When we have a [[Multivariate Gaussian|Gaussian Vector]] $X \in \mathbb R^d$, and we have $n$ copies of that, then $\mu$\bar X_n$ denotes the average of each element of $X$ over all $n$ copies. - $\bar X_n \in \mathbb R^d$ is a random vector, which converges to a multivariate Gaussian. - $\mu \in \mathbb R^d$ is a deterministic vector as it holds the population means. When we subtract the vector by its mean, we are centering the multivariate Gaussian around zero on each of the $d$ dimensions. $ \sqrt n \,(\bar X_n-\mu) \xrightarrow[n \to \infty]{(d)}\mathcal N_d(0, \Sigma) $ To arrive at a statement of convergence to a multivariate standard Gaussian, we take following transformations: $ \begin{align} \sqrt n \,(\bar X_n-\mu) &\xrightarrow[n \to \infty]{(d)}\mathcal N_d(0, \Sigma) \tag{1}\\[4pt] \Sigma^{-1/2}\sqrt n \,(\bar X_n-\mu) &\xrightarrow[n \to \infty]{(d)} \Sigma^{-1/2} \mathcal N_d(0, \Sigma) \tag{2} \\[4pt] \Sigma^{-1/2}\sqrt n \,(\bar X_n-\mu) &\xrightarrow[n \to \infty]{(d)} \mathcal N_d \big(0, \Sigma^{-1/2} \, \Sigma (\Sigma^{-1/2})^T \big) \tag{3} \\[4pt] \Sigma^{-1/2} \sqrt n \,(\bar X_n-\mu) &\xrightarrow[n \to \infty]{(d)}\mathcal N_d(0, \mathbf I_d) \tag{4} \end{align} $ where: - (2) Multiply both sides with $\Sigma^{-1/2}$. - (3) Apply [[Covariance Matrix#Affine Transformation|Affine Transformation to Covariance Matrix]]. - (4) Since a covariance matrix is always positive definite, the variance of $\mathcal N_d$ simplifies to the identity $\mathbf I_d$.