# hw3sol - Stat 310A/Math 230A Theory of Probability Homework...

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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). Cauchy-Schwarz 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 sub-additivity 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 right-most 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 ( ω ) := sup m ≥ n | X m ( ω ) | noting that Z n ↓ Z for every ω , with Z ≥ 0. Since X n ( ω ) → 0 if and only if Z ( ω ) = 0, the a.s. convergence of X n to 0 is equivalent to P ( Z > ) = 0 for each...
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hw3sol - Stat 310A/Math 230A Theory of Probability Homework...

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