How to find complete log likelihood for mixture of PPCA












1












$begingroup$


In Appendix C of a paper by Michael E. Tipping and Christopher M. Bishop about mixture models for probabilistic PCA, the probability of a single data vector $mathbf{t}$ is expressed as a mixture of PCA models (equation 69):



$$
p(mathbf{t}) = sum_{i=1}^Mpi_i p(mathbf{t}|i)
$$



where $pi$ is the mixing proportion and $p(mathbf{t}|i)$ is a single probabilistic PCA model.



The model underlying the probabilistic PCA method is (equation 2)



$$
mathbf{t} = mathbf{Wx} + boldsymbolmu + boldsymbolepsilon.
$$
Where $mathbf{x}$ is a latent variable. By introducing a new set of variables $z_{ni}$ "labelling which model is responsible for generating each data point $mathbf{t}_n$", Bishop formulates the complete log likelihood as (equation 70):



$$
mathcal{L}_C = sum_{n=1}^Nsum_{i=1}^Mz_{ni}ln{pi_ip(mathbf{t}_n, mathbf{x}_{ni})}.
$$
I would like to understand how he derives this expression as he doesn't provide a solution himself. How is this expression for the complete log likelihood found?






probability-theory machine-learning







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    In Appendix C of a paper by Michael E. Tipping and Christopher M. Bishop about mixture models for probabilistic PCA, the probability of a single data vector $mathbf{t}$ is expressed as a mixture of PCA models (equation 69):



    $$
    p(mathbf{t}) = sum_{i=1}^Mpi_i p(mathbf{t}|i)
    $$



    where $pi$ is the mixing proportion and $p(mathbf{t}|i)$ is a single probabilistic PCA model.



    The model underlying the probabilistic PCA method is (equation 2)



    $$
    mathbf{t} = mathbf{Wx} + boldsymbolmu + boldsymbolepsilon.
    $$
    Where $mathbf{x}$ is a latent variable. By introducing a new set of variables $z_{ni}$ "labelling which model is responsible for generating each data point $mathbf{t}_n$", Bishop formulates the complete log likelihood as (equation 70):



    $$
    mathcal{L}_C = sum_{n=1}^Nsum_{i=1}^Mz_{ni}ln{pi_ip(mathbf{t}_n, mathbf{x}_{ni})}.
    $$
    I would like to understand how he derives this expression as he doesn't provide a solution himself. How is this expression for the complete log likelihood found?






    probability-theory machine-learning







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      In Appendix C of a paper by Michael E. Tipping and Christopher M. Bishop about mixture models for probabilistic PCA, the probability of a single data vector $mathbf{t}$ is expressed as a mixture of PCA models (equation 69):



      $$
      p(mathbf{t}) = sum_{i=1}^Mpi_i p(mathbf{t}|i)
      $$



      where $pi$ is the mixing proportion and $p(mathbf{t}|i)$ is a single probabilistic PCA model.



      The model underlying the probabilistic PCA method is (equation 2)



      $$
      mathbf{t} = mathbf{Wx} + boldsymbolmu + boldsymbolepsilon.
      $$
      Where $mathbf{x}$ is a latent variable. By introducing a new set of variables $z_{ni}$ "labelling which model is responsible for generating each data point $mathbf{t}_n$", Bishop formulates the complete log likelihood as (equation 70):



      $$
      mathcal{L}_C = sum_{n=1}^Nsum_{i=1}^Mz_{ni}ln{pi_ip(mathbf{t}_n, mathbf{x}_{ni})}.
      $$
      I would like to understand how he derives this expression as he doesn't provide a solution himself. How is this expression for the complete log likelihood found?






      probability-theory machine-learning







      share|cite|improve this question









      $endgroup$




      In Appendix C of a paper by Michael E. Tipping and Christopher M. Bishop about mixture models for probabilistic PCA, the probability of a single data vector $mathbf{t}$ is expressed as a mixture of PCA models (equation 69):



      $$
      p(mathbf{t}) = sum_{i=1}^Mpi_i p(mathbf{t}|i)
      $$



      where $pi$ is the mixing proportion and $p(mathbf{t}|i)$ is a single probabilistic PCA model.



      The model underlying the probabilistic PCA method is (equation 2)



      $$
      mathbf{t} = mathbf{Wx} + boldsymbolmu + boldsymbolepsilon.
      $$
      Where $mathbf{x}$ is a latent variable. By introducing a new set of variables $z_{ni}$ "labelling which model is responsible for generating each data point $mathbf{t}_n$", Bishop formulates the complete log likelihood as (equation 70):



      $$
      mathcal{L}_C = sum_{n=1}^Nsum_{i=1}^Mz_{ni}ln{pi_ip(mathbf{t}_n, mathbf{x}_{ni})}.
      $$
      I would like to understand how he derives this expression as he doesn't provide a solution himself. How is this expression for the complete log likelihood found?






      probability-theory machine-learning




      probability-theory machine-learning






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 4 at 11:07









      SandiSandi

      255112




      255112






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          Let's concentrate for the time being on the $n^{rm th}$ datapoint, $mathbf t_n$. Suppose this datapoint is generated from the $i_n^{rm th}$ model. Then $$ z_{ni} = begin{cases} 1 & {rm if } i = i_n \ 0 & {rm otherwise}end{cases}.$$Thus we have
          $$sum_{i=1}^M z_{ni} ln left(pi_i p(mathbf t_n ,mathbf x_{ni}) right) = ln left( pi_{i_n} p(mathbf t_n , mathbf x_{n{i_n}})right).$$
          The expression on the right-hand side is log-likelihood for the $n$th datapoint. To spell it out:





          • $pi_{i_n}$ is the probability that the $n^{rm th}$ datapoint is generated by the $i_n^{rm th}$ model.


          • $p(mathbf t_n, mathbf x_{ni_n})$ is the probability of encountering this particular latent vector $mathbf x_{ni_n}$ and this particular visible vector $mathbf t_n$ for the $n^{rm th}$ datapoint, given that this datapoint is generated from the the $i_n^{rm th}$ model. [In fact, $$p(mathbf t_n, mathbf x_{ni_n}) = mathcal N(mathbf x_{ni_n} | mathbf 0, mathbf I) times mathcal N(mathbf t_n - mathbf W mathbf x_{ni_n} - mathbf mu_n |mathbf 0, sigma_n^2 mathbf I),$$ assuming that $mathbf epsilon_{n} sim mathcal N(mathbf 0, sigma_n^2 mathbf I)$.]


          Since the datapoints are generated independently, the log-likelihood for the entire dataset is a sum over the log-likelihoods for the individual datapoints, giving the desired result.






          share|cite|improve this answer









          $endgroup$













            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: "69"
            };
            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: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            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
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061531%2fhow-to-find-complete-log-likelihood-for-mixture-of-ppca%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            Let's concentrate for the time being on the $n^{rm th}$ datapoint, $mathbf t_n$. Suppose this datapoint is generated from the $i_n^{rm th}$ model. Then $$ z_{ni} = begin{cases} 1 & {rm if } i = i_n \ 0 & {rm otherwise}end{cases}.$$Thus we have
            $$sum_{i=1}^M z_{ni} ln left(pi_i p(mathbf t_n ,mathbf x_{ni}) right) = ln left( pi_{i_n} p(mathbf t_n , mathbf x_{n{i_n}})right).$$
            The expression on the right-hand side is log-likelihood for the $n$th datapoint. To spell it out:





            • $pi_{i_n}$ is the probability that the $n^{rm th}$ datapoint is generated by the $i_n^{rm th}$ model.


            • $p(mathbf t_n, mathbf x_{ni_n})$ is the probability of encountering this particular latent vector $mathbf x_{ni_n}$ and this particular visible vector $mathbf t_n$ for the $n^{rm th}$ datapoint, given that this datapoint is generated from the the $i_n^{rm th}$ model. [In fact, $$p(mathbf t_n, mathbf x_{ni_n}) = mathcal N(mathbf x_{ni_n} | mathbf 0, mathbf I) times mathcal N(mathbf t_n - mathbf W mathbf x_{ni_n} - mathbf mu_n |mathbf 0, sigma_n^2 mathbf I),$$ assuming that $mathbf epsilon_{n} sim mathcal N(mathbf 0, sigma_n^2 mathbf I)$.]


            Since the datapoints are generated independently, the log-likelihood for the entire dataset is a sum over the log-likelihoods for the individual datapoints, giving the desired result.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Let's concentrate for the time being on the $n^{rm th}$ datapoint, $mathbf t_n$. Suppose this datapoint is generated from the $i_n^{rm th}$ model. Then $$ z_{ni} = begin{cases} 1 & {rm if } i = i_n \ 0 & {rm otherwise}end{cases}.$$Thus we have
              $$sum_{i=1}^M z_{ni} ln left(pi_i p(mathbf t_n ,mathbf x_{ni}) right) = ln left( pi_{i_n} p(mathbf t_n , mathbf x_{n{i_n}})right).$$
              The expression on the right-hand side is log-likelihood for the $n$th datapoint. To spell it out:





              • $pi_{i_n}$ is the probability that the $n^{rm th}$ datapoint is generated by the $i_n^{rm th}$ model.


              • $p(mathbf t_n, mathbf x_{ni_n})$ is the probability of encountering this particular latent vector $mathbf x_{ni_n}$ and this particular visible vector $mathbf t_n$ for the $n^{rm th}$ datapoint, given that this datapoint is generated from the the $i_n^{rm th}$ model. [In fact, $$p(mathbf t_n, mathbf x_{ni_n}) = mathcal N(mathbf x_{ni_n} | mathbf 0, mathbf I) times mathcal N(mathbf t_n - mathbf W mathbf x_{ni_n} - mathbf mu_n |mathbf 0, sigma_n^2 mathbf I),$$ assuming that $mathbf epsilon_{n} sim mathcal N(mathbf 0, sigma_n^2 mathbf I)$.]


              Since the datapoints are generated independently, the log-likelihood for the entire dataset is a sum over the log-likelihoods for the individual datapoints, giving the desired result.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Let's concentrate for the time being on the $n^{rm th}$ datapoint, $mathbf t_n$. Suppose this datapoint is generated from the $i_n^{rm th}$ model. Then $$ z_{ni} = begin{cases} 1 & {rm if } i = i_n \ 0 & {rm otherwise}end{cases}.$$Thus we have
                $$sum_{i=1}^M z_{ni} ln left(pi_i p(mathbf t_n ,mathbf x_{ni}) right) = ln left( pi_{i_n} p(mathbf t_n , mathbf x_{n{i_n}})right).$$
                The expression on the right-hand side is log-likelihood for the $n$th datapoint. To spell it out:





                • $pi_{i_n}$ is the probability that the $n^{rm th}$ datapoint is generated by the $i_n^{rm th}$ model.


                • $p(mathbf t_n, mathbf x_{ni_n})$ is the probability of encountering this particular latent vector $mathbf x_{ni_n}$ and this particular visible vector $mathbf t_n$ for the $n^{rm th}$ datapoint, given that this datapoint is generated from the the $i_n^{rm th}$ model. [In fact, $$p(mathbf t_n, mathbf x_{ni_n}) = mathcal N(mathbf x_{ni_n} | mathbf 0, mathbf I) times mathcal N(mathbf t_n - mathbf W mathbf x_{ni_n} - mathbf mu_n |mathbf 0, sigma_n^2 mathbf I),$$ assuming that $mathbf epsilon_{n} sim mathcal N(mathbf 0, sigma_n^2 mathbf I)$.]


                Since the datapoints are generated independently, the log-likelihood for the entire dataset is a sum over the log-likelihoods for the individual datapoints, giving the desired result.






                share|cite|improve this answer









                $endgroup$



                Let's concentrate for the time being on the $n^{rm th}$ datapoint, $mathbf t_n$. Suppose this datapoint is generated from the $i_n^{rm th}$ model. Then $$ z_{ni} = begin{cases} 1 & {rm if } i = i_n \ 0 & {rm otherwise}end{cases}.$$Thus we have
                $$sum_{i=1}^M z_{ni} ln left(pi_i p(mathbf t_n ,mathbf x_{ni}) right) = ln left( pi_{i_n} p(mathbf t_n , mathbf x_{n{i_n}})right).$$
                The expression on the right-hand side is log-likelihood for the $n$th datapoint. To spell it out:





                • $pi_{i_n}$ is the probability that the $n^{rm th}$ datapoint is generated by the $i_n^{rm th}$ model.


                • $p(mathbf t_n, mathbf x_{ni_n})$ is the probability of encountering this particular latent vector $mathbf x_{ni_n}$ and this particular visible vector $mathbf t_n$ for the $n^{rm th}$ datapoint, given that this datapoint is generated from the the $i_n^{rm th}$ model. [In fact, $$p(mathbf t_n, mathbf x_{ni_n}) = mathcal N(mathbf x_{ni_n} | mathbf 0, mathbf I) times mathcal N(mathbf t_n - mathbf W mathbf x_{ni_n} - mathbf mu_n |mathbf 0, sigma_n^2 mathbf I),$$ assuming that $mathbf epsilon_{n} sim mathcal N(mathbf 0, sigma_n^2 mathbf I)$.]


                Since the datapoints are generated independently, the log-likelihood for the entire dataset is a sum over the log-likelihoods for the individual datapoints, giving the desired result.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 20:56









                Kenny WongKenny Wong

                18.5k21438




                18.5k21438






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • 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%2fmath.stackexchange.com%2fquestions%2f3061531%2fhow-to-find-complete-log-likelihood-for-mixture-of-ppca%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

                    'app-layout' is not a known element: how to share Component with different Modules

                    Image
                    0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule ...
                    Read more

                    android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d ...
                    Read more

                    WPF add header to Image with URL pettitions [duplicate]

                    Image
                    up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
                    Read more