multidimensional Gaussian integral MultidimensionalGaussianIntegral 2002-02-13 00:40:49 2006-10-14 16:07:05 Theorem % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newcommand{\mv}[1]{\mathbf{#1}} % matrix or vector \newcommand{\cov}{\mathrm{cov}} \newcommand{\mvt}[1]{\mv{#1}^{\mathrm{T}}} \newcommand{\mvi}[1]{\mv{#1}^{-1}} \newcommand{\mpderiv}[1]{\frac{\partial}{\partial {#1}}} Let $\mv{x} = [x_1\ x_2\ \ldots\ x_n]^T$ and $d^n \mv{x} \equiv \prod_{i=1}^{n} d x_i$. \newtheorem{thm}{Theorem} \begin{thm}[] Let $K$ be a symmetric \PMlinkname{positive definite}{PositiveDefinite} 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 $|\mv{K}| = \det{\mv{K}}$. \end{thm} \emph{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: \begin{align} \int e^{-\frac{1}{2} \mvt{x} \mv{T} \mv{\Lambda} \mvi{T} \mv{x}} d^n \mv{x} &= \int e^{-\frac{1}{2} \mvt{x} \mv{T} \mv{\Lambda} \mvt{T} \mv{x}} d^n \mv{x}\\ &= \int e^{-\frac{1}{2} \mvt{y} \mv{\Lambda}\mv{y}} |\mv{J}| d^n \mv{y}\\ \end{align} 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 \begin{align} \int e^{-\frac{1}{2} \mvt{y} \mv{\Lambda}\mv{y}} d^n \mv{y} &= \prod_{k=1}^{n} \int e^{-\frac{1}{2} \lambda_k y_k^2} d y_k\\ &= \prod_{k=1}^{n} \left(\frac{2 \pi}{\lambda_k}\right)^{\frac{1}{2}} \\ &= \left(\frac{(2 \pi)^n}{\prod_{k=1}^{n}\lambda_k}\right)^{\frac{1}{2}} \\ &= \left(\frac{(2 \pi)^n}{|\mv{\Lambda}|}\right)^{\frac{1}{2}}. \\ \end{align} 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.