PDF of Mahalanobis distance of a multivariate random normal variable

Question

Suppose $x$ is a multivariate normal random variable with mean $\mu$ and covariance matrix $C$. Let $h(x)$ be a the Mahalanobis distance of $x$ given by $h(x)=\sqrt{(x-\mu)^TC^{-1}(x-\mu)}$. Then how do we prove that the PDF of $h(x)$, is given by $p(h(x))=\frac{2}{\sqrt{2\pi}}exp(-\frac{1}{2}h(x)^2)$? I found this result mentioned in a research paper (https://pdfs.semanticscholar.org/b81a/a3046e7f9213949fb37e0c59cacdca4572c4.pdf) but I am unable to understand how this holds.

Answer 1

Let the eigendecomposition of $C$ be,

$$C = U\Lambda U^T = U\Lambda^{1/2}(U\Lambda^{1/2})^T$$

where $\Lambda$ is the diagonal matrix of eigenvalues and the columns of $U$ are unit eigenvectors. Note that $U^{-1} = U^T$ and therefore, $C^{-1} = U\Lambda^{-1}U^T$.

$$X \sim \mathcal{N}(\mu, C) \implies X = \mu + U\Lambda^{1/2}Z, \ \text{where } Z \sim \mathcal{N}(0,1)$$

$$Z = (U\Lambda^{1/2})^{-1}(X-\mu) = \Lambda^{-1/2}U^T(X-\mu)$$ $$ Z^{T}Z = (X-\mu)^{T}U\Lambda^{-1/2}\Lambda^{-1/2}U^T(X-\mu) = (X-\mu)^{T}U\Lambda^{-1}U^T(X-\mu)$$

$$Z^TZ = (X-\mu)^TC^{-1}(X-\mu)$$

Hope this helps.