TITLE: Multivariate Gaussian, why divide by determinant of covariance matrix?
QUESTION [6 upvotes]: Given the formula for the density of the multivariate gaussian:
$$f_Y(x)=\frac{1}{\sqrt{(2\pi)^n|\boldsymbol\Sigma|}}
\exp\left(-\frac{1}{2}({x}-{m})^T{\boldsymbol\Sigma}^{-1}({x}-{m})
\right)$$
Can anybody tell me what the rationale is behind using the determinant in the square root term in the denominator? In the univariate case you divide by the standard deviation, why do you have to use the determinant in the multivariate case? What relation is there between some measure of deviation and the determinant of the covariance matrix?

REPLY [2 votes]: Here's one way to look at it:
\begin{align}
f_Y(x) \, dx & =\frac{1}{\sqrt{(2\pi)^n}}
\exp\left(-\frac{1}{2}\Big({\boldsymbol\Sigma}^{-1/2}(x-m)^T\Big)^T \Big( {\boldsymbol\Sigma}^{-1/2} (x-m)
\right) \, \frac{dx}{|\boldsymbol{\Sigma}^{-1/2}|} \\[12pt]
& = \frac{1}{\sqrt{(2\pi)^n}} \exp\left( -\frac 1 2 (u-\ell)^T(u-\ell) \right) \, du
\end{align}
An infinitely small volume $dx = dx_1\cdots dx_n$ is transformed by multiplication by the matrix $\boldsymbol{\Sigma}^{-1/2}$ to a region of volume $|\Sigma^{-1/2}|\,du_1\cdots du_n$.  That is what determinants of matrices do.
For this to make sense it is necessary to know that $\boldsymbol{\Sigma}$ and its inverse are positive-definite symmetric matrices and that every such matrix has a positive-definite symmetric square root.  That follows from the spectral theorem of linear algebra.