I have a $n$-dimensional multivariate Gaussian random variable $x \sim N(0, \Sigma_{n})$ where $\Sigma_{n} \in \mathbb{R}^{n \times n}$ is a valid covariance matrix. Assume a "whitened" representation $z = \Sigma_{n}^{-1/2} x$. Presumably now $z \sim N(0, I_{n})$.
Question: For multivariate CDFs of the form $F_X(x) = P(X_0 \leq x_0, X_1 \leq x_1, ... X_{n-1} \leq x_{n-1})$, what is the relationship between $F_X(x)$ and $F_Z(z)$ ?
For the 1-dimensional case, it is clear that $F_X(x) = P(X \leq x) = P(\Sigma^{1/2} Z \leq x) = P(Z \leq \Sigma^{-1/2}x) = F_Z(z)$. The same logic holds when $\Sigma$ is a diagonal matrix as each co-ordinate can be treated independently in a similar manner, but it is not clear to me what is the relationship for a general $n$-dimensional covariance matrix $\Sigma$. Any pointers?
It is a special case for linear transformations of a random vector. Generally, let $\bf{x}$ be a random vector with pdf $F_{\bf{x}}(\bf x')$ and let $\bf{z}=Ax$. If $\bf A$ is a positive-semidefinite matrix $\bf{x}\leq x' \iff \bf{Ax}\leq Ax' \iff \bf{z}\leq z' $. Therefore, $F_{\bf{x}}(\bf x')=F_{\bf{z}}(\bf z')$. If $\bf A$ is not a positive-semidefinite matrix it's complicated since the inequality is not satisfied. Since $\Sigma_n$ is a valid covariance matrix it is a positive-semidefinite matrix.