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Good day to everyone!

My question concerns Hidden Markov Models and is pretty basic. In one of the books ("Introduction to Machine Learning" by Ethem Alpaydin, 2nd Edition, p.373), I get the following transition:

$P(O_1, \ldots, O_T \mid q_t = S) = P(O_1, \ldots, O_t \mid q_t = S) \cdot P(O_{t+1}, \ldots, O_T \mid q_t = S)$,

where $O_i$ are the public observations for HMM, and $q_t$ is the $t$-th hidden state, $S$ is some state. My question is why is the joint probability equals to product? Intuitively I understand that: we have conditioning on $q_t = S$, and with respect to it, the observations after $t$ are independent of the ones before $t$. But I can't prove that rigorously, having already wasted lots of paper :(

Thanks a lot for any hint!

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1 Answer 1

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You are right that this equation amounts to conditional independence of the future and the past provided the present is fixed (i.e., the Markov property). However, one has to use the right chain. In this case the easiest is to consider the "compound" Markov chain on $S\times V$, where $S$ is the set of states of the original chain, and $V$ is the set of observables. The transition probabilities are $$ \tilde p ( (s,v), (s',v') ) = p(s,s') b(s',v') \;, $$ where $p$ are the transition probabilities of the base chain on $S$, and $b$ are the observation probabilities. Then one just has to right down the Markov property for this chain and notice that conditioned by $s_0$ both its future and past are independent of $v_0$.

PS I find the notation used in this book really confusing and making things more complicated than they really are.

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Thanks a lot, R W, I found this explanation sufficiently reasonable for me. P.S. Had the same feeling about the text of the book. –  AGro Jan 9 '13 at 15:27

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