Take the derivative of a product of sequence












2














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 $frac{d}{d x}logF(x)=frac{d logF(x)}{d F(x)}cdot frac{d F(x)}{d x}$,



@Ankit Kumar help me with:



$$frac{d(logF(x))}{dx}=sum_{i=1}^nfrac{d(log(1-e^{-lambda_ix}))}{dx}$$,
then $$frac{1}{F(x)}frac{d(F(x))}{dx}=sum_{i=1}^nfrac{1}{log(1-e^{-lambda_ix})}{lambda_ie^{-lambda_ix}}$$, but I don't know how to move on with $frac{d(F(x))}{dx}=sum_{i=1}^nfrac{1}{log(1-e^{-lambda_ix})}{lambda_ie^{-lambda_ix}}cdot prod_{i=1}^{n}(1-e^{-lambda _{i}cdot x})$, how to simplify it? since it is a multiplication of a sum of sequence and a product of sequence.






calculus probability derivatives







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    2














    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 $frac{d}{d x}logF(x)=frac{d logF(x)}{d F(x)}cdot frac{d F(x)}{d x}$,



    @Ankit Kumar help me with:



    $$frac{d(logF(x))}{dx}=sum_{i=1}^nfrac{d(log(1-e^{-lambda_ix}))}{dx}$$,
    then $$frac{1}{F(x)}frac{d(F(x))}{dx}=sum_{i=1}^nfrac{1}{log(1-e^{-lambda_ix})}{lambda_ie^{-lambda_ix}}$$, but I don't know how to move on with $frac{d(F(x))}{dx}=sum_{i=1}^nfrac{1}{log(1-e^{-lambda_ix})}{lambda_ie^{-lambda_ix}}cdot prod_{i=1}^{n}(1-e^{-lambda _{i}cdot x})$, how to simplify it? since it is a multiplication of a sum of sequence and a product of sequence.






    calculus probability derivatives







    share|cite|improve this question

























      2












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      2


      0





      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 $frac{d}{d x}logF(x)=frac{d logF(x)}{d F(x)}cdot frac{d F(x)}{d x}$,



      @Ankit Kumar help me with:



      $$frac{d(logF(x))}{dx}=sum_{i=1}^nfrac{d(log(1-e^{-lambda_ix}))}{dx}$$,
      then $$frac{1}{F(x)}frac{d(F(x))}{dx}=sum_{i=1}^nfrac{1}{log(1-e^{-lambda_ix})}{lambda_ie^{-lambda_ix}}$$, but I don't know how to move on with $frac{d(F(x))}{dx}=sum_{i=1}^nfrac{1}{log(1-e^{-lambda_ix})}{lambda_ie^{-lambda_ix}}cdot prod_{i=1}^{n}(1-e^{-lambda _{i}cdot x})$, how to simplify it? since it is a multiplication of a sum of sequence and a product of sequence.






      calculus probability derivatives







      share|cite|improve this question













      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 $frac{d}{d x}logF(x)=frac{d logF(x)}{d F(x)}cdot frac{d F(x)}{d x}$,



      @Ankit Kumar help me with:



      $$frac{d(logF(x))}{dx}=sum_{i=1}^nfrac{d(log(1-e^{-lambda_ix}))}{dx}$$,
      then $$frac{1}{F(x)}frac{d(F(x))}{dx}=sum_{i=1}^nfrac{1}{log(1-e^{-lambda_ix})}{lambda_ie^{-lambda_ix}}$$, but I don't know how to move on with $frac{d(F(x))}{dx}=sum_{i=1}^nfrac{1}{log(1-e^{-lambda_ix})}{lambda_ie^{-lambda_ix}}cdot prod_{i=1}^{n}(1-e^{-lambda _{i}cdot x})$, how to simplify it? since it is a multiplication of a sum of sequence and a product of sequence.






      calculus probability derivatives




      calculus probability derivatives






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      asked Dec 9 at 15:43









      Crisp

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          Here is another approach. Recall the product formula $(fcdot g)^{prime}=f^{prime}cdot g+fcdot g^{prime}$ of the product of two differentiable functions $f,g$. We can generalise this (and for instance prove it with induction) and obtain
          begin{align*}
          color{blue}{left(prod_{j=1}^n f_j(x)right)^{prime}}
          &=left(f_1(x)right)^prime prod_{j=2}^nf_j(x)+f_1(x)left(prod_{j=2}^n f_j(x)right)^prime\
          &=left(f_1(x)right)^prime prod_{j=2}^nf_j(x)+f_1(x)left(left(f_2(x)right)^prime prod_{j=3}^nf_j(x)+f_2(x)left(prod_{j=3}^n f_j(x)right)^primeright)\
          &=left(f_1(x)right)^prime prod_{{j=1}atop{jne 1}}^nf_j(x)+left(f_2(x)right)^prime prod_{{j=1}atop{jne 2}}^nf_j(x)+f_1(x)f_2(x)left(prod_{j=3}^n f_j(x)right)^prime\
          & vdots\
          &=left(f_1(x)right)^prime prod_{{j=1}atop{jne 1}}^nf_j(x)+left(f_2(x)right)^prime prod_{{j=1}atop{jne 2}}^nf_j(x)
          +cdots+left(f_n(x)right)^prime prod_{{j=1}atop{jne n}}^nf_j(x)\
          &,,color{blue}{=sum_{k=1}^nleft(f_k(x)right)^{prime} prod_{{j=1}atop{jne k}}^nf_j(x)}tag{1}
          end{align*}




          We obtain according to (1)
          begin{align*}
          frac{d}{dx}left(prod_{j=1}^nleft(1-e^{-lambda_j x}right)right)
          &=sum_{k=1}^nleft(frac{d}{dx}left(1-e^{-lambda_k x}right)right)prod_{{j=1}atop{jne k}}^nleft(1-e^{-lambda_j x}right)\
          &=sum_{k=1}^nlambda_k e^{-lambda_k x}prod_{{j=1}atop{jne k}}^nleft(1-e^{-lambda_j x}right)
          end{align*}







          share|cite|improve this answer





















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            1 Answer
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            active

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            active

            oldest

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            2














            Here is another approach. Recall the product formula $(fcdot g)^{prime}=f^{prime}cdot g+fcdot g^{prime}$ of the product of two differentiable functions $f,g$. We can generalise this (and for instance prove it with induction) and obtain
            begin{align*}
            color{blue}{left(prod_{j=1}^n f_j(x)right)^{prime}}
            &=left(f_1(x)right)^prime prod_{j=2}^nf_j(x)+f_1(x)left(prod_{j=2}^n f_j(x)right)^prime\
            &=left(f_1(x)right)^prime prod_{j=2}^nf_j(x)+f_1(x)left(left(f_2(x)right)^prime prod_{j=3}^nf_j(x)+f_2(x)left(prod_{j=3}^n f_j(x)right)^primeright)\
            &=left(f_1(x)right)^prime prod_{{j=1}atop{jne 1}}^nf_j(x)+left(f_2(x)right)^prime prod_{{j=1}atop{jne 2}}^nf_j(x)+f_1(x)f_2(x)left(prod_{j=3}^n f_j(x)right)^prime\
            & vdots\
            &=left(f_1(x)right)^prime prod_{{j=1}atop{jne 1}}^nf_j(x)+left(f_2(x)right)^prime prod_{{j=1}atop{jne 2}}^nf_j(x)
            +cdots+left(f_n(x)right)^prime prod_{{j=1}atop{jne n}}^nf_j(x)\
            &,,color{blue}{=sum_{k=1}^nleft(f_k(x)right)^{prime} prod_{{j=1}atop{jne k}}^nf_j(x)}tag{1}
            end{align*}




            We obtain according to (1)
            begin{align*}
            frac{d}{dx}left(prod_{j=1}^nleft(1-e^{-lambda_j x}right)right)
            &=sum_{k=1}^nleft(frac{d}{dx}left(1-e^{-lambda_k x}right)right)prod_{{j=1}atop{jne k}}^nleft(1-e^{-lambda_j x}right)\
            &=sum_{k=1}^nlambda_k e^{-lambda_k x}prod_{{j=1}atop{jne k}}^nleft(1-e^{-lambda_j x}right)
            end{align*}







            share|cite|improve this answer


























              2














              Here is another approach. Recall the product formula $(fcdot g)^{prime}=f^{prime}cdot g+fcdot g^{prime}$ of the product of two differentiable functions $f,g$. We can generalise this (and for instance prove it with induction) and obtain
              begin{align*}
              color{blue}{left(prod_{j=1}^n f_j(x)right)^{prime}}
              &=left(f_1(x)right)^prime prod_{j=2}^nf_j(x)+f_1(x)left(prod_{j=2}^n f_j(x)right)^prime\
              &=left(f_1(x)right)^prime prod_{j=2}^nf_j(x)+f_1(x)left(left(f_2(x)right)^prime prod_{j=3}^nf_j(x)+f_2(x)left(prod_{j=3}^n f_j(x)right)^primeright)\
              &=left(f_1(x)right)^prime prod_{{j=1}atop{jne 1}}^nf_j(x)+left(f_2(x)right)^prime prod_{{j=1}atop{jne 2}}^nf_j(x)+f_1(x)f_2(x)left(prod_{j=3}^n f_j(x)right)^prime\
              & vdots\
              &=left(f_1(x)right)^prime prod_{{j=1}atop{jne 1}}^nf_j(x)+left(f_2(x)right)^prime prod_{{j=1}atop{jne 2}}^nf_j(x)
              +cdots+left(f_n(x)right)^prime prod_{{j=1}atop{jne n}}^nf_j(x)\
              &,,color{blue}{=sum_{k=1}^nleft(f_k(x)right)^{prime} prod_{{j=1}atop{jne k}}^nf_j(x)}tag{1}
              end{align*}




              We obtain according to (1)
              begin{align*}
              frac{d}{dx}left(prod_{j=1}^nleft(1-e^{-lambda_j x}right)right)
              &=sum_{k=1}^nleft(frac{d}{dx}left(1-e^{-lambda_k x}right)right)prod_{{j=1}atop{jne k}}^nleft(1-e^{-lambda_j x}right)\
              &=sum_{k=1}^nlambda_k e^{-lambda_k x}prod_{{j=1}atop{jne k}}^nleft(1-e^{-lambda_j x}right)
              end{align*}







              share|cite|improve this answer
























                2












                2








                2






                Here is another approach. Recall the product formula $(fcdot g)^{prime}=f^{prime}cdot g+fcdot g^{prime}$ of the product of two differentiable functions $f,g$. We can generalise this (and for instance prove it with induction) and obtain
                begin{align*}
                color{blue}{left(prod_{j=1}^n f_j(x)right)^{prime}}
                &=left(f_1(x)right)^prime prod_{j=2}^nf_j(x)+f_1(x)left(prod_{j=2}^n f_j(x)right)^prime\
                &=left(f_1(x)right)^prime prod_{j=2}^nf_j(x)+f_1(x)left(left(f_2(x)right)^prime prod_{j=3}^nf_j(x)+f_2(x)left(prod_{j=3}^n f_j(x)right)^primeright)\
                &=left(f_1(x)right)^prime prod_{{j=1}atop{jne 1}}^nf_j(x)+left(f_2(x)right)^prime prod_{{j=1}atop{jne 2}}^nf_j(x)+f_1(x)f_2(x)left(prod_{j=3}^n f_j(x)right)^prime\
                & vdots\
                &=left(f_1(x)right)^prime prod_{{j=1}atop{jne 1}}^nf_j(x)+left(f_2(x)right)^prime prod_{{j=1}atop{jne 2}}^nf_j(x)
                +cdots+left(f_n(x)right)^prime prod_{{j=1}atop{jne n}}^nf_j(x)\
                &,,color{blue}{=sum_{k=1}^nleft(f_k(x)right)^{prime} prod_{{j=1}atop{jne k}}^nf_j(x)}tag{1}
                end{align*}




                We obtain according to (1)
                begin{align*}
                frac{d}{dx}left(prod_{j=1}^nleft(1-e^{-lambda_j x}right)right)
                &=sum_{k=1}^nleft(frac{d}{dx}left(1-e^{-lambda_k x}right)right)prod_{{j=1}atop{jne k}}^nleft(1-e^{-lambda_j x}right)\
                &=sum_{k=1}^nlambda_k e^{-lambda_k x}prod_{{j=1}atop{jne k}}^nleft(1-e^{-lambda_j x}right)
                end{align*}







                share|cite|improve this answer












                Here is another approach. Recall the product formula $(fcdot g)^{prime}=f^{prime}cdot g+fcdot g^{prime}$ of the product of two differentiable functions $f,g$. We can generalise this (and for instance prove it with induction) and obtain
                begin{align*}
                color{blue}{left(prod_{j=1}^n f_j(x)right)^{prime}}
                &=left(f_1(x)right)^prime prod_{j=2}^nf_j(x)+f_1(x)left(prod_{j=2}^n f_j(x)right)^prime\
                &=left(f_1(x)right)^prime prod_{j=2}^nf_j(x)+f_1(x)left(left(f_2(x)right)^prime prod_{j=3}^nf_j(x)+f_2(x)left(prod_{j=3}^n f_j(x)right)^primeright)\
                &=left(f_1(x)right)^prime prod_{{j=1}atop{jne 1}}^nf_j(x)+left(f_2(x)right)^prime prod_{{j=1}atop{jne 2}}^nf_j(x)+f_1(x)f_2(x)left(prod_{j=3}^n f_j(x)right)^prime\
                & vdots\
                &=left(f_1(x)right)^prime prod_{{j=1}atop{jne 1}}^nf_j(x)+left(f_2(x)right)^prime prod_{{j=1}atop{jne 2}}^nf_j(x)
                +cdots+left(f_n(x)right)^prime prod_{{j=1}atop{jne n}}^nf_j(x)\
                &,,color{blue}{=sum_{k=1}^nleft(f_k(x)right)^{prime} prod_{{j=1}atop{jne k}}^nf_j(x)}tag{1}
                end{align*}




                We obtain according to (1)
                begin{align*}
                frac{d}{dx}left(prod_{j=1}^nleft(1-e^{-lambda_j x}right)right)
                &=sum_{k=1}^nleft(frac{d}{dx}left(1-e^{-lambda_k x}right)right)prod_{{j=1}atop{jne k}}^nleft(1-e^{-lambda_j x}right)\
                &=sum_{k=1}^nlambda_k e^{-lambda_k x}prod_{{j=1}atop{jne k}}^nleft(1-e^{-lambda_j x}right)
                end{align*}








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                answered Dec 9 at 17:28









                Markus Scheuer

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