{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HO5-sol-09

# HO5-sol-09 - Math 236 Homework 5 Solutions Problem 1 From...

This preview shows pages 1–3. Sign up to view the full content.

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 π Δ ( x k ) exp ( - ( x k +1 - x k - b ( x k t ) 2 2 σ 2 ( x k t ) and p ( k, x k , x k +1 ) = 1 2 π Δ ( x k ) exp ( - ( x k +1 - x k ) 2 2 σ 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 σ 2 ( x k t ) = n - 1 X k =0 b ( x k ) σ 2 ( x k ) ( x k +1 - x k ) - 1 2 n - 1 X k =0 ( b ( x k ) σ ( x k ) ) 2 Δ t. By direct calculation we have E P [ M n | X 0 , X 1 , . . . , X n - 1 ] = Z Ω 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 Ω 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 Ω 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 0 = x 0 , Y 1 = x 1 , . . . , Y n = x n ) = P ( Y 0 = x 0 ) P ( Y 1 = x 1 | Y 0 = x 0 ) · · · P ( Y n = x n | Y n - 1 = x n - 1 ) = P ( Y 0 = x 0 ) n - 1 Y k =0 p * ( k, x k , x k +1 ) . Because b ( x ) = - γx and σ ( x ) = σ 0 , we have P ( Y 0 = x 0 , Y 1 = x 1 , . . . , Y n = x n ) = P ( Y 0 = x 0 ) (2 π Δ t ) n/ 2 σ n 0 exp - 1 2 σ 2 0 Δ t n - 1 X k =0 ( x k +1 - x k (1 - γ Δ t )) 2 ! . Taking log and derivative w.r.t. γ , we have - 1 2 σ 2 0 Δ t n - 1 X k =0 2( x k +1 - x k + x k γ Δ t )( - x k Δ t ) = 0 ˆ γ = - n - 1 k =0 x k ( x k +1 - x k ) Δ t n - 1 k =0 x 2 k . 1

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

View Full Document
Problem 2. As given in Problem 1 , we know P { max 0 k n | Y ( k Δ t ) - Y k | > δ } → 0 . Thus Y n converges to Y ( n Δ t ) in probability when n → ∞ and n Δ t fixed. When b ( x ) = - γx and σ ( x ) = σ , i.e. Y ( t ) satisfies d Y ( t ) = - γY ( t )d t + σ d B ( t ) , we know Y ( t ) is an OU process. When Δ t 0 and n Δ t = t , we have n - 1 X k =0 y k ( y k +1 - y k ) Z t 0 Y ( s )d Y ( s ) , Δ t n - 1 X k =0 y 2 k Z t 0 Y 2 ( s )d s.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 4

HO5-sol-09 - Math 236 Homework 5 Solutions Problem 1 From...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online