Prove that $E (\overline{X} - \mu)^2 = \frac{1}{n}\sigma^2$ How to prove $E (\overline{X} - \mu)^2  = \frac{1}{n}\sigma^2$ (from wiki),
where $\overline{X}$ - is the sample mean ?
What I have so far:
\begin{align}
E (\overline{X} - \mu)^2  = \frac{1}{n}\sigma^2 &=\\ 
&= E (\overline{X}^2 + \mu^2 - 2\mu\overline{X}) = E(\overline{X}^2) - \mu^2 =\\
&= \frac{\sum_{i=1}^n\sum_{j=1}^n E (X_iX_j)}{n^2} - \mu^2
\end{align}
Could you give me some tips how to proceed?
 A: Let $\{X_i\}_{1\le i\le n}$ be $n$ independent random variables with mean $\mu$ and variance $\sigma^2$. We need 2 results
1) $var(aX_i) = a^2 \times var(X_i)$
2) $var(\sum_{i=1}^n X_i) = \sum_{i=1}^n var(X_i)$.
Also, Expectation is a linear operator. That is
${E}[aX+bY] = a{E}[X] + b{E}[Y]$
Hence, $E[\bar{X}] = E[\frac{\sum_{i=1}^n X_i}{n}] = \frac{1}{n}\sum_{i=1}^nE[X_i] = \frac{1}{n}\sum_{i=1}^n \mu = \mu\,.$  
Hence, 
\begin{align}
E(\bar{X}-\mu)^2 &= E(\bar{X} - E[\bar{X}])^2 = var(\bar{X}) \\
&=var\left(\frac{\sum_{i=1}^n X_i}{n}\right) \\
&=\frac{1}{n^2}var\left(\sum_{i=1}^n X_i\right) \\
&=\frac{1}{n^2}\sum_{i=1}^n var\left( X_i\right) \\
&=\frac{n\sigma^2}{n^2}\\
&=\frac{\sigma^2}{n}
\end{align}
A: We know that $E(X^2) = [E(X)]^2 + \text{Var}(X)$, so
\begin{align*}
E((\overline{X} - \mu)^2) &= [E(\overline{X} - \mu)]^2 + \text{Var}(\overline{X} - \mu) \\
&= \text{Var}(\overline{X}) \\
&= \text{Var}\left(\frac{1}{n}(X_1 + \cdots + X_n) \right) \\
&= \frac{n\sigma^2}{n^2},
\end{align*}
where we use the following:
$$E(\overline{X}) = \mu, \qquad Var(a + X) = Var(X), \qquad \text{Var}(aX) = a^2\text{Var}(X), $$
and the independence of the $X_i$ so that the variance adds linearly.
