Let $X$ be some random variable; we will take different measurements of this same variable $(X_1, X_2, ... X_N)$; distribution of each random variable then is identical, but variables are not independent; I am now trying to show that:
$$\mathbb{V}[\bar{X}] = \frac{1}{N^2}\sum_{n,m=1}^N Cov (X_n,X_m)$$
The above is the result from the book, whereas I have:
$$\mathbb{E}[\bar{X}^2]-\mathbb{E}[\bar{X}]^2 = \frac{1}{N^2} \left[ \mathbb{E}\left[ \sum_{i=1}^N X_i^2 \right] +2\mathbb{E}\left[ \sum_{i=1}^{N-1} \sum_{i<m} X_i X_m \right] - \mathbb{E} \left[\sum_{i=1}^N X_i\right]^2 \right]$$
Note that $$\mbox{Var}[\overline{X}] = \frac{1}{N^2} \mbox{Var}\left[\sum_{i=1}^{N} X_i \right] = \frac{1}{N^2} \mbox{Cov}\left(\sum_{i=1}^{N} X_i, \sum_{j=1}^{N} X_j\right).$$