This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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  > } ....
View
Full
Document
This note was uploaded on 11/08/2009 for the course MATH 236 taught by Professor Papanicolaou during the Winter '08 term at Stanford.
 Winter '08
 Papanicolaou
 Math, Differential Equations, Equations

Click to edit the document details