Mixing
Extend this integral to higher dimension
$begingroup$
I'm solving a high dimensional integral where $x in mathbb{R}^D$; where $A in mathbb{R}^{DxD}$ positive definite and $b in mathbb{R}^D$; $c in mathbb{R}$
$$int^{infty}_{-infty} boldsymbol{x}^2exp(- frac{1}{2}boldsymbol{x}^TAboldsymbol{x} + b^Tboldsymbol{x} + c)dboldsymbol{x}$$.
So basically, the variance of some unnormalized probability distribution. I first start with the one dimensional case:
begin{align}
&int x^2exp(-ax^2+bx+c)\
&=exp(c) int x^2 exp(-ax^2 + bx)\
&=exp(c) int x^2 exp(-a(x - frac{b}{2a})^2 + frac{b^2}{4a})\
&= exp(c)exp(frac{b^2}{4a})int x^2 exp(-a(x-frac{b}{2a})^2)dx\
end{align}
By change of variable
let $u = x-frac{b}{2a} implies du = dx$ and $x = u + frac{b}{2a}$
$x^2 = u^2 + 2ufrac{b}{2a} + frac{b^2}{4a^2}$
we have
begin{align}
&=exp(c)exp(frac{b^2}{4a})int (u^2 + 2ufrac{b}{2a} + frac{b^2}{4a^2}) exp(-a(u)^2)du\
&= exp(c + frac{b^2}{4a}) big(int u^2 exp(-a(u)^2)du + int 2ufrac{b}{2a}exp(-a(u)^2)du + int frac{b^2}{4a^2} exp(-a(u)^2)du big)
end{align}
for each three terms in the big bracket;
$$int^{infty}_{infty} u^2 exp(-a(u)^2)du = dfrac{sqrt{{pi}}}{2a^frac{3}{2}}$$
$$int^{infty}_{infty} 2ufrac{b}{2a} exp(-a(u)^2)du = 0 $$
$$int^{infty}_{infty} frac{b^2}{4a^2} exp(-a(u)^2)du = dfrac{sqrt{{pi}}b^2}{4a^frac{5}{2}}$$
Finally; leads me to the final solution of
$$ exp(c + frac{b^2}{4a}) sqrt{{pi}} (dfrac{1}{2a^frac{3}{2}} + dfrac{b^2}{4a^frac{5}{2}} )$$
However, I'm having trouble to generalize it to high dimension. The answer should a vector in $mathbb{R}^{DxD}$
My intuition based on Normal distribution in high dimension tells my the answer should be look like
$$exp(c + frac{b^Tb}{4det(A)})pi^{frac{D}{2}} Big( dots Big)$$
But I couldn't make the bracket term in one dimension to higher dimension while still keeps the final solution a vector with dimension $DxD$ (it looks like it will be ends up with a scalar). Any helps Thanks
probability integration
asked Jan 28 at 1:40
ElleryLElleryL
725519
$endgroup$
add a comment |
$begingroup$
I'm solving a high dimensional integral where $x in mathbb{R}^D$; where $A in mathbb{R}^{DxD}$ positive definite and $b in mathbb{R}^D$; $c in mathbb{R}$
$$int^{infty}_{-infty} boldsymbol{x}^2exp(- frac{1}{2}boldsymbol{x}^TAboldsymbol{x} + b^Tboldsymbol{x} + c)dboldsymbol{x}$$.
So basically, the variance of some unnormalized probability distribution. I first start with the one dimensional case:
begin{align}
&int x^2exp(-ax^2+bx+c)\
&=exp(c) int x^2 exp(-ax^2 + bx)\
&=exp(c) int x^2 exp(-a(x - frac{b}{2a})^2 + frac{b^2}{4a})\
&= exp(c)exp(frac{b^2}{4a})int x^2 exp(-a(x-frac{b}{2a})^2)dx\
end{align}
By change of variable
let $u = x-frac{b}{2a} implies du = dx$ and $x = u + frac{b}{2a}$
$x^2 = u^2 + 2ufrac{b}{2a} + frac{b^2}{4a^2}$
we have
begin{align}
&=exp(c)exp(frac{b^2}{4a})int (u^2 + 2ufrac{b}{2a} + frac{b^2}{4a^2}) exp(-a(u)^2)du\
&= exp(c + frac{b^2}{4a}) big(int u^2 exp(-a(u)^2)du + int 2ufrac{b}{2a}exp(-a(u)^2)du + int frac{b^2}{4a^2} exp(-a(u)^2)du big)
end{align}
for each three terms in the big bracket;
$$int^{infty}_{infty} u^2 exp(-a(u)^2)du = dfrac{sqrt{{pi}}}{2a^frac{3}{2}}$$
$$int^{infty}_{infty} 2ufrac{b}{2a} exp(-a(u)^2)du = 0 $$
$$int^{infty}_{infty} frac{b^2}{4a^2} exp(-a(u)^2)du = dfrac{sqrt{{pi}}b^2}{4a^frac{5}{2}}$$
Finally; leads me to the final solution of
$$ exp(c + frac{b^2}{4a}) sqrt{{pi}} (dfrac{1}{2a^frac{3}{2}} + dfrac{b^2}{4a^frac{5}{2}} )$$
However, I'm having trouble to generalize it to high dimension. The answer should a vector in $mathbb{R}^{DxD}$
My intuition based on Normal distribution in high dimension tells my the answer should be look like
$$exp(c + frac{b^Tb}{4det(A)})pi^{frac{D}{2}} Big( dots Big)$$
But I couldn't make the bracket term in one dimension to higher dimension while still keeps the final solution a vector with dimension $DxD$ (it looks like it will be ends up with a scalar). Any helps Thanks
probability integration
asked Jan 28 at 1:40
ElleryLElleryL
725519
$endgroup$
add a comment |
$begingroup$
I'm solving a high dimensional integral where $x in mathbb{R}^D$; where $A in mathbb{R}^{DxD}$ positive definite and $b in mathbb{R}^D$; $c in mathbb{R}$
$$int^{infty}_{-infty} boldsymbol{x}^2exp(- frac{1}{2}boldsymbol{x}^TAboldsymbol{x} + b^Tboldsymbol{x} + c)dboldsymbol{x}$$.
So basically, the variance of some unnormalized probability distribution. I first start with the one dimensional case:
begin{align}
&int x^2exp(-ax^2+bx+c)\
&=exp(c) int x^2 exp(-ax^2 + bx)\
&=exp(c) int x^2 exp(-a(x - frac{b}{2a})^2 + frac{b^2}{4a})\
&= exp(c)exp(frac{b^2}{4a})int x^2 exp(-a(x-frac{b}{2a})^2)dx\
end{align}
By change of variable
let $u = x-frac{b}{2a} implies du = dx$ and $x = u + frac{b}{2a}$
$x^2 = u^2 + 2ufrac{b}{2a} + frac{b^2}{4a^2}$
we have
begin{align}
&=exp(c)exp(frac{b^2}{4a})int (u^2 + 2ufrac{b}{2a} + frac{b^2}{4a^2}) exp(-a(u)^2)du\
&= exp(c + frac{b^2}{4a}) big(int u^2 exp(-a(u)^2)du + int 2ufrac{b}{2a}exp(-a(u)^2)du + int frac{b^2}{4a^2} exp(-a(u)^2)du big)
end{align}
for each three terms in the big bracket;
$$int^{infty}_{infty} u^2 exp(-a(u)^2)du = dfrac{sqrt{{pi}}}{2a^frac{3}{2}}$$
$$int^{infty}_{infty} 2ufrac{b}{2a} exp(-a(u)^2)du = 0 $$
$$int^{infty}_{infty} frac{b^2}{4a^2} exp(-a(u)^2)du = dfrac{sqrt{{pi}}b^2}{4a^frac{5}{2}}$$
Finally; leads me to the final solution of
$$ exp(c + frac{b^2}{4a}) sqrt{{pi}} (dfrac{1}{2a^frac{3}{2}} + dfrac{b^2}{4a^frac{5}{2}} )$$
However, I'm having trouble to generalize it to high dimension. The answer should a vector in $mathbb{R}^{DxD}$
My intuition based on Normal distribution in high dimension tells my the answer should be look like
$$exp(c + frac{b^Tb}{4det(A)})pi^{frac{D}{2}} Big( dots Big)$$
But I couldn't make the bracket term in one dimension to higher dimension while still keeps the final solution a vector with dimension $DxD$ (it looks like it will be ends up with a scalar). Any helps Thanks
probability integration
asked Jan 28 at 1:40
ElleryLElleryL
725519
$endgroup$
I'm solving a high dimensional integral where $x in mathbb{R}^D$; where $A in mathbb{R}^{DxD}$ positive definite and $b in mathbb{R}^D$; $c in mathbb{R}$
$$int^{infty}_{-infty} boldsymbol{x}^2exp(- frac{1}{2}boldsymbol{x}^TAboldsymbol{x} + b^Tboldsymbol{x} + c)dboldsymbol{x}$$.
So basically, the variance of some unnormalized probability distribution. I first start with the one dimensional case:
begin{align}
&int x^2exp(-ax^2+bx+c)\
&=exp(c) int x^2 exp(-ax^2 + bx)\
&=exp(c) int x^2 exp(-a(x - frac{b}{2a})^2 + frac{b^2}{4a})\
&= exp(c)exp(frac{b^2}{4a})int x^2 exp(-a(x-frac{b}{2a})^2)dx\
end{align}
By change of variable
let $u = x-frac{b}{2a} implies du = dx$ and $x = u + frac{b}{2a}$
$x^2 = u^2 + 2ufrac{b}{2a} + frac{b^2}{4a^2}$
we have
begin{align}
&=exp(c)exp(frac{b^2}{4a})int (u^2 + 2ufrac{b}{2a} + frac{b^2}{4a^2}) exp(-a(u)^2)du\
&= exp(c + frac{b^2}{4a}) big(int u^2 exp(-a(u)^2)du + int 2ufrac{b}{2a}exp(-a(u)^2)du + int frac{b^2}{4a^2} exp(-a(u)^2)du big)
end{align}
for each three terms in the big bracket;
$$int^{infty}_{infty} u^2 exp(-a(u)^2)du = dfrac{sqrt{{pi}}}{2a^frac{3}{2}}$$
$$int^{infty}_{infty} 2ufrac{b}{2a} exp(-a(u)^2)du = 0 $$
$$int^{infty}_{infty} frac{b^2}{4a^2} exp(-a(u)^2)du = dfrac{sqrt{{pi}}b^2}{4a^frac{5}{2}}$$
Finally; leads me to the final solution of
$$ exp(c + frac{b^2}{4a}) sqrt{{pi}} (dfrac{1}{2a^frac{3}{2}} + dfrac{b^2}{4a^frac{5}{2}} )$$
However, I'm having trouble to generalize it to high dimension. The answer should a vector in $mathbb{R}^{DxD}$
My intuition based on Normal distribution in high dimension tells my the answer should be look like
$$exp(c + frac{b^Tb}{4det(A)})pi^{frac{D}{2}} Big( dots Big)$$
But I couldn't make the bracket term in one dimension to higher dimension while still keeps the final solution a vector with dimension $DxD$ (it looks like it will be ends up with a scalar). Any helps Thanks
probability integration
probability integration
asked Jan 28 at 1:40
ElleryLElleryL
725519
asked Jan 28 at 1:40
ElleryLElleryL
725519
edited Jan 29 at 1:57
ElleryL
asked Jan 28 at 1:40
ElleryLElleryL
725519
asked Jan 28 at 1:40
ElleryLElleryL
725519
asked Jan 28 at 1:40
ElleryLElleryL
725519
725519
add a comment |
add a comment |
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