Mahalanobis distance

Remarks

Example: let $\mu = 0$ and $\Sigma = \operatorname{diag}(4, 1)$. Then $\Sigma^{-1} = \operatorname{diag}(1/4, 1)$, so $D_M(2, 0) = \sqrt{4/4} = 1$ and $D_M(0, 1) = 1$, while $D_M(0, 2) = 2$. Non-example: if $\Sigma = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, then $\Sigma$ is not invertible, so the formula is not defined. If $\Sigma = A A^T$ with $A$ invertible, then $d_M(x, y) = \abs{A^{-1} x - A^{-1} y}$, the Euclidean distance after the change of variables $x \mapsto A^{-1} x$. So $d_M$ is a metric on $\R^p$. If $\Sigma$ is the identity, $d_M$ is the Euclidean distance. If $X$ is multivariate normal, then $D_M(X)^2$ has the chi-squared distribution with $p$ degrees of freedom.