Degenerate Gaussian IntegralWhat are the origin and applications of this result?Expectation under a...
Degenerate Gaussian Integral
What are the origin and applications of this result?Expectation under a t-distributionEquivalence between choosing a subspace and choosing its orthogonalRegression with correlation structureExpectation involving maximum of Gaussian variablesQuantifying the effect of noise on the posterior variance in Gaussian processes / multivariate Gaussian vectorsApproximating the mathematical expectation of the argmax of a Gaussian random vector
$begingroup$
I have an integral over a subspace of $mathbb{R}^n times mathbb{R}^n$ with an integrand of the form
$$expleft(-frac{1}{2}left[||u^2|| + langle u, v rangle + ||v||^2right]right)$$
The subspace is exactly the space for which $u_{i} = u_{n-i}$ (assume $n$ is even). In other words, $v$ is a true $n$-dimensional vector, whereas $u$ is two copies of an $n/2$ dimensional vector.
In order to evaluate this integral, I have been thinking about it as a density over ALL of $mathbb{R}^n times mathbb{R}^n$ of two correlated Gaussians with covariance given by the block matrix
$$begin{bmatrix}
A & B \ B^t & C
end{bmatrix}
$$
where $A,C$ are $n times n$ identity matrices and $B$ is the block matrix
$$begin{bmatrix}
I_{n/2} & I_{n/2} \ 0 & 0
end{bmatrix}
$$
However, given that this is degenerate, I am having trouble finishing the computation. Would greatly appreciate any tips!
pr.probability st.statistics probability-distributions
asked 8 hours ago
DJADJA
854 bronze badges
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add a comment
|
$begingroup$
I have an integral over a subspace of $mathbb{R}^n times mathbb{R}^n$ with an integrand of the form
$$expleft(-frac{1}{2}left[||u^2|| + langle u, v rangle + ||v||^2right]right)$$
The subspace is exactly the space for which $u_{i} = u_{n-i}$ (assume $n$ is even). In other words, $v$ is a true $n$-dimensional vector, whereas $u$ is two copies of an $n/2$ dimensional vector.
In order to evaluate this integral, I have been thinking about it as a density over ALL of $mathbb{R}^n times mathbb{R}^n$ of two correlated Gaussians with covariance given by the block matrix
$$begin{bmatrix}
A & B \ B^t & C
end{bmatrix}
$$
where $A,C$ are $n times n$ identity matrices and $B$ is the block matrix
$$begin{bmatrix}
I_{n/2} & I_{n/2} \ 0 & 0
end{bmatrix}
$$
However, given that this is degenerate, I am having trouble finishing the computation. Would greatly appreciate any tips!
pr.probability st.statistics probability-distributions
asked 8 hours ago
DJADJA
854 bronze badges
$endgroup$
$begingroup$
Just want to make sure you don't actually mean inner product of u and v to have a coefficient of 2--that's how these integrals usually look because inner product of u+v with itself yields a coefficient of 2 on the middle term. This would make Iosef's answer even simpler.
$endgroup$
– Sheridan Grant
4 hours ago
add a comment
|
$begingroup$
I have an integral over a subspace of $mathbb{R}^n times mathbb{R}^n$ with an integrand of the form
$$expleft(-frac{1}{2}left[||u^2|| + langle u, v rangle + ||v||^2right]right)$$
The subspace is exactly the space for which $u_{i} = u_{n-i}$ (assume $n$ is even). In other words, $v$ is a true $n$-dimensional vector, whereas $u$ is two copies of an $n/2$ dimensional vector.
In order to evaluate this integral, I have been thinking about it as a density over ALL of $mathbb{R}^n times mathbb{R}^n$ of two correlated Gaussians with covariance given by the block matrix
$$begin{bmatrix}
A & B \ B^t & C
end{bmatrix}
$$
where $A,C$ are $n times n$ identity matrices and $B$ is the block matrix
$$begin{bmatrix}
I_{n/2} & I_{n/2} \ 0 & 0
end{bmatrix}
$$
However, given that this is degenerate, I am having trouble finishing the computation. Would greatly appreciate any tips!
pr.probability st.statistics probability-distributions
asked 8 hours ago
DJADJA
854 bronze badges
$endgroup$
I have an integral over a subspace of $mathbb{R}^n times mathbb{R}^n$ with an integrand of the form
$$expleft(-frac{1}{2}left[||u^2|| + langle u, v rangle + ||v||^2right]right)$$
The subspace is exactly the space for which $u_{i} = u_{n-i}$ (assume $n$ is even). In other words, $v$ is a true $n$-dimensional vector, whereas $u$ is two copies of an $n/2$ dimensional vector.
In order to evaluate this integral, I have been thinking about it as a density over ALL of $mathbb{R}^n times mathbb{R}^n$ of two correlated Gaussians with covariance given by the block matrix
$$begin{bmatrix}
A & B \ B^t & C
end{bmatrix}
$$
where $A,C$ are $n times n$ identity matrices and $B$ is the block matrix
$$begin{bmatrix}
I_{n/2} & I_{n/2} \ 0 & 0
end{bmatrix}
$$
However, given that this is degenerate, I am having trouble finishing the computation. Would greatly appreciate any tips!
pr.probability st.statistics probability-distributions
pr.probability st.statistics probability-distributions
asked 8 hours ago
DJADJA
854 bronze badges
asked 8 hours ago
DJADJA
854 bronze badges
asked 8 hours ago
DJADJA
854 bronze badges
asked 8 hours ago
DJADJA
854 bronze badges
asked 8 hours ago
DJADJA
854 bronze badges
854 bronze badges
$begingroup$
Just want to make sure you don't actually mean inner product of u and v to have a coefficient of 2--that's how these integrals usually look because inner product of u+v with itself yields a coefficient of 2 on the middle term. This would make Iosef's answer even simpler.
$endgroup$
– Sheridan Grant
4 hours ago
add a comment
|
$begingroup$
Just want to make sure you don't actually mean inner product of u and v to have a coefficient of 2--that's how these integrals usually look because inner product of u+v with itself yields a coefficient of 2 on the middle term. This would make Iosef's answer even simpler.
$endgroup$
– Sheridan Grant
4 hours ago
$begingroup$
Just want to make sure you don't actually mean inner product of u and v to have a coefficient of 2--that's how these integrals usually look because inner product of u+v with itself yields a coefficient of 2 on the middle term. This would make Iosef's answer even simpler.
$endgroup$
– Sheridan Grant
4 hours ago
$begingroup$
Just want to make sure you don't actually mean inner product of u and v to have a coefficient of 2--that's how these integrals usually look because inner product of u+v with itself yields a coefficient of 2 on the middle term. This would make Iosef's answer even simpler.
$endgroup$
– Sheridan Grant
4 hours ago
add a comment
|
1 Answer
1
active
oldest
votes
$begingroup$
$newcommand{R}{mathbb{R}}$
Let $U:={uinR^ncolon u_i=u_{n-i} forall i}$ be your $n/2$-dimensional subspace. I am assuming that your integral is with respect to the product of the Lebesgue measures on $U$ and $R^n$, and I will denote those measures by $du$ and $dv$, respectively. So, if $cdot$ denotes the dot product, your integral is
begin{align}
I&:=int_U du,int_{R^n}dv,expbig(-(|u|^2+ucdot v+|v|^2)/2big) \
&=int_U du,int_{R^n}dv,expbig(-(3|u|^2/4+|v+u/2|^2)/2big) \
&=int_U du,int_{R^n}dw,expbig(-(3|u|^2/4+|w|^2)/2big) \
&=(2pi)^{n/2}int_U du,expbig(-3|u|^2/8big) \
&=(2pi)^{n/2}int_{R^{n/2}} dt,expbig(-3|t|^2/8big) \
&=(2pi)^{n/2}(2pi)^{n/4}(4/3)^{n/4}.
end{align}
The penultimate equality here holds because both the Euclidean norm and the Lebesgue measure are rotation invariant, whereas the dimension of $U$ is $n/2$; in fact, this is how the Lebesgue measure on $U$ can/should be defined: by the condition that
begin{equation}
int_U du,f(u)=int_{R^{n/2}} dt,f(Tt)
end{equation}
for all nonnegative Borel-measurable functions $fcolon UtoR$, where $TcolonR^{n/2}to U$ is a linear isomorphism.
answered 7 hours ago
Iosif PinelisIosif Pinelis
25.5k3 gold badges30 silver badges69 bronze badges
$endgroup$
1
$begingroup$
Ah, I see. It is as simple as just factoring and completing the square? Thanks for your quick + helpful answers Iosif!
$endgroup$
– DJA
6 hours ago
1
$begingroup$
@DJA : From my viewpoint, the main thing here is to properly/conveniently define the Lebesgue measure on a linear subspace.
$endgroup$
– Iosif Pinelis
5 hours ago
add a comment
|
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1 Answer
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1 Answer
1
active
oldest
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oldest
votes
$begingroup$
$newcommand{R}{mathbb{R}}$
Let $U:={uinR^ncolon u_i=u_{n-i} forall i}$ be your $n/2$-dimensional subspace. I am assuming that your integral is with respect to the product of the Lebesgue measures on $U$ and $R^n$, and I will denote those measures by $du$ and $dv$, respectively. So, if $cdot$ denotes the dot product, your integral is
begin{align}
I&:=int_U du,int_{R^n}dv,expbig(-(|u|^2+ucdot v+|v|^2)/2big) \
&=int_U du,int_{R^n}dv,expbig(-(3|u|^2/4+|v+u/2|^2)/2big) \
&=int_U du,int_{R^n}dw,expbig(-(3|u|^2/4+|w|^2)/2big) \
&=(2pi)^{n/2}int_U du,expbig(-3|u|^2/8big) \
&=(2pi)^{n/2}int_{R^{n/2}} dt,expbig(-3|t|^2/8big) \
&=(2pi)^{n/2}(2pi)^{n/4}(4/3)^{n/4}.
end{align}
The penultimate equality here holds because both the Euclidean norm and the Lebesgue measure are rotation invariant, whereas the dimension of $U$ is $n/2$; in fact, this is how the Lebesgue measure on $U$ can/should be defined: by the condition that
begin{equation}
int_U du,f(u)=int_{R^{n/2}} dt,f(Tt)
end{equation}
for all nonnegative Borel-measurable functions $fcolon UtoR$, where $TcolonR^{n/2}to U$ is a linear isomorphism.
answered 7 hours ago
Iosif PinelisIosif Pinelis
25.5k3 gold badges30 silver badges69 bronze badges
$endgroup$
1
$begingroup$
Ah, I see. It is as simple as just factoring and completing the square? Thanks for your quick + helpful answers Iosif!
$endgroup$
– DJA
6 hours ago
1
$begingroup$
@DJA : From my viewpoint, the main thing here is to properly/conveniently define the Lebesgue measure on a linear subspace.
$endgroup$
– Iosif Pinelis
5 hours ago
add a comment
|
$begingroup$
$newcommand{R}{mathbb{R}}$
Let $U:={uinR^ncolon u_i=u_{n-i} forall i}$ be your $n/2$-dimensional subspace. I am assuming that your integral is with respect to the product of the Lebesgue measures on $U$ and $R^n$, and I will denote those measures by $du$ and $dv$, respectively. So, if $cdot$ denotes the dot product, your integral is
begin{align}
I&:=int_U du,int_{R^n}dv,expbig(-(|u|^2+ucdot v+|v|^2)/2big) \
&=int_U du,int_{R^n}dv,expbig(-(3|u|^2/4+|v+u/2|^2)/2big) \
&=int_U du,int_{R^n}dw,expbig(-(3|u|^2/4+|w|^2)/2big) \
&=(2pi)^{n/2}int_U du,expbig(-3|u|^2/8big) \
&=(2pi)^{n/2}int_{R^{n/2}} dt,expbig(-3|t|^2/8big) \
&=(2pi)^{n/2}(2pi)^{n/4}(4/3)^{n/4}.
end{align}
The penultimate equality here holds because both the Euclidean norm and the Lebesgue measure are rotation invariant, whereas the dimension of $U$ is $n/2$; in fact, this is how the Lebesgue measure on $U$ can/should be defined: by the condition that
begin{equation}
int_U du,f(u)=int_{R^{n/2}} dt,f(Tt)
end{equation}
for all nonnegative Borel-measurable functions $fcolon UtoR$, where $TcolonR^{n/2}to U$ is a linear isomorphism.
answered 7 hours ago
Iosif PinelisIosif Pinelis
25.5k3 gold badges30 silver badges69 bronze badges
$endgroup$
1
$begingroup$
Ah, I see. It is as simple as just factoring and completing the square? Thanks for your quick + helpful answers Iosif!
$endgroup$
– DJA
6 hours ago
1
$begingroup$
@DJA : From my viewpoint, the main thing here is to properly/conveniently define the Lebesgue measure on a linear subspace.
$endgroup$
– Iosif Pinelis
5 hours ago
add a comment
|
$begingroup$
$newcommand{R}{mathbb{R}}$
Let $U:={uinR^ncolon u_i=u_{n-i} forall i}$ be your $n/2$-dimensional subspace. I am assuming that your integral is with respect to the product of the Lebesgue measures on $U$ and $R^n$, and I will denote those measures by $du$ and $dv$, respectively. So, if $cdot$ denotes the dot product, your integral is
begin{align}
I&:=int_U du,int_{R^n}dv,expbig(-(|u|^2+ucdot v+|v|^2)/2big) \
&=int_U du,int_{R^n}dv,expbig(-(3|u|^2/4+|v+u/2|^2)/2big) \
&=int_U du,int_{R^n}dw,expbig(-(3|u|^2/4+|w|^2)/2big) \
&=(2pi)^{n/2}int_U du,expbig(-3|u|^2/8big) \
&=(2pi)^{n/2}int_{R^{n/2}} dt,expbig(-3|t|^2/8big) \
&=(2pi)^{n/2}(2pi)^{n/4}(4/3)^{n/4}.
end{align}
The penultimate equality here holds because both the Euclidean norm and the Lebesgue measure are rotation invariant, whereas the dimension of $U$ is $n/2$; in fact, this is how the Lebesgue measure on $U$ can/should be defined: by the condition that
begin{equation}
int_U du,f(u)=int_{R^{n/2}} dt,f(Tt)
end{equation}
for all nonnegative Borel-measurable functions $fcolon UtoR$, where $TcolonR^{n/2}to U$ is a linear isomorphism.
answered 7 hours ago
Iosif PinelisIosif Pinelis
25.5k3 gold badges30 silver badges69 bronze badges
$endgroup$
$newcommand{R}{mathbb{R}}$
Let $U:={uinR^ncolon u_i=u_{n-i} forall i}$ be your $n/2$-dimensional subspace. I am assuming that your integral is with respect to the product of the Lebesgue measures on $U$ and $R^n$, and I will denote those measures by $du$ and $dv$, respectively. So, if $cdot$ denotes the dot product, your integral is
begin{align}
I&:=int_U du,int_{R^n}dv,expbig(-(|u|^2+ucdot v+|v|^2)/2big) \
&=int_U du,int_{R^n}dv,expbig(-(3|u|^2/4+|v+u/2|^2)/2big) \
&=int_U du,int_{R^n}dw,expbig(-(3|u|^2/4+|w|^2)/2big) \
&=(2pi)^{n/2}int_U du,expbig(-3|u|^2/8big) \
&=(2pi)^{n/2}int_{R^{n/2}} dt,expbig(-3|t|^2/8big) \
&=(2pi)^{n/2}(2pi)^{n/4}(4/3)^{n/4}.
end{align}
The penultimate equality here holds because both the Euclidean norm and the Lebesgue measure are rotation invariant, whereas the dimension of $U$ is $n/2$; in fact, this is how the Lebesgue measure on $U$ can/should be defined: by the condition that
begin{equation}
int_U du,f(u)=int_{R^{n/2}} dt,f(Tt)
end{equation}
for all nonnegative Borel-measurable functions $fcolon UtoR$, where $TcolonR^{n/2}to U$ is a linear isomorphism.
answered 7 hours ago
Iosif PinelisIosif Pinelis
25.5k3 gold badges30 silver badges69 bronze badges
edited 7 hours ago
answered 7 hours ago
Iosif PinelisIosif Pinelis
25.5k3 gold badges30 silver badges69 bronze badges
answered 7 hours ago
Iosif PinelisIosif Pinelis
25.5k3 gold badges30 silver badges69 bronze badges
answered 7 hours ago
Iosif PinelisIosif Pinelis
25.5k3 gold badges30 silver badges69 bronze badges
25.5k3 gold badges30 silver badges69 bronze badges
1
$begingroup$
Ah, I see. It is as simple as just factoring and completing the square? Thanks for your quick + helpful answers Iosif!
$endgroup$
– DJA
6 hours ago
1
$begingroup$
@DJA : From my viewpoint, the main thing here is to properly/conveniently define the Lebesgue measure on a linear subspace.
$endgroup$
– Iosif Pinelis
5 hours ago
add a comment
|
1
$begingroup$
Ah, I see. It is as simple as just factoring and completing the square? Thanks for your quick + helpful answers Iosif!
$endgroup$
– DJA
6 hours ago
1
$begingroup$
@DJA : From my viewpoint, the main thing here is to properly/conveniently define the Lebesgue measure on a linear subspace.
$endgroup$
– Iosif Pinelis
5 hours ago
1
1
$begingroup$
Ah, I see. It is as simple as just factoring and completing the square? Thanks for your quick + helpful answers Iosif!
$endgroup$
– DJA
6 hours ago
$begingroup$
Ah, I see. It is as simple as just factoring and completing the square? Thanks for your quick + helpful answers Iosif!
$endgroup$
– DJA
6 hours ago
1
1
$begingroup$
@DJA : From my viewpoint, the main thing here is to properly/conveniently define the Lebesgue measure on a linear subspace.
$endgroup$
– Iosif Pinelis
5 hours ago
$begingroup$
@DJA : From my viewpoint, the main thing here is to properly/conveniently define the Lebesgue measure on a linear subspace.
$endgroup$
– Iosif Pinelis
5 hours ago
add a comment
|
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$begingroup$
Just want to make sure you don't actually mean inner product of u and v to have a coefficient of 2--that's how these integrals usually look because inner product of u+v with itself yields a coefficient of 2 on the middle term. This would make Iosef's answer even simpler.
$endgroup$
– Sheridan Grant
4 hours ago