Suppose that $X_1,\ldots,X_n$ are iid $N(0,1)$ random variables. Consider the random variable given by $$ \xi_n =\Bigl|\frac1{\sqrt{n}}\sum_{t=1}^nX_t\Bigr|^2-\frac1n\sum_{t=1}^nX_t^2 =\frac1n\sum_{s\ne t}X_sX_t. $$ What is the distribution of $\xi_n$?

My observations are as follow.

- We have that $n^{-1/2}\sum_{t=1}^nX_t\sim N(0,1)$ and $|n^{-1/2}\sum_{t=1}^nX_t|^2\sim\chi^2(1)$, where $\chi^2(k)$ is the chi-square distribution with $k$ degrees of freedom.
- $\sum_{t=1}^nX_t^2\sim\chi^2(n)$ and $n^{-1}\sum_{t=1}^nX_t^2\sim\Gamma(k=n/2,\theta = 2n^{-1})$ (see here).
- $|n^{-1/2}\sum_{t=1}^nX_t|^2$ and $n^{-1}\sum_{t=1}^nX_t^2$ are not independent since the covariance between them is equal to $2/n$ (although the covariance goes to $0$ as $n\to\infty$).
- $n^{-1}\sum_{t=1}^nX_t^2\to 1$ in probability as $n\to\infty$ using the law of large numbers and hence it follows that $\xi_n\to \chi^2(1)-1$ in distribution as $n\to\infty$ using Slutsky's theorem. However, I am interested if the exact distribution has some manageable form when $n$ is some fixed positive integer.

If $|n^{-1/2}\sum_{t=1}^nX_t|^2$ and $n^{-1}\sum_{t=1}^nX_t^2$ were independent, then we could try to use the answers to this question to derive the exact distribution. But they are not independent. Maybe Cochran's theorem might be of use here.

Does the exact distribution of $\xi_n$ have some manageable expression?

Any help is much appreciated!