I am looking at the derivation of $f_{\vec{Y}}(\vec{y})$ where $\vec{Y}=A \vec{X}$ and $\vec{X}$ is a vector of i.i.d standard normal random variables. $A$ is an $n \times n$ non-singular matrix.
The multivariate standard normal density is given by :
$$f_{\vec{X}}(\vec{x})=\left(\frac{1}{\sqrt{2 \pi}}\right)^{n} \exp \left[-\frac{1}{2} \vec{x}^{\top} \vec{x}\right]$$
And by the change of variables theorem ;
$$f_{Y}(y)=\left\{\begin{array}{ll}f_{X}(h(Y))\left|J_{h}(y)\right| & \text { for } y \in R \\ 0 & \text { otherwise }\end{array}\right.$$
$$h(y)=g^{-1}(y)$$
Then :
$$\begin{aligned} f_{\vec{Y}}(\vec{y}) &=\left(\frac{1}{\sqrt{2 \pi}}\right)^{n}\|A\|^{-1} \exp \left[-\frac{1}{2} \vec{y}^{\top}\left(A^{-1}\right)^{\top} A^{-1} \vec{y}\right] \\ &=\left(\frac{1}{\sqrt{2 \pi}}\right)^{n}\|\Sigma\|^{-1 / 2} \exp \left[-\frac{1}{2} \vec{y}^{\top} \Sigma^{-1} \vec{y}\right].\end{aligned}$$
I am having trouble figuring out why $\left\|\sum\right\|^{2}=\|A\|$, or why $\left(A^{-1}\right)^{\top} A^{-1}=\sum^{-1}$. That is, how do we get the covariance matrix $\sum$ into the formulation for the multivariate standard normal of this transformation.
Or simply, why is $Var(\vec{Y}) = \sum=A A^{T}$ ... ? Where does this result come from ?
This derivation may be what you are looking for:
$$ \Sigma_{Y}= E\left[YY^{T}\right]=E\left[AX\left(AX\right)^{T}\right] $$ $$ \Sigma_{Y}= E\left[AXX^{T}A^{T}\right]=AE\left[XX^{T}\right]A^{T}=A\Sigma_{X} A^{T} $$
I hope this helps.