# Find the distribution of $Y=X'\Sigma^{-1}X$ using mgf

Let $\bf{X}\sim N_p(\bf{0},\bf{\Sigma})$, find the distribution of $\bf{Y=X'\Sigma^{-1}X}$ using the moment generating function.

I know that $$M_X(t)=E[\exp(t'X]=\exp(\frac{1}{2}t'\Sigma t)$$ then $$M_Y(t)=E[\exp(t'Y)]=E[\exp(t'X'\Sigma^{-1}X)]$$ Here I think about use the fact that $\Sigma=LL'$ where $L$ is a lower triangular matrix, then $$M_Y(t)=E[\exp(t'X'(LL')^{-1}X)]=E[\exp(t'X'(L')^{-1}L^{-1}X]=E[\exp(t'X'(L^{-1})'L^{-1}X]$$

$$=E[\exp(t'(L^{-1}X)'(L^{-1}X))]$$

but now I don't know what to do. Any hint?

First, note that $X'\Sigma^{-1}X$ is a number, not a vector, so you don't really need that prime on the $t$ of your mgf for Y.
I suppose we're assuming $\Sigma$ is invertible based on the premise of he question, so we can take the expectation $E(\exp(t(X'\Sigma^{-1}X)))$ using the density for $N(0,\Sigma)$ $$f(x) = \frac{1}{\sqrt{\det(2\pi\Sigma)}} \exp(-\frac{1}{2}x'\Sigma^{-1}x).$$ So we have $$E(\exp(t(X'\Sigma^{-1}X))) = \frac{1}{\sqrt{\det(2\pi\Sigma)}} \int \exp(tx'\Sigma^{-1}x)\exp\left(-\frac{1}{2}x'\Sigma^{-1}x\right)\,d^p x\\= \frac{1}{\sqrt{\det(2\pi\Sigma)}} \int \exp\left(-\frac{1}{2}(1-2t)x'\Sigma^{-1}x\right)\,d^p x \\= \frac{\sqrt{\det\left(\frac{2\pi}{1-2t}\Sigma\right)}}{\sqrt{\det(2\pi\Sigma)}} \\= \frac{1}{(1-2t)^{p/2}}.$$
• if $c$ is a constant and $A$ is a $p\times p$ matrix, $\det (cA) = c^p\det(A)$ Sep 10, 2017 at 1:24
• used the fact that $$\int e^{-\frac{1}{2}x'\sigma^{-1}x} d^px = \sqrt{\det(2\pi\Sigma)}$$ (that's why the normalization factor is what it is) and substituted to rescale $x$ by $\sqrt{1-2t}$. It might have actually been more clear to have $\frac{1}{(1-2t)^{p/2}}$ on the outside on the second to last line. Sep 10, 2017 at 2:10