I found this post somewhere on the net. I am also getting a similar problem. Can you all please help?(adsbygoogle = window.adsbygoogle || []).push({});

+++++++++++++++

I have a problem, I try to solve but still get stuck.

Can you please read my solution and point out what my wrong points are?

Xn is a Markov Chain . n=1....M. with P(X(0)=i)=a_i. and P(X(n+1)=i|X(n)=j)=Pji.

Yn n=1..M are discrete random variables which are conditionally independent

given X with

P(Y(n)=i|X(n)=j)=f(i,j)

I need to find a recursion for alpha_N(i)= P(Y(N),Y(N-1),...,Y(0), and X(N)=i).

where P(.) is the probability of (.)

My solution is that:

alpha_N(i)= P(Y(N),Y(N-1),...,Y(0), and X(N)=i)

=P(Y(N),Y(N-1),...,Y(0)|X(N)=i)P(X(N)=i)

=P(Y(N)|X(N)=1)*P(Y(N-1),..,Y(0)|X(N)=i)P(X(N)=i) (1)

***question*******

Is (1) correct? I am confused because as the provided information

Yn n=1..M are discrete random variables which are conditionally independent

given X. That means P(Y(0),...Y(N)|X)=P(Y(0)|X)*...P(Y(N)|X)

where X is a whole chain, not only one sample X(n).

Does it still guarantee P(Y(0),...Y(N)|X(N))=P(Y(0)|X(N))*...P(Y(N)|X(N))? ?

***end question****

=f(y(n),i)*Sum_k{P(X(N)=i|P(X(N-1)=k)P(X(N-1)=k}*P(Y(N-1),....,Y(0)|X(N)=i)

=f(y(n),i)*Sum_k{P_ki*P(X(N-1)=k}*P(Y(N-1),....,Y(0)|X(N)=i)

I get stuck here. The answer is wrong While the correct answer should be as:

alpha_N(i)=f(y(N),i)*Sum_k{P_ki*P(X(N-1)=k*P(Y(N-1),....,Y(0)|X(N-1)=k)}

=Sum_k {f(Y(N),i)*P_ki*alpha_(N-1)(k); where Sum_k means summary for k=1 to M.

I think the problem that I can not get the correct answer is the step (1).

Can you please help me out?

Thanks

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# Conditionally independent with a Random Chain.

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