Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conditionally independent with a Random Chain.

  1. Oct 22, 2006 #1
    I found this post somewhere on the net. I am also getting a similar problem. Can you all please help?


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

    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
     
  2. jcsd
  3. Oct 23, 2006 #2
    anyone can help please!!!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Conditionally independent with a Random Chain.
Loading...