$(n-1)S\sim W_p(n-1,\Sigma)$, so $S\sim W_p(n-1,\Sigma/(n-1))$

If $\underline x_1,\dots,\underline x_n$ ($n\ge p+1$) are distributed independently each according to $N_p(\underline\mu,\Sigma)$, then the distribution of $S=\frac1{n-1}\sum_{\alpha=1}^n(\underline x_\alpha-\bar{\underline x})(\underline x_\alpha-\bar{\underline x})'$ is: