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: CME308: Assignment 5 Due: Thursday, May 27, 2010 Due Date: This assignment is due on Thursday, May 27, 2010, by 5pm under the door of 380383V. See the course website for the policy on incentives for L A T E X solutions. Problem 1 (10 pts): This problem is concerned with computing recursive filtering equations in the Markov chain context. Let X = ( X n : n 0) be a Markov chain in discrete statespace S , and let Z = ( Z n : n 0) be a noise corrupted version of X defined via P { Z dz ,Z 1 dz 1 ,...,Z n dz n  X } = n Y i =0 f ( z i ; X i ) dz i , where ( f ( ; x ) : x S ) is a family of given density functions. Let = ( ( x ) : x S ) be a prior distribution of X (i.e ( x ) = P ( X = x )). Put n ( x ) = P ( X n = x  Z , ,Z n ). Show how n +1 ( x ) can be recursively computed from n Solution: For simplicity we assume the function f is strictly positive. We first establish the base case of our recursion ( x ) = P { X = x  Z } = P { X = x,Z dz } P { Z dz } = P { Z dz  X = x } P { X = x } y S f ( Z ; y ) dz ( y ) = f ( Z ; x ) dz ( x ) y S f ( Z ; y ) dz ( y ) = f ( Z ; x ) ( x ) y S f ( Z ; y ) ( y ) We now provide a recursive formula to compute n +1 n +1 ( x ) = P { X n +1 = y  Z ,...,Z n +1 } = P { Z n +1 dz n +1 ,X n +1 = y  Z , ,Z n } P { Z n +1 dz n +1 } = P { Z n +1 dz n +1  X n +1 = y,Z , ,Z n } P { X n +1 = y  Z , ,Z n } P { Z n +1 dz n +1  Z , ,Z n } = A ( y ) B Note that B = y S A ( y ). We now focus on computing A ( y ): A ( y ) = P { Z n +1 dz n +1  X n +1 = y,Z , ,Z n } P { X n +1 = y  Z , ,Z n } = P { Z n +1 dz n +1  X n +1 = y,Z , ,Z n } X x S P { X n +1 = y  X n = x,Z , ,Z n } P { X n = x  Z ,...,Z n } by the Markov porperty and the assumed kowledge of n ( x ) this simplifies to: = P { Z n +1 dz n +1  X n +1 = y,Z , ,Z n } X x S P { X n +1 = y  X n = x } n ( x ) now assuming we know the transition probabilities of the underlying Markov chain we have = P { Z n +1 dz n +1  X n +1 = y,Z , ,Z n } X x S P xy n ( x ) 1 CME308: Assignment 5 Due: Thursday, May 27, 2010 We now need to compute: P { Z n +1 dz n +1  X n +1 = y,Z , ,Z n } . Note that P { Z n +1 dz n +1  X n +1 = y,Z , ,Z n } = X paths P { path } P { Z n +1 dz n +1  X n +1 ,X n , ,X ,Z n , ,Z } We now show that the last term is independent of the path X , ,X n P { Z n +1 dz n +1  X n +1 ,X n , ,X ,Z n , ,Z } = P { Z n +1 dz n +1 , ,Z dz  X n +1 ,X n , ,X } P { Z n dz n +1 , ,Z dz  X n , ,X } = Q n +1 j =0 f ( dz i ; X i ) dz i Q n j =0 f ( dz i ; X i ) dz i = f ( dz n +1 ; y ) dz n +1 and thus P { Z n +1 dz n +1  X n +1 = y,Z , ,Z n } = X pathsto X n +1 = y P { path } f ( dz n +1 ,y ) dz n +1 = f ( dz...
View
Full
Document
This note was uploaded on 06/17/2010 for the course CME 308 taught by Professor Peterglynn during the Spring '08 term at Stanford.
 Spring '08
 PETERGLYNN

Click to edit the document details