How to take the derivative of product of a sequence?












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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







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    $begingroup$


    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 $frac{partial}{partial x}logF(x)$






    calculus probability derivatives







    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 $frac{partial}{partial x}logF(x)$






      calculus probability derivatives







      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




      calculus probability derivatives






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Dec 9 '18 at 4:20









      CrispCrisp

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      133






















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          $begingroup$

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






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            1 Answer
            1






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            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

                5,3462724




                5,3462724






























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