Let $\psi(x) = \mathbb{P}(X>\mathbf{x})$, where $X\sim N_{p}(\mathbf{\mu}, \mathbf{\Sigma})$. Make the transformation $Y = \mathbf{\Sigma}^{-1/2}(X - \mathbf{\mu})$. Let $\phi(x) = \mathbb{P}(Y>\mathbf{\Sigma}^{-1/2}(x - \mathbf{\mu}))$, where $Y\sim N_{p}(\mathbf{0}, \mathbf{I})$. Idealy $\psi(x) = \phi(x)$ for all $\mathbf{\Sigma}^{-1/2}$. However, the factorization $\mathbf{\Sigma}^{-1} = \mathbf{\Sigma}^{-1/2}\mathbf{\Sigma}^{-1/2}$, is not unique. My question is that, which factorization i.e., $\mathbf{\Sigma}^{-1/2}$ should one take?
I did some computations and found that $\psi(x) \neq \phi(x)$ for different choices of $\mathbf{\Sigma}^{-1/2}$. May be I am making some mistakes.