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I have been trying to understand hidden Markov models with observational probability but I often find myself confused. I have discussed with my tutor for further help however, he is often rude and does not help and so I have decided to turn to the community.

I am trying to determine the probability of observing the following sequences: AABGA, ABBGT.

From what I understand, you would first need to find probabilities of the sequence occurring without observations and then begin looking for probabilities of the sequence with observational probability. I think I am often confused on questions like this because I do not know how I could check my answer to ensure I have obtained all the possible probabilities of the sequence. Would appreciate on insight into whether my approach was correct and if I have correctly obtained the probability of observing the sequences. Thanks in advance.

enter image description here

States with their initial probabilities

$A(0.3) $

$A(0.4) $

$B(0.2) $

$G(0.0) $

$T(0.1) $

Markov model - Updated based on suggestion: enter image description here

AABGA

$A,A,A,G,G = 0.3 * 0.7 * 0.1 * 0.5 * 0.2 = 0.21%$

$A,A,A,A,A = 0.3 * 0.7 * 0.1 * 0.3 * 0.5 = 0.315%$

$A,A,A,A,A = 0.3 * 0.7 * 0.1 * 0.3 * 0.1 = 0.063%$

$A,A,A,G,G = 0.4 * 0.5 * 0.1 * 0.5 * 0.2 = 0.2%$

$A,A,A,G,G = 0.4 * 0.1 * 0.1 * 0.5 * 0.2 = 0.04%$

$A,A,A,A,A = 0.4 * 0.1 * 0.1 * 0.3 * 0.5 = 0.06%$

$A,A,A,A,A = 0.4 * 0.1 * 0.1 * 0.3 * 0.1 = 0.012%$

$P(AABGA)$ = 0.21% + 0.315% + 0.063% + 0.2% + 0.04% + 0.06% + 0.012% = $0.9%$

ABBGT

$A,A,A,G,G = 0.4 * 0.1 * 0.1 * 0.5 * 0.2 = 0.04%$

$A,A,A,A,T = 0.4 * 0.1 * 0.1 * 0.3 * 0.5 = 0.06%$

$P(ABBGT)$ = 0.04% + 0.06% = $0.1%$

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  • $\begingroup$ Looks like a hidden Markov model. $\endgroup$
    – Wuestenfux
    Commented Apr 11, 2022 at 8:15

1 Answer 1

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Not yet an answer, but too long for the comments section.

Updated in light of added diagram

It would appear from the added diagram that the underlying Markov chain has $5$ states $\ 1,2,\dots,5\ $, with transition matrix $$ \pmatrix{0.2&0.7&0&0.1&0\\ 0&0&0.5&0.5&0\\ 0.7&0&0.3&0&0\\ 0&0&0&0.2&0.8\\ 0.4&0.2&0.4&0&0}\ , $$ and initial state probabilities $\ \pmatrix{0.3&\color{red}{0.4}&0.1&0&0.2}\ $.

It's not clear what the emission probabilities are, however. Is $\ P(A|1)=1.0\ $ or is it $\ 0\ $, for instance? It can't be both, but you've given both as its value. The same inconsistency occurs for $\ P(A|i)\ $ for all the other states. The labels on the diagram seem to suggest that the output is always $\ A\ $ whenever the state is $\ 1\ $ or $\ 2\ $ and always $\ T, G\ $ or $\ B\ $ whenever the state is $\ 3,4\ $ or $\ 5\ $ respectively. But if that were the case, what would be the meaning of the quantities $\ P(A|i), P(B|i)\ $ etc. given in the question?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Aloizio Macedo
    Commented Apr 15, 2022 at 15:04

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