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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 zeromean random variables converges w.p.1. Using Kronecker’s lemma for b n = n and x n = Y n ( ω ) E Y n , it follows that almost surely, 1 n n X k =1 Y k ( ω ) 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 ( ω ) → 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 BorelCantelli 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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 '11
 MONTANARI,A.
 Central Limit Theorem, Law Of Large Numbers, Probability, Probability theory, Convergence, Trigraph, yk, Dominated convergence theorem

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