Let $X_1,\dots,X_n$ be i.i.d $N(0,1)$ random variables. Let $\overline{X_k}=\frac{1}{k}\sum_{i=1}^{k}X_i$, $\overline{X_{n-k}}=\frac{1}{n-k}\sum_{i=k+1}^{n}X_i.$
What is the distribution of $\frac{1}{2}(\overline{X_k}+\overline{X_{n-k}})$?
My attempt:
$\sum_{i=1}^{k}X_i\sim N(0,k)$
$\sum_{i=k+1}^{n}X_i\sim N(0,n-k-1)$
$\overline{X_k}\sim N(0,\frac{1}{k})$
$\overline{X_{n-k}}\sim N(0,\frac{n-k-1}{(n-k)^{2}})=N(0,\frac{1}{n-k}-\frac{1}{(n-k)^2})$
$(\overline{X_k}+\overline{X_{n-k}})\sim N(0,\frac{1}{k}+\frac{1}{n-k}-\frac{1}{(n-k)^2})=N(0,\frac{n}{k(n-k)}-\frac{1}{(n-k)^2})=N(0,\frac{n(n-k)-k}{k(n-k)^2}) $
$\frac{1}{2}(\overline{X_k}+\overline{X_{n-k}})\sim N(0,\frac{n}{k(n-k)}-\frac{1}{(n-k)^2})=N(0,\frac{n(n-k)-k}{4k(n-k)^2}) $
Is this correct? Is it possible to express the answer in terms of another distribution (gamma, t, F, $\chi^2$, etc.)?
Note that $$\frac{X_{1}+\ldots+X_{k}}{k} \sim N \left( 0,\frac{1}{k} \right)$$
and
$$\frac{X_{k+1}+\ldots+X_{n}}{n-k} \sim N \left( 0,\frac{1}{n-k} \right)$$
So
$$\frac{X_{1}+\ldots+X_{k}}{k}+\frac{X_{k+1}+\ldots+X_{n}}{n-k} \sim N \left( 0,\frac{1}{k}+\frac{1}{n-k} \right)$$
In particular when $n=2k$,
$$\frac{1}{2} \left( \frac{X_{1}+\ldots+X_{k}}{k}+\frac{X_{k+1}+\ldots+X_{n}}{n-k} \right) \sim N \left( 0,\frac{1}{2k} \right)$$
that is
$$\frac{X_{1}+\ldots+X_{n}}{n} \sim N \left( 0,\frac{1}{n} \right)$$