Integrate: $int_{-infty}^infty exp(-||vec x||^2) ||vec x||^{-m}mathrm{d}vec{x}$
$begingroup$
Let $m,n$ be two positive integers with $0 < m < n$.
Can we integrate this:
$$I = int_{-infty}^infty mathrm{d}x_1 dots int_{-infty}^infty mathrm{d}x_n left(sum_{i=1}^n x_i^2right)^{-m/2} expleft(-sum_{i=1}^n x_i^2right)$$
If a closed analytical expression is not possible, what I really need is to evaluate a large deviation limit of the form:
$$lim_{nrightarrowinfty} frac{1}{n} log I$$
where it is assumed that the ratio $m/n$ remains fixed.
Note that $I$ resembles a negative moment of the radious of a multivariate Gaussian.
I asked a very similar question here: Integrate: $int_0^1 ||vec x||^{-m}mathrm{d}vec{x}$. But note that this integral has different limits and a decaying quadratic exponential. Maybe I have better luck with this variant ;)
normal-distribution large-deviation-theory
asked Dec 6 '18 at 19:20
beckobecko
2,34931942
$endgroup$
add a comment |
$begingroup$
Let $m,n$ be two positive integers with $0 < m < n$.
Can we integrate this:
$$I = int_{-infty}^infty mathrm{d}x_1 dots int_{-infty}^infty mathrm{d}x_n left(sum_{i=1}^n x_i^2right)^{-m/2} expleft(-sum_{i=1}^n x_i^2right)$$
If a closed analytical expression is not possible, what I really need is to evaluate a large deviation limit of the form:
$$lim_{nrightarrowinfty} frac{1}{n} log I$$
where it is assumed that the ratio $m/n$ remains fixed.
Note that $I$ resembles a negative moment of the radious of a multivariate Gaussian.
I asked a very similar question here: Integrate: $int_0^1 ||vec x||^{-m}mathrm{d}vec{x}$. But note that this integral has different limits and a decaying quadratic exponential. Maybe I have better luck with this variant ;)
normal-distribution large-deviation-theory
asked Dec 6 '18 at 19:20
beckobecko
2,34931942
$endgroup$
add a comment |
$begingroup$
Let $m,n$ be two positive integers with $0 < m < n$.
Can we integrate this:
$$I = int_{-infty}^infty mathrm{d}x_1 dots int_{-infty}^infty mathrm{d}x_n left(sum_{i=1}^n x_i^2right)^{-m/2} expleft(-sum_{i=1}^n x_i^2right)$$
If a closed analytical expression is not possible, what I really need is to evaluate a large deviation limit of the form:
$$lim_{nrightarrowinfty} frac{1}{n} log I$$
where it is assumed that the ratio $m/n$ remains fixed.
Note that $I$ resembles a negative moment of the radious of a multivariate Gaussian.
I asked a very similar question here: Integrate: $int_0^1 ||vec x||^{-m}mathrm{d}vec{x}$. But note that this integral has different limits and a decaying quadratic exponential. Maybe I have better luck with this variant ;)
normal-distribution large-deviation-theory
asked Dec 6 '18 at 19:20
beckobecko
2,34931942
$endgroup$
Let $m,n$ be two positive integers with $0 < m < n$.
Can we integrate this:
$$I = int_{-infty}^infty mathrm{d}x_1 dots int_{-infty}^infty mathrm{d}x_n left(sum_{i=1}^n x_i^2right)^{-m/2} expleft(-sum_{i=1}^n x_i^2right)$$
If a closed analytical expression is not possible, what I really need is to evaluate a large deviation limit of the form:
$$lim_{nrightarrowinfty} frac{1}{n} log I$$
where it is assumed that the ratio $m/n$ remains fixed.
Note that $I$ resembles a negative moment of the radious of a multivariate Gaussian.
I asked a very similar question here: Integrate: $int_0^1 ||vec x||^{-m}mathrm{d}vec{x}$. But note that this integral has different limits and a decaying quadratic exponential. Maybe I have better luck with this variant ;)
normal-distribution large-deviation-theory
normal-distribution large-deviation-theory
asked Dec 6 '18 at 19:20
beckobecko
2,34931942
asked Dec 6 '18 at 19:20
beckobecko
2,34931942
edited Dec 6 '18 at 19:24
becko
asked Dec 6 '18 at 19:20
beckobecko
2,34931942
asked Dec 6 '18 at 19:20
beckobecko
2,34931942
asked Dec 6 '18 at 19:20
beckobecko
2,34931942
2,34931942
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider the homomorphism $x mapsto left( |x|, dfrac{x}{|x|} right)$ between $mathbf{R}^d setminus {0}$ and $(0, infty) times mathbf{S}_{d-1}.$ By the change of measures formula, the integral becomes
$$begin{align*}
intlimits_0^infty dr intlimits_{mathbf{S}_{d-1}} dsigma_d e^{-r^2}r^{-m} r^{d-1} &= sigma_{d-1}(mathbf{S}_{d-1}) intlimits_{0}^infty dr e^{-r^2} r^{d-m-1} \
&= dfrac{sigma_{d-1}(mathbf{S}_{d-1})}{2} intlimits_0^infty dr e^{-r} r^{frac{d-m-2}{2}} \
&= dfrac{sigma_{d-1}(mathbf{S}_{d-1})}{2} Gammaleft(dfrac{d-m}{2}right). square
end{align*}$$
answered Dec 6 '18 at 19:53
Will M.Will M.
2,440314
$endgroup$
$begingroup$
Thanks, but the notation is a bit obscure to me. This is the same as what I got, right?
$endgroup$
– becko
Dec 6 '18 at 20:40
$begingroup$
Yours is wrong, I think. To begin with, $N = d$ and $M = m,$ and in the integral you made the change of variables $u = r^2,$ so that $u' = 2r$ and then $int e^{-r^2} r^alpha dr = dfrac{1}{2}int e^{-u} u^{frac{alpha - 1}{2}} du.$ I think you factored out a $2^{frac{d - m}{2}-1}$ wrongly.
$endgroup$
– Will M.
Dec 7 '18 at 1:54
$begingroup$
I see what's wrong. In the original question the exponent was not divided by 2, but in my answer for some reason I divided by 2. If you correct for that we are getting equivalent results.
$endgroup$
– becko
Dec 7 '18 at 15:33
add a comment |
$begingroup$
Introduce the spherical coordinates $r, theta, xi_1, ldots, xi_{N - 2}$,
$$ x_1 = r cos (xi_1), quad x_2 = r sin (xi_1) cos (xi_2), quad
ldots, quad x_N = r sin (xi_1) ldots sin (xi_{N - 2}) sin
(theta) $$
where $0 leqslant xi_i leqslant pi$, $0 leqslant theta leqslant 2 pi$
and $0 leqslant r < infty$. The volume element is given by $mathrm d x_1
ldots mathrm d x_N = r^{N - 1} mathrm d r mathrm d sigma$, where $mathrm d
sigma$ is the unit-sphere surface element,
$$mathrm d sigma = sin^{N - 2} (xi_1) ldots sin (xi_{n - 2}) mathrm d theta
mathrm d xi_1 ldots mathrm d xi_{N - 2}$$
It follows that:
$$begin{eqnarray*}
I & = & int_{- infty}^{infty} | {vec x} |^{- M} mathrm e^{- |
{vec x} |^2 / 2} mathrm d vec x\
& = & left( int_0^{infty} r^{N - M - 1} mathrm e^{- r^2/2}
mathrm d r right) left( int_{mathcal{S}} mathrm d sigma right)\
& = & 2^{(N - M) / 2 - 1} Gamma left( frac{N - M}{2}
right) frac{2 pi^{N / 2}}{Gamma (N / 2)}
end{eqnarray*}$$
where $mathcal S$ is the unit sphere.
answered Dec 6 '18 at 19:24
beckobecko
2,34931942
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the homomorphism $x mapsto left( |x|, dfrac{x}{|x|} right)$ between $mathbf{R}^d setminus {0}$ and $(0, infty) times mathbf{S}_{d-1}.$ By the change of measures formula, the integral becomes
$$begin{align*}
intlimits_0^infty dr intlimits_{mathbf{S}_{d-1}} dsigma_d e^{-r^2}r^{-m} r^{d-1} &= sigma_{d-1}(mathbf{S}_{d-1}) intlimits_{0}^infty dr e^{-r^2} r^{d-m-1} \
&= dfrac{sigma_{d-1}(mathbf{S}_{d-1})}{2} intlimits_0^infty dr e^{-r} r^{frac{d-m-2}{2}} \
&= dfrac{sigma_{d-1}(mathbf{S}_{d-1})}{2} Gammaleft(dfrac{d-m}{2}right). square
end{align*}$$
answered Dec 6 '18 at 19:53
Will M.Will M.
2,440314
$endgroup$
$begingroup$
Thanks, but the notation is a bit obscure to me. This is the same as what I got, right?
$endgroup$
– becko
Dec 6 '18 at 20:40
$begingroup$
Yours is wrong, I think. To begin with, $N = d$ and $M = m,$ and in the integral you made the change of variables $u = r^2,$ so that $u' = 2r$ and then $int e^{-r^2} r^alpha dr = dfrac{1}{2}int e^{-u} u^{frac{alpha - 1}{2}} du.$ I think you factored out a $2^{frac{d - m}{2}-1}$ wrongly.
$endgroup$
– Will M.
Dec 7 '18 at 1:54
$begingroup$
I see what's wrong. In the original question the exponent was not divided by 2, but in my answer for some reason I divided by 2. If you correct for that we are getting equivalent results.
$endgroup$
– becko
Dec 7 '18 at 15:33
add a comment |
$begingroup$
Consider the homomorphism $x mapsto left( |x|, dfrac{x}{|x|} right)$ between $mathbf{R}^d setminus {0}$ and $(0, infty) times mathbf{S}_{d-1}.$ By the change of measures formula, the integral becomes
$$begin{align*}
intlimits_0^infty dr intlimits_{mathbf{S}_{d-1}} dsigma_d e^{-r^2}r^{-m} r^{d-1} &= sigma_{d-1}(mathbf{S}_{d-1}) intlimits_{0}^infty dr e^{-r^2} r^{d-m-1} \
&= dfrac{sigma_{d-1}(mathbf{S}_{d-1})}{2} intlimits_0^infty dr e^{-r} r^{frac{d-m-2}{2}} \
&= dfrac{sigma_{d-1}(mathbf{S}_{d-1})}{2} Gammaleft(dfrac{d-m}{2}right). square
end{align*}$$
answered Dec 6 '18 at 19:53
Will M.Will M.
2,440314
$endgroup$
$begingroup$
Thanks, but the notation is a bit obscure to me. This is the same as what I got, right?
$endgroup$
– becko
Dec 6 '18 at 20:40
$begingroup$
Yours is wrong, I think. To begin with, $N = d$ and $M = m,$ and in the integral you made the change of variables $u = r^2,$ so that $u' = 2r$ and then $int e^{-r^2} r^alpha dr = dfrac{1}{2}int e^{-u} u^{frac{alpha - 1}{2}} du.$ I think you factored out a $2^{frac{d - m}{2}-1}$ wrongly.
$endgroup$
– Will M.
Dec 7 '18 at 1:54
$begingroup$
I see what's wrong. In the original question the exponent was not divided by 2, but in my answer for some reason I divided by 2. If you correct for that we are getting equivalent results.
$endgroup$
– becko
Dec 7 '18 at 15:33
add a comment |
$begingroup$
Consider the homomorphism $x mapsto left( |x|, dfrac{x}{|x|} right)$ between $mathbf{R}^d setminus {0}$ and $(0, infty) times mathbf{S}_{d-1}.$ By the change of measures formula, the integral becomes
$$begin{align*}
intlimits_0^infty dr intlimits_{mathbf{S}_{d-1}} dsigma_d e^{-r^2}r^{-m} r^{d-1} &= sigma_{d-1}(mathbf{S}_{d-1}) intlimits_{0}^infty dr e^{-r^2} r^{d-m-1} \
&= dfrac{sigma_{d-1}(mathbf{S}_{d-1})}{2} intlimits_0^infty dr e^{-r} r^{frac{d-m-2}{2}} \
&= dfrac{sigma_{d-1}(mathbf{S}_{d-1})}{2} Gammaleft(dfrac{d-m}{2}right). square
end{align*}$$
answered Dec 6 '18 at 19:53
Will M.Will M.
2,440314
$endgroup$
Consider the homomorphism $x mapsto left( |x|, dfrac{x}{|x|} right)$ between $mathbf{R}^d setminus {0}$ and $(0, infty) times mathbf{S}_{d-1}.$ By the change of measures formula, the integral becomes
$$begin{align*}
intlimits_0^infty dr intlimits_{mathbf{S}_{d-1}} dsigma_d e^{-r^2}r^{-m} r^{d-1} &= sigma_{d-1}(mathbf{S}_{d-1}) intlimits_{0}^infty dr e^{-r^2} r^{d-m-1} \
&= dfrac{sigma_{d-1}(mathbf{S}_{d-1})}{2} intlimits_0^infty dr e^{-r} r^{frac{d-m-2}{2}} \
&= dfrac{sigma_{d-1}(mathbf{S}_{d-1})}{2} Gammaleft(dfrac{d-m}{2}right). square
end{align*}$$
answered Dec 6 '18 at 19:53
Will M.Will M.
2,440314
answered Dec 6 '18 at 19:53
Will M.Will M.
2,440314
answered Dec 6 '18 at 19:53
Will M.Will M.
2,440314
answered Dec 6 '18 at 19:53
Will M.Will M.
2,440314
2,440314
$begingroup$
Thanks, but the notation is a bit obscure to me. This is the same as what I got, right?
$endgroup$
– becko
Dec 6 '18 at 20:40
$begingroup$
Yours is wrong, I think. To begin with, $N = d$ and $M = m,$ and in the integral you made the change of variables $u = r^2,$ so that $u' = 2r$ and then $int e^{-r^2} r^alpha dr = dfrac{1}{2}int e^{-u} u^{frac{alpha - 1}{2}} du.$ I think you factored out a $2^{frac{d - m}{2}-1}$ wrongly.
$endgroup$
– Will M.
Dec 7 '18 at 1:54
$begingroup$
I see what's wrong. In the original question the exponent was not divided by 2, but in my answer for some reason I divided by 2. If you correct for that we are getting equivalent results.
$endgroup$
– becko
Dec 7 '18 at 15:33
add a comment |
$begingroup$
Thanks, but the notation is a bit obscure to me. This is the same as what I got, right?
$endgroup$
– becko
Dec 6 '18 at 20:40
$begingroup$
Yours is wrong, I think. To begin with, $N = d$ and $M = m,$ and in the integral you made the change of variables $u = r^2,$ so that $u' = 2r$ and then $int e^{-r^2} r^alpha dr = dfrac{1}{2}int e^{-u} u^{frac{alpha - 1}{2}} du.$ I think you factored out a $2^{frac{d - m}{2}-1}$ wrongly.
$endgroup$
– Will M.
Dec 7 '18 at 1:54
$begingroup$
I see what's wrong. In the original question the exponent was not divided by 2, but in my answer for some reason I divided by 2. If you correct for that we are getting equivalent results.
$endgroup$
– becko
Dec 7 '18 at 15:33
$begingroup$
Thanks, but the notation is a bit obscure to me. This is the same as what I got, right?
$endgroup$
– becko
Dec 6 '18 at 20:40
$begingroup$
Thanks, but the notation is a bit obscure to me. This is the same as what I got, right?
$endgroup$
– becko
Dec 6 '18 at 20:40
$begingroup$
Yours is wrong, I think. To begin with, $N = d$ and $M = m,$ and in the integral you made the change of variables $u = r^2,$ so that $u' = 2r$ and then $int e^{-r^2} r^alpha dr = dfrac{1}{2}int e^{-u} u^{frac{alpha - 1}{2}} du.$ I think you factored out a $2^{frac{d - m}{2}-1}$ wrongly.
$endgroup$
– Will M.
Dec 7 '18 at 1:54
$begingroup$
Yours is wrong, I think. To begin with, $N = d$ and $M = m,$ and in the integral you made the change of variables $u = r^2,$ so that $u' = 2r$ and then $int e^{-r^2} r^alpha dr = dfrac{1}{2}int e^{-u} u^{frac{alpha - 1}{2}} du.$ I think you factored out a $2^{frac{d - m}{2}-1}$ wrongly.
$endgroup$
– Will M.
Dec 7 '18 at 1:54
$begingroup$
I see what's wrong. In the original question the exponent was not divided by 2, but in my answer for some reason I divided by 2. If you correct for that we are getting equivalent results.
$endgroup$
– becko
Dec 7 '18 at 15:33
$begingroup$
I see what's wrong. In the original question the exponent was not divided by 2, but in my answer for some reason I divided by 2. If you correct for that we are getting equivalent results.
$endgroup$
– becko
Dec 7 '18 at 15:33
add a comment |
$begingroup$
Introduce the spherical coordinates $r, theta, xi_1, ldots, xi_{N - 2}$,
$$ x_1 = r cos (xi_1), quad x_2 = r sin (xi_1) cos (xi_2), quad
ldots, quad x_N = r sin (xi_1) ldots sin (xi_{N - 2}) sin
(theta) $$
where $0 leqslant xi_i leqslant pi$, $0 leqslant theta leqslant 2 pi$
and $0 leqslant r < infty$. The volume element is given by $mathrm d x_1
ldots mathrm d x_N = r^{N - 1} mathrm d r mathrm d sigma$, where $mathrm d
sigma$ is the unit-sphere surface element,
$$mathrm d sigma = sin^{N - 2} (xi_1) ldots sin (xi_{n - 2}) mathrm d theta
mathrm d xi_1 ldots mathrm d xi_{N - 2}$$
It follows that:
$$begin{eqnarray*}
I & = & int_{- infty}^{infty} | {vec x} |^{- M} mathrm e^{- |
{vec x} |^2 / 2} mathrm d vec x\
& = & left( int_0^{infty} r^{N - M - 1} mathrm e^{- r^2/2}
mathrm d r right) left( int_{mathcal{S}} mathrm d sigma right)\
& = & 2^{(N - M) / 2 - 1} Gamma left( frac{N - M}{2}
right) frac{2 pi^{N / 2}}{Gamma (N / 2)}
end{eqnarray*}$$
where $mathcal S$ is the unit sphere.
answered Dec 6 '18 at 19:24
beckobecko
2,34931942
$endgroup$
add a comment |
$begingroup$
Introduce the spherical coordinates $r, theta, xi_1, ldots, xi_{N - 2}$,
$$ x_1 = r cos (xi_1), quad x_2 = r sin (xi_1) cos (xi_2), quad
ldots, quad x_N = r sin (xi_1) ldots sin (xi_{N - 2}) sin
(theta) $$
where $0 leqslant xi_i leqslant pi$, $0 leqslant theta leqslant 2 pi$
and $0 leqslant r < infty$. The volume element is given by $mathrm d x_1
ldots mathrm d x_N = r^{N - 1} mathrm d r mathrm d sigma$, where $mathrm d
sigma$ is the unit-sphere surface element,
$$mathrm d sigma = sin^{N - 2} (xi_1) ldots sin (xi_{n - 2}) mathrm d theta
mathrm d xi_1 ldots mathrm d xi_{N - 2}$$
It follows that:
$$begin{eqnarray*}
I & = & int_{- infty}^{infty} | {vec x} |^{- M} mathrm e^{- |
{vec x} |^2 / 2} mathrm d vec x\
& = & left( int_0^{infty} r^{N - M - 1} mathrm e^{- r^2/2}
mathrm d r right) left( int_{mathcal{S}} mathrm d sigma right)\
& = & 2^{(N - M) / 2 - 1} Gamma left( frac{N - M}{2}
right) frac{2 pi^{N / 2}}{Gamma (N / 2)}
end{eqnarray*}$$
where $mathcal S$ is the unit sphere.
answered Dec 6 '18 at 19:24
beckobecko
2,34931942
$endgroup$
add a comment |
$begingroup$
Introduce the spherical coordinates $r, theta, xi_1, ldots, xi_{N - 2}$,
$$ x_1 = r cos (xi_1), quad x_2 = r sin (xi_1) cos (xi_2), quad
ldots, quad x_N = r sin (xi_1) ldots sin (xi_{N - 2}) sin
(theta) $$
where $0 leqslant xi_i leqslant pi$, $0 leqslant theta leqslant 2 pi$
and $0 leqslant r < infty$. The volume element is given by $mathrm d x_1
ldots mathrm d x_N = r^{N - 1} mathrm d r mathrm d sigma$, where $mathrm d
sigma$ is the unit-sphere surface element,
$$mathrm d sigma = sin^{N - 2} (xi_1) ldots sin (xi_{n - 2}) mathrm d theta
mathrm d xi_1 ldots mathrm d xi_{N - 2}$$
It follows that:
$$begin{eqnarray*}
I & = & int_{- infty}^{infty} | {vec x} |^{- M} mathrm e^{- |
{vec x} |^2 / 2} mathrm d vec x\
& = & left( int_0^{infty} r^{N - M - 1} mathrm e^{- r^2/2}
mathrm d r right) left( int_{mathcal{S}} mathrm d sigma right)\
& = & 2^{(N - M) / 2 - 1} Gamma left( frac{N - M}{2}
right) frac{2 pi^{N / 2}}{Gamma (N / 2)}
end{eqnarray*}$$
where $mathcal S$ is the unit sphere.
answered Dec 6 '18 at 19:24
beckobecko
2,34931942
$endgroup$
Introduce the spherical coordinates $r, theta, xi_1, ldots, xi_{N - 2}$,
$$ x_1 = r cos (xi_1), quad x_2 = r sin (xi_1) cos (xi_2), quad
ldots, quad x_N = r sin (xi_1) ldots sin (xi_{N - 2}) sin
(theta) $$
where $0 leqslant xi_i leqslant pi$, $0 leqslant theta leqslant 2 pi$
and $0 leqslant r < infty$. The volume element is given by $mathrm d x_1
ldots mathrm d x_N = r^{N - 1} mathrm d r mathrm d sigma$, where $mathrm d
sigma$ is the unit-sphere surface element,
$$mathrm d sigma = sin^{N - 2} (xi_1) ldots sin (xi_{n - 2}) mathrm d theta
mathrm d xi_1 ldots mathrm d xi_{N - 2}$$
It follows that:
$$begin{eqnarray*}
I & = & int_{- infty}^{infty} | {vec x} |^{- M} mathrm e^{- |
{vec x} |^2 / 2} mathrm d vec x\
& = & left( int_0^{infty} r^{N - M - 1} mathrm e^{- r^2/2}
mathrm d r right) left( int_{mathcal{S}} mathrm d sigma right)\
& = & 2^{(N - M) / 2 - 1} Gamma left( frac{N - M}{2}
right) frac{2 pi^{N / 2}}{Gamma (N / 2)}
end{eqnarray*}$$
where $mathcal S$ is the unit sphere.
answered Dec 6 '18 at 19:24
beckobecko
2,34931942
answered Dec 6 '18 at 19:24
beckobecko
2,34931942
answered Dec 6 '18 at 19:24
beckobecko
2,34931942
answered Dec 6 '18 at 19:24
beckobecko
2,34931942
2,34931942
add a comment |
add a comment |
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
