Confusion in modelling finite mixture model
Clash Royale CLAN TAG#URR8PPP
$begingroup$
From the book "Machine Learning a probabilistic Perspective",
I'm reading about finite/infinite mixture models. Particularly at paragraph 25.2.1 it's stated:
The usual representation (of a finite mixture model) is as follows:
$p(x_i|z_i = k, boldsymboltheta) = p(x_i|boldsymboltheta_k)$
$p(z_i = k| boldsymbolpi = pi_k) = pi_k$
$p(boldsymbolpi|alpha) =textDir(boldsymbolpi|(alpha/K)boldsymbol1_K)$
The form of $p(boldsymboltheta_k|lambda)$ is chosen to be be
conjugate to $p(x_i|boldsymboltheta_k)$. We can write
$p(x_i|boldsymboltheta_k)$ as $boldsymbolx_i sim F(boldsymboltheta_z_i)$ where F is the observation distribution.
Similarly, we can write $boldsymboltheta_k sim H(lambda)$, where H
is the prior.
Now this modelling is quite confusing to me. What is the difference between $boldsymboltheta_k$ and $boldsymboltheta_z_i$? What is meant by "Observation distribution"? Can we apply EM algorithm to this model, how?
machine-learning probability unsupervised-learning mixture finite-mixture-model
$endgroup$
add a comment |
$begingroup$
From the book "Machine Learning a probabilistic Perspective",
I'm reading about finite/infinite mixture models. Particularly at paragraph 25.2.1 it's stated:
The usual representation (of a finite mixture model) is as follows:
$p(x_i|z_i = k, boldsymboltheta) = p(x_i|boldsymboltheta_k)$
$p(z_i = k| boldsymbolpi = pi_k) = pi_k$
$p(boldsymbolpi|alpha) =textDir(boldsymbolpi|(alpha/K)boldsymbol1_K)$
The form of $p(boldsymboltheta_k|lambda)$ is chosen to be be
conjugate to $p(x_i|boldsymboltheta_k)$. We can write
$p(x_i|boldsymboltheta_k)$ as $boldsymbolx_i sim F(boldsymboltheta_z_i)$ where F is the observation distribution.
Similarly, we can write $boldsymboltheta_k sim H(lambda)$, where H
is the prior.
Now this modelling is quite confusing to me. What is the difference between $boldsymboltheta_k$ and $boldsymboltheta_z_i$? What is meant by "Observation distribution"? Can we apply EM algorithm to this model, how?
machine-learning probability unsupervised-learning mixture finite-mixture-model
$endgroup$
add a comment |
$begingroup$
From the book "Machine Learning a probabilistic Perspective",
I'm reading about finite/infinite mixture models. Particularly at paragraph 25.2.1 it's stated:
The usual representation (of a finite mixture model) is as follows:
$p(x_i|z_i = k, boldsymboltheta) = p(x_i|boldsymboltheta_k)$
$p(z_i = k| boldsymbolpi = pi_k) = pi_k$
$p(boldsymbolpi|alpha) =textDir(boldsymbolpi|(alpha/K)boldsymbol1_K)$
The form of $p(boldsymboltheta_k|lambda)$ is chosen to be be
conjugate to $p(x_i|boldsymboltheta_k)$. We can write
$p(x_i|boldsymboltheta_k)$ as $boldsymbolx_i sim F(boldsymboltheta_z_i)$ where F is the observation distribution.
Similarly, we can write $boldsymboltheta_k sim H(lambda)$, where H
is the prior.
Now this modelling is quite confusing to me. What is the difference between $boldsymboltheta_k$ and $boldsymboltheta_z_i$? What is meant by "Observation distribution"? Can we apply EM algorithm to this model, how?
machine-learning probability unsupervised-learning mixture finite-mixture-model
$endgroup$
From the book "Machine Learning a probabilistic Perspective",
I'm reading about finite/infinite mixture models. Particularly at paragraph 25.2.1 it's stated:
The usual representation (of a finite mixture model) is as follows:
$p(x_i|z_i = k, boldsymboltheta) = p(x_i|boldsymboltheta_k)$
$p(z_i = k| boldsymbolpi = pi_k) = pi_k$
$p(boldsymbolpi|alpha) =textDir(boldsymbolpi|(alpha/K)boldsymbol1_K)$
The form of $p(boldsymboltheta_k|lambda)$ is chosen to be be
conjugate to $p(x_i|boldsymboltheta_k)$. We can write
$p(x_i|boldsymboltheta_k)$ as $boldsymbolx_i sim F(boldsymboltheta_z_i)$ where F is the observation distribution.
Similarly, we can write $boldsymboltheta_k sim H(lambda)$, where H
is the prior.
Now this modelling is quite confusing to me. What is the difference between $boldsymboltheta_k$ and $boldsymboltheta_z_i$? What is meant by "Observation distribution"? Can we apply EM algorithm to this model, how?
machine-learning probability unsupervised-learning mixture finite-mixture-model
machine-learning probability unsupervised-learning mixture finite-mixture-model
edited Jan 10 at 12:14
Xi'an
55.1k792353
55.1k792353
asked Jan 10 at 10:36
Tommaso BendinelliTommaso Bendinelli
265
265
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you consider a mixture model,$$Xsimsum_k pi_k p(x|theta_k)$$it can be expressed as the marginal of the joint model$$(X,Z)sim underbracepi_z_p_pi(z)timesunderbracetheta_z)_p_theta(x$$where $Z$ is an integer valued random variable. This implies that$$X|Z=ksim p(x|theta_k)$$which can also be written as $$X|Zsim p(x|theta_Z)$$The random variable $Z$ is also latent in that (a) it is not observed and (b) it does not necessarily "exist" in the original experiment modelled by the mixture. But the EM algorithm that returns the MLE of the parameters $pi_k$ and $theta_k$ takes advantage of this joint representation by iteratively
- calculating $mathbbE[Z_i|X_i,theta,pi]$
- maximising the expected log-likelihood in $(theta,pi)$
$endgroup$
$begingroup$
Thank you! And what is the difference between $boldsymboltheta_k$ and $boldsymboltheta_z_i$? And what is meant by "We can write > $p(x_i|boldsymboltheta_k)$ as $boldsymbolx_i sim F(boldsymboltheta_z_i)$ where F is the observation distribution"?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:23
$begingroup$
If $ X|Z=ksim p(x|theta_k)$ and $X|Zsim p(x|theta_Z)$ are identical, because both represent the random varialbe $X|Z$ why in one case $theta$ is $theta_k$ and in the other $theta_z$ ?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:42
$begingroup$
I don't get both your point, (i) where have you seen $X|Z = Z|X$, (ii) What you "by is indexed".
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:49
$begingroup$
Also do we know the distribution $F(theta_z_i)$?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:54
add a comment |
$begingroup$
What is the difference between $theta_k$ and $theta_z_i$?
The only difference is the subscript. The value $theta_z_i$ refers to the value $theta_k$ where $k = z_i$. So the bit which says $boldsymbolx sim F(boldsymboltheta_z_i)$ simply means that $boldsymbolx sim F(boldsymboltheta_k)$ where the subscript $k$ is the latent random variable $z_i$.
$endgroup$
$begingroup$
Thanks for answering, do we know the distribution $(F(boldsymboltheta_k))$ or is it unknown? Intuitively I would think that we don't know it, because if we do there is no point in modeling a Gaussian mixture right?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 13:29
$begingroup$
It is not specified in the problem, since this is a general description of the finite mixture model, which can take on all sorts of distributions. Usually it would be a distribution with an assumed particular parametric form, but with unknown parameters.
$endgroup$
– Ben
Jan 10 at 13:31
$begingroup$
Okey, so just to be sure I understood. $theta_z_i$ are random variable, because $z_i$ is random variable. While $theta_k$, once sampled from H it's a value. Correct?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 13:33
$begingroup$
Yes, but in a mixture model you will usually only observe the $x$s. The rest are unobserved "latent variables" which are effectively parameter values that you infer via Bayes theorem.
$endgroup$
– Ben
Jan 10 at 13:36
$begingroup$
Please have a look here stats.stackexchange.com/questions/386681/…
$endgroup$
– Tommaso Bendinelli
Jan 11 at 9:48
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you consider a mixture model,$$Xsimsum_k pi_k p(x|theta_k)$$it can be expressed as the marginal of the joint model$$(X,Z)sim underbracepi_z_p_pi(z)timesunderbracetheta_z)_p_theta(x$$where $Z$ is an integer valued random variable. This implies that$$X|Z=ksim p(x|theta_k)$$which can also be written as $$X|Zsim p(x|theta_Z)$$The random variable $Z$ is also latent in that (a) it is not observed and (b) it does not necessarily "exist" in the original experiment modelled by the mixture. But the EM algorithm that returns the MLE of the parameters $pi_k$ and $theta_k$ takes advantage of this joint representation by iteratively
- calculating $mathbbE[Z_i|X_i,theta,pi]$
- maximising the expected log-likelihood in $(theta,pi)$
$endgroup$
$begingroup$
Thank you! And what is the difference between $boldsymboltheta_k$ and $boldsymboltheta_z_i$? And what is meant by "We can write > $p(x_i|boldsymboltheta_k)$ as $boldsymbolx_i sim F(boldsymboltheta_z_i)$ where F is the observation distribution"?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:23
$begingroup$
If $ X|Z=ksim p(x|theta_k)$ and $X|Zsim p(x|theta_Z)$ are identical, because both represent the random varialbe $X|Z$ why in one case $theta$ is $theta_k$ and in the other $theta_z$ ?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:42
$begingroup$
I don't get both your point, (i) where have you seen $X|Z = Z|X$, (ii) What you "by is indexed".
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:49
$begingroup$
Also do we know the distribution $F(theta_z_i)$?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:54
add a comment |
$begingroup$
If you consider a mixture model,$$Xsimsum_k pi_k p(x|theta_k)$$it can be expressed as the marginal of the joint model$$(X,Z)sim underbracepi_z_p_pi(z)timesunderbracetheta_z)_p_theta(x$$where $Z$ is an integer valued random variable. This implies that$$X|Z=ksim p(x|theta_k)$$which can also be written as $$X|Zsim p(x|theta_Z)$$The random variable $Z$ is also latent in that (a) it is not observed and (b) it does not necessarily "exist" in the original experiment modelled by the mixture. But the EM algorithm that returns the MLE of the parameters $pi_k$ and $theta_k$ takes advantage of this joint representation by iteratively
- calculating $mathbbE[Z_i|X_i,theta,pi]$
- maximising the expected log-likelihood in $(theta,pi)$
$endgroup$
$begingroup$
Thank you! And what is the difference between $boldsymboltheta_k$ and $boldsymboltheta_z_i$? And what is meant by "We can write > $p(x_i|boldsymboltheta_k)$ as $boldsymbolx_i sim F(boldsymboltheta_z_i)$ where F is the observation distribution"?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:23
$begingroup$
If $ X|Z=ksim p(x|theta_k)$ and $X|Zsim p(x|theta_Z)$ are identical, because both represent the random varialbe $X|Z$ why in one case $theta$ is $theta_k$ and in the other $theta_z$ ?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:42
$begingroup$
I don't get both your point, (i) where have you seen $X|Z = Z|X$, (ii) What you "by is indexed".
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:49
$begingroup$
Also do we know the distribution $F(theta_z_i)$?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:54
add a comment |
$begingroup$
If you consider a mixture model,$$Xsimsum_k pi_k p(x|theta_k)$$it can be expressed as the marginal of the joint model$$(X,Z)sim underbracepi_z_p_pi(z)timesunderbracetheta_z)_p_theta(x$$where $Z$ is an integer valued random variable. This implies that$$X|Z=ksim p(x|theta_k)$$which can also be written as $$X|Zsim p(x|theta_Z)$$The random variable $Z$ is also latent in that (a) it is not observed and (b) it does not necessarily "exist" in the original experiment modelled by the mixture. But the EM algorithm that returns the MLE of the parameters $pi_k$ and $theta_k$ takes advantage of this joint representation by iteratively
- calculating $mathbbE[Z_i|X_i,theta,pi]$
- maximising the expected log-likelihood in $(theta,pi)$
$endgroup$
If you consider a mixture model,$$Xsimsum_k pi_k p(x|theta_k)$$it can be expressed as the marginal of the joint model$$(X,Z)sim underbracepi_z_p_pi(z)timesunderbracetheta_z)_p_theta(x$$where $Z$ is an integer valued random variable. This implies that$$X|Z=ksim p(x|theta_k)$$which can also be written as $$X|Zsim p(x|theta_Z)$$The random variable $Z$ is also latent in that (a) it is not observed and (b) it does not necessarily "exist" in the original experiment modelled by the mixture. But the EM algorithm that returns the MLE of the parameters $pi_k$ and $theta_k$ takes advantage of this joint representation by iteratively
- calculating $mathbbE[Z_i|X_i,theta,pi]$
- maximising the expected log-likelihood in $(theta,pi)$
edited Jan 10 at 12:35
answered Jan 10 at 12:19
Xi'anXi'an
55.1k792353
55.1k792353
$begingroup$
Thank you! And what is the difference between $boldsymboltheta_k$ and $boldsymboltheta_z_i$? And what is meant by "We can write > $p(x_i|boldsymboltheta_k)$ as $boldsymbolx_i sim F(boldsymboltheta_z_i)$ where F is the observation distribution"?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:23
$begingroup$
If $ X|Z=ksim p(x|theta_k)$ and $X|Zsim p(x|theta_Z)$ are identical, because both represent the random varialbe $X|Z$ why in one case $theta$ is $theta_k$ and in the other $theta_z$ ?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:42
$begingroup$
I don't get both your point, (i) where have you seen $X|Z = Z|X$, (ii) What you "by is indexed".
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:49
$begingroup$
Also do we know the distribution $F(theta_z_i)$?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:54
add a comment |
$begingroup$
Thank you! And what is the difference between $boldsymboltheta_k$ and $boldsymboltheta_z_i$? And what is meant by "We can write > $p(x_i|boldsymboltheta_k)$ as $boldsymbolx_i sim F(boldsymboltheta_z_i)$ where F is the observation distribution"?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:23
$begingroup$
If $ X|Z=ksim p(x|theta_k)$ and $X|Zsim p(x|theta_Z)$ are identical, because both represent the random varialbe $X|Z$ why in one case $theta$ is $theta_k$ and in the other $theta_z$ ?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:42
$begingroup$
I don't get both your point, (i) where have you seen $X|Z = Z|X$, (ii) What you "by is indexed".
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:49
$begingroup$
Also do we know the distribution $F(theta_z_i)$?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:54
$begingroup$
Thank you! And what is the difference between $boldsymboltheta_k$ and $boldsymboltheta_z_i$? And what is meant by "We can write > $p(x_i|boldsymboltheta_k)$ as $boldsymbolx_i sim F(boldsymboltheta_z_i)$ where F is the observation distribution"?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:23
$begingroup$
Thank you! And what is the difference between $boldsymboltheta_k$ and $boldsymboltheta_z_i$? And what is meant by "We can write > $p(x_i|boldsymboltheta_k)$ as $boldsymbolx_i sim F(boldsymboltheta_z_i)$ where F is the observation distribution"?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:23
$begingroup$
If $ X|Z=ksim p(x|theta_k)$ and $X|Zsim p(x|theta_Z)$ are identical, because both represent the random varialbe $X|Z$ why in one case $theta$ is $theta_k$ and in the other $theta_z$ ?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:42
$begingroup$
If $ X|Z=ksim p(x|theta_k)$ and $X|Zsim p(x|theta_Z)$ are identical, because both represent the random varialbe $X|Z$ why in one case $theta$ is $theta_k$ and in the other $theta_z$ ?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:42
$begingroup$
I don't get both your point, (i) where have you seen $X|Z = Z|X$, (ii) What you "by is indexed".
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:49
$begingroup$
I don't get both your point, (i) where have you seen $X|Z = Z|X$, (ii) What you "by is indexed".
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:49
$begingroup$
Also do we know the distribution $F(theta_z_i)$?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:54
$begingroup$
Also do we know the distribution $F(theta_z_i)$?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 12:54
add a comment |
$begingroup$
What is the difference between $theta_k$ and $theta_z_i$?
The only difference is the subscript. The value $theta_z_i$ refers to the value $theta_k$ where $k = z_i$. So the bit which says $boldsymbolx sim F(boldsymboltheta_z_i)$ simply means that $boldsymbolx sim F(boldsymboltheta_k)$ where the subscript $k$ is the latent random variable $z_i$.
$endgroup$
$begingroup$
Thanks for answering, do we know the distribution $(F(boldsymboltheta_k))$ or is it unknown? Intuitively I would think that we don't know it, because if we do there is no point in modeling a Gaussian mixture right?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 13:29
$begingroup$
It is not specified in the problem, since this is a general description of the finite mixture model, which can take on all sorts of distributions. Usually it would be a distribution with an assumed particular parametric form, but with unknown parameters.
$endgroup$
– Ben
Jan 10 at 13:31
$begingroup$
Okey, so just to be sure I understood. $theta_z_i$ are random variable, because $z_i$ is random variable. While $theta_k$, once sampled from H it's a value. Correct?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 13:33
$begingroup$
Yes, but in a mixture model you will usually only observe the $x$s. The rest are unobserved "latent variables" which are effectively parameter values that you infer via Bayes theorem.
$endgroup$
– Ben
Jan 10 at 13:36
$begingroup$
Please have a look here stats.stackexchange.com/questions/386681/…
$endgroup$
– Tommaso Bendinelli
Jan 11 at 9:48
add a comment |
$begingroup$
What is the difference between $theta_k$ and $theta_z_i$?
The only difference is the subscript. The value $theta_z_i$ refers to the value $theta_k$ where $k = z_i$. So the bit which says $boldsymbolx sim F(boldsymboltheta_z_i)$ simply means that $boldsymbolx sim F(boldsymboltheta_k)$ where the subscript $k$ is the latent random variable $z_i$.
$endgroup$
$begingroup$
Thanks for answering, do we know the distribution $(F(boldsymboltheta_k))$ or is it unknown? Intuitively I would think that we don't know it, because if we do there is no point in modeling a Gaussian mixture right?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 13:29
$begingroup$
It is not specified in the problem, since this is a general description of the finite mixture model, which can take on all sorts of distributions. Usually it would be a distribution with an assumed particular parametric form, but with unknown parameters.
$endgroup$
– Ben
Jan 10 at 13:31
$begingroup$
Okey, so just to be sure I understood. $theta_z_i$ are random variable, because $z_i$ is random variable. While $theta_k$, once sampled from H it's a value. Correct?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 13:33
$begingroup$
Yes, but in a mixture model you will usually only observe the $x$s. The rest are unobserved "latent variables" which are effectively parameter values that you infer via Bayes theorem.
$endgroup$
– Ben
Jan 10 at 13:36
$begingroup$
Please have a look here stats.stackexchange.com/questions/386681/…
$endgroup$
– Tommaso Bendinelli
Jan 11 at 9:48
add a comment |
$begingroup$
What is the difference between $theta_k$ and $theta_z_i$?
The only difference is the subscript. The value $theta_z_i$ refers to the value $theta_k$ where $k = z_i$. So the bit which says $boldsymbolx sim F(boldsymboltheta_z_i)$ simply means that $boldsymbolx sim F(boldsymboltheta_k)$ where the subscript $k$ is the latent random variable $z_i$.
$endgroup$
What is the difference between $theta_k$ and $theta_z_i$?
The only difference is the subscript. The value $theta_z_i$ refers to the value $theta_k$ where $k = z_i$. So the bit which says $boldsymbolx sim F(boldsymboltheta_z_i)$ simply means that $boldsymbolx sim F(boldsymboltheta_k)$ where the subscript $k$ is the latent random variable $z_i$.
answered Jan 10 at 13:20
BenBen
23.4k224113
23.4k224113
$begingroup$
Thanks for answering, do we know the distribution $(F(boldsymboltheta_k))$ or is it unknown? Intuitively I would think that we don't know it, because if we do there is no point in modeling a Gaussian mixture right?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 13:29
$begingroup$
It is not specified in the problem, since this is a general description of the finite mixture model, which can take on all sorts of distributions. Usually it would be a distribution with an assumed particular parametric form, but with unknown parameters.
$endgroup$
– Ben
Jan 10 at 13:31
$begingroup$
Okey, so just to be sure I understood. $theta_z_i$ are random variable, because $z_i$ is random variable. While $theta_k$, once sampled from H it's a value. Correct?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 13:33
$begingroup$
Yes, but in a mixture model you will usually only observe the $x$s. The rest are unobserved "latent variables" which are effectively parameter values that you infer via Bayes theorem.
$endgroup$
– Ben
Jan 10 at 13:36
$begingroup$
Please have a look here stats.stackexchange.com/questions/386681/…
$endgroup$
– Tommaso Bendinelli
Jan 11 at 9:48
add a comment |
$begingroup$
Thanks for answering, do we know the distribution $(F(boldsymboltheta_k))$ or is it unknown? Intuitively I would think that we don't know it, because if we do there is no point in modeling a Gaussian mixture right?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 13:29
$begingroup$
It is not specified in the problem, since this is a general description of the finite mixture model, which can take on all sorts of distributions. Usually it would be a distribution with an assumed particular parametric form, but with unknown parameters.
$endgroup$
– Ben
Jan 10 at 13:31
$begingroup$
Okey, so just to be sure I understood. $theta_z_i$ are random variable, because $z_i$ is random variable. While $theta_k$, once sampled from H it's a value. Correct?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 13:33
$begingroup$
Yes, but in a mixture model you will usually only observe the $x$s. The rest are unobserved "latent variables" which are effectively parameter values that you infer via Bayes theorem.
$endgroup$
– Ben
Jan 10 at 13:36
$begingroup$
Please have a look here stats.stackexchange.com/questions/386681/…
$endgroup$
– Tommaso Bendinelli
Jan 11 at 9:48
$begingroup$
Thanks for answering, do we know the distribution $(F(boldsymboltheta_k))$ or is it unknown? Intuitively I would think that we don't know it, because if we do there is no point in modeling a Gaussian mixture right?
$endgroup$
– Tommaso Bendinelli
Jan 10 at 13:29
$begingroup$
Thanks for answering, do we know the distribution $(F(boldsymboltheta_k))$ or is it unknown? Intuitively I would think that we don't know it, because if we do there is no point in modeling a Gaussian mixture right?
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– Tommaso Bendinelli
Jan 10 at 13:29
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It is not specified in the problem, since this is a general description of the finite mixture model, which can take on all sorts of distributions. Usually it would be a distribution with an assumed particular parametric form, but with unknown parameters.
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– Ben
Jan 10 at 13:31
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It is not specified in the problem, since this is a general description of the finite mixture model, which can take on all sorts of distributions. Usually it would be a distribution with an assumed particular parametric form, but with unknown parameters.
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– Ben
Jan 10 at 13:31
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Okey, so just to be sure I understood. $theta_z_i$ are random variable, because $z_i$ is random variable. While $theta_k$, once sampled from H it's a value. Correct?
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– Tommaso Bendinelli
Jan 10 at 13:33
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Okey, so just to be sure I understood. $theta_z_i$ are random variable, because $z_i$ is random variable. While $theta_k$, once sampled from H it's a value. Correct?
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– Tommaso Bendinelli
Jan 10 at 13:33
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Yes, but in a mixture model you will usually only observe the $x$s. The rest are unobserved "latent variables" which are effectively parameter values that you infer via Bayes theorem.
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– Ben
Jan 10 at 13:36
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Yes, but in a mixture model you will usually only observe the $x$s. The rest are unobserved "latent variables" which are effectively parameter values that you infer via Bayes theorem.
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– Ben
Jan 10 at 13:36
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Please have a look here stats.stackexchange.com/questions/386681/…
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– Tommaso Bendinelli
Jan 11 at 9:48
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Please have a look here stats.stackexchange.com/questions/386681/…
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– Tommaso Bendinelli
Jan 11 at 9:48
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