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Unformatted text preview: Stat 310A/Math 230A Theory of Probability Homework 3 Solutions Andrea Montanari Due on 10/14/2010 Exercises on inequalities and convergence Exercise [1.3.21] (a). CauchySchwarz implies ( E Y I ( Y >a ) ) 2 E Y 2 P ( Y > a ) For E Y > a 0 the left hand side is larger than ( E Y a ) 2 so rearranging gives the desired result. (b). Let B 1 = and B i = S i 1 j =1 A j for i = 2 ,...,n . Noting that C i = A i B c i are disjoint sets, such that S n i =1 A i = S n i =1 C i , we have by the additivity of probability measures that P ( n [ i =1 A i ) = P ( n [ i =1 C i ) = n X i =1 P ( C i ) = n X i =1 P ( A i ) n X i =2 P ( A i B i ) (omitting the zero probability of A 1 B 1 = on the right side). Since A i B i S j<i A i A j , we get by subadditivity of probability measures that for i = 2 ,...,n , P ( A i B i ) i 1 X j =1 P ( A i A j ) , which by the preceeding identity results with the second Bonferroni inequality. For Y = n i =1 I A i we have that { Y > } = S n i =1 A i whereas m = E Y = n X i =1 P ( A i ) , v = E Y 2 = n X i,j =1 P ( A i A j ) . Excluding the trivial case where P ( A i ) = 0 for all i , we have from part (a) that P ( Y > 0) m 2 /v 2 m v . That is, P ( n [ i =1 A i ) n X i =1 P ( A i ) 2 X 1 j<i n P ( A i A j ) . Part (b) improves on the latter bound by removing the factor 2 in the rightmost correlation term. However, this improvement is somewhat irrelevant if one uses such bounds to approximate P ( Y > 0) by E Y = n i =1 P ( A i ) upon showing that the correlation term is much smaller than E Y . Exercise [1.3.36] Let Z n ( ) := sup m n  X m ( )  noting that Z n Z for every , with Z 0. Since X n ( ) 0 if and only if Z ( ) = 0, the a.s. convergence of X n to 0 is equivalent to P ( Z > ) = 0 for each...
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 MONTANARI,A.
 Probability

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