Derivation of a closed-form solution for the integral of a 3D Gaussian over the *positive* reals
$begingroup$
in a post from over four years ago, Przemo gave the following formula for the integral over a Gaussian function over the positive reals in three dimensions (denoted here as $mathbb{R}^3_+$) with inverse covariance $A=left(begin{array}{rrr}a & a_{12} & a_{13}\a_{12}& b&a_{23}\a_{13}&a_{23}&cend{array}right)$:
begin{multline}
int_{mathbb{R}^3_+} mathrm{d}x ; expleft(-frac{1}{2}x^top A xright) = \
frac{pi}{sqrt{2 det A}}intlimits_0^infty mathrm{erfc}left(u sqrt{frac{-2 a_{12}a_{13}a_{23}+a_{13}^2 b+a_{12}^2 c}{det(A)}}right)e^{-u^2}cdot \
left[-mathrm{erfc}left(sqrt{a} frac{a_{12} c-a_{13}a_{23}}{a_{13} sqrt{det(A)}} uright)+mathrm{erfc}left(sqrt{a} frac{a_{12}a_{23}-a_{13} b}{a_{12} sqrt{det(A)}} uright)right] du
end{multline}
This integral can be done using the techniques outlined here, yielding a closed-form solution to the original problem.
I would like to understand the derivation of the intermediate result given above. In his original post, Przemo details the calculation in two dimensions: complete the square in the quadratic form; do the first integral, which yields an error function; expand the error function and do the remaining integral term by term. How can you adapt this approach to 3D?
Steps undertaken so far:
Completing the square in one variable, say $x$, leaves me with
$$int_{mathbb{R}_+^2} mathrm{d}ymathrm{d}z expleft(-frac{1}{2} frac{mathrm{det},A_3}{mathrm{det},A_2}z^2right) expleft(-frac{1}{2} frac{mathrm{det}, A_2}{a}(y-m z)^2right) left[1 - mathrm{erf}left(frac{a_{12}y+a_{13}z}{sqrt{2a}}right) right] $$
where $A_2=begin{pmatrix} a & a_{12}\ & bend{pmatrix}$, $A_3$ as defined above, and $m$ is a function of the coefficients of the matrices. However, I do not know how to proceed from there: expanding the error function to do the integral in y, say, is a nightmare due to the constant term in z; I also did not find a way to do a coordinate transform à la $s=a_{12}y+a_{13}z$ or something similar.Indeed, the formula above looks more like one completed the square in two of the variables independently; but what happened to the cross-term? I cannot find a factorisation of the exponent that would allow me to complete two integrals over the half-line with only one variable left in the error function yielded by the integral.
Any help / hint would be greatly appreciated! Thank you in advance for your time :)
definite-integrals gaussian-integral
asked Dec 21 '18 at 9:05
workandheatworkandheat
536
$endgroup$
add a comment |
$begingroup$
in a post from over four years ago, Przemo gave the following formula for the integral over a Gaussian function over the positive reals in three dimensions (denoted here as $mathbb{R}^3_+$) with inverse covariance $A=left(begin{array}{rrr}a & a_{12} & a_{13}\a_{12}& b&a_{23}\a_{13}&a_{23}&cend{array}right)$:
begin{multline}
int_{mathbb{R}^3_+} mathrm{d}x ; expleft(-frac{1}{2}x^top A xright) = \
frac{pi}{sqrt{2 det A}}intlimits_0^infty mathrm{erfc}left(u sqrt{frac{-2 a_{12}a_{13}a_{23}+a_{13}^2 b+a_{12}^2 c}{det(A)}}right)e^{-u^2}cdot \
left[-mathrm{erfc}left(sqrt{a} frac{a_{12} c-a_{13}a_{23}}{a_{13} sqrt{det(A)}} uright)+mathrm{erfc}left(sqrt{a} frac{a_{12}a_{23}-a_{13} b}{a_{12} sqrt{det(A)}} uright)right] du
end{multline}
This integral can be done using the techniques outlined here, yielding a closed-form solution to the original problem.
I would like to understand the derivation of the intermediate result given above. In his original post, Przemo details the calculation in two dimensions: complete the square in the quadratic form; do the first integral, which yields an error function; expand the error function and do the remaining integral term by term. How can you adapt this approach to 3D?
Steps undertaken so far:
Completing the square in one variable, say $x$, leaves me with
$$int_{mathbb{R}_+^2} mathrm{d}ymathrm{d}z expleft(-frac{1}{2} frac{mathrm{det},A_3}{mathrm{det},A_2}z^2right) expleft(-frac{1}{2} frac{mathrm{det}, A_2}{a}(y-m z)^2right) left[1 - mathrm{erf}left(frac{a_{12}y+a_{13}z}{sqrt{2a}}right) right] $$
where $A_2=begin{pmatrix} a & a_{12}\ & bend{pmatrix}$, $A_3$ as defined above, and $m$ is a function of the coefficients of the matrices. However, I do not know how to proceed from there: expanding the error function to do the integral in y, say, is a nightmare due to the constant term in z; I also did not find a way to do a coordinate transform à la $s=a_{12}y+a_{13}z$ or something similar.Indeed, the formula above looks more like one completed the square in two of the variables independently; but what happened to the cross-term? I cannot find a factorisation of the exponent that would allow me to complete two integrals over the half-line with only one variable left in the error function yielded by the integral.
Any help / hint would be greatly appreciated! Thank you in advance for your time :)
definite-integrals gaussian-integral
asked Dec 21 '18 at 9:05
workandheatworkandheat
536
$endgroup$
$begingroup$
@ workandheat I have provided detailed explanations in the original post if you wanna see them. Unfortunately for the time being I am unable to generalize tyhe solution to dimensions higher than three. My question to you is therefore why are you interested in this very problem? Have you managed to to make any progress? I have some ideas which I can share with you if you wish.
$endgroup$
– Przemo
Jan 25 at 16:43
add a comment |
$begingroup$
in a post from over four years ago, Przemo gave the following formula for the integral over a Gaussian function over the positive reals in three dimensions (denoted here as $mathbb{R}^3_+$) with inverse covariance $A=left(begin{array}{rrr}a & a_{12} & a_{13}\a_{12}& b&a_{23}\a_{13}&a_{23}&cend{array}right)$:
begin{multline}
int_{mathbb{R}^3_+} mathrm{d}x ; expleft(-frac{1}{2}x^top A xright) = \
frac{pi}{sqrt{2 det A}}intlimits_0^infty mathrm{erfc}left(u sqrt{frac{-2 a_{12}a_{13}a_{23}+a_{13}^2 b+a_{12}^2 c}{det(A)}}right)e^{-u^2}cdot \
left[-mathrm{erfc}left(sqrt{a} frac{a_{12} c-a_{13}a_{23}}{a_{13} sqrt{det(A)}} uright)+mathrm{erfc}left(sqrt{a} frac{a_{12}a_{23}-a_{13} b}{a_{12} sqrt{det(A)}} uright)right] du
end{multline}
This integral can be done using the techniques outlined here, yielding a closed-form solution to the original problem.
I would like to understand the derivation of the intermediate result given above. In his original post, Przemo details the calculation in two dimensions: complete the square in the quadratic form; do the first integral, which yields an error function; expand the error function and do the remaining integral term by term. How can you adapt this approach to 3D?
Steps undertaken so far:
Completing the square in one variable, say $x$, leaves me with
$$int_{mathbb{R}_+^2} mathrm{d}ymathrm{d}z expleft(-frac{1}{2} frac{mathrm{det},A_3}{mathrm{det},A_2}z^2right) expleft(-frac{1}{2} frac{mathrm{det}, A_2}{a}(y-m z)^2right) left[1 - mathrm{erf}left(frac{a_{12}y+a_{13}z}{sqrt{2a}}right) right] $$
where $A_2=begin{pmatrix} a & a_{12}\ & bend{pmatrix}$, $A_3$ as defined above, and $m$ is a function of the coefficients of the matrices. However, I do not know how to proceed from there: expanding the error function to do the integral in y, say, is a nightmare due to the constant term in z; I also did not find a way to do a coordinate transform à la $s=a_{12}y+a_{13}z$ or something similar.Indeed, the formula above looks more like one completed the square in two of the variables independently; but what happened to the cross-term? I cannot find a factorisation of the exponent that would allow me to complete two integrals over the half-line with only one variable left in the error function yielded by the integral.
Any help / hint would be greatly appreciated! Thank you in advance for your time :)
definite-integrals gaussian-integral
asked Dec 21 '18 at 9:05
workandheatworkandheat
536
$endgroup$
in a post from over four years ago, Przemo gave the following formula for the integral over a Gaussian function over the positive reals in three dimensions (denoted here as $mathbb{R}^3_+$) with inverse covariance $A=left(begin{array}{rrr}a & a_{12} & a_{13}\a_{12}& b&a_{23}\a_{13}&a_{23}&cend{array}right)$:
begin{multline}
int_{mathbb{R}^3_+} mathrm{d}x ; expleft(-frac{1}{2}x^top A xright) = \
frac{pi}{sqrt{2 det A}}intlimits_0^infty mathrm{erfc}left(u sqrt{frac{-2 a_{12}a_{13}a_{23}+a_{13}^2 b+a_{12}^2 c}{det(A)}}right)e^{-u^2}cdot \
left[-mathrm{erfc}left(sqrt{a} frac{a_{12} c-a_{13}a_{23}}{a_{13} sqrt{det(A)}} uright)+mathrm{erfc}left(sqrt{a} frac{a_{12}a_{23}-a_{13} b}{a_{12} sqrt{det(A)}} uright)right] du
end{multline}
This integral can be done using the techniques outlined here, yielding a closed-form solution to the original problem.
I would like to understand the derivation of the intermediate result given above. In his original post, Przemo details the calculation in two dimensions: complete the square in the quadratic form; do the first integral, which yields an error function; expand the error function and do the remaining integral term by term. How can you adapt this approach to 3D?
Steps undertaken so far:
Completing the square in one variable, say $x$, leaves me with
$$int_{mathbb{R}_+^2} mathrm{d}ymathrm{d}z expleft(-frac{1}{2} frac{mathrm{det},A_3}{mathrm{det},A_2}z^2right) expleft(-frac{1}{2} frac{mathrm{det}, A_2}{a}(y-m z)^2right) left[1 - mathrm{erf}left(frac{a_{12}y+a_{13}z}{sqrt{2a}}right) right] $$
where $A_2=begin{pmatrix} a & a_{12}\ & bend{pmatrix}$, $A_3$ as defined above, and $m$ is a function of the coefficients of the matrices. However, I do not know how to proceed from there: expanding the error function to do the integral in y, say, is a nightmare due to the constant term in z; I also did not find a way to do a coordinate transform à la $s=a_{12}y+a_{13}z$ or something similar.Indeed, the formula above looks more like one completed the square in two of the variables independently; but what happened to the cross-term? I cannot find a factorisation of the exponent that would allow me to complete two integrals over the half-line with only one variable left in the error function yielded by the integral.
Any help / hint would be greatly appreciated! Thank you in advance for your time :)
definite-integrals gaussian-integral
definite-integrals gaussian-integral
asked Dec 21 '18 at 9:05
workandheatworkandheat
536
asked Dec 21 '18 at 9:05
workandheatworkandheat
536
edited Dec 21 '18 at 15:55
workandheat
asked Dec 21 '18 at 9:05
workandheatworkandheat
536
asked Dec 21 '18 at 9:05
workandheatworkandheat
536
asked Dec 21 '18 at 9:05
workandheatworkandheat
536
536
$begingroup$
@ workandheat I have provided detailed explanations in the original post if you wanna see them. Unfortunately for the time being I am unable to generalize tyhe solution to dimensions higher than three. My question to you is therefore why are you interested in this very problem? Have you managed to to make any progress? I have some ideas which I can share with you if you wish.
$endgroup$
– Przemo
Jan 25 at 16:43
add a comment |
$begingroup$
@ workandheat I have provided detailed explanations in the original post if you wanna see them. Unfortunately for the time being I am unable to generalize tyhe solution to dimensions higher than three. My question to you is therefore why are you interested in this very problem? Have you managed to to make any progress? I have some ideas which I can share with you if you wish.
$endgroup$
– Przemo
Jan 25 at 16:43
$begingroup$
@ workandheat I have provided detailed explanations in the original post if you wanna see them. Unfortunately for the time being I am unable to generalize tyhe solution to dimensions higher than three. My question to you is therefore why are you interested in this very problem? Have you managed to to make any progress? I have some ideas which I can share with you if you wish.
$endgroup$
– Przemo
Jan 25 at 16:43
$begingroup$
@ workandheat I have provided detailed explanations in the original post if you wanna see them. Unfortunately for the time being I am unable to generalize tyhe solution to dimensions higher than three. My question to you is therefore why are you interested in this very problem? Have you managed to to make any progress? I have some ideas which I can share with you if you wish.
$endgroup$
– Przemo
Jan 25 at 16:43
add a comment |
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$begingroup$
@ workandheat I have provided detailed explanations in the original post if you wanna see them. Unfortunately for the time being I am unable to generalize tyhe solution to dimensions higher than three. My question to you is therefore why are you interested in this very problem? Have you managed to to make any progress? I have some ideas which I can share with you if you wish.
$endgroup$
– Przemo
Jan 25 at 16:43