hw6sol - Stat 310A/Math 230A Theory of Probability Homework...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

Page1 / 4

hw6sol - Stat 310A/Math 230A Theory of Probability Homework...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online