How to take the derivative of product of a sequence?
$begingroup$
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
asked Dec 9 '18 at 4:20
CrispCrisp
133
$endgroup$
add a comment |
$begingroup$
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
asked Dec 9 '18 at 4:20
CrispCrisp
133
$endgroup$
add a comment |
$begingroup$
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
asked Dec 9 '18 at 4:20
CrispCrisp
133
$endgroup$
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
asked Dec 9 '18 at 4:20
CrispCrisp
133
asked Dec 9 '18 at 4:20
CrispCrisp
133
asked Dec 9 '18 at 4:20
CrispCrisp
133
asked Dec 9 '18 at 4:20
CrispCrisp
133
asked Dec 9 '18 at 4:20
CrispCrisp
133
133
add a comment |
add a comment |
1 Answer
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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
$$
frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)=frac{d}{dx}e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}=e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}sum_{i=1}^n frac{d}{dx}log(1-e^{-lambda _{i}x})
$$
$$
=left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)sum_{j=1}^nfrac{lambda_j e^{-lambda_j x}}{1-e^{-lambda_j x}} .
$$
answered Dec 24 '18 at 15:30
Pierpaolo VivoPierpaolo Vivo
5,3562724
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1 Answer
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1 Answer
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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
$$
frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)=frac{d}{dx}e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}=e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}sum_{i=1}^n frac{d}{dx}log(1-e^{-lambda _{i}x})
$$
$$
=left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)sum_{j=1}^nfrac{lambda_j e^{-lambda_j x}}{1-e^{-lambda_j x}} .
$$
answered Dec 24 '18 at 15:30
Pierpaolo VivoPierpaolo Vivo
5,3562724
$endgroup$
add a comment |
$begingroup$
$$
frac{d}{dx}prod_{i=1}^{n}(1-e^{-lambda _{i}x})I(x>0)=frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)I(x>0)+prod_{i=1}^{n}(1-e^{-lambda _{i}x})frac{d}{dx}I(x>0) .
$$
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
$$
frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)=frac{d}{dx}e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}=e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}sum_{i=1}^n frac{d}{dx}log(1-e^{-lambda _{i}x})
$$
$$
=left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)sum_{j=1}^nfrac{lambda_j e^{-lambda_j x}}{1-e^{-lambda_j x}} .
$$
answered Dec 24 '18 at 15:30
Pierpaolo VivoPierpaolo Vivo
5,3562724
$endgroup$
add a comment |
$begingroup$
$$
frac{d}{dx}prod_{i=1}^{n}(1-e^{-lambda _{i}x})I(x>0)=frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)I(x>0)+prod_{i=1}^{n}(1-e^{-lambda _{i}x})frac{d}{dx}I(x>0) .
$$
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
$$
frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)=frac{d}{dx}e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}=e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}sum_{i=1}^n frac{d}{dx}log(1-e^{-lambda _{i}x})
$$
$$
=left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)sum_{j=1}^nfrac{lambda_j e^{-lambda_j x}}{1-e^{-lambda_j x}} .
$$
answered Dec 24 '18 at 15:30
Pierpaolo VivoPierpaolo Vivo
5,3562724
$endgroup$
$$
frac{d}{dx}prod_{i=1}^{n}(1-e^{-lambda _{i}x})I(x>0)=frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)I(x>0)+prod_{i=1}^{n}(1-e^{-lambda _{i}x})frac{d}{dx}I(x>0) .
$$
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
$$
frac{d}{dx}left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)=frac{d}{dx}e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}=e^{sum_{i=1}^n log(1-e^{-lambda _{i}x})}sum_{i=1}^n frac{d}{dx}log(1-e^{-lambda _{i}x})
$$
$$
=left(prod_{i=1}^{n}(1-e^{-lambda _{i}x})right)sum_{j=1}^nfrac{lambda_j e^{-lambda_j x}}{1-e^{-lambda_j x}} .
$$
answered Dec 24 '18 at 15:30
Pierpaolo VivoPierpaolo Vivo
5,3562724
answered Dec 24 '18 at 15:30
Pierpaolo VivoPierpaolo Vivo
5,3562724
answered Dec 24 '18 at 15:30
Pierpaolo VivoPierpaolo Vivo
5,3562724
answered Dec 24 '18 at 15:30
Pierpaolo VivoPierpaolo Vivo
5,3562724
5,3562724
add a comment |
add a comment |
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