Let $\mathbf{X}=(X_1,\dots,X_p)\sim\mathcal{N}(\mu,\Sigma)$ be a Gaussian random vector. We all know that
$$d^2(\mathbf{X},\mu) = (\mathbf{X}-\mu)^T\Sigma^{-1}(\mathbf{X}-\mu) $$
has a $\chi^2_p-$distribution. We also know how the distribution changes when we don't know $\mu$ and $\Sigma$, but we have sample estimates $\bar{\mathbf{X}}$ and $S$. I wonder what is the distribution of this other quantity: if $\mathbf{X_1}$ and $\mathbf{X_2}$ are two independent draws from $\mathcal{N}(\mu,\Sigma)$, then how is
$$d^2(\mathbf{X_1},\mathbf{X_2}) = (\mathbf{X_1}-\mathbf{X_2})^T\Sigma^{-1}(\mathbf{X_1}-\mathbf{X_2}) $$
distributed? Is it $\chi^2_{2p}-$distributed?
Finally, if $D=\{\mathbf{X_1},\dots,\mathbf{X_n}\}$ is a random sample of size $n$ from $\mathcal{N}(\mu,\Sigma)$, and $\mathbf{X_i}$ and $\mathbf{X_j}$ are observations in the sample $D$, how is
$$d^2(\mathbf{X_i},\mathbf{X_j}) = (\mathbf{X_i}-\mathbf{X_j})^TS^{-1}(\mathbf{X_i}-\mathbf{X_j}) $$
distributed ($S$ is obviously the sample covariance matrix)?
