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I am having a hard time understanding how to use the observation matrix B for continuous HMM assuming the observation of each hidden state Normal.

So far I defined the matrix B as an Nx2 matrix where N is the number of states and 2 is mean and variance of the Normal dist. Question 1: is this correct?

Now, I want to use the values inside the matrix B in the forward algorithm (let's call b the terms inside the matrix B). The forward algorithm I am referring is the following one:

enter image description here

In particular, in step (b), I have to multiply the bracket by bj(Ot) and I am not sure how to define it.

Online I found the following definition in terms of pdf: enter image description here

Question 2: Is this formula correct to be used in the forward alg.? In this formula I know the mean and variance for each hidden state that are contained in the matrix B. By the way, I do not know how to define Ot. In fact, my initial idea was to use the single observation at time t. By the way, knowing that the continuous pdf of a single number is 0, this makes no sense. Question 3: Am I missing something? How can I formulate this?

Thank you so much in advance!

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Question 1: Yes, you will have one mean and one variance for each hidden state.

Question 2 and 3: $o_t$ is your observation, the realization of your observed random variable. In case of continuous observations it is then a scalar. Yes, the formula of the gaussian density function is the correct formula to plug in your observations. It is the quantity to be used in the forward-backward algorithm. It is not $0$ as you will see when you will implement the algorithm. This density has to be thought as a likelihood of the observations which gives a score at seeing this observation, not a probability in the strict sense.

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  • $\begingroup$ The problem I have is when I implement the algorithm (Forward). Every iteration on t I obtain always a very small value of bj(t) (close to 10^-4/10^-5). For this reason, every iteration I obtain smaller and smaller values of alpha that lead to a final probability equal to 0. I am implementing the code in MATLAB and I am using the function normdist. $\endgroup$ Dec 8, 2020 at 22:35
  • $\begingroup$ For instance, b1(o1)=normpdf(786,mu=1151.6, std=181)=2.8660e-04. Is there a way to overcome this problem? Can normalizing the bj(ot) values be an idea? Thank you so much in advance! $\endgroup$ Dec 8, 2020 at 22:37
  • $\begingroup$ Yes renormalization at each time step of you forward and backward probabilities is permitted and is a way to handle this problem (this article deals with the topic) $\endgroup$
    – TheCG
    Dec 9, 2020 at 6:34

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