Consider the sequence $(X_n)_{n \in \mathbb{N}}$ of random variables all with $\mu = 0$ and $\sigma^2 = 1$. Further, we have that $$\text{Cov}(X_i, X_j) = \begin{cases} 0 & \text{if} \: \: |i-j| > 1 \\ 1 & \text{otherwise}\end{cases}$$
Now I want to calculate the variance of $$\overline{X_N} = \frac{1}{N}\sum_{i=1}^N X_i$$
My first thought was that its as simple as calculating $$\text{Var}(\overline{X_N}) = \text{Var}(\frac{1}{N}\sum_{i=1}^N X_i) = \frac{1}{N^2} \cdot N \cdot \text{Var}(X_i) = \frac{1}{N}$$
However, the solution says since we have $\mathbb{E}(X_i) = 0$ and thus $$\text{Var}(X_i) = \mathbb{E}(X_i^2) - \mathbb{E}(X_i)^2 = \mathbb{E}(X_i^2) = 1$$ that we get $$\text{Var}(\overline{X_N}) = \frac{1}{N^2} \mathbb{E}(\overline{X_N}^2)$$ $$= \mathbb{E}(X_1^2 + \cdots + X_N^2 + X_1 \cdot X_2 + X_2 \cdot X_1 + X_2 \cdot X_3 + \cdots + X_{N-1} \cdot X_{N-2} + X_{N-1} \cdot X_{N} + X_{N} \cdot X_{N-1} + Y)$$ where $Y$ is the sum of all products $X_i \cdot X_j$ which do not contribute due to covariance condition. Then finally, $$\text{Var}(\overline{X_N}) = \frac{1}{N^2} \cdot (N + 2(N-2) + 2) = \frac{3N-2}{N^2}$$
I don't really understand the solution, since I don't know how they got this long sum and how they evaluated that. Can someone explain the solution?
Use the definition of Variance of sum of correlated rvs: $$ \mathbf{Var}[X_i + X_j] = \mathbf{Var}X_i + \mathbf{Var}X_j + 2 Cov(X_i,X_j) $$ In your case, a total number of covariance terms is $n-1$, so ther contribution is $2(n-1)$.