I'm trying to study hidden Markov models by solving some exercises, and stumbled upon this one:
Given HMM 2, what is $P(O_{100} = A, O_{101} = A, O_{102} = A)$?
I think that this can be solved using the forward scoring algorithm and by assuming that the previous states can be anything (I'm not sure however if I'm allowed to make this assumption?). This means that the previous steps in the recursion of the forward algorithm are $\alpha_A(99) = 1$ and $\alpha_B(99) = 1$ with $\alpha_i(t)$ where $i$ is one of the states of the HMM and $t$ is the current step in the algorithm.
By applying $\alpha_j(t+1) = (\sum_{i=1}^{n}{\alpha_i(t) * a_{ij}}) * b_j(O_{t+1})$ up until $O_{102}$ I can find the desired probability.
My question is now, if I'm making the right assumptions and follow the correct way of solving this, or am I missing something?
Given: $P(O_1=A)=P(O_1=B)=0.5$
By symmetry: $P(O_{100}=A)=P(O_{100}=B)=0.5$
Thus: $P(O_{100}=A,O_{101}=A,O_{102}=A)=0.5\times0.8\times0.8=0.32$