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Unformatted text preview: MATH 507b TAKEHOME MIDTERM SOLUTIONS SPRING 2008 Prof. Alexander (1)(a) Since E ( Y k  F k 1 ) Y k 1 we have for k > n that E ( Y k  F n ) = E ( E ( Y k  F k 1 )  F n ) E ( Y k 1  F n ) , so { E ( Y k  F n ) ,k n } is monotone nondecreasing a.s., so lim k E ( Y k  F n ) exists a.s. (possibly + .) Let M = sup k EY k < . By Monotone Convergence, E (lim k E ( Y k  F n )) = lim k E ( E ( Y k  F n )) = lim k EY k M , we must have lim k E ( Y k  F n ) < a.s. (b) Let Y n be the size of the n th generation in a branching process with mean family size = 1 (with family size not a.s. equal to 1.) We showed in lecture that Y n is a martingale with Y n 0 a.s., i.e. Y 0. Y n is L 1bounded since it is nonnegative. Letting F = { , } we have lim k E ( Y k  F ) = lim k EY k = lim k 1 = 1 6 = 0 = EY = E ( Y  F ) . (c) From (a), EZ n M for all n , so Z n is L 1bounded. By the Monotone Convergence Theorem for conditional expectation (Chapter 4 (1.1c) p. 223) we have E ( Z n +1  F n ) = E lim k E ( Y k  F n +1 ) F n = lim k E ( E ( Y k  F n +1 ) F n ) = lim k E ( Y k  F n ) = Z n , so { Z n } is a martingale. (d) Let Y k = X + k and Y 00 k = X k . These are both L 1bounded submartingales (since the positive and negative parts are convex functions), so by (c), Z n = lim...
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 Spring '08
 Alexander
 Math, Probability

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