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Suppose that we have two models for a 2-state HMM and both have two output symbols: $A$ and $B$.

Model 1:

  • Transition probabilities: $a_{11}=0.6$, $a_{12}=0.4$, $a_{21}=0.0$, $π‘Ž_{22}=1.0$.
  • Output probabilities: $𝑏_1(𝐴)=0.45$, $𝑏_1(𝐡)=0.55$, $𝑏_2(𝐴)=0.5$, $𝑏_2(𝐡)=0.5$.
  • Initial probabilities: $πœ‹_1=0.4$, $πœ‹_2=0.6$.

Model 2:

  • Transition probabilities: $a_{11}=0.2$, $a_{12}=0.8$, $a_{21}=0.0$, $π‘Ž_{22}=1.0$.
  • Output probabilities: $𝑏_1(𝐴)=0.2$, $𝑏_1(𝐡)=0.8$, $𝑏_2(𝐴)=0.6$, $𝑏_2(𝐡)=0.4$.
  • Initial probabilities: $πœ‹_1=0.7$, $πœ‹_2=0.3$.

Which model is more likely to produce the observation sequence $\{A, B, A\}$?

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  • $\begingroup$ This seems doable, if a bit tedious. What's your problem? $\endgroup$
    – Gijs
    Apr 23 '19 at 14:41
  • $\begingroup$ I am new to Markov Chains and I want to know how can I produce the probability of the sequence in both the models. $\endgroup$
    – Ali J.
    Apr 23 '19 at 14:57
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You can answer this samll problem with the basic version of forward algorithm, described in the famous tutorial paper A tutorial on hidden Markov models and selected applications in speech recognition by Lawrence Rabiner. The interesting part of the paper for you is on page 6 :

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Using an old code i had on my computer I get : $$p({A, B, A}|\mathrm{Model} 1) \approx 0.1291, $$ while $$p({A, B, A}|\mathrm{Model} 2) \approx 0.0817. $$

Thus Model 1 is more likely to generate this sequence.

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