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Unformatted text preview: Math 136  Stochastic Processes Homework Set 3, Autumn 2008, Due: October 15 Questions? See Bo Shen. 1. Exercise 1.4.30 . Use Monotone Convergence to show that E ( summationdisplay n =1 Y n ) = summationdisplay n =1 E Y n , for any sequence of nonnegative R.V. Y n . Deduce that if X 0 and A n are disjoint sets with P ( n A n ) = 1, then E ( X ) = summationdisplay n =1 E ( XI A n ) . Further, show that this applies also for any X L 1. ANS: For each m let X m = m n =1 Y n . Since the Y n are nonnegative it follows that { X m } is a non negative nondecreasing sequence with (possibly infinite) limit n =1 Y n . Hence by monotone conver gence (Theorem 1.4.29) and the linearity of the expectation, E ( summationdisplay n =1 Y n ) = E ( lim m X m ) = lim m E ( X m ) = lim m parenleftBigg m summationdisplay n =1 E ( Y n ) parenrightBigg = summationdisplay n =1 E ( Y n ) . Suppose that X 0 and A n are disjoint with P ( n A n ) = 1. Then the random variables Y n = XI A n satisfy the criterion of the first part of the problem. Using that P ( n A n ) = 1, we have E ( X ) = E ( XI n A n ) = E ( X summationdisplay n =1 I A n ) = E ( summationdisplay n =1 XI A n ) = summationdisplay n =1 E ( XI A n ) . Finally, suppose X L 1. Let X + = max( X, 0) and X = min( X, 0) = max( X, 0) denote the positive and negative parts of X , respectively. Applying the previous part to the nonnegative random variables X + and X , we get E X = E X + E X = summationdisplay n =1 E X + I A n summationdisplay n =1 E X I A n = summationdisplay n =1 E ( X + X ) I A n = summationdisplay n =1 E XI A n . Note that this could also have been accomplished just as easily by applying dominated convergence to the sequence X n = n k =1 XI A k (with  X n   X  for all n )....
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 Fall '08
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