# How do we know that $m^T\Sigma^{-1}m$ is the sum of independent squared standard normal variables?

Say we have a vector of $k$ random variables $$m\sim N(0,\Sigma)$$ Then it is the case that $m^T\Sigma^{-1}m$ has a $\chi^2_k$-distribution, because it is the sum of $k$ independent standard normal variables.

Now, if $\Sigma$ is a diagonal matrix, so that the $m_i$ are independent, then I understand why this is the case. We simply get: $$m^T\Sigma^{-1}m=\frac {m_1^2}{\sigma_1^2}+...+\frac {m_k^2}{\sigma_k^2}$$

However, when the variables are correlated, then I can't prove the result. How do we show the general case?

• Hint: Can you find a matrix $A$ such that $Am$ is a vector of $k$ independent, standard normal random variables? Jan 9, 2018 at 17:57
• @MatthewGunn, :O, you mean $A=\Sigma^{-\frac 1 2}$? That must work, but I'm not at all sure what the elements of that matrix are, or how to show that this $Am$ is a vector of standard indep. normal variables. Jan 9, 2018 at 18:02
• You're on the right track. If $m$ is multivariate normal then what distribution is $Am$? You can also use zero correlation $\Leftrightarrow$ independence for multivariate normal distribution. (Another comment, you need $\Sigma$ is rank $k$. Think about what happens if rank $<k$.) Jan 9, 2018 at 18:13
• Thanks to my linear models professor, here's the result you need. See slide 31 of dnett.github.io/S510/01Preliminaries.pdf. I can't remember how to prove this currently. Jan 9, 2018 at 18:15
• @MatthewGunn $Var(Am)=AVar(m)A^T=I_K$..... lol, that was easy. if $\Sigma$ is not rank $k$, then it is not invertible, so the whole thing doesn't work. Jan 9, 2018 at 18:15