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For usual HMM problems, observations are generated for all time steps. But let's suppose that we only observe a subset of the outputs: $x_{t1},...x_{tk}$ at the time steps $t1,...tk$. How can we modify the forward algorithm to calculate P(X) (ie P(x_{t1},...x_{tk}))?

MrT77
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    This is an interesting problem, but more suitable for stats.stackexchange.com. That said, it seems like you can handle this as an ordinary HMM with non-uniform transition matrix. Just roll up the transitions for the missing observations into a single transition between one non-missing to the next non-missing. – Robert Dodier Jan 29 '19 at 22:16
  • Thanks for the commentand the answer. About the first, I'll be putting this in stats.stackexchange.com. About the answer, sorry, but could you be more explicit (with formula if possible)? It's really unclear for me what you mean. – MrT77 Jan 30 '19 at 05:53
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    What I mean about rolling up the transitions: a Markov chain with constant transition matrix is (x0, x1, x2, x3, ...) = (x0, A.x0, A.A.x0, A.A.A.x0, ...) where A is the transition matrix and . means matrix multiplication. You can group that differently: x(tk) = (A^t1).(A^(t2 - t1)).(A^(t3 - t2))...(A^(tk - t(k-1))).x0 where A^m is the m-fold product A.A.A... A, m times. Now that's a Markov chain with non-constant transitions. I've omitted the emission or observation part of the HMM but you can work out the rest, this is the basic idea. – Robert Dodier Jan 30 '19 at 20:13

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