Proof of w obeys Chi-square distribution with r degrees of freedom

Question

In my econometrics book, it has a footnote saying:

Let $\mathbf{x}$ be an $m$ dimensional random vector, if $\mathbf{x} \sim N(\mu, \Sigma)$ where $\Sigma$ is nonsingular, then $ (\mathbf{x} - \mu)'\Sigma^{-1} (\mathbf{x} - \mu) \sim \chi^2(m)$ with $m$ degrees of freedom.

How to prove this?

Thanks a lot.

Answer 1

It is not correct and should be: $$(\mathbf x-\mu)^T\Sigma^{-1}(\mathbf x-\mu)\sim\chi^2(m)$$

You can write $\mathbf x-\mu=A\mathbf u$ where $\mathbf u$ has standard normal distribution and $\Sigma=AA^T$.

Then: $$(\mathbf x-\mu)^T\Sigma^{-1}(\mathbf x-\mu)=\mathbf u^T\mathbf u$$

Here $\mathbf u^T\mathbf u$ has chi-square distribution.