability of X 1 (c.f. proof of Proposition ?? ), so by the first Borel-Cantelli lemma P ( c k X k 6 = Y k i.o. ) = 0, and hence almost surely n- 1 n k =1 c k X k 0 for n . Exercise [3.1.10] 1. By independence, b n = Var( R n ) = n X k =1 Var( B k ) = n X k =1 k- 1 (1- k- 1 ) = n X k =1 k- 1- n X k =1 k- 2 ....

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