Videoclip: Probability of a sequence – worked example

[Cecilia C]

In the videoclip on learn “Developing the HMM: probability of a sequence – worked example” I don’t understand how should we calculate the PDF and how the most probable state sequence has been found. Should we substitute each value of the sequence to the x in the function? And how we determine how many values each “step” of a model will produce?

[Simon]

When you say “calculate the PDF” do you mean how do we estimate the parameters of the Gaussians in an HMM? That will be coming up in lectures shortly.

As for finding the most probable state sequence, that is the Viterbi algorithm and is coming up in lectures even sooner.

[Norbert G]

Cecilia/Simon,

your x values should be the observations, so 5 4 2 3 3 3,
then you have given most likely state sequence for each model.
Hence, I took each observation and calculated emission prob using mean and sd of the most likely model in the sequence. However, not all my results exactly match the answers. I blame the calculator :).
Simon, was my way of doing it right? I am confused, since some calculations are like in your answer and some different? Are we going to talk about it tomorrow?

[Simon]

Please post your observation probabilities here and I’ll compare them to mine – it’s possible there are errors in my calculations.

[Norbert G]

model A emission probs: 0.18,0.12,0.12,0.2,0.2,0.2
model B emission probs: 0.24,0.05, 0.09, 0.1,0.1,0.18

[Simon]

I think we both have errors 🙂 I’ve attached a spreadsheet that calculates this worked example.

model A: 0.22, 0.10, 0.19, 0.28, 0.28, 0.28 – total is 0.00010

model B: 0.24, 0.05, 0.18, 0.20, 0.20, 0.22 – total is 0.00002

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