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multidimensional Gaussian integral (Theorem)

Let $\mv{x} = [x_1\ x_2\ \ldots\ x_n]^T$ and $d^n \mv{x} \equiv \prod_{i=1}^{n} d x_i$ .

Theorem 1   Let $K$ be a symmetric positive definite matrix and $f: R^n \to R$ , where $f(x) = \exp{(- \frac{1}{2} \mv{x}^T \mv{K}^{-1} \mv{x})}$ . Then \begin{equation} \int e^{-\frac{1}{2} \mv{x}^T \mv{K}^{-1} \mv{x}} d^n \mv{x} = \left((2\pi)^n |\mv{K}| \right)^{\frac{1}{2}} \end{equation}where $|\mv{K}| = \det{\mv{K}}$ .
Proof. $\mvi{K}$ is real and symmetric (since $(\mvi{K})^{\mathrm{T}} = (\mvt{K})^{-1} = \mv{K}^{-1})$ . For convenience, let $\mv{A} = \mvi{K}$ . We can decompose $\mv{A}$ into $\mv{A} = \mv{T} \mv{\Lambda} \mvi{T}$ , where $\mv{T}$ is an orthonormal ($\mvt{T} \mv{T} = \mv{I}$ ) matrix of the eigenvectors of $\mv{A}$ and $\mv{\Lambda}$ is a diagonal matrix of the eigenvalues of $\mv{A}$ . Then \begin{equation} \int e^{-\frac{1}{2} \mvt{x} \mv{A} \mv{x}} d^n \mv{x} = \int e^{-\frac{1}{2} \mvt{x} \mv{T} \mv{\Lambda} \mvi{T} \mv{x}} d^n \mv{x}. \end{equation} Because $\mv{T}$ is orthonormal, we have $\mvi{T} = \mvt{T}$ . Now define a new vector variable $\mv{y} \equiv \mvt{T} \mv{x}$ , and substitute:
$\displaystyle \int e^{-\frac{1}{2} \mathbf{x}^{\mathrm{T}} \mathbf{T} \mathbf{\Lambda} \mathbf{T}^{-1} \mathbf{x}} d^n \mathbf{x}$ $\displaystyle = \int e^{-\frac{1}{2} \mathbf{x}^{\mathrm{T}} \mathbf{T} \mathbf{\Lambda} \mathbf{T}^{\mathrm{T}} \mathbf{x}} d^n \mathbf{x}$ (1)
  $\displaystyle = \int e^{-\frac{1}{2} \mathbf{y}^{\mathrm{T}} \mathbf{\Lambda}\mathbf{y}} \vert\mathbf{J}\vert d^n \mathbf{y}$ (2)

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

Since $\mv{\Lambda}$ 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

\begin{equation} \int e^{-\frac{1}{2} a t^2} dt = \left(\frac{2 \pi}{a}\right)^{\frac{1}{2}}. \end{equation} Carrying out this program, we get

$\displaystyle \int e^{-\frac{1}{2} \mathbf{y}^{\mathrm{T}} \mathbf{\Lambda}\mathbf{y}} d^n \mathbf{y}$ $\displaystyle = \prod_{k=1}^{n} \int e^{-\frac{1}{2} \lambda_k y_k^2} d y_k$ (3)
  $\displaystyle = \prod_{k=1}^{n} \left(\frac{2 \pi}{\lambda_k}\right)^{\frac{1}{2}}$ (4)
  $\displaystyle = \left(\frac{(2 \pi)^n}{\prod_{k=1}^{n}\lambda_k}\right)^{\frac{1}{2}}$ (5)
  $\displaystyle = \left(\frac{(2 \pi)^n}{\vert\mathbf{\Lambda}\vert}\right)^{\frac{1}{2}}.$ (6)

Now, we have $|\mv{A}| = |\mv{T} \mv{\Lambda} \mvi{T}| = |\mv{T}| |\mv{\Lambda}| |\mvi{T}| = |\mv{\Lambda}|$ , so this becomes

\begin{equation} \int e^{-\frac{1}{2} \mvt{x} \mv{A} \mv{x}} d^n \mv{x} = \left(\frac{(2 \pi)^n}{|\mv{A}|}\right)^{\frac{1}{2}}. \end{equation} Substituting back in for $\mvi{K}$ , we get

\begin{equation} \int e^{-\frac{1}{2} \mvt{x} \mvi{K} \mv{x}} d^n \mv{x} = \left(\frac{(2 \pi)^n}{|\mvi{K}|}\right)^{\frac{1}{2}} = \left((2 \pi)^n |\mv{K}|\right)^{\frac{1}{2}}, \end{equation}as promised.




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See Also: Jacobi determinant, area under Gaussian curve, proof of Gaussian maximizes entropy for given covariance

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Cross-references: formula, integrate, distributions, Gaussian, independent, product, separated, integral, diagonal, Jacobian matrix, determinant, variable, vector, eigenvalues, diagonal matrix, eigenvectors, orthonormal, real, proof, matrix, symmetric
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This is version 19 of multidimensional Gaussian integral, born on 2002-02-13, modified 2006-10-14.
Object id is 1914, canonical name is MultidimensionalGaussianIntegral.
Accessed 13648 times total.

Classification:
AMS MSC60B11 (Probability theory and stochastic processes :: Probability theory on algebraic and topological structures :: Probability theory on linear topological spaces)
 62H10 (Statistics :: Multivariate analysis :: Distribution of statistics)
 62H99 (Statistics :: Multivariate analysis :: Miscellaneous)

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