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# sol2 - STA3007 Applied probability(08-09 Chapter 2 Solution...

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STA3007 Applied probability (08-09) Solution to Assignment 2 Chapter 2 Problem 5.1 Let X, Y, Z denote X n +2 , X n +1 , ( X 0 , · · · , X n ) T , then, according to the given formula E[E( X | Y, Z ) | Z ] = E[ X | Z ] , we have E[ X n +2 | X 0 , · · · , X n ] = E[E( X n +2 | X 0 , · · · , X n +1 ) | X 0 , · · · , X n ] . By the definition of martingale, we have E( X n +2 | X 0 , · · · , X n +1 ) = X n +1 . Hence E[ X n +2 | X 0 , · · · , X n ] = E[ X n +1 | X 0 , · · · , X n ] = X n . By using induction, it is easy to generalize the above formula to E[ X n + m | X 0 , · · · , X n ] = X n , for any m > 0 . Problem 5.3 Since ε 1 , · · · , ε n are identically and independently distributed as exponential distribu- tion with mean 1, thus E( | X n | ) = E( X n ) = n i =1 0 2 exp( - ε i ) · exp( - ε i ) i = 1 < . Observe S n +1 = S n + ε n +1 , we have X n +1 = 2 n +1 exp( - S n +1 ) = 2 · 2 n exp( - S n - ε n +1 ) = X n · 2 exp( - ε n +1 ) , hence, E( X n +1 | X 0 , · · · , X n ) = E( X n · 2 exp( - ε n +1 ) | X n ) = X n E(2 exp( - ε n +1 )) = X n 0 2 exp( - 2 ε n +1 ) n +1 = X n 1

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Therefore, X n is a martingale.
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sol2 - STA3007 Applied probability(08-09 Chapter 2 Solution...

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