How to take the derivative of product of a sequence?












0












$begingroup$


How to take the derivative of $prod_{i=1}^{n}(1-e^{-lambda _{i}cdot x})I(x>0)$ with regard to x?



Here $F(X)=prod_{i=1}^{n}(1-e^{-lambda _{i}cdot x})I(x>0)$ is a CDF, and I want to take the derivative of it and get the pdf of X.



I try to take the log of F(X), so $frac{partial}{partial x}logF(x)=frac{partial logF(x)}{partial F(x)}cdot frac{partial F(x)}{partial x}$, then I can get $f(x)= frac{partial F(x)}{partial x} $, but I don't know how the move on with $frac{partial}{partial x}logF(x)$










share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    How to take the derivative of $prod_{i=1}^{n}(1-e^{-lambda _{i}cdot x})I(x>0)$ with regard to x?



    Here $F(X)=prod_{i=1}^{n}(1-e^{-lambda _{i}cdot x})I(x>0)$ is a CDF, and I want to take the derivative of it and get the pdf of X.



    I try to take the log of F(X), so $frac{partial}{partial x}logF(x)=frac{partial logF(x)}{partial F(x)}cdot frac{partial F(x)}{partial x}$, then I can get $f(x)= frac{partial F(x)}{partial x} $, but I don't know how the move on with $frac{partial}{partial x}logF(x)$










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      How to take the derivative of $prod_{i=1}^{n}(1-e^{-lambda _{i}cdot x})I(x>0)$ with regard to x?



      Here $F(X)=prod_{i=1}^{n}(1-e^{-lambda _{i}cdot x})I(x>0)$ is a CDF, and I want to take the derivative of it and get the pdf of X.



      I try to take the log of F(X), so $frac{partial}{partial x}logF(x)=frac{partial logF(x)}{partial F(x)}cdot frac{partial F(x)}{partial x}$, then I can get $f(x)= frac{partial F(x)}{partial x} $, but I don't know how the move on with $frac{partial}{partial x}logF(x)$










      share|cite|improve this question









      $endgroup$




      How to take the derivative of $prod_{i=1}^{n}(1-e^{-lambda _{i}cdot x})I(x>0)$ with regard to x?



      Here $F(X)=prod_{i=1}^{n}(1-e^{-lambda _{i}cdot x})I(x>0)$ is a CDF, and I want to take the derivative of it and get the pdf of X.



      I try to take the log of F(X), so $frac{partial}{partial x}logF(x)=frac{partial logF(x)}{partial F(x)}cdot frac{partial F(x)}{partial x}$, then I can get $f(x)= frac{partial F(x)}{partial x} $, but I don't know how the move on with $frac{partial}{partial x}logF(x)$







      calculus probability derivatives






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 9 '18 at 4:20









      CrispCrisp

      133




      133






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          $$
          frac{d}{dx}prod_{i=1}^{n}(1-e^{-lambda _{i}x})I(x>0)=frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)I(x>0)+prod_{i=1}^{n}(1-e^{-lambda _{i}x})frac{d}{dx}I(x>0) .
          $$

          Note that
          $$
          frac{d}{dx}I(x>0)=delta(x) ,
          $$

          therefore in principle your pdf would have a Dirac mass at $x=0$, whose coefficient is $prod_{i=1}^{n}(1-e^{-lambda _{i}x})Big|_{x=0}=0$ (so no point mass after all...).



          Evaluating $frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)$ is not hard, just rewrite
          $$
          frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)=frac{d}{dx}e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}=e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}sum_{i=1}^n frac{d}{dx}log(1-e^{-lambda _{i}x})
          $$

          $$
          =left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)sum_{j=1}^nfrac{lambda_j e^{-lambda_j x}}{1-e^{-lambda_j x}} .
          $$






          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%2f3032002%2fhow-to-take-the-derivative-of-product-of-a-sequence%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












            $begingroup$

            $$
            frac{d}{dx}prod_{i=1}^{n}(1-e^{-lambda _{i}x})I(x>0)=frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)I(x>0)+prod_{i=1}^{n}(1-e^{-lambda _{i}x})frac{d}{dx}I(x>0) .
            $$

            Note that
            $$
            frac{d}{dx}I(x>0)=delta(x) ,
            $$

            therefore in principle your pdf would have a Dirac mass at $x=0$, whose coefficient is $prod_{i=1}^{n}(1-e^{-lambda _{i}x})Big|_{x=0}=0$ (so no point mass after all...).



            Evaluating $frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)$ is not hard, just rewrite
            $$
            frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)=frac{d}{dx}e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}=e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}sum_{i=1}^n frac{d}{dx}log(1-e^{-lambda _{i}x})
            $$

            $$
            =left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)sum_{j=1}^nfrac{lambda_j e^{-lambda_j x}}{1-e^{-lambda_j x}} .
            $$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              $$
              frac{d}{dx}prod_{i=1}^{n}(1-e^{-lambda _{i}x})I(x>0)=frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)I(x>0)+prod_{i=1}^{n}(1-e^{-lambda _{i}x})frac{d}{dx}I(x>0) .
              $$

              Note that
              $$
              frac{d}{dx}I(x>0)=delta(x) ,
              $$

              therefore in principle your pdf would have a Dirac mass at $x=0$, whose coefficient is $prod_{i=1}^{n}(1-e^{-lambda _{i}x})Big|_{x=0}=0$ (so no point mass after all...).



              Evaluating $frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)$ is not hard, just rewrite
              $$
              frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)=frac{d}{dx}e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}=e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}sum_{i=1}^n frac{d}{dx}log(1-e^{-lambda _{i}x})
              $$

              $$
              =left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)sum_{j=1}^nfrac{lambda_j e^{-lambda_j x}}{1-e^{-lambda_j x}} .
              $$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                $$
                frac{d}{dx}prod_{i=1}^{n}(1-e^{-lambda _{i}x})I(x>0)=frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)I(x>0)+prod_{i=1}^{n}(1-e^{-lambda _{i}x})frac{d}{dx}I(x>0) .
                $$

                Note that
                $$
                frac{d}{dx}I(x>0)=delta(x) ,
                $$

                therefore in principle your pdf would have a Dirac mass at $x=0$, whose coefficient is $prod_{i=1}^{n}(1-e^{-lambda _{i}x})Big|_{x=0}=0$ (so no point mass after all...).



                Evaluating $frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)$ is not hard, just rewrite
                $$
                frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)=frac{d}{dx}e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}=e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}sum_{i=1}^n frac{d}{dx}log(1-e^{-lambda _{i}x})
                $$

                $$
                =left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)sum_{j=1}^nfrac{lambda_j e^{-lambda_j x}}{1-e^{-lambda_j x}} .
                $$






                share|cite|improve this answer









                $endgroup$



                $$
                frac{d}{dx}prod_{i=1}^{n}(1-e^{-lambda _{i}x})I(x>0)=frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)I(x>0)+prod_{i=1}^{n}(1-e^{-lambda _{i}x})frac{d}{dx}I(x>0) .
                $$

                Note that
                $$
                frac{d}{dx}I(x>0)=delta(x) ,
                $$

                therefore in principle your pdf would have a Dirac mass at $x=0$, whose coefficient is $prod_{i=1}^{n}(1-e^{-lambda _{i}x})Big|_{x=0}=0$ (so no point mass after all...).



                Evaluating $frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)$ is not hard, just rewrite
                $$
                frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)=frac{d}{dx}e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}=e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}sum_{i=1}^n frac{d}{dx}log(1-e^{-lambda _{i}x})
                $$

                $$
                =left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)sum_{j=1}^nfrac{lambda_j e^{-lambda_j x}}{1-e^{-lambda_j x}} .
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 24 '18 at 15:30









                Pierpaolo VivoPierpaolo Vivo

                5,3562724




                5,3562724






























                    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%2f3032002%2fhow-to-take-the-derivative-of-product-of-a-sequence%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

                    Probability when a professor distributes a quiz and homework assignment to a class of n students.

                    Aardman Animations

                    Are they similar matrix