I have an HMM with $N$ states and $T$ possible obsevations where $A \in \mathbb{R}^{N \times N}$ is transition probability matrix and $B \in \mathbb{R}^{N \times T}$ is emission probability matrix. I follow the introductory text https://web.stanford.edu/~jurafsky/slp3/A.pdf and understand the concepts and formulas.

However, in my case observations are also probabilistic. It means instead of observing only one state, I have a probability vector of observed states at each discrete time. I can modify the likelihood and decoding algorithms by summing over the observations with probabilities as weights. So, if the probability of being in state $j$ after the first $t$ observations for a particular observation $o_i$ at time $t$ is $\alpha_t(j)$, then for a vector of observation probabilities $o_t$ I can write

$$ \bar{\alpha}_t(j) := \sum_{i=1}^T o_{ti} \alpha_t(j) $$

Hopefully that makes sense, because I'm quite new to the topic. However, I'm not sure how to do the training with this setting. Is there a name for this kind of problem? Can you suggest me a reference if there is? Alternatively, is there an easy way to modify the training formulation to be used in this problem?

  • $\begingroup$ Aren't probabilistic observations the same as emission probabilities? i.e. a probability distribution over possible observations? $\endgroup$
    – Adam Kells
    Commented Oct 18, 2021 at 13:29
  • $\begingroup$ @AdamKells as I understand given the observation sequence we try to find the hidden state sequence depending on the transition and emission probabilities. But the given observation sequence is absolute, i.e. only one state is observed at each discrete step. However, in my case observations are probabilistic, so in a sequence $o_1o_2 \dots o_n$ each $o_t$ is a vector with probabilities of observations instead of a scalar number. $\endgroup$
    – obareey
    Commented Oct 19, 2021 at 7:00

1 Answer 1


I'm pretty sure you can just update the emission probability matrix to take the probablistic observations into account.

There are a lot of ways we could add probablistic "measurement error" to the observations. Let's say that $\gamma$ is the probability that the observation accurately reflects the emission, and $1 - \gamma$ the probability that the observation is a random sample from the $T$ possible observations, and $T=3$ in this example.

$B_i = [b_i^1, b_i^2, b_i^3]$ is the vector of emission probabilities from state $i$ before measurement error, and so are also the observations probabilities in cases where no measurement error occurs), and $E = [\frac{1}{3}, \frac{1}{3}, \frac{1}{3}]$ are the observation probabilities on cases where measurement error does occur. Since the probability of measurement error occurring is $1 - \gamma$, the marginal observation probabilities are therefore

$$ \hat B_i = \gamma B_i + (1-\gamma)E \\ = [ \gamma b_i^1 + (1-\gamma)\frac{1}{3}, \ \gamma b_i^2 + (1-\gamma)\frac{1}{3}, \ \gamma b_i^3 + (1-\gamma)\frac{1}{3} ] \\ = [ \gamma b_i^1 + \frac{(1-\gamma)}{3}, \ \gamma b_i^2 + \frac{(1-\gamma)}{3}, \ \gamma b_i^3 + \frac{(1-\gamma)}{3} ] $$

  • $\begingroup$ So this is one way of incorporating the measurement errors into the model and there could be others. However, I'm still not clear how this can be used in the training of HMM since marginal emission probabilities are changing at each discrete time step. $\endgroup$
    – obareey
    Commented Nov 3, 2021 at 12:45
  • $\begingroup$ I'm sorry, you've lost me. Emission/observation probabilities don't change over time, they're just the probabilities of observing each outcome if the system is in a particular state. $\endgroup$
    – Eoin
    Commented Nov 3, 2021 at 14:00
  • $\begingroup$ Yes, I meant the marginal emission probabilities, $\hat{B}_i$. Because $E$ changes over time (at each measurement). $\endgroup$
    – obareey
    Commented Nov 3, 2021 at 14:57
  • $\begingroup$ Ah, OK. To be honest, you've gone beyond my knowledge in that case, but I suspect that if the observation probabilities change over time, this isn't a Markov model any more, and it probably can't be estimated. $\endgroup$
    – Eoin
    Commented Nov 3, 2021 at 17:03

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