hw5soln - CME308: Assignment 5 Due: Thursday, May 27, 2010...

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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 380-383V. 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 state-space 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...
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This note was uploaded on 06/17/2010 for the course CME 308 taught by Professor Peterglynn during the Spring '08 term at Stanford.

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hw5soln - CME308: Assignment 5 Due: Thursday, May 27, 2010...

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