Re-arranging of formulation for Generative Adversarial Networks (GAN) formulation for optimal $D$












1












$begingroup$


I am following This paper. In section 3.1. Justification of regularizer I am trying to duplicate their transition from $D(x)=frac{p_x(x)}{p_x(x)+p_G(x)}$ to the form of $log(p_X(x))= log(D(x)) − log(1 − D(x))
+ log(p_Z(z)) + log(|frac{partial Z}{partial x}|)$.



I am aware that the when the GAN converges $p_x=p_G$ hence $D(x)=0.5$ though I am not sure if it helps.



So far I got:



$$log(D(x))=log(p_x(x))-log(p_x(x)+p_G(x))$$
$$log(D(x))=log(p_x(x))-log(p_x(x)+p_z(z)|frac{partial Z}{partial x}|)$$



How do I proceed?






neural-networks







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am following This paper. In section 3.1. Justification of regularizer I am trying to duplicate their transition from $D(x)=frac{p_x(x)}{p_x(x)+p_G(x)}$ to the form of $log(p_X(x))= log(D(x)) − log(1 − D(x))
    + log(p_Z(z)) + log(|frac{partial Z}{partial x}|)$    .



    I am aware that the when the GAN converges $p_x=p_G$ hence $D(x)=0.5$ though I am not sure if it helps.



    So far I got:



    $$log(D(x))=log(p_x(x))-log(p_x(x)+p_G(x))$$
    $$log(D(x))=log(p_x(x))-log(p_x(x)+p_z(z)|frac{partial Z}{partial x}|)$$



    How do I proceed?






    neural-networks







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am following This paper. In section 3.1. Justification of regularizer I am trying to duplicate their transition from $D(x)=frac{p_x(x)}{p_x(x)+p_G(x)}$ to the form of $log(p_X(x))= log(D(x)) − log(1 − D(x))
      + log(p_Z(z)) + log(|frac{partial Z}{partial x}|)$      .



      I am aware that the when the GAN converges $p_x=p_G$ hence $D(x)=0.5$ though I am not sure if it helps.



      So far I got:



      $$log(D(x))=log(p_x(x))-log(p_x(x)+p_G(x))$$
      $$log(D(x))=log(p_x(x))-log(p_x(x)+p_z(z)|frac{partial Z}{partial x}|)$$



      How do I proceed?






      neural-networks







      share|cite|improve this question









      $endgroup$




      I am following This paper. In section 3.1. Justification of regularizer I am trying to duplicate their transition from $D(x)=frac{p_x(x)}{p_x(x)+p_G(x)}$ to the form of $log(p_X(x))= log(D(x)) − log(1 − D(x))
      + log(p_Z(z)) + log(|frac{partial Z}{partial x}|)$      .



      I am aware that the when the GAN converges $p_x=p_G$ hence $D(x)=0.5$ though I am not sure if it helps.



      So far I got:



      $$log(D(x))=log(p_x(x))-log(p_x(x)+p_G(x))$$
      $$log(D(x))=log(p_x(x))-log(p_x(x)+p_z(z)|frac{partial Z}{partial x}|)$$



      How do I proceed?






      neural-networks




      neural-networks






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 4 '18 at 10:40









      havakokhavakok

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      510215






















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          $begingroup$

          The log of a sum is an indicator that your first equation won't be helpful, because you won't be able to simplify it.
          Replace $p_G(x)$ by $p_Z(z) left| frac{partial z}{partial x}right|$ in the original equation, multiply both sides by the denominator, and rearrange to group the $p_X(x)$s together:
          $$ p_X(x)(1-D(x)) = D(x) p_Z(z) left| frac{partial z}{partial x}right|.$$
          Now take logs to get the result you want.






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            The log of a sum is an indicator that your first equation won't be helpful, because you won't be able to simplify it.
            Replace $p_G(x)$ by $p_Z(z) left| frac{partial z}{partial x}right|$ in the original equation, multiply both sides by the denominator, and rearrange to group the $p_X(x)$s together:
            $$ p_X(x)(1-D(x)) = D(x) p_Z(z) left| frac{partial z}{partial x}right|.$$
            Now take logs to get the result you want.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The log of a sum is an indicator that your first equation won't be helpful, because you won't be able to simplify it.
              Replace $p_G(x)$ by $p_Z(z) left| frac{partial z}{partial x}right|$ in the original equation, multiply both sides by the denominator, and rearrange to group the $p_X(x)$s together:
              $$ p_X(x)(1-D(x)) = D(x) p_Z(z) left| frac{partial z}{partial x}right|.$$
              Now take logs to get the result you want.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The log of a sum is an indicator that your first equation won't be helpful, because you won't be able to simplify it.
                Replace $p_G(x)$ by $p_Z(z) left| frac{partial z}{partial x}right|$ in the original equation, multiply both sides by the denominator, and rearrange to group the $p_X(x)$s together:
                $$ p_X(x)(1-D(x)) = D(x) p_Z(z) left| frac{partial z}{partial x}right|.$$
                Now take logs to get the result you want.






                share|cite|improve this answer









                $endgroup$



                The log of a sum is an indicator that your first equation won't be helpful, because you won't be able to simplify it.
                Replace $p_G(x)$ by $p_Z(z) left| frac{partial z}{partial x}right|$ in the original equation, multiply both sides by the denominator, and rearrange to group the $p_X(x)$s together:
                $$ p_X(x)(1-D(x)) = D(x) p_Z(z) left| frac{partial z}{partial x}right|.$$
                Now take logs to get the result you want.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 4 '18 at 11:17









                Matthew TowersMatthew Towers

                7,47622344




                7,47622344






























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