Hidden Markov Model probability of producing a sequence

Question

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

Model 1:

• Transition probabilities: $a_{11}=0.6$, $a_{12}=0.4$, $a_{21}=0.0$, $𝑎_{22}=1.0$.
• Output probabilities: $𝑏_1(𝐴)=0.45$, $𝑏_1(𝐵)=0.55$, $𝑏_2(𝐴)=0.5$, $𝑏_2(𝐵)=0.5$.
• Initial probabilities: $𝜋_1=0.4$, $𝜋_2=0.6$.

Model 2:

• Transition probabilities: $a_{11}=0.2$, $a_{12}=0.8$, $a_{21}=0.0$, $𝑎_{22}=1.0$.
• Output probabilities: $𝑏_1(𝐴)=0.2$, $𝑏_1(𝐵)=0.8$, $𝑏_2(𝐴)=0.6$, $𝑏_2(𝐵)=0.4$.
• Initial probabilities: $𝜋_1=0.7$, $𝜋_2=0.3$.

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

Answer 1

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 :

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.