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The backward algorithm computes suffix likelihoods in a Hidden Markov model. With ${\beta }_t(i)=P(o_{t+1},\dots ,o_T\mid Z_t=i)$, its recurrence is ${\beta }_t(i)={\sum }_j{a}_{ij}{b}_j(o_{t+1}){\beta }_{t+1}(j)$. Together with the forward algorithm, it gives posterior state probabilities proportional to ${\alpha }_t(i){\beta }_t(i)$. The terminal condition must include any explicit End-state transition.

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.