hw5soln

# hw5soln - CME308 Assignment 5 Due Thursday Due Date This...

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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 μ = ( μ ( x ) : x ∈ S ) be a prior distribution of X (i.e μ ( x ) = P ( X = x )). Put μ n ( x ) = P ( X n = x | Z , ··· ,Z n ). Show how μ n +1 ( x ) can be recursively computed from μ n Solution: For simplicity we assume the function f is strictly positive. We first establish the base case of our recursion μ ( 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 μ ( y ) = f ( Z ; x ) dz μ ( x ) ∑ y ∈ S f ( Z ; y ) dz μ ( y ) = f ( Z ; x ) μ ( x ) ∑ y ∈ S f ( Z ; y ) μ ( y ) We now provide a recursive formula to compute μ n +1 μ 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 μ 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 } μ 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 μ 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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hw5soln - CME308 Assignment 5 Due Thursday Due Date This...

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