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
2,35431942
$endgroup$
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
2,35431942
$endgroup$
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
2,35431942
$endgroup$
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
2,35431942
asked Dec 6 '18 at 19:20
beckobecko
2,35431942
edited Dec 6 '18 at 19:24
becko
asked Dec 6 '18 at 19:20
beckobecko
2,35431942
asked Dec 6 '18 at 19:20
beckobecko
2,35431942
asked Dec 6 '18 at 19:20
beckobecko
2,35431942
2,35431942
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.
2,630315
$endgroup$
$begingroup$
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
2,35431942
$endgroup$
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.
2,630315
$endgroup$
$begingroup$
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.
2,630315
$endgroup$
$begingroup$
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.
2,630315
$endgroup$
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.
2,630315
answered Dec 6 '18 at 19:53
Will M.Will M.
2,630315
answered Dec 6 '18 at 19:53
Will M.Will M.
2,630315
answered Dec 6 '18 at 19:53
Will M.Will M.
2,630315
2,630315
$begingroup$
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
2,35431942
$endgroup$
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
2,35431942
$endgroup$
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
2,35431942
$endgroup$
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
2,35431942
answered Dec 6 '18 at 19:24
beckobecko
2,35431942
answered Dec 6 '18 at 19:24
beckobecko
2,35431942
answered Dec 6 '18 at 19:24
beckobecko
2,35431942
2,35431942
add a comment |
add a comment |
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