Simplify double sum
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1
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Does the following expression has a closed form
begin{align}
E left[ | Z|^k exp(t |Z|^2)right]
end{align}
for $k$ is even and $Z$ is standard normal vector.
For the case of $k=2$ the computation is very easy. For $k=1/2$ see here Compute $E[ sqrt{Q} e^{-tQ} ]$ where $Q$ is non-central chi-square random variable..
However, this question is about even $k$. I there a way to use a binomina theorem to solve this ?
normal-distribution gaussian-integral
edited Nov 17 at 1:26
David G. Stork
9,30321232
asked Nov 17 at 1:13
Lisa
631213
add a comment |
up vote
1
down vote
favorite
Does the following expression has a closed form
begin{align}
E left[ | Z|^k exp(t |Z|^2)right]
end{align}
for $k$ is even and $Z$ is standard normal vector.
For the case of $k=2$ the computation is very easy. For $k=1/2$ see here Compute $E[ sqrt{Q} e^{-tQ} ]$ where $Q$ is non-central chi-square random variable..
However, this question is about even $k$. I there a way to use a binomina theorem to solve this ?
normal-distribution gaussian-integral
edited Nov 17 at 1:26
David G. Stork
9,30321232
asked Nov 17 at 1:13
Lisa
631213
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Does the following expression has a closed form
begin{align}
E left[ | Z|^k exp(t |Z|^2)right]
end{align}
for $k$ is even and $Z$ is standard normal vector.
For the case of $k=2$ the computation is very easy. For $k=1/2$ see here Compute $E[ sqrt{Q} e^{-tQ} ]$ where $Q$ is non-central chi-square random variable..
However, this question is about even $k$. I there a way to use a binomina theorem to solve this ?
normal-distribution gaussian-integral
edited Nov 17 at 1:26
David G. Stork
9,30321232
asked Nov 17 at 1:13
Lisa
631213
Does the following expression has a closed form
begin{align}
E left[ | Z|^k exp(t |Z|^2)right]
end{align}
for $k$ is even and $Z$ is standard normal vector.
For the case of $k=2$ the computation is very easy. For $k=1/2$ see here Compute $E[ sqrt{Q} e^{-tQ} ]$ where $Q$ is non-central chi-square random variable..
However, this question is about even $k$. I there a way to use a binomina theorem to solve this ?
normal-distribution gaussian-integral
normal-distribution gaussian-integral
edited Nov 17 at 1:26
David G. Stork
9,30321232
asked Nov 17 at 1:13
Lisa
631213
edited Nov 17 at 1:26
David G. Stork
9,30321232
asked Nov 17 at 1:13
Lisa
631213
edited Nov 17 at 1:26
David G. Stork
9,30321232
edited Nov 17 at 1:26
David G. Stork
9,30321232
edited Nov 17 at 1:26
David G. Stork
9,30321232
9,30321232
asked Nov 17 at 1:13
Lisa
631213
asked Nov 17 at 1:13
Lisa
631213
asked Nov 17 at 1:13
Lisa
631213
631213
add a comment |
add a comment |
1 Answer
1
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0
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In hyperspherical coordinates in $mathbb R^n$, the integral becomes
$$operatorname E left[ lVert Z rVert^k e^{t lVert Z rVert^2} right] =
(2 pi)^{-n/2} S_{n - 1} int_0^infty
rho^{k + n - 1} e^{t rho^2 - rho^2/2} drho = \
2^{k/2} (1 - 2 t)^{(-k - n)/2} left( frac n 2 right)_{k/2},$$
where $S_n$ is the surface area of a unit $n$-sphere and $(n)_k$ is the rising factorial.
answered Nov 17 at 23:58
Maxim
4,346219
Thanks. Can you talk more about the change of variable?
– Lisa
Nov 18 at 16:06
We integrate $lVert x rVert^k e^{t lVert x rVert^2} f_Z(x)$ over $mathbb R^n$, $x = (x_1, dots, x_n)$. The Jacobian of the coordinate transformation to spherical coordinates also gives the area element of a sphere of radius $rho$, therefore integrating the Jacobian over the spherical angles gives the area of the sphere; $lVert x rVert$ and $f_Z(x)$ depend only on $rho$.
– Maxim
Nov 18 at 18:56
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
In hyperspherical coordinates in $mathbb R^n$, the integral becomes
$$operatorname E left[ lVert Z rVert^k e^{t lVert Z rVert^2} right] =
(2 pi)^{-n/2} S_{n - 1} int_0^infty
rho^{k + n - 1} e^{t rho^2 - rho^2/2} drho = \
2^{k/2} (1 - 2 t)^{(-k - n)/2} left( frac n 2 right)_{k/2},$$
where $S_n$ is the surface area of a unit $n$-sphere and $(n)_k$ is the rising factorial.
answered Nov 17 at 23:58
Maxim
4,346219
Thanks. Can you talk more about the change of variable?
– Lisa
Nov 18 at 16:06
We integrate $lVert x rVert^k e^{t lVert x rVert^2} f_Z(x)$ over $mathbb R^n$, $x = (x_1, dots, x_n)$. The Jacobian of the coordinate transformation to spherical coordinates also gives the area element of a sphere of radius $rho$, therefore integrating the Jacobian over the spherical angles gives the area of the sphere; $lVert x rVert$ and $f_Z(x)$ depend only on $rho$.
– Maxim
Nov 18 at 18:56
add a comment |
up vote
0
down vote
In hyperspherical coordinates in $mathbb R^n$, the integral becomes
$$operatorname E left[ lVert Z rVert^k e^{t lVert Z rVert^2} right] =
(2 pi)^{-n/2} S_{n - 1} int_0^infty
rho^{k + n - 1} e^{t rho^2 - rho^2/2} drho = \
2^{k/2} (1 - 2 t)^{(-k - n)/2} left( frac n 2 right)_{k/2},$$
where $S_n$ is the surface area of a unit $n$-sphere and $(n)_k$ is the rising factorial.
answered Nov 17 at 23:58
Maxim
4,346219
Thanks. Can you talk more about the change of variable?
– Lisa
Nov 18 at 16:06
We integrate $lVert x rVert^k e^{t lVert x rVert^2} f_Z(x)$ over $mathbb R^n$, $x = (x_1, dots, x_n)$. The Jacobian of the coordinate transformation to spherical coordinates also gives the area element of a sphere of radius $rho$, therefore integrating the Jacobian over the spherical angles gives the area of the sphere; $lVert x rVert$ and $f_Z(x)$ depend only on $rho$.
– Maxim
Nov 18 at 18:56
add a comment |
up vote
0
down vote
up vote
0
down vote
In hyperspherical coordinates in $mathbb R^n$, the integral becomes
$$operatorname E left[ lVert Z rVert^k e^{t lVert Z rVert^2} right] =
(2 pi)^{-n/2} S_{n - 1} int_0^infty
rho^{k + n - 1} e^{t rho^2 - rho^2/2} drho = \
2^{k/2} (1 - 2 t)^{(-k - n)/2} left( frac n 2 right)_{k/2},$$
where $S_n$ is the surface area of a unit $n$-sphere and $(n)_k$ is the rising factorial.
answered Nov 17 at 23:58
Maxim
4,346219
In hyperspherical coordinates in $mathbb R^n$, the integral becomes
$$operatorname E left[ lVert Z rVert^k e^{t lVert Z rVert^2} right] =
(2 pi)^{-n/2} S_{n - 1} int_0^infty
rho^{k + n - 1} e^{t rho^2 - rho^2/2} drho = \
2^{k/2} (1 - 2 t)^{(-k - n)/2} left( frac n 2 right)_{k/2},$$
where $S_n$ is the surface area of a unit $n$-sphere and $(n)_k$ is the rising factorial.
answered Nov 17 at 23:58
Maxim
4,346219
answered Nov 17 at 23:58
Maxim
4,346219
answered Nov 17 at 23:58
Maxim
4,346219
answered Nov 17 at 23:58
Maxim
4,346219
4,346219
Thanks. Can you talk more about the change of variable?
– Lisa
Nov 18 at 16:06
We integrate $lVert x rVert^k e^{t lVert x rVert^2} f_Z(x)$ over $mathbb R^n$, $x = (x_1, dots, x_n)$. The Jacobian of the coordinate transformation to spherical coordinates also gives the area element of a sphere of radius $rho$, therefore integrating the Jacobian over the spherical angles gives the area of the sphere; $lVert x rVert$ and $f_Z(x)$ depend only on $rho$.
– Maxim
Nov 18 at 18:56
add a comment |
Thanks. Can you talk more about the change of variable?
– Lisa
Nov 18 at 16:06
We integrate $lVert x rVert^k e^{t lVert x rVert^2} f_Z(x)$ over $mathbb R^n$, $x = (x_1, dots, x_n)$. The Jacobian of the coordinate transformation to spherical coordinates also gives the area element of a sphere of radius $rho$, therefore integrating the Jacobian over the spherical angles gives the area of the sphere; $lVert x rVert$ and $f_Z(x)$ depend only on $rho$.
– Maxim
Nov 18 at 18:56
Thanks. Can you talk more about the change of variable?
– Lisa
Nov 18 at 16:06
Thanks. Can you talk more about the change of variable?
– Lisa
Nov 18 at 16:06
We integrate $lVert x rVert^k e^{t lVert x rVert^2} f_Z(x)$ over $mathbb R^n$, $x = (x_1, dots, x_n)$. The Jacobian of the coordinate transformation to spherical coordinates also gives the area element of a sphere of radius $rho$, therefore integrating the Jacobian over the spherical angles gives the area of the sphere; $lVert x rVert$ and $f_Z(x)$ depend only on $rho$.
– Maxim
Nov 18 at 18:56
We integrate $lVert x rVert^k e^{t lVert x rVert^2} f_Z(x)$ over $mathbb R^n$, $x = (x_1, dots, x_n)$. The Jacobian of the coordinate transformation to spherical coordinates also gives the area element of a sphere of radius $rho$, therefore integrating the Jacobian over the spherical angles gives the area of the sphere; $lVert x rVert$ and $f_Z(x)$ depend only on $rho$.
– Maxim
Nov 18 at 18:56
add a comment |
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