Mixing
Is a GMM-HMM equivalent to a no-mixture HMM enriched with more states?
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I'm trying to model sequence data that has 5 hidden states. Observation data conditional to each state is gaussian except for one state for which mixture of 2 gaussians seems more appropriate. Unfortunately, the R package that I'm using (depmix) does not seem to support (without extending the package) a GMM as a possible response distribution. So I was considering the possibility of adding a 6th state so I could interpret one of this enriched set of states as a state for which observation distribution is the first gaussian in my above mixture and another one as a state for which observation distribution is the second gaussian.
Am I wrong thinking that the two approaches are equivalent?
hidden-markov-model gaussian-mixture
asked 1 hour ago
Patrick
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up vote
2
down vote
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I'm trying to model sequence data that has 5 hidden states. Observation data conditional to each state is gaussian except for one state for which mixture of 2 gaussians seems more appropriate. Unfortunately, the R package that I'm using (depmix) does not seem to support (without extending the package) a GMM as a possible response distribution. So I was considering the possibility of adding a 6th state so I could interpret one of this enriched set of states as a state for which observation distribution is the first gaussian in my above mixture and another one as a state for which observation distribution is the second gaussian.
Am I wrong thinking that the two approaches are equivalent?
hidden-markov-model gaussian-mixture
asked 1 hour ago
Patrick
1093
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm trying to model sequence data that has 5 hidden states. Observation data conditional to each state is gaussian except for one state for which mixture of 2 gaussians seems more appropriate. Unfortunately, the R package that I'm using (depmix) does not seem to support (without extending the package) a GMM as a possible response distribution. So I was considering the possibility of adding a 6th state so I could interpret one of this enriched set of states as a state for which observation distribution is the first gaussian in my above mixture and another one as a state for which observation distribution is the second gaussian.
Am I wrong thinking that the two approaches are equivalent?
hidden-markov-model gaussian-mixture
asked 1 hour ago
Patrick
1093
I'm trying to model sequence data that has 5 hidden states. Observation data conditional to each state is gaussian except for one state for which mixture of 2 gaussians seems more appropriate. Unfortunately, the R package that I'm using (depmix) does not seem to support (without extending the package) a GMM as a possible response distribution. So I was considering the possibility of adding a 6th state so I could interpret one of this enriched set of states as a state for which observation distribution is the first gaussian in my above mixture and another one as a state for which observation distribution is the second gaussian.
Am I wrong thinking that the two approaches are equivalent?
hidden-markov-model gaussian-mixture
hidden-markov-model gaussian-mixture
asked 1 hour ago
Patrick
1093
asked 1 hour ago
Patrick
1093
asked 1 hour ago
Patrick
1093
asked 1 hour ago
Patrick
1093
asked 1 hour ago
Patrick
1093
1093
add a comment |Â
add a comment |Â
2 Answers
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No you are not wrong thinking that.
If $Y mid X_1 sim alpha f_1(y) + (1-alpha)f_2(y)$, then you can also let $X_2 sim textBernoulli(alpha)$ independently and say
$$
Y mid X_1, X_2 = 1 sim f_1(y)
$$
and
$$
Y mid X_1, X_2 = 0 sim f_2(y).
$$
This is because
$$
f_Y(y mid x_1) = sum_i=1^2f_Y(y mid x_1, x_2) f(x_2) = alpha f_1(y) + (1-alpha)f_2(y).
$$
Keep in mind the sequence through time $X_2^t_t$ is iid, and so the Markov structure is overkill (but still perfectly fine).
answered 52 mins ago
Taylor
10.6k11642
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up vote
2
down vote
It is not exactly equivalent: the 6-state HMM can model everything the GMM-HMM can, but not the other way around.
Suppose you start with the GMM-HMM, with $s_5$ being the GMM state, and turn it into the 6-state HMM with states $s_6$ and $s_7$ instead of $s_5$.
Let $p_6$ and $p_7$ be the prior probabilities of the two components of the GMM (that are then transformed into states $s_6$ and $s_7$).
For every transition from a state $s_i$ to $s_5$ in the GMM-HMM (with probability $t$), create two transition probabilities in the 6-state HMM:
$s_i$ to $s_6$ with probability $t cdot p_6$
$s_i$ to $s_7$ with probability $t cdot p_7$
For every transition from $s_5$ to a state $s_i$ in the GMM-HMM (with probability $t$), create two transition probabilities, respectively from $s_6$ and $s_7$, going to $s_i$, both with the same probability $t$.
If I am not mistaken, the resulting 6-state HMM is equivalent to the GMM-HMM.
However, the other way around doesn't always work. Imagine you are starting the the 6-state HMM.
Suppose that the transition probabilities for $s_i rightarrow s_6$ and $s_i rightarrow s_7$ are not equal do not have the same ratio as $p_6$ and $p_7$ (EDIT). You could not carry this information into the GMM-HMM.
In short, the 6-state HMM should be able to represent everything the GMM-HMM can, and more.
answered 52 mins ago
Vincent B. Lortie
212
New contributor
Vincent B. Lortie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I'm not sure that last "suppose" is correct. What if the mixture probabilities are, say, 0.3 and 0.7? Wouldn't that imply that the transition probabilities into $s_6$ and $s_7$ weren't equal?
– jbowman
26 mins ago
1
On the other hand, there is an implication of the GMM that $p_i,6/p_i,7 = p_j,6/p_j,7$ for all $i,j$, well except the zero transition probabilities of course, which doesn't have to be enforced by the 6-state HMM model, so your fundamental point is correct.
– jbowman
25 mins ago
@jbowman you are correct, I've changed it.
– Vincent B. Lortie
5 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
No you are not wrong thinking that.
If $Y mid X_1 sim alpha f_1(y) + (1-alpha)f_2(y)$, then you can also let $X_2 sim textBernoulli(alpha)$ independently and say
$$
Y mid X_1, X_2 = 1 sim f_1(y)
$$
and
$$
Y mid X_1, X_2 = 0 sim f_2(y).
$$
This is because
$$
f_Y(y mid x_1) = sum_i=1^2f_Y(y mid x_1, x_2) f(x_2) = alpha f_1(y) + (1-alpha)f_2(y).
$$
Keep in mind the sequence through time $X_2^t_t$ is iid, and so the Markov structure is overkill (but still perfectly fine).
answered 52 mins ago
Taylor
10.6k11642
add a comment |Â
up vote
2
down vote
No you are not wrong thinking that.
If $Y mid X_1 sim alpha f_1(y) + (1-alpha)f_2(y)$, then you can also let $X_2 sim textBernoulli(alpha)$ independently and say
$$
Y mid X_1, X_2 = 1 sim f_1(y)
$$
and
$$
Y mid X_1, X_2 = 0 sim f_2(y).
$$
This is because
$$
f_Y(y mid x_1) = sum_i=1^2f_Y(y mid x_1, x_2) f(x_2) = alpha f_1(y) + (1-alpha)f_2(y).
$$
Keep in mind the sequence through time $X_2^t_t$ is iid, and so the Markov structure is overkill (but still perfectly fine).
answered 52 mins ago
Taylor
10.6k11642
add a comment |Â
up vote
2
down vote
up vote
2
down vote
No you are not wrong thinking that.
If $Y mid X_1 sim alpha f_1(y) + (1-alpha)f_2(y)$, then you can also let $X_2 sim textBernoulli(alpha)$ independently and say
$$
Y mid X_1, X_2 = 1 sim f_1(y)
$$
and
$$
Y mid X_1, X_2 = 0 sim f_2(y).
$$
This is because
$$
f_Y(y mid x_1) = sum_i=1^2f_Y(y mid x_1, x_2) f(x_2) = alpha f_1(y) + (1-alpha)f_2(y).
$$
Keep in mind the sequence through time $X_2^t_t$ is iid, and so the Markov structure is overkill (but still perfectly fine).
answered 52 mins ago
Taylor
10.6k11642
No you are not wrong thinking that.
If $Y mid X_1 sim alpha f_1(y) + (1-alpha)f_2(y)$, then you can also let $X_2 sim textBernoulli(alpha)$ independently and say
$$
Y mid X_1, X_2 = 1 sim f_1(y)
$$
and
$$
Y mid X_1, X_2 = 0 sim f_2(y).
$$
This is because
$$
f_Y(y mid x_1) = sum_i=1^2f_Y(y mid x_1, x_2) f(x_2) = alpha f_1(y) + (1-alpha)f_2(y).
$$
Keep in mind the sequence through time $X_2^t_t$ is iid, and so the Markov structure is overkill (but still perfectly fine).
answered 52 mins ago
Taylor
10.6k11642
edited 39 mins ago
answered 52 mins ago
Taylor
10.6k11642
answered 52 mins ago
Taylor
10.6k11642
answered 52 mins ago
Taylor
10.6k11642
10.6k11642
add a comment |Â
add a comment |Â
up vote
2
down vote
It is not exactly equivalent: the 6-state HMM can model everything the GMM-HMM can, but not the other way around.
Suppose you start with the GMM-HMM, with $s_5$ being the GMM state, and turn it into the 6-state HMM with states $s_6$ and $s_7$ instead of $s_5$.
Let $p_6$ and $p_7$ be the prior probabilities of the two components of the GMM (that are then transformed into states $s_6$ and $s_7$).
For every transition from a state $s_i$ to $s_5$ in the GMM-HMM (with probability $t$), create two transition probabilities in the 6-state HMM:
$s_i$ to $s_6$ with probability $t cdot p_6$
$s_i$ to $s_7$ with probability $t cdot p_7$
For every transition from $s_5$ to a state $s_i$ in the GMM-HMM (with probability $t$), create two transition probabilities, respectively from $s_6$ and $s_7$, going to $s_i$, both with the same probability $t$.
If I am not mistaken, the resulting 6-state HMM is equivalent to the GMM-HMM.
However, the other way around doesn't always work. Imagine you are starting the the 6-state HMM.
Suppose that the transition probabilities for $s_i rightarrow s_6$ and $s_i rightarrow s_7$ are not equal do not have the same ratio as $p_6$ and $p_7$ (EDIT). You could not carry this information into the GMM-HMM.
In short, the 6-state HMM should be able to represent everything the GMM-HMM can, and more.
answered 52 mins ago
Vincent B. Lortie
212
New contributor
Vincent B. Lortie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I'm not sure that last "suppose" is correct. What if the mixture probabilities are, say, 0.3 and 0.7? Wouldn't that imply that the transition probabilities into $s_6$ and $s_7$ weren't equal?
– jbowman
26 mins ago
1
On the other hand, there is an implication of the GMM that $p_i,6/p_i,7 = p_j,6/p_j,7$ for all $i,j$, well except the zero transition probabilities of course, which doesn't have to be enforced by the 6-state HMM model, so your fundamental point is correct.
– jbowman
25 mins ago
@jbowman you are correct, I've changed it.
– Vincent B. Lortie
5 mins ago
add a comment |Â
up vote
2
down vote
It is not exactly equivalent: the 6-state HMM can model everything the GMM-HMM can, but not the other way around.
Suppose you start with the GMM-HMM, with $s_5$ being the GMM state, and turn it into the 6-state HMM with states $s_6$ and $s_7$ instead of $s_5$.
Let $p_6$ and $p_7$ be the prior probabilities of the two components of the GMM (that are then transformed into states $s_6$ and $s_7$).
For every transition from a state $s_i$ to $s_5$ in the GMM-HMM (with probability $t$), create two transition probabilities in the 6-state HMM:
$s_i$ to $s_6$ with probability $t cdot p_6$
$s_i$ to $s_7$ with probability $t cdot p_7$
For every transition from $s_5$ to a state $s_i$ in the GMM-HMM (with probability $t$), create two transition probabilities, respectively from $s_6$ and $s_7$, going to $s_i$, both with the same probability $t$.
If I am not mistaken, the resulting 6-state HMM is equivalent to the GMM-HMM.
However, the other way around doesn't always work. Imagine you are starting the the 6-state HMM.
Suppose that the transition probabilities for $s_i rightarrow s_6$ and $s_i rightarrow s_7$ are not equal do not have the same ratio as $p_6$ and $p_7$ (EDIT). You could not carry this information into the GMM-HMM.
In short, the 6-state HMM should be able to represent everything the GMM-HMM can, and more.
answered 52 mins ago
Vincent B. Lortie
212
New contributor
Vincent B. Lortie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I'm not sure that last "suppose" is correct. What if the mixture probabilities are, say, 0.3 and 0.7? Wouldn't that imply that the transition probabilities into $s_6$ and $s_7$ weren't equal?
– jbowman
26 mins ago
1
On the other hand, there is an implication of the GMM that $p_i,6/p_i,7 = p_j,6/p_j,7$ for all $i,j$, well except the zero transition probabilities of course, which doesn't have to be enforced by the 6-state HMM model, so your fundamental point is correct.
– jbowman
25 mins ago
@jbowman you are correct, I've changed it.
– Vincent B. Lortie
5 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It is not exactly equivalent: the 6-state HMM can model everything the GMM-HMM can, but not the other way around.
Suppose you start with the GMM-HMM, with $s_5$ being the GMM state, and turn it into the 6-state HMM with states $s_6$ and $s_7$ instead of $s_5$.
Let $p_6$ and $p_7$ be the prior probabilities of the two components of the GMM (that are then transformed into states $s_6$ and $s_7$).
For every transition from a state $s_i$ to $s_5$ in the GMM-HMM (with probability $t$), create two transition probabilities in the 6-state HMM:
$s_i$ to $s_6$ with probability $t cdot p_6$
$s_i$ to $s_7$ with probability $t cdot p_7$
For every transition from $s_5$ to a state $s_i$ in the GMM-HMM (with probability $t$), create two transition probabilities, respectively from $s_6$ and $s_7$, going to $s_i$, both with the same probability $t$.
If I am not mistaken, the resulting 6-state HMM is equivalent to the GMM-HMM.
However, the other way around doesn't always work. Imagine you are starting the the 6-state HMM.
Suppose that the transition probabilities for $s_i rightarrow s_6$ and $s_i rightarrow s_7$ are not equal do not have the same ratio as $p_6$ and $p_7$ (EDIT). You could not carry this information into the GMM-HMM.
In short, the 6-state HMM should be able to represent everything the GMM-HMM can, and more.
answered 52 mins ago
Vincent B. Lortie
212
New contributor
Vincent B. Lortie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
It is not exactly equivalent: the 6-state HMM can model everything the GMM-HMM can, but not the other way around.
Suppose you start with the GMM-HMM, with $s_5$ being the GMM state, and turn it into the 6-state HMM with states $s_6$ and $s_7$ instead of $s_5$.
Let $p_6$ and $p_7$ be the prior probabilities of the two components of the GMM (that are then transformed into states $s_6$ and $s_7$).
For every transition from a state $s_i$ to $s_5$ in the GMM-HMM (with probability $t$), create two transition probabilities in the 6-state HMM:
$s_i$ to $s_6$ with probability $t cdot p_6$
$s_i$ to $s_7$ with probability $t cdot p_7$
For every transition from $s_5$ to a state $s_i$ in the GMM-HMM (with probability $t$), create two transition probabilities, respectively from $s_6$ and $s_7$, going to $s_i$, both with the same probability $t$.
If I am not mistaken, the resulting 6-state HMM is equivalent to the GMM-HMM.
However, the other way around doesn't always work. Imagine you are starting the the 6-state HMM.
Suppose that the transition probabilities for $s_i rightarrow s_6$ and $s_i rightarrow s_7$ are not equal do not have the same ratio as $p_6$ and $p_7$ (EDIT). You could not carry this information into the GMM-HMM.
In short, the 6-state HMM should be able to represent everything the GMM-HMM can, and more.
answered 52 mins ago
Vincent B. Lortie
212
New contributor
Vincent B. Lortie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 mins ago
answered 52 mins ago
Vincent B. Lortie
212
New contributor
Vincent B. Lortie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 52 mins ago
Vincent B. Lortie
212
answered 52 mins ago
Vincent B. Lortie
212
212
New contributor
Vincent B. Lortie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Vincent B. Lortie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Vincent B. Lortie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I'm not sure that last "suppose" is correct. What if the mixture probabilities are, say, 0.3 and 0.7? Wouldn't that imply that the transition probabilities into $s_6$ and $s_7$ weren't equal?
– jbowman
26 mins ago
1
On the other hand, there is an implication of the GMM that $p_i,6/p_i,7 = p_j,6/p_j,7$ for all $i,j$, well except the zero transition probabilities of course, which doesn't have to be enforced by the 6-state HMM model, so your fundamental point is correct.
– jbowman
25 mins ago
@jbowman you are correct, I've changed it.
– Vincent B. Lortie
5 mins ago
add a comment |Â
1
I'm not sure that last "suppose" is correct. What if the mixture probabilities are, say, 0.3 and 0.7? Wouldn't that imply that the transition probabilities into $s_6$ and $s_7$ weren't equal?
– jbowman
26 mins ago
1
On the other hand, there is an implication of the GMM that $p_i,6/p_i,7 = p_j,6/p_j,7$ for all $i,j$, well except the zero transition probabilities of course, which doesn't have to be enforced by the 6-state HMM model, so your fundamental point is correct.
– jbowman
25 mins ago
@jbowman you are correct, I've changed it.
– Vincent B. Lortie
5 mins ago
1
1
I'm not sure that last "suppose" is correct. What if the mixture probabilities are, say, 0.3 and 0.7? Wouldn't that imply that the transition probabilities into $s_6$ and $s_7$ weren't equal?
– jbowman
26 mins ago
I'm not sure that last "suppose" is correct. What if the mixture probabilities are, say, 0.3 and 0.7? Wouldn't that imply that the transition probabilities into $s_6$ and $s_7$ weren't equal?
– jbowman
26 mins ago
1
1
On the other hand, there is an implication of the GMM that $p_i,6/p_i,7 = p_j,6/p_j,7$ for all $i,j$, well except the zero transition probabilities of course, which doesn't have to be enforced by the 6-state HMM model, so your fundamental point is correct.
– jbowman
25 mins ago
On the other hand, there is an implication of the GMM that $p_i,6/p_i,7 = p_j,6/p_j,7$ for all $i,j$, well except the zero transition probabilities of course, which doesn't have to be enforced by the 6-state HMM model, so your fundamental point is correct.
– jbowman
25 mins ago
@jbowman you are correct, I've changed it.
– Vincent B. Lortie
5 mins ago
@jbowman you are correct, I've changed it.
– Vincent B. Lortie
5 mins ago
add a comment |Â
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