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CNET zhengh HW2

# CNET zhengh HW2 - CNETzhengh HW2 Zheng Hong Contents 1...

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CNETzhengh HW2 Zheng Hong October 12, 2009 Contents 1 Problem 1 2 2 Problem 2 3 3 Problem 3 4 4 Problem 4 5 5 Problem 5 7 6 Problem 6 8 7 Problem 7 9 1

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1 Problem 1 (a) Proof: E [ W 3 ( t ) | W ( s ) , s 6 t ] = E [( W ( t ) - W ( s ) + W ( s )) 3 | W ( s ) , s 6 t ] use the deposition into indep. increments = E [( W ( t ) - W ( s )) 3 + 3( W ( t ) - W ( s )) 2 W ( s ) + 3( W ( t ) - W ( s )) W ( s ) 2 + W ( s ) 3 | W ( s ) , s 6 t ] expand = E [( W ( t ) - W ( s )) 3 ] + 3 W ( s ) E [( W ( t ) - W ( s )) 2 ] + 3 W ( s ) 2 E [ W ( t ) - W ( s )] + W ( s ) 3 indep. of the incre. = 0 + 3 W ( s )( t - s ) + 0 + W ( s ) 3 table1.1 = 3 W ( s )( t - s ) + W ( s ) 3 (1) (b) Proof: Martingale M ( t ) satisfies E [ M ( t ) | M ( s ) , 0 6 s < t ] = M ( s ). To make M (3) W ( t ) = W ( t ) 3 + a 2 ( t ) W ( t ) 2 + a 1 ( t ) W ( t ) + a 0 ( t ) a Martingale, we should let E [ M (3) W ( t ) | W ( s )] = M (3) W ( s ); that means: E [ W ( t ) 3 + a 2 ( t ) W ( t ) 2 + a 1 ( t ) W ( t ) + a 0 ( t ) | W ( s ) , 0 6 s < t ] = W ( s ) 3 + 3 W ( s )( t - s ) + a 2 ( t ) E [( W ( t ) - W ( s ) + W ( s )) 2 | W ( s ) , 0 6 s < t ] + a 1 ( t ) W ( s ) + a 0 ( t ) = W ( s ) 3 + 3 W ( s )( t - s ) + a 2 ( t ) W ( s ) 2 + a 2 ( t )( t - s ) + a 1 ( t ) W ( s ) + a 0 ( t ) = W ( s ) 3 + a 2 ( t ) W ( s ) 2 + (3( t - s ) + a 1 ( t )) W ( s ) + a 2 ( t )( t - s ) + a 0 ( t ) = W ( s ) 3 + a 2 ( s ) W ( s ) 2 + a 1 ( s ) W ( s ) + a 0 ( s ) (2) Compare W ( m ) ( s )’s coefficients, a 2 ( t ) = a 2 ( s ) for arbitrary s,t, 0 6 s < t ; so a 2 ( t ) = 0; 3( t - s ) + a 1 ( t ) = a 1 ( s ), so a 1 ( t ) = - 3 t ; a 2 ( t )( t - s ) + a 0 ( t ) = a 0 ( s ), so a 0 ( t ) = 0; That is to say, E [ W ( t ) 3 - 3 tW ( t ) | W ( s ) , 0 6 s < t ] = W ( s ) 3 - 3 sW ( s ). 2
2 Problem 2 (a) Proof: E [ P 2 ( t ) | P ( s ) , s 6 t ] = E [( P ( t ) - P ( s ) + P ( s )) 2 | P ( s ) , s 6 t ] use the deposition into indep. increments = E [( P ( t ) - P ( s )) 2 + 2( P ( t ) - P ( s )) P ( s ) + P ( s ) 2 | P ( s ) , s 6 t ] expand = E [( P ( t ) - P ( s )) 2 ] + 2 P ( s ) E [ P ( t ) - P ( s )] + P ( s ) 2 indep. of the incre. = λ 0 ( t - s )(1 + λ 0 ( t - s )) + 2 P ( s ) λ 0 ( t - s ) + P ( s ) 2 table1.2 (3) (b) Proof: Martingale M ( t ) satisfies E [ M ( t ) | M ( s ) , 0 6 s < t ] = M ( s ). To make M (2) P ( t ) = P ( t ) 2 + a 1 ( t ) P ( t ) + a 0 ( t ) a Martingale, we should let E [ M (2) P ( t ) | P ( s )] = M (2) P ( s ); that means: E [ P ( t ) 2 + a 1 ( t ) P ( t ) + a 0 ( t ) | P ( s ) , 0 6 s < t ] = λ 0 ( t - s )(1 + λ 0 ( t - s )) + 2 P ( s ) λ 0 ( t - s ) + P ( s ) 2 + a 1 ( t ) λ 0 ( t - s ) + a 1 ( t ) P ( s ) + a 0 ( t ) = P ( s ) 2 + (2 λ 0 ( t - s ) + a 1 ( t )) P ( s ) + λ 0 ( t - s )(1 + λ 0 ( t - s ) + a 1 ( t )) + a 0 ( t ) = P ( s ) 2 + a 1 ( s )

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