multidimensional Gaussian integral

Let $𝐱={[x_1{x}_2\mathrm{\dots }{x}_n]}^T$ and $d^n𝐱\equiv {\prod }_{i=1}^{n}d{x}_i$.

Theorem 1

Let $K$ be a symmetric positive definite (http://planetmath.org/PositiveDefinite) matrix and$f\mathrm{:}{R}^n\mathrm{\to }R$, where$f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{=}\exp \mathit{}\mathrm{(}\mathrm{-}\frac{\mathrm{1}}{\mathrm{2}}\mathit{}{\mathrm{x}}^T\mathit{}{\mathrm{K}}^{\mathrm{-}\mathrm{1}}\mathit{}\mathrm{x}\mathrm{)}$.Then

$$\int e^{-\frac{1}{2}{𝐱}^T{𝐊}^{-1}𝐱}{d}^n𝐱={\left({(2\pi )}^n|𝐊|\right)}^{\frac{1}{2}}$$(1)

where $\mathrm{|}\mathrm{K}\mathrm{|}\mathrm{=}\det \mathit{}\mathrm{K}$.

Proof. ${𝐊}^{-1}$ is real and symmetric (since ${({𝐊}^{-1})}^{\mathrm{T}}={({𝐊}^{\mathrm{T}})}^{-1}={𝐊}^{-1})$. For convenience, let $𝐀={𝐊}^{-1}$. We can decompose $𝐀$ into $𝐀=𝐓𝚲{𝐓}^{-1}$, where $𝐓$ is an orthonormal (${𝐓}^{\mathrm{T}}𝐓=𝐈$) matrix of the eigenvectors of $𝐀$ and $𝚲$ is a diagonal matrix of the eigenvalues of $𝐀$. Then

$$\int e^{-\frac{1}{2}{𝐱}^{\mathrm{T}}\mathrm{𝐀𝐱}}{d}^n𝐱=\int e^{-\frac{1}{2}{𝐱}^{\mathrm{T}}𝐓𝚲{𝐓}^{-1}𝐱}{d}^n𝐱.$$(2)

Because $𝐓$ is orthonormal, we have ${𝐓}^{-1}={𝐓}^{\mathrm{T}}$. Now define a new vector variable $𝐲\equiv {𝐓}^{\mathrm{T}}𝐱$, and substitute:

${\displaystyle \int e^{-\frac{1}{2}{𝐱}^{\mathrm{T}}𝐓𝚲{𝐓}^{-1}𝐱}{d}^n𝐱}$$={\displaystyle \int e^{-\frac{1}{2}{𝐱}^{\mathrm{T}}𝐓𝚲{𝐓}^{\mathrm{T}}𝐱}{d}^n𝐱}$(3)

$={\displaystyle \int e^{-\frac{1}{2}{𝐲}^{\mathrm{T}}𝚲𝐲}|𝐉|d^n𝐲}$(4)

where $|𝐉|$ is the determinant of the Jacobian matrix $J_{mn}=\frac{\partial x_m}{\partial y_n}$. In this case, $𝐉=𝐓$ and thus $|𝐉|=1$.

Since $𝚲$ is diagonal, the integral may be separated into the product of $n$ independent Gaussian distributions, each of which we can integrate separately using the well-known formula

$$\int e^{-\frac{1}{2}a{t}^2}𝑑t={\left(\frac{2\pi }{a}\right)}^{\frac{1}{2}}.$$(6)

Carrying out this program, we get

${\displaystyle \int e^{-\frac{1}{2}{𝐲}^{\mathrm{T}}𝚲𝐲}{d}^n𝐲}$$={\displaystyle \prod _{k=1}^n}{\displaystyle \int e^{-\frac{1}{2}{\lambda }_k{y}_k^2}𝑑y_k}$(7)

$={\displaystyle \prod _{k=1}^n}{\left({\displaystyle \frac{2\pi }{{\lambda }_k}}\right)}^{\frac{1}{2}}$(8)

$={\left({\displaystyle \frac{{(2\pi )}^n}{{\prod }_{k=1}^n{\lambda }_k}}\right)}^{\frac{1}{2}}$(9)

$={\left({\displaystyle \frac{{(2\pi )}^n}{|𝚲|}}\right)}^{\frac{1}{2}}.$(10)

Now, we have $|𝐀|=|𝐓𝚲{𝐓}^{-1}|=|𝐓||𝚲||{𝐓}^{-1}|=|𝚲|$, so this becomes

$$\int e^{-\frac{1}{2}{𝐱}^{\mathrm{T}}\mathrm{𝐀𝐱}}{d}^n𝐱={\left(\frac{{(2\pi )}^n}{|𝐀|}\right)}^{\frac{1}{2}}.$$(12)

Substituting back in for ${𝐊}^{-1}$, we get

$$\int e^{-\frac{1}{2}{𝐱}^{\mathrm{T}}{𝐊}^{-1}𝐱}{d}^n𝐱={\left(\frac{{(2\pi )}^n}{|{𝐊}^{-1}|}\right)}^{\frac{1}{2}}={\left({(2\pi )}^n|𝐊|\right)}^{\frac{1}{2}},$$(13)

as promised.