How to find complete log likelihood for mixture of PPCA












1














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







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    1














    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

























      1












      1








      1







      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













      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













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      asked Jan 4 at 11:07









      SandiSandi

      255112




      255112






















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          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





















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            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


























              0














              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
























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                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












                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 yesterday









                Kenny WongKenny Wong

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