Laplace transform of normal distribution function?$k$-dimensional normal distribution functionNormal...
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Laplace transform of normal distribution function?
$k$-dimensional normal distribution functionNormal Distribution Calculating ProbabilityUnilateral Laplace transform calculationLaplace transform of convolution integralLaplace transform of a normal distributionHow to compute the Laplace transform of a normally distributed density function?Laplace Transform gets complicatedRegion of Convergence of Bilateral Laplace TransformProduct distribution of independent Normal and Exponential random variablesSolve loss function for a normal distribution by integration
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In my notes this was left as an exercise and I am a bit rusty with my calculus.
Starting with the definitions:
$$mathcal{L}_X(t) = mathbb{E}[e^{-tX}] = int_0^infty e^{-Xt}f(t)dt ;;text{ and };;Xsimmathcal{N}(mu,sigma);;text{ i.e. };;f(t) = frac{1}{sqrt{2pi}sigma}e^{-frac{1}{2}frac{(x-mu)^2}{sigma^2}}$$ so
$$mathcal{L}_X(t) = int_0^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}frac{(x-mu)^2}{sigma^2}} e^{-xt}dx$$
$$= frac{1}{sqrt{2pi}sigma} int_0^infty e^{-frac{1}{2}big(frac{(x-mu)^2}{sigma^2}big)-tx }dx$$
suppose I now say $u = frac{x-mu}{sigma}$ so that $x = usigma+mu$ I get the following but have run out of ideas for how to continue:
$$ = frac{1}{sqrt{2pi}sigma} int_{-frac{mu}{sigma}}^infty e^{-frac{1}{2}u^2-t(usigma+mu) }du$$
I considered trying to complete the square but don't think it helped: the exponent becomes $-frac{1}{2}(u^2-2tusigma-2tmu)$ giving $-frac{1}{2}((u-tsigma)^2 -t^2sigma^2+2tmu)$, which doesn't seem to be any simpler to integrate.
EDIT
After thinking about the comments I think this should be a double sided integral, as the expectation would be an integral over the probability distribution's domain (?)
So now I get
$$ mathcal{L}_X(t) = frac{1}{sqrt{2pi}sigma} int_{-infty}^infty e^{-frac{1}{2}u^2-t(usigma+mu) }du$$
Now I try completing the square as above and get
$$ frac{1}{sqrt{2pi}sigma} int_{-infty}^infty e^{-frac{1}{2}((u-tsigma)^2 -t^2sigma^2+2tmu) }du = frac{1}{sqrt{2pi}sigma} e^{-t^2sigma^2-2tmu} int_{-infty}^infty
e^{-frac{1}{2}(u-tsigma)^2}du$$
Now I make a substitution to say $z = frac{1}{sqrt{2}}(u-tsigma)$ and then we get $u = sqrt{2}z+tsigma$, and $du = sqrt{2}dz$ so finally:
$$ frac{1}{sqrt{2pi}sigma}e^{-t^2sigma^2-2tmu} int_{-infty}^infty e^{-z^2} sqrt{2}dz =
frac{1}{sigma}e^{-t^2sigma^2-2tmu}$$
OK, so that is my attempt. I am very not confident about it, so any help/corrections are very welcome.
normal-distribution laplace-transform
asked Nov 22 '15 at 10:54
LuskentyrianLuskentyrian
147210
$endgroup$
add a comment |
$begingroup$
In my notes this was left as an exercise and I am a bit rusty with my calculus.
Starting with the definitions:
$$mathcal{L}_X(t) = mathbb{E}[e^{-tX}] = int_0^infty e^{-Xt}f(t)dt ;;text{ and };;Xsimmathcal{N}(mu,sigma);;text{ i.e. };;f(t) = frac{1}{sqrt{2pi}sigma}e^{-frac{1}{2}frac{(x-mu)^2}{sigma^2}}$$ so
$$mathcal{L}_X(t) = int_0^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}frac{(x-mu)^2}{sigma^2}} e^{-xt}dx$$
$$= frac{1}{sqrt{2pi}sigma} int_0^infty e^{-frac{1}{2}big(frac{(x-mu)^2}{sigma^2}big)-tx }dx$$
suppose I now say $u = frac{x-mu}{sigma}$ so that $x = usigma+mu$ I get the following but have run out of ideas for how to continue:
$$ = frac{1}{sqrt{2pi}sigma} int_{-frac{mu}{sigma}}^infty e^{-frac{1}{2}u^2-t(usigma+mu) }du$$
I considered trying to complete the square but don't think it helped: the exponent becomes $-frac{1}{2}(u^2-2tusigma-2tmu)$ giving $-frac{1}{2}((u-tsigma)^2 -t^2sigma^2+2tmu)$, which doesn't seem to be any simpler to integrate.
EDIT
After thinking about the comments I think this should be a double sided integral, as the expectation would be an integral over the probability distribution's domain (?)
So now I get
$$ mathcal{L}_X(t) = frac{1}{sqrt{2pi}sigma} int_{-infty}^infty e^{-frac{1}{2}u^2-t(usigma+mu) }du$$
Now I try completing the square as above and get
$$ frac{1}{sqrt{2pi}sigma} int_{-infty}^infty e^{-frac{1}{2}((u-tsigma)^2 -t^2sigma^2+2tmu) }du = frac{1}{sqrt{2pi}sigma} e^{-t^2sigma^2-2tmu} int_{-infty}^infty
e^{-frac{1}{2}(u-tsigma)^2}du$$
Now I make a substitution to say $z = frac{1}{sqrt{2}}(u-tsigma)$ and then we get $u = sqrt{2}z+tsigma$, and $du = sqrt{2}dz$ so finally:
$$ frac{1}{sqrt{2pi}sigma}e^{-t^2sigma^2-2tmu} int_{-infty}^infty e^{-z^2} sqrt{2}dz =
frac{1}{sigma}e^{-t^2sigma^2-2tmu}$$
OK, so that is my attempt. I am very not confident about it, so any help/corrections are very welcome.
normal-distribution laplace-transform
asked Nov 22 '15 at 10:54
LuskentyrianLuskentyrian
147210
$endgroup$
$begingroup$
Typically for probability we do moment generating functions which is the 2-sided Laplace transform, integrate on $(-infty, infty)$. Are you sure it is the one-tailed Laplace you need?
$endgroup$
– Nero
Nov 22 '15 at 11:00
$begingroup$
@Nero I think one-sided as that is the only transform we defined in the course. The one example worked was for the exponential distribution, which was one sided. Is it much easier to do it two-sided?
$endgroup$
– Luskentyrian
Nov 22 '15 at 11:47
$begingroup$
Yes, 2 sided is much easier because the range is unchanged by the transformation.
$endgroup$
– Nero
Nov 22 '15 at 12:28
$begingroup$
@Nero On second thoughts I think it should be double sided as the expectation would run from $-infty$ to $infty$. I have made a new attempt at this in an edit.
$endgroup$
– Luskentyrian
Nov 30 '15 at 12:42
add a comment |
$begingroup$
In my notes this was left as an exercise and I am a bit rusty with my calculus.
Starting with the definitions:
$$mathcal{L}_X(t) = mathbb{E}[e^{-tX}] = int_0^infty e^{-Xt}f(t)dt ;;text{ and };;Xsimmathcal{N}(mu,sigma);;text{ i.e. };;f(t) = frac{1}{sqrt{2pi}sigma}e^{-frac{1}{2}frac{(x-mu)^2}{sigma^2}}$$ so
$$mathcal{L}_X(t) = int_0^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}frac{(x-mu)^2}{sigma^2}} e^{-xt}dx$$
$$= frac{1}{sqrt{2pi}sigma} int_0^infty e^{-frac{1}{2}big(frac{(x-mu)^2}{sigma^2}big)-tx }dx$$
suppose I now say $u = frac{x-mu}{sigma}$ so that $x = usigma+mu$ I get the following but have run out of ideas for how to continue:
$$ = frac{1}{sqrt{2pi}sigma} int_{-frac{mu}{sigma}}^infty e^{-frac{1}{2}u^2-t(usigma+mu) }du$$
I considered trying to complete the square but don't think it helped: the exponent becomes $-frac{1}{2}(u^2-2tusigma-2tmu)$ giving $-frac{1}{2}((u-tsigma)^2 -t^2sigma^2+2tmu)$, which doesn't seem to be any simpler to integrate.
EDIT
After thinking about the comments I think this should be a double sided integral, as the expectation would be an integral over the probability distribution's domain (?)
So now I get
$$ mathcal{L}_X(t) = frac{1}{sqrt{2pi}sigma} int_{-infty}^infty e^{-frac{1}{2}u^2-t(usigma+mu) }du$$
Now I try completing the square as above and get
$$ frac{1}{sqrt{2pi}sigma} int_{-infty}^infty e^{-frac{1}{2}((u-tsigma)^2 -t^2sigma^2+2tmu) }du = frac{1}{sqrt{2pi}sigma} e^{-t^2sigma^2-2tmu} int_{-infty}^infty
e^{-frac{1}{2}(u-tsigma)^2}du$$
Now I make a substitution to say $z = frac{1}{sqrt{2}}(u-tsigma)$ and then we get $u = sqrt{2}z+tsigma$, and $du = sqrt{2}dz$ so finally:
$$ frac{1}{sqrt{2pi}sigma}e^{-t^2sigma^2-2tmu} int_{-infty}^infty e^{-z^2} sqrt{2}dz =
frac{1}{sigma}e^{-t^2sigma^2-2tmu}$$
OK, so that is my attempt. I am very not confident about it, so any help/corrections are very welcome.
normal-distribution laplace-transform
asked Nov 22 '15 at 10:54
LuskentyrianLuskentyrian
147210
$endgroup$
In my notes this was left as an exercise and I am a bit rusty with my calculus.
Starting with the definitions:
$$mathcal{L}_X(t) = mathbb{E}[e^{-tX}] = int_0^infty e^{-Xt}f(t)dt ;;text{ and };;Xsimmathcal{N}(mu,sigma);;text{ i.e. };;f(t) = frac{1}{sqrt{2pi}sigma}e^{-frac{1}{2}frac{(x-mu)^2}{sigma^2}}$$ so
$$mathcal{L}_X(t) = int_0^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}frac{(x-mu)^2}{sigma^2}} e^{-xt}dx$$
$$= frac{1}{sqrt{2pi}sigma} int_0^infty e^{-frac{1}{2}big(frac{(x-mu)^2}{sigma^2}big)-tx }dx$$
suppose I now say $u = frac{x-mu}{sigma}$ so that $x = usigma+mu$ I get the following but have run out of ideas for how to continue:
$$ = frac{1}{sqrt{2pi}sigma} int_{-frac{mu}{sigma}}^infty e^{-frac{1}{2}u^2-t(usigma+mu) }du$$
I considered trying to complete the square but don't think it helped: the exponent becomes $-frac{1}{2}(u^2-2tusigma-2tmu)$ giving $-frac{1}{2}((u-tsigma)^2 -t^2sigma^2+2tmu)$, which doesn't seem to be any simpler to integrate.
EDIT
After thinking about the comments I think this should be a double sided integral, as the expectation would be an integral over the probability distribution's domain (?)
So now I get
$$ mathcal{L}_X(t) = frac{1}{sqrt{2pi}sigma} int_{-infty}^infty e^{-frac{1}{2}u^2-t(usigma+mu) }du$$
Now I try completing the square as above and get
$$ frac{1}{sqrt{2pi}sigma} int_{-infty}^infty e^{-frac{1}{2}((u-tsigma)^2 -t^2sigma^2+2tmu) }du = frac{1}{sqrt{2pi}sigma} e^{-t^2sigma^2-2tmu} int_{-infty}^infty
e^{-frac{1}{2}(u-tsigma)^2}du$$
Now I make a substitution to say $z = frac{1}{sqrt{2}}(u-tsigma)$ and then we get $u = sqrt{2}z+tsigma$, and $du = sqrt{2}dz$ so finally:
$$ frac{1}{sqrt{2pi}sigma}e^{-t^2sigma^2-2tmu} int_{-infty}^infty e^{-z^2} sqrt{2}dz =
frac{1}{sigma}e^{-t^2sigma^2-2tmu}$$
OK, so that is my attempt. I am very not confident about it, so any help/corrections are very welcome.
normal-distribution laplace-transform
normal-distribution laplace-transform
asked Nov 22 '15 at 10:54
LuskentyrianLuskentyrian
147210
asked Nov 22 '15 at 10:54
LuskentyrianLuskentyrian
147210
edited Nov 30 '15 at 12:41
Luskentyrian
asked Nov 22 '15 at 10:54
LuskentyrianLuskentyrian
147210
asked Nov 22 '15 at 10:54
LuskentyrianLuskentyrian
147210
asked Nov 22 '15 at 10:54
LuskentyrianLuskentyrian
147210
147210
$begingroup$
Typically for probability we do moment generating functions which is the 2-sided Laplace transform, integrate on $(-infty, infty)$. Are you sure it is the one-tailed Laplace you need?
$endgroup$
– Nero
Nov 22 '15 at 11:00
$begingroup$
@Nero I think one-sided as that is the only transform we defined in the course. The one example worked was for the exponential distribution, which was one sided. Is it much easier to do it two-sided?
$endgroup$
– Luskentyrian
Nov 22 '15 at 11:47
$begingroup$
Yes, 2 sided is much easier because the range is unchanged by the transformation.
$endgroup$
– Nero
Nov 22 '15 at 12:28
$begingroup$
@Nero On second thoughts I think it should be double sided as the expectation would run from $-infty$ to $infty$. I have made a new attempt at this in an edit.
$endgroup$
– Luskentyrian
Nov 30 '15 at 12:42
add a comment |
$begingroup$
Typically for probability we do moment generating functions which is the 2-sided Laplace transform, integrate on $(-infty, infty)$. Are you sure it is the one-tailed Laplace you need?
$endgroup$
– Nero
Nov 22 '15 at 11:00
$begingroup$
@Nero I think one-sided as that is the only transform we defined in the course. The one example worked was for the exponential distribution, which was one sided. Is it much easier to do it two-sided?
$endgroup$
– Luskentyrian
Nov 22 '15 at 11:47
$begingroup$
Yes, 2 sided is much easier because the range is unchanged by the transformation.
$endgroup$
– Nero
Nov 22 '15 at 12:28
$begingroup$
@Nero On second thoughts I think it should be double sided as the expectation would run from $-infty$ to $infty$. I have made a new attempt at this in an edit.
$endgroup$
– Luskentyrian
Nov 30 '15 at 12:42
$begingroup$
Typically for probability we do moment generating functions which is the 2-sided Laplace transform, integrate on $(-infty, infty)$. Are you sure it is the one-tailed Laplace you need?
$endgroup$
– Nero
Nov 22 '15 at 11:00
$begingroup$
Typically for probability we do moment generating functions which is the 2-sided Laplace transform, integrate on $(-infty, infty)$. Are you sure it is the one-tailed Laplace you need?
$endgroup$
– Nero
Nov 22 '15 at 11:00
$begingroup$
@Nero I think one-sided as that is the only transform we defined in the course. The one example worked was for the exponential distribution, which was one sided. Is it much easier to do it two-sided?
$endgroup$
– Luskentyrian
Nov 22 '15 at 11:47
$begingroup$
@Nero I think one-sided as that is the only transform we defined in the course. The one example worked was for the exponential distribution, which was one sided. Is it much easier to do it two-sided?
$endgroup$
– Luskentyrian
Nov 22 '15 at 11:47
$begingroup$
Yes, 2 sided is much easier because the range is unchanged by the transformation.
$endgroup$
– Nero
Nov 22 '15 at 12:28
$begingroup$
Yes, 2 sided is much easier because the range is unchanged by the transformation.
$endgroup$
– Nero
Nov 22 '15 at 12:28
$begingroup$
@Nero On second thoughts I think it should be double sided as the expectation would run from $-infty$ to $infty$. I have made a new attempt at this in an edit.
$endgroup$
– Luskentyrian
Nov 30 '15 at 12:42
$begingroup$
@Nero On second thoughts I think it should be double sided as the expectation would run from $-infty$ to $infty$. I have made a new attempt at this in an edit.
$endgroup$
– Luskentyrian
Nov 30 '15 at 12:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Instead of doing the $u$-substitution:
$$
begin{align}
mathcal{L}_X (t) & = int_{-infty}^infty e^{-tx} frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}right)^2} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}right)^2-tx} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2txright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2t(x-mu)+2tmuright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2(tsigma)left(frac{x-mu}{sigma}right)+2tmuright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2(tsigma)left(frac{x-mu}{sigma}right)+t^2sigma^2-t^2sigma^2+2tmuright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2(tsigma)left(frac{x-mu}{sigma}right)+t^2sigma^2right)+frac{1}{2}t^2sigma^2-tmu} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}+tsigmaright)^2+frac{1}{2}t^2sigma^2-tmu} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}+tsigmaright)^2}e^{frac{1}{2}t^2sigma^2-tmu} mathrm{d}x \
& = e^{frac{1}{2}t^2sigma^2-tmu} int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}+tsigmaright)^2} mathrm{d}x \
& = e^{frac{1}{2}t^2sigma^2-tmu} int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{(x+tsigma^2)-mu}{sigma}right)^2} mathrm{d}x
end{align}
$$
Let $y=x+tsigma^2$. $mathrm{d}y=mathrm{d}x$, $lim_{xtoinfty}ytoinfty$, $lim_{xto-infty}yto-infty$.
$$
begin{align}
mathcal{L}_X (t) & = e^{frac{1}{2}t^2sigma^2-tmu} int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{y-mu}{sigma}right)^2} mathrm{d}y \
& = e^{frac{1}{2}t^2sigma^2-tmu}
end{align}
$$
answered Mar 1 at 3:59
Vibius Vibidius ZosimusVibius Vibidius Zosimus
365
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Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
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– dantopa
Mar 1 at 4:16
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1 Answer
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1 Answer
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$begingroup$
Instead of doing the $u$-substitution:
$$
begin{align}
mathcal{L}_X (t) & = int_{-infty}^infty e^{-tx} frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}right)^2} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}right)^2-tx} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2txright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2t(x-mu)+2tmuright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2(tsigma)left(frac{x-mu}{sigma}right)+2tmuright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2(tsigma)left(frac{x-mu}{sigma}right)+t^2sigma^2-t^2sigma^2+2tmuright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2(tsigma)left(frac{x-mu}{sigma}right)+t^2sigma^2right)+frac{1}{2}t^2sigma^2-tmu} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}+tsigmaright)^2+frac{1}{2}t^2sigma^2-tmu} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}+tsigmaright)^2}e^{frac{1}{2}t^2sigma^2-tmu} mathrm{d}x \
& = e^{frac{1}{2}t^2sigma^2-tmu} int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}+tsigmaright)^2} mathrm{d}x \
& = e^{frac{1}{2}t^2sigma^2-tmu} int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{(x+tsigma^2)-mu}{sigma}right)^2} mathrm{d}x
end{align}
$$
Let $y=x+tsigma^2$. $mathrm{d}y=mathrm{d}x$, $lim_{xtoinfty}ytoinfty$, $lim_{xto-infty}yto-infty$.
$$
begin{align}
mathcal{L}_X (t) & = e^{frac{1}{2}t^2sigma^2-tmu} int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{y-mu}{sigma}right)^2} mathrm{d}y \
& = e^{frac{1}{2}t^2sigma^2-tmu}
end{align}
$$
answered Mar 1 at 3:59
Vibius Vibidius ZosimusVibius Vibidius Zosimus
365
$endgroup$
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
$endgroup$
– dantopa
Mar 1 at 4:16
add a comment |
$begingroup$
Instead of doing the $u$-substitution:
$$
begin{align}
mathcal{L}_X (t) & = int_{-infty}^infty e^{-tx} frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}right)^2} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}right)^2-tx} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2txright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2t(x-mu)+2tmuright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2(tsigma)left(frac{x-mu}{sigma}right)+2tmuright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2(tsigma)left(frac{x-mu}{sigma}right)+t^2sigma^2-t^2sigma^2+2tmuright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2(tsigma)left(frac{x-mu}{sigma}right)+t^2sigma^2right)+frac{1}{2}t^2sigma^2-tmu} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}+tsigmaright)^2+frac{1}{2}t^2sigma^2-tmu} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}+tsigmaright)^2}e^{frac{1}{2}t^2sigma^2-tmu} mathrm{d}x \
& = e^{frac{1}{2}t^2sigma^2-tmu} int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}+tsigmaright)^2} mathrm{d}x \
& = e^{frac{1}{2}t^2sigma^2-tmu} int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{(x+tsigma^2)-mu}{sigma}right)^2} mathrm{d}x
end{align}
$$
Let $y=x+tsigma^2$. $mathrm{d}y=mathrm{d}x$, $lim_{xtoinfty}ytoinfty$, $lim_{xto-infty}yto-infty$.
$$
begin{align}
mathcal{L}_X (t) & = e^{frac{1}{2}t^2sigma^2-tmu} int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{y-mu}{sigma}right)^2} mathrm{d}y \
& = e^{frac{1}{2}t^2sigma^2-tmu}
end{align}
$$
answered Mar 1 at 3:59
Vibius Vibidius ZosimusVibius Vibidius Zosimus
365
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– dantopa
Mar 1 at 4:16
add a comment |
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Instead of doing the $u$-substitution:
$$
begin{align}
mathcal{L}_X (t) & = int_{-infty}^infty e^{-tx} frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}right)^2} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}right)^2-tx} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2txright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2t(x-mu)+2tmuright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2(tsigma)left(frac{x-mu}{sigma}right)+2tmuright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2(tsigma)left(frac{x-mu}{sigma}right)+t^2sigma^2-t^2sigma^2+2tmuright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2(tsigma)left(frac{x-mu}{sigma}right)+t^2sigma^2right)+frac{1}{2}t^2sigma^2-tmu} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}+tsigmaright)^2+frac{1}{2}t^2sigma^2-tmu} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}+tsigmaright)^2}e^{frac{1}{2}t^2sigma^2-tmu} mathrm{d}x \
& = e^{frac{1}{2}t^2sigma^2-tmu} int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}+tsigmaright)^2} mathrm{d}x \
& = e^{frac{1}{2}t^2sigma^2-tmu} int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{(x+tsigma^2)-mu}{sigma}right)^2} mathrm{d}x
end{align}
$$
Let $y=x+tsigma^2$. $mathrm{d}y=mathrm{d}x$, $lim_{xtoinfty}ytoinfty$, $lim_{xto-infty}yto-infty$.
$$
begin{align}
mathcal{L}_X (t) & = e^{frac{1}{2}t^2sigma^2-tmu} int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{y-mu}{sigma}right)^2} mathrm{d}y \
& = e^{frac{1}{2}t^2sigma^2-tmu}
end{align}
$$
answered Mar 1 at 3:59
Vibius Vibidius ZosimusVibius Vibidius Zosimus
365
$endgroup$
Instead of doing the $u$-substitution:
$$
begin{align}
mathcal{L}_X (t) & = int_{-infty}^infty e^{-tx} frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}right)^2} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}right)^2-tx} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2txright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2t(x-mu)+2tmuright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2(tsigma)left(frac{x-mu}{sigma}right)+2tmuright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2(tsigma)left(frac{x-mu}{sigma}right)+t^2sigma^2-t^2sigma^2+2tmuright)} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(left(frac{x-mu}{sigma}right)^2+2(tsigma)left(frac{x-mu}{sigma}right)+t^2sigma^2right)+frac{1}{2}t^2sigma^2-tmu} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}+tsigmaright)^2+frac{1}{2}t^2sigma^2-tmu} mathrm{d}x \
& = int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}+tsigmaright)^2}e^{frac{1}{2}t^2sigma^2-tmu} mathrm{d}x \
& = e^{frac{1}{2}t^2sigma^2-tmu} int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{x-mu}{sigma}+tsigmaright)^2} mathrm{d}x \
& = e^{frac{1}{2}t^2sigma^2-tmu} int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{(x+tsigma^2)-mu}{sigma}right)^2} mathrm{d}x
end{align}
$$
Let $y=x+tsigma^2$. $mathrm{d}y=mathrm{d}x$, $lim_{xtoinfty}ytoinfty$, $lim_{xto-infty}yto-infty$.
$$
begin{align}
mathcal{L}_X (t) & = e^{frac{1}{2}t^2sigma^2-tmu} int_{-infty}^infty frac{1}{sqrt{2pi}sigma} e^{-frac{1}{2}left(frac{y-mu}{sigma}right)^2} mathrm{d}y \
& = e^{frac{1}{2}t^2sigma^2-tmu}
end{align}
$$
answered Mar 1 at 3:59
Vibius Vibidius ZosimusVibius Vibidius Zosimus
365
edited Mar 18 at 19:19
answered Mar 1 at 3:59
Vibius Vibidius ZosimusVibius Vibidius Zosimus
365
answered Mar 1 at 3:59
Vibius Vibidius ZosimusVibius Vibidius Zosimus
365
answered Mar 1 at 3:59
Vibius Vibidius ZosimusVibius Vibidius Zosimus
365
365
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– dantopa
Mar 1 at 4:16
add a comment |
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
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– dantopa
Mar 1 at 4:16
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
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– dantopa
Mar 1 at 4:16
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
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– dantopa
Mar 1 at 4:16
add a comment |
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Typically for probability we do moment generating functions which is the 2-sided Laplace transform, integrate on $(-infty, infty)$. Are you sure it is the one-tailed Laplace you need?
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– Nero
Nov 22 '15 at 11:00
$begingroup$
@Nero I think one-sided as that is the only transform we defined in the course. The one example worked was for the exponential distribution, which was one sided. Is it much easier to do it two-sided?
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– Luskentyrian
Nov 22 '15 at 11:47
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Yes, 2 sided is much easier because the range is unchanged by the transformation.
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– Nero
Nov 22 '15 at 12:28
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@Nero On second thoughts I think it should be double sided as the expectation would run from $-infty$ to $infty$. I have made a new attempt at this in an edit.
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– Luskentyrian
Nov 30 '15 at 12:42