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?










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?










    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?










      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






      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

      510215




      510215






















          1 Answer
          1






          active

          oldest

          votes


















          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









          $endgroup$













            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%2f3025411%2fre-arranging-of-formulation-for-generative-adversarial-networks-gan-formulatio%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









            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












              $begingroup$

              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






























                    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%2f3025411%2fre-arranging-of-formulation-for-generative-adversarial-networks-gan-formulatio%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

                    Ellipse (mathématiques)

                    Quarter-circle Tiles

                    Mont Emei