Mixing
Probability of observation sequence not knowing previous observations
$begingroup$
I'm trying to study hidden Markov models by solving some exercises, and stumbled upon this one:
Given HMM 2, what is $P(O_{100} = A, O_{101} = A, O_{102} = A)$?
I think that this can be solved using the forward scoring algorithm and by assuming that the previous states can be anything (I'm not sure however if I'm allowed to make this assumption?). This means that the previous steps in the recursion of the forward algorithm are $alpha_A(99) = 1$ and $alpha_B(99) = 1$ with $alpha_i(t)$ where $i$ is one of the states of the HMM and $t$ is the current step in the algorithm.
By applying $alpha_j(t+1) = (sum_{i=1}^{n}{alpha_i(t) * a_{ij}}) * b_j(O_{t+1})$ up until $O_{102}$ I can find the desired probability.
My question is now, if I'm making the right assumptions and follow the correct way of solving this, or am I missing something?
probability probability-theory hidden-markov-models
asked Jan 18 at 8:06
Pieter VerschaffeltPieter Verschaffelt
277110
$endgroup$
add a comment |
$begingroup$
I'm trying to study hidden Markov models by solving some exercises, and stumbled upon this one:
Given HMM 2, what is $P(O_{100} = A, O_{101} = A, O_{102} = A)$?
I think that this can be solved using the forward scoring algorithm and by assuming that the previous states can be anything (I'm not sure however if I'm allowed to make this assumption?). This means that the previous steps in the recursion of the forward algorithm are $alpha_A(99) = 1$ and $alpha_B(99) = 1$ with $alpha_i(t)$ where $i$ is one of the states of the HMM and $t$ is the current step in the algorithm.
By applying $alpha_j(t+1) = (sum_{i=1}^{n}{alpha_i(t) * a_{ij}}) * b_j(O_{t+1})$ up until $O_{102}$ I can find the desired probability.
My question is now, if I'm making the right assumptions and follow the correct way of solving this, or am I missing something?
probability probability-theory hidden-markov-models
asked Jan 18 at 8:06
Pieter VerschaffeltPieter Verschaffelt
277110
$endgroup$
add a comment |
$begingroup$
I'm trying to study hidden Markov models by solving some exercises, and stumbled upon this one:
Given HMM 2, what is $P(O_{100} = A, O_{101} = A, O_{102} = A)$?
I think that this can be solved using the forward scoring algorithm and by assuming that the previous states can be anything (I'm not sure however if I'm allowed to make this assumption?). This means that the previous steps in the recursion of the forward algorithm are $alpha_A(99) = 1$ and $alpha_B(99) = 1$ with $alpha_i(t)$ where $i$ is one of the states of the HMM and $t$ is the current step in the algorithm.
By applying $alpha_j(t+1) = (sum_{i=1}^{n}{alpha_i(t) * a_{ij}}) * b_j(O_{t+1})$ up until $O_{102}$ I can find the desired probability.
My question is now, if I'm making the right assumptions and follow the correct way of solving this, or am I missing something?
probability probability-theory hidden-markov-models
asked Jan 18 at 8:06
Pieter VerschaffeltPieter Verschaffelt
277110
$endgroup$
I'm trying to study hidden Markov models by solving some exercises, and stumbled upon this one:
Given HMM 2, what is $P(O_{100} = A, O_{101} = A, O_{102} = A)$?
I think that this can be solved using the forward scoring algorithm and by assuming that the previous states can be anything (I'm not sure however if I'm allowed to make this assumption?). This means that the previous steps in the recursion of the forward algorithm are $alpha_A(99) = 1$ and $alpha_B(99) = 1$ with $alpha_i(t)$ where $i$ is one of the states of the HMM and $t$ is the current step in the algorithm.
By applying $alpha_j(t+1) = (sum_{i=1}^{n}{alpha_i(t) * a_{ij}}) * b_j(O_{t+1})$ up until $O_{102}$ I can find the desired probability.
My question is now, if I'm making the right assumptions and follow the correct way of solving this, or am I missing something?
probability probability-theory hidden-markov-models
probability probability-theory hidden-markov-models
asked Jan 18 at 8:06
Pieter VerschaffeltPieter Verschaffelt
277110
asked Jan 18 at 8:06
Pieter VerschaffeltPieter Verschaffelt
277110
edited Jan 18 at 8:12
Pieter Verschaffelt
asked Jan 18 at 8:06
Pieter VerschaffeltPieter Verschaffelt
277110
asked Jan 18 at 8:06
Pieter VerschaffeltPieter Verschaffelt
277110
asked Jan 18 at 8:06
Pieter VerschaffeltPieter Verschaffelt
277110
277110
add a comment |
add a comment |
1 Answer
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$begingroup$
Given: $P(O_1=A)=P(O_1=B)=0.5$
By symmetry: $P(O_{100}=A)=P(O_{100}=B)=0.5$
Thus: $P(O_{100}=A,O_{101}=A,O_{102}=A)=0.5times0.8times0.8=0.32$
answered Jan 18 at 15:00
Daniel MathiasDaniel Mathias
1,30518
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$begingroup$
Given: $P(O_1=A)=P(O_1=B)=0.5$
By symmetry: $P(O_{100}=A)=P(O_{100}=B)=0.5$
Thus: $P(O_{100}=A,O_{101}=A,O_{102}=A)=0.5times0.8times0.8=0.32$
answered Jan 18 at 15:00
Daniel MathiasDaniel Mathias
1,30518
$endgroup$
add a comment |
$begingroup$
Given: $P(O_1=A)=P(O_1=B)=0.5$
By symmetry: $P(O_{100}=A)=P(O_{100}=B)=0.5$
Thus: $P(O_{100}=A,O_{101}=A,O_{102}=A)=0.5times0.8times0.8=0.32$
answered Jan 18 at 15:00
Daniel MathiasDaniel Mathias
1,30518
$endgroup$
add a comment |
$begingroup$
Given: $P(O_1=A)=P(O_1=B)=0.5$
By symmetry: $P(O_{100}=A)=P(O_{100}=B)=0.5$
Thus: $P(O_{100}=A,O_{101}=A,O_{102}=A)=0.5times0.8times0.8=0.32$
answered Jan 18 at 15:00
Daniel MathiasDaniel Mathias
1,30518
$endgroup$
Given: $P(O_1=A)=P(O_1=B)=0.5$
By symmetry: $P(O_{100}=A)=P(O_{100}=B)=0.5$
Thus: $P(O_{100}=A,O_{101}=A,O_{102}=A)=0.5times0.8times0.8=0.32$
answered Jan 18 at 15:00
Daniel MathiasDaniel Mathias
1,30518
answered Jan 18 at 15:00
Daniel MathiasDaniel Mathias
1,30518
answered Jan 18 at 15:00
Daniel MathiasDaniel Mathias
1,30518
answered Jan 18 at 15:00
Daniel MathiasDaniel Mathias
1,30518
1,30518
add a comment |
add a comment |
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