HO5-sol-09

HO5-sol-09 - Math 236 Homework 5 Solutions March 17, 2009...

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Unformatted text preview: Math 236 Homework 5 Solutions March 17, 2009 Problem 1. From the definition of Gaussian, we have p * ( k, x k , x k +1 ) = 1 2 t ( x k ) exp (- ( x k +1- x k- b ( x k ) t ) 2 2 2 ( x k ) t ) and p ( k, x k , x k +1 ) = 1 2 t ( x k ) exp (- ( x k +1- x k ) 2 2 2 ( x k ) t ) . Then log M n = n- 1 X k =0 log exp (- ( x k +1- x k ) 2 + b 2 ( x x ) t 2- 2( x k +1- x k ) b ( x k ) t- ( x k +1- x k ) 2 2 2 ( x k ) t ) = n- 1 X k =0 b ( x k ) 2 ( x k ) ( x k +1- x k )- 1 2 n- 1 X k =0 ( b ( x k ) ( x k ) ) 2 t. By direct calculation we have E P [ M n | X , X 1 , . . . , X n- 1 ] = Z n- 1 Y k =0 p * ( k, x k , x k +1 ) p ( k, x k , x k +1 ) p ( n- 1 , x n- 1 , x n )d x n = Z n- 2 Y k =0 p * ( k, x k , x k +1 ) p ( k, x k , x k +1 ) p * ( n- 1 , x n- 1 , x n )d x n = M n- 1 Z p * ( n- 1 , x n- 1 , x n )d x n = M n- 1 . Notice that in the recursive formula of Y n , the value of next iterate only depends on the most current iterate. Hence P ( Y = x , Y 1 = x 1 , . . . , Y n = x n ) = P ( Y = x ) P ( Y 1 = x 1 | Y = x ) P ( Y n = x n | Y n- 1 = x n- 1 ) = P ( Y = x ) n- 1 Y k =0 p * ( k, x k , x k +1 ) . Because b ( x ) =- x and ( x ) = , we have P ( Y = x , Y 1 = x 1 , . . . , Y n = x n ) = P ( Y = x ) (2 t ) n/ 2 n exp- 1 2 2 t n- 1 X k =0 ( x k +1- x k (1- t )) 2 ! . Taking log and derivative w.r.t. , we have- 1 2 2 t n- 1 X k =0 2( x k +1- x k + x k t )(- x k t ) = 0 =- n- 1 k =0 x k ( x k +1- x k ) t n- 1 k =0 x 2 k . 1 Problem 2. As given in Problem 1 , we know P { max k n | Y ( k t )- Y k | > } ....
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This note was uploaded on 11/08/2009 for the course MATH 236 taught by Professor Papanicolaou during the Winter '08 term at Stanford.

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HO5-sol-09 - Math 236 Homework 5 Solutions March 17, 2009...

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