multidimensional Gaussian integral

Let $\mathbf{x}=[x_{1}\ x_{2}\ \ldots\ x_{n}]^{T}$ and $d^{n}\mathbf{x}\equiv\prod_{i=1}^{n}dx_{i}$ .

Theorem 1

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

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

(1)

where $|\mathbf{K}|=\det{\mathbf{K}}$ .

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

$\int e^{-\frac{1}{2}\mathbf{x}^{\mathrm{T}}\mathbf{A}\mathbf{x}}d^{n}\mathbf{x% }=\int e^{-\frac{1}{2}\mathbf{x}^{\mathrm{T}}\mathbf{T}\mathbf{\Lambda}\mathbf% {T}^{-1}\mathbf{x}}d^{n}\mathbf{x}.$

(2)

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

	$\displaystyle\int e^{-\frac{1}{2}\mathbf{x}^{\mathrm{T}}\mathbf{T}\mathbf{% \Lambda}\mathbf{T}^{-1}\mathbf{x}}d^{n}\mathbf{x}$	$\displaystyle=\int e^{-\frac{1}{2}\mathbf{x}^{\mathrm{T}}\mathbf{T}\mathbf{% \Lambda}\mathbf{T}^{\mathrm{T}}\mathbf{x}}d^{n}\mathbf{x}$		(3)
		$\displaystyle=\int e^{-\frac{1}{2}\mathbf{y}^{\mathrm{T}}\mathbf{\Lambda}% \mathbf{y}}\|\mathbf{J}\|d^{n}\mathbf{y}$		(4)

where $|\mathbf{J}|$ is the determinant of the Jacobian matrix $J_{mn}=\frac{\partial{x_{m}}}{\partial{y_{n}}}$ . In this case, $\mathbf{J}=\mathbf{T}$ and thus $|\mathbf{J}|=1$ .

Since $\mathbf{\Lambda}$ 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

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

(6)

Carrying out this program, we get

$\displaystyle\int e^{-\frac{1}{2}\mathbf{y}^{\mathrm{T}}\mathbf{\Lambda}% \mathbf{y}}d^{n}\mathbf{y}$	$\displaystyle=\prod_{k=1}^{n}\int e^{-\frac{1}{2}\lambda_{k}y_{k}^{2}}dy_{k}$	(7)
	$\displaystyle=\prod_{k=1}^{n}\left(\frac{2\pi}{\lambda_{k}}\right)^{\frac{1}{2}}$	(8)
	$\displaystyle=\left(\frac{(2\pi)^{n}}{\prod_{k=1}^{n}\lambda_{k}}\right)^{% \frac{1}{2}}$	(9)
	$\displaystyle=\left(\frac{(2\pi)^{n}}{\|\mathbf{\Lambda}\|}\right)^{\frac{1}{2}}.$	(10)

Now, we have $|\mathbf{A}|=|\mathbf{T}\mathbf{\Lambda}\mathbf{T}^{-1}|=|\mathbf{T}||\mathbf{% \Lambda}||\mathbf{T}^{-1}|=|\mathbf{\Lambda}|$ , so this becomes

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

(12)

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

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

(13)

as promised.

Title	multidimensional Gaussian integral
Canonical name	MultidimensionalGaussianIntegral
Date of creation	2013-03-22 12:18:44
Last modified on	2013-03-22 12:18:44
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	22
Author	Mathprof (13753)
Entry type	Theorem
Classification	msc 60B11
Classification	msc 62H99
Classification	msc 62H10
Related topic	JacobiDeterminant
Related topic	AreaUnderGaussianCurve
Related topic	ProofOfGaussianMaximizesEntropyForGivenCovariance