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Suppose I have an observation $Y_t$ that is conditionally dependent on $X_t$. (More specifically, Y is a series of observations emitted by an underlying hidden Markov state sequence X.)

I can describe the distribution of $p(Y_t)$ as

$\sum_{x_t} p(Y_t|X_t)p(X_t),$

that is to say, the sum of the emission probabilities for all possible hidden states.

My question is, suppose I also have a third variable Z. Can I express $p(Y_t|z)$ as:

$\sum_{x_t} p(Y_t|X_t)p(X_t|z)?$

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Yes – generally one would have to conditionalize the entire expression,

$ p(Y_t|z)=\sum_{x_t} p(Y_t|X_t,z)p(X_t|z)\;, $

but as you rightly pointed out in a comment, in this case $Y_t$ depends only on $X_t$, so $p(Y_t|X_t,z)=p(Y_t|X_t)$ and the sum reduces to your sum,

$ p(Y_t|z)=\sum_{x_t} p(Y_t|X_t)p(X_t|z)\;. $

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    OK that clears up the last of my confusion, so accepted! :)2012-10-22