{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Background image of page 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 )
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}