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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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up vote 1 down vote accepted

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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Thank you. What I don't quite understand about this is the following. Since we are dealing with a HMM, we know observation $Y_t$ is dependent on $X_t$. So if we know the value for a particular $X_t$, wouldn't $Y_t$ then be independent of Z? So isn't it safe to say $$p(Y_t|X_t, z) = p(Y_t|X_t)$$ if we know X? Thanks for your kind help :) – davec Oct 22 '12 at 18:21
@davec: Sorry, I hadn't read the question carefully enough -- you're right, I edited the answer accordingly. – joriki Oct 22 '12 at 19:22
OK that clears up the last of my confusion, so accepted! :) – davec Oct 22 '12 at 20:35

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