The Gaussian sample mean is independent of the Wishart sample scatter

Define $Y_i={\mathrm{\Sigma }}^{-1\mathrm{/}2}(X_i-\mu )$. The positive root exists by [F10], and if ${\mathrm{\Sigma }}^{1\mathrm{/}2}v=0$ then $v^T\mathrm{\Sigma }v=\mathrm{\parallel }{\mathrm{\Sigma }}^{1\mathrm{/}2}v{\mathrm{\parallel }}^2=0$, so $\mathrm{\Sigma }\succ 0$ makes the root invertible. Whitening is a continuous, hence Borel, map by [F9]; [F6] therefore preserves independence of the sample vectors. By [F2], each $Y_i\sim N_p(0,I_p)$ has independent standard-normal coordinates, and [F11] gives the product joint law across $i$. Thus all $np$ entries $Y_{ia}$ are independent standard normals. Let $u=n^{-1\mathrm{/}2}(1,\dots ,1{)}^T\in {\mathbb{R}}^n$. By [F3], complete $u$ to an orthonormal basis $u_1=u,u_2,\dots ,u_n$ of ${\mathbb{R}}^n$. For each coordinate $a\in \{1,\dots ,p\}$, the sample-coordinate vector $(Y_{1a},\dots ,Y_{na}{)}^T$ is therefore standard normal in ${\mathbb{R}}^n$, and these $p$ vectors are independent.

Let $Y=[Y_1\mathrm{\mid }\cdots \mathrm{\mid }{Y}_n]$ and, for each $j,a$, define $Y_{ja}^{\mathrm{\prime }}=u_j^T(Y_{1a},\dots ,Y_{na}{)}^T$. Applying the orthogonal invariance [F4] separately to the $p$ independent coordinate vectors gives independent standard-normal entries within each transformed coordinate vector; [F6] preserves independence between those $p$ vectors under their separate orthogonal maps. Thus all $np$ entries $Y_{ja}^{\mathrm{\prime }}$ are independent standard normals, and the $p$-vectors $Y_1^{\mathrm{\prime }},\dots ,Y_n^{\mathrm{\prime }}$ formed from their coordinates are independent $N_p(0,I_p)$ vectors.

Since $u_1=u$, $Y_1^{\mathrm{\prime }}=\sqrt{n}{n}^{-1}{\sum }_i{Y}_i=\sqrt{n}{\mathrm{\Sigma }}^{-1\mathrm{/}2}(\bar{X}-\mu )$. Its law and [F2] give $\sqrt{n}(\bar{X}-\mu )\sim N_p(0,\mathrm{\Sigma })$. The vector $Y_1^{\mathrm{\prime }}$ is independent of $(Y_2^{\mathrm{\prime }},\dots ,Y_n^{\mathrm{\prime }})$ by [F5].

The matrix $U=[u_1\mathrm{\mid }\cdots \mathrm{\mid }{u}_n]$ is orthogonal and $u_1=u$. Hence $Y{Y}^T$ splits as $Y{Y}^T=Yu{u}^T{Y}^T+{\sum }_{j=2}^{n}Y{u}_j{u}_j^T{Y}^T$. The first term is $n\bar{Y}{\bar{Y}}^T$, and the remaining terms are $Y_j^{\mathrm{\prime }}{Y}_j^{\mathrm{\prime }T}$, so ${\sum }_i(Y_i-\bar{Y})(Y_i-\bar{Y}{)}^T={\sum }_{j=2}^n{Y}_j^{\mathrm{\prime }}{Y}_j^{\mathrm{\prime }T}$. Since $X_i-\bar{X}={\mathrm{\Sigma }}^{1\mathrm{/}2}(Y_i-\bar{Y})$, it follows that $(n-1)S={\mathrm{\Sigma }}^{1\mathrm{/}2}({\sum }_{j=2}^n{Y}_j^{\mathrm{\prime }}{Y}_j^{\mathrm{\prime }T}){\mathrm{\Sigma }}^{1\mathrm{/}2}$. The independent vectors in [F5] and Wishart definition [F7] give $(n-1)S\sim W_p(\mathrm{\Sigma },n-1)$. The sample mean is a function of $Y_1^{\mathrm{\prime }}$, while $S$ is a function of $Y_2^{\mathrm{\prime }},\dots ,Y_n^{\mathrm{\prime }}$; the maps are Borel by [F9], so [F6] gives $\bar{X}\perp S$.

If $n>p$, then $n-1\ge p$ and the residual vectors $Y_2^{\mathrm{\prime }},\dots ,Y_n^{\mathrm{\prime }}$ are independent standard-normal columns. By [F8], their $p\times (n-1)$ matrix has row rank $p$ almost surely. Its sum of column outer products is therefore positive definite; the congruence in step 4.1 and $n-1>0$ show $S\succ 0$ almost surely.