 Solution

Let $Z_t$ be the hidden state. With the usual convention ${\alpha }_t(i)=P(o_1,\dots ,o_t,Z_t=i)$ and ${\beta }_t(i)=P(o_{t+1},\dots ,o_T\mid Z_t=i)$, the Markov property and the conditional independence of emissions factor the joint probability asThe requested state-path probability after observing the complete data is the corresponding conditional probability:Here the forward algorithm and backward algorithm give the same normalizer at any interior time,There is no extra factor $b_i(o_t)$: it is already included in ${\alpha }_t(i)$. If an unconditioned joint probability with the observations is intended, it is the numerator alone. Summing the boxed expression over all $i,j,k$ gives one.