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Unformatted text preview: Math 136  Stochastic Processes Homework Set 6, Autumn 2009, Due: Nov 4 1. Exercise 4.1.5 . Suppose ( X n , F n ) is a martingale. Show that then { X n } is also a martingale with respect to its canonical filtration and that a.s. E [ X l F n ] = X n for all l > n . ANS: Its easy to check the integrability condition, and suppose the canonical filtration is G n , by tower property, we have that E ( X n G n 1 ) = E ( E ( X n F n 1 ) G n 1 ) = E ( X n 1 G n 1 ) = X n 1 , where the second to last equality we have used the condition that ( X n , F n ) is a martingale. Hence, { X n } is also a martingale w.r.t. its canonical filtration. For the second part, again iteratively using the tower property, we have: E ( X l F n ) = E ( E ( X l F l 1 ) F n ) = E ( X l 1 F n ) = ... = X n where all equality holds in the a.s. sense since they are random variables. 2. Exercise 4.1.6 . Provide an example of a probability space ( , F , P ), a filtration {F n } and a stochastic process { X n } adapted to {F n } such that: (a) { X n } is a martingale with respect to its canonical filtration but ( X n , F n ) is not a martingale. ANS: Take = { a,b } , F = F = 2 , X = 0, X 1 = 1 with probability 1 / 2 and X n = X 1 for all n 2. Then { X n } is a martingale with respect to its canonical filtration since: X = 0 = E ( X 1 ) = E ( X 1  X ) and X n = E ( X n  X n ) = E ( X n +1  X n ) = E ( X n +1  ( X ,...,X n )) for all n 1. Now consider the filtration {F n } where F n = 2 for all n . Then, X = 0 6 = X 1 = E ( X 1 F ) , so that ( X n , F n ) is not a martingale. (b) Provide a probability measure Q on ( , F ) under which { X n } is not a martingale even with respect to its canonical filtration. ANS: Let Q be a probability measure on ( , F ) such that X 1 = 1 with probability p > 1 / 2 and X 1 = 1 with probability 1 p < 1 / 2. Then E X 1 = (2 p 1) > 6 = 0 = X so that { X n } is not a martingale with respect to its canonical filtration. 1 3. Exercise 4.1.23 . Let 1 , 2 ,... be independent with E i = 0 and E 2 i = 2 i . (a) Let S n = n i =1 i and s 2 n = n i =1 2 i . Show that { S 2 n } is a submartingale and { S 2 n s 2 n } is a martingale. ANS: Using the same argument of Example 4.1.8 we know that { S n } is a martingale with respect to its canonical filtration. Moreover, from the fact that S 2 n = n i =1 2 i + 2 1 i<j n i j it is clear that E  S 2 n  < for all n . Thus since x 7 x 2 is a convex function it follows from the conditional Jensen inequality that S 2 n is a submartingale. Letting F n = ( S 1 ,...,S n ) and using that n +1 is independent of F n , we have E [ S 2 n +1 F n ] = E [( S n + n +1 ) 2 F n ] = E [ S 2 n + 2 n +1 S n + 2 n +1...
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This document was uploaded on 04/12/2010.
 Spring '09
 Math

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