Given that $X_1,...,X_n$ is an i.i.d sample and its sample mean is $\overline X_n$, I have to prove the following equation:
\begin{equation*} \frac{n-1}{n} \sum_{i=1}^n(\overline{X}_{n-1,i}^2 - \overline{X}_n^2)^2 - \frac{\hat{\alpha}_2^2}{n^2(n-1)} = \frac{4 \overline X_n^2 \hat \alpha_2}{n} - \frac{4 \overline X_n \hat \alpha_3}{n(n-1)} + \frac{\hat \alpha_4}{n(n-1)^2}- \frac{\hat{\alpha}_2^2}{n^2(n-1)}, \end{equation*}
where $\overline X_{n-1,i} = \frac{1}{n-1} \sum_{j\neq i}^{n} X_j$, $\overline{X}_n^2$ is the square of the sample mean, and $\hat \alpha_k = \frac{1}{n-1} \sum_{i=1}^n (X_i - \overline X_n)^k$.
Can you help me proving this? Thanks