Variance of sample mean (problems with proof) Assuming that I have $\{x_1,\ldots, x_N\}$ - an iid (independent identically distributed) sample size $N$ of observations of random variable $\xi$ with unknown mean $m_1$, variance (second central moment) $m_{c_2}$ and second raw moment $m_2$. I try to use sample mean $\overline{x}=\frac{1}{N}\sum_{i=1}^Nx_i$ as an estimator of the true mean. 
So I want to find its’ bias and the variance. Everything is simple with bias but not with the variance.
First I use the fact of independency: 
$$\mathbb{var}[\overline{x}]=
\mathbb{var}\left[ \frac{1}{N}\sum_{i=1}^Nx_i \right]\stackrel{\text{iid}}{=}
 \frac{1}{N^2}\sum_{i=1}^N\mathbb{var}\left[x_i \right]=
\frac{m_{c_2}}{N}$$
At the same time I can use the representation of variance in terms of raw first and second moments:
$$\mathbb{var}[\overline{x}]=
\mathbb{E}[(\overline{x}-\mathbb{E}[\overline{x}])^2]=
\mathbb{E}[\overline{x}^2]-\mathbb{E}[\overline{x}]^2=
\mathbb{E}[\overline{x}^2]-m_1^2$$
Then
$$
\begin{eqnarray}
\mathbb{E}[\overline{x}^2]&=&\mathbb{E}\left[\left( \frac{1}{N}\sum_{i=1}^Nx_i \right)\left( \frac{1}{N}\sum_{j=1}^Nx_j \right)\right]= 
\mathbb{E}\left[\frac{1}{N^2}\sum_{i=1}^N\sum_{j=1}^Nx_i x_j \right]=
\\
&=&\mathbb{E}\left[\frac{1}{N^2}\sum_{i=1}^Nx^2_i+\frac{1}{N^2}\sum_{i=1,i\neq j}^N\sum_{j=1}^Nx_i x_j \right]\stackrel{\text{iid}}{=}\frac{m_2}{N}
\end{eqnarray}
$$
So $$\mathbb{var}[\overline{x}]=\frac{m_2}{N}-m_1^2$$ 
Which clearly doesn’t match what I’ve got earlier. So where’s the catch?
 A: $$
[\frac{1}{N^2}\sum_{i = 1}^n\sum_{j \neq i}x_ix_j]
=
\frac{1}{N^2}\sum_{i = 1}^n\sum_{j \neq i}[x_i][x_j]
$$
And this is not equal to zero.
A: Corrigendum
$$\mathbb{var}[\overline{x}]=\mathbb{var}\left[ \frac{1}{N}\sum_{i=1}^Nx_i \right]\stackrel{\text{iid}}{=} \frac{1}{N^2}\sum_{i=1}^N\mathbb{var}\left[x_i \right]=N \cdot \frac{m_{c_2}}{N^2}=\frac{m_{c_2}}{N}$$
$Cov({x_i},{x_j}) = E\left[ {{x_i}{x_j}} \right] - E\left[ {{x_i}} \right]E\left[ {{x_j}} \right] = 0$ Since $\{x_1,\ldots, x_N\}$ iid :
$$\begin{eqnarray}
\mathbb{E}[\overline{x}^2]&=&\mathbb{E}\left[\left( \frac{1}{N}\sum_{i=1}^Nx_i \right)\left( \frac{1}{N}\sum_{j=1}^Nx_j \right)\right]= \mathbb{E}\left[\frac{1}{N^2}\sum_{i=1}^N\sum_{j=1}^Nx_i x_j \right]=\\
&=&\mathbb{E}\left[\frac{1}{N^2}\sum_{i=1}^Nx^2_i+\frac{1}{N^2}\sum_{i=1,i\neq j}^N\sum_{j=1}^Nx_i x_j \right]\stackrel{\text{iid}}{=}\frac{m_2}{N}+\frac{N^2-N}{N^2}m_1^2
\end{eqnarray}$$
$$\mathbb{var}[\overline{x}]=\mathbb{E}[\overline{x}^2]-\mathbb{E}[\overline{x}]^2=\frac{m_2-m_1^2}{N}=\frac{m_{c2}}{N}$$
Updated: reduced to the final form according to the comments.
