$X_1,...,X_n\sim N(0,\theta)$, define: $$ V= \frac{(n-2)\sum_{i=1}^nX_i^2+(n\bar{X})^2}{n(n-1)} $$
I realized that $Y=\sum_{i=1}^{n} X_{i}^{2} \sim \theta \chi^{2}(n), \quad Z= \bar{X}^{2} \sim \frac{\theta}{n} \chi^{2}(1)$, then the question can be simplified: $$ Cov(V)=\frac{(n-2)^2}{(n(n-1))^2}Var(\theta Y) +\frac{n^4}{(n(n-1))^2)}Var(\frac{\theta}{n}Z) +\frac{n^2(n-2)}{(n(n-1))^2}\cdot \theta \cdot \frac{\theta}{n}Cov(Y,Z). $$
Two variances $Var(\theta Y)$ and $Var(\frac{\theta}{n}Z)$ are easy to calculate, but I can't figure out how to calculate $Cov(Y,Z)$.
As I know, $Cov(Y,Z)=E(YZ)-E(Y)E(Z)$, it's easy to get $E(Y)E(Z)$, but what is troubling me most is $E(YZ)$.
Can anyone help me? Thanks a lot !!
Sketch (almost solution):
$cov(\sum_{i=1}^{n} X_{i}^{2}, \bar{X}^2) = \sum_{i=1}^{n} cov(X_{i}^{2}, \bar{X}^2) = n \cdot cov(X_{1}^{2}, \bar{X}^2)$ because $X_i$ are i.i.d.
Further $$cov(X_{1}^{2}, \bar{X}^2) = cov(X_{1}^{2}, (\sum_{i=1}^{n} X_{i}^{2} )^2 ) = cov(X_{1}^{2}, (\sum_{i=1}^{n} X_{i}^{2} )\cdot ( \sum_{j=1}^{n} X_{j}^{2} )) $$ $$= cov(X_{1}^{2}, \sum_{i,j=1}^{n} X_{i}^{2}X_j^2 ) = \sum_{i,j=1}^{n} cov(X_{1}^{2}, X_{i}^{2}X_j^2 )$$ Hence it's sufficient to find $cov(X_{1}^{2}, X_{i}^{2}X_j^2 )$ If $i, j \ge 2$ then $cov(X_{1}^{2}, X_{i}^{2}X_j^2 ) = 0$ because $X_{1}^{2}$ and $X_{i}^{2}X_j^2$ are independent. If $i=1, j>1$ (similarly $j=1<i$) we have $cov(X_{1}^{2}, X_{1}^{2}X_i^2 ) = EX_1^4 X_i^2 - EX_1^2 E(X_{1}^{2}X_i^2 ) = EX_1^4 EX_i^2 - EX_i^2 (EX_1^2)^2$ by independence. If $i=j=1$ then it's easy to find $cov(X_{1}^{2}, X_{1}^{2}X_1^2 )$.