This is rather basic question. I was going through Speech and Language Processing by Jurafsky and Martin. In the book, they define a Hidden Markov Model (HMM) as follows:

An HMM is specified by the following components:

  • $Q = q_1q_2 ...q_N$ : a set of N states
  • $A = a_{11} ...a_{i j} ...a_{NN}$ : a transition probability matrix $A$ , each > $a_{ij}$ representing the probability of moving from state $i$ to state $j$, s.t. $\sum_{j=1}^Na_{ij}=1 \quad ∀i$
  • $O = o_1o_2 ...o_T$ : a sequence of $T$ observations, each one drawn from a vocabulary $V = v_1, v_2,..., v_V$
  • $B = b_i(o_t)$ : a sequence of observation likelihoods, also called emission probabilities, each expressing the probability of an observation $o_t$ being generated from a state $q_i$
  • $π = π_1,π_2,...,π_N$: an initial probability distribution over states. $π_i$ is the probability that the Markov chain will start in state $i$. Some states $j$ may have $π_j = 0$, meaning that they cannot be initial states. Also, $\sum_{i=1}^n\pi_i=1$

My doubt is shouldn't emission probabilities $B$ sum to 1? That is, shouldnt it be the case that $\sum_{i=1}^n b_i(o_t)=1$ (or maybe $\sum_{t=1}^{n_t} b_i(o_t)=1$). If not, why? If yes, why doesn't the book specify either of these?


1 Answer 1


You're right that a probability distribution should sum to 1, but not in the way that you wrote it. The sum of the probability mass over all events should be 1.

In other words, $\sum_{k=1}^{V} b_i\left(v_k\right) = 1$. At every position in the sequence, the probability of emitting a given symbol given that you're in state $i$ is what's summed up to make a normalized distribution. This is true, whether you're at time $t=1$ or at time $t=T$ or any time in between.

For each possible state $q_i$, you'll have a different summing-to-one distribution, which is conditioned on you being in state $q_i$.

  • $\begingroup$ Very sorry, I took long to ponder. Want to confirm (for all ?s) my understandings: Lets consider transition matrix A & emission matrix B in this diagram. So each row in A and B should sum to 1? Definition talks about A's rows, but forgot to talk about B's rows? Since each row corresponds to different state/position in sequence, it makes no sense for any column (of either A or B) sum to be 1? Can there be any condition / constraint (for example, on sequence formation) in which column may need to sum to 1? $\endgroup$
    – Mahesha999
    Commented Aug 25, 2021 at 15:50
  • $\begingroup$ Your understanding is right. It’s also impossible for all columns of B to sum to 1, unless the number of states equals the number of possible outcomes at 1 time. (Pigeonhole principle.) If all columns of A sum to 1, the matrix is doubly stochastic and the limiting distribution of the Markov chain is uniform. Can’t imagine a compelling reason to include that constraint. $\endgroup$ Commented Aug 25, 2021 at 15:58

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