Confusion in modelling finite mixture model

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP












2












$begingroup$


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?










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    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?










    share|cite|improve this question











    $endgroup$














      2












      2








      2


      0



      $begingroup$


      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?










      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 10 at 12:14









      Xi'an

      55.1k792353




      55.1k792353










      asked Jan 10 at 10:36









      Tommaso BendinelliTommaso Bendinelli

      265




      265




















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          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



          1. calculating $mathbbE[Z_i|X_i,theta,pi]$

          2. maximising the expected log-likelihood in $(theta,pi)$





          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            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



















          1












          $begingroup$


          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 $z_i$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            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











          Your Answer





          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "65"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: false,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          imageUploader:
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          ,
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













          draft saved

          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f386513%2fconfusion-in-modelling-finite-mixture-model%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          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



          1. calculating $mathbbE[Z_i|X_i,theta,pi]$

          2. maximising the expected log-likelihood in $(theta,pi)$





          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            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
















          2












          $begingroup$

          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



          1. calculating $mathbbE[Z_i|X_i,theta,pi]$

          2. maximising the expected log-likelihood in $(theta,pi)$





          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            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














          2












          2








          2





          $begingroup$

          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



          1. calculating $mathbbE[Z_i|X_i,theta,pi]$

          2. maximising the expected log-likelihood in $(theta,pi)$





          share|cite|improve this answer











          $endgroup$



          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



          1. calculating $mathbbE[Z_i|X_i,theta,pi]$

          2. maximising the expected log-likelihood in $(theta,pi)$






          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          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

















          • $begingroup$
            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














          1












          $begingroup$


          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 $z_i$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            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
















          1












          $begingroup$


          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 $z_i$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            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














          1












          1








          1





          $begingroup$


          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 $z_i$.






          share|cite|improve this answer









          $endgroup$




          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 $z_i$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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

















          • $begingroup$
            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?
          $endgroup$
          – 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




          $begingroup$
          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





          $begingroup$
          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




          $begingroup$
          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





          $begingroup$
          Please have a look here stats.stackexchange.com/questions/386681/…
          $endgroup$
          – Tommaso Bendinelli
          Jan 11 at 9:48


















          draft saved

          draft discarded
















































          Thanks for contributing an answer to Cross Validated!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid


          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.

          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f386513%2fconfusion-in-modelling-finite-mixture-model%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown






          Popular posts from this blog

          How to check contact read email or not when send email to Individual?

          Bahrain

          Postfix configuration issue with fips on centos 7; mailgun relay