# 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 MultidimensionalGaussianIntegral 2013-03-22 12:18:44 2013-03-22 12:18:44 Mathprof (13753) Mathprof (13753) 22 Mathprof (13753) Theorem msc 60B11 msc 62H99 msc 62H10 JacobiDeterminant AreaUnderGaussianCurve ProofOfGaussianMaximizesEntropyForGivenCovariance