Compute $Var[\frac{1}{n}\sum_{i=1}^n (X_i-E[X])^2]$ Let $X_1,\ldots, X_n$ be positive random variables in the range $[a, b]$ and $\sum_i^n X_i = 1$.
I want to compute
\begin{eqnarray}
Var[\frac{1}{n}\sum_{i=1}^n (X_i-E[X])^2] 
&=& \frac{1}{n^2}Var[\sum_{i=1}^n (X_i-E[X])^2]\\
&=& \frac{1}{n^2}\sum_{i=1}^n Var[X_i^2] - 4Var[X_i E[X]] + Var[E[X]^2]\\
\end{eqnarray}
Here I got stuck. Are there any approximations I can use to go further or to at least estimate an upper bound for the variance?

I simplify the term as far I can think of
$$\frac{1}{n^2}Var[\sum_{i=1}^n (X_i-E[X])^2]$$
Next you multiply out the square brackets and leaving variance opterator  still out of the brackets.
$$\frac{1}{n^2}Var[\sum_{i=1}^n (X_i^2-2E[X]X_i+E[X]^2)]$$
$$\frac{1}{n^2}Var[\sum_{i=1}^n X_i^2-2E[X]\sum_{i=1}^n X_i+\sum_{i=1}^nE[X]^2)]$$
$E[X]^2$ is a constant and its variance is 0.
$$\frac{1}{n^2}Var[\sum_{i=1}^n X_i^2-2E[X]\sum_{i=1}^n X_i+n\cdot E[X]^2)]$$
$$\frac{1}{n^2}[Var\left(\sum_{i=1}^n X_i^2\right)-4E[X]^2Var\left(\sum_{i=1}^n X_i\right)]$$
Since $\sum_{i=1}^n X_i=1$, we have a variance of a constant again.
$$\frac{1}{n^2}\cdot Var\left(\sum_{i=1}^n X_i^2\right)$$
To calculate/estimate it explicity further informations are needed.
