Variance of $X_{i}+X_{j}$ if $X_{1} + \ldots + X_{N}=1$ I have random variables $X_{1}, X_{2}, \ldots, X_{N}$, where $X_{i} \in \{0,1\}$ and
$$X_{1} + \ldots + X_{N}=1$$
I.e. exactly one of the $X_{i}$'s are $1$, and the rest are $0$. 
In addition, denote $P(X_{i} = 1) = p_{i}$, where $\sum_{i} p_{i}=1$. I do not want to assume that the $X_{i}$'s are identically distributed, meaning we may have $p_{i} \neq p_{j}$ for some $i\neq j$. 
I now want to calculate the variance of $X_{i}+X_{j}$ for any $i \neq j$, but am having some difficulty. Can anyone provide a formula and an explanation of how to do so for this example? Much appreciated!

*

*$X_i + X_j$ takes values in $\{0,1\}$, so is a Bernoulli random variable.


*

*This is because at most one of the $X_k$ is equal to $1$.


*$\mathbb{P}(X_i + X_j = 1) = \mathbb{P}(X_i =1 \text{ or } X_j = 1) = \mathbb{P}(X_i = 1) + \mathbb{P}(X_j = 1) = p_i + p_j$


*

*The second equality holds because the events $\{ X_i = 1\}, \{ X_j = 1\}$ are disjoint.


*Thus, $X_i + X_j \sim \text{Ber}(p_i + p_j)$

*If $Y \sim \text{Ber}(q)$, then $\text{Var}(Y)=q(1-q)$

*Thus, the variance of $(X_i + X_j)$ is $(p_i + p_j) \cdot (1 - p_i - p_j)$

