Covariance of multinomial distribution Let $X = (X_1,\ldots, X_k)$ be multinomially distributed based upon $n$ trials with parameters $p_1,\ldots,p_k$ such that the sum of the parameters is equal to $1$. I am trying to find, for $i \neq j$, $\operatorname{Var}(X_i + X_j)$. Knowing this will be sufficient to find the $\operatorname{Cov}(X_i,X_j)$.
Now $X_i \sim \text{Bin}(n, p_i)$. The natural thing to say would be that $X_i + X_j\sim \text{Bin}(n, p_i+p_j)$ (and this would, indeed, yield the right result), but I m not sure if this is indeed so.
Suggestions for how to go about this are greatly appreciated!
UPDATE:  @grand_chat very nicely answered the question about the distribution of $X_i + X_j$. How would we go about computing the variance of $X_i - X_j$? As @grand_chat correctly points out, this cannot be binomial because it is not guaranteed to be positive. How, then, should one go about computing the variance of this random variable?
UPDATE 2: The answer in this link answers the question in my UPDATE.
 A: $X_i+X_j$ is indeed a binomial variable because it counts the number of trials that land in either bin $i$ or bin $j$. The $n$ trials are independent, and the probability of "success" is $$P(\text{trial lands in $i$}) + P(\text{trial lands in $j$}) = p_i+p_j.$$
A: There are several ways to do this, but one neat proof of the covariance of a multinomial uses the property you mention that $X_i + X_j \sim \text{Bin}(n, p_i + p_j)$ which some people call the "lumping" property
Covariance in a Multinomial
Given $(X_1,...,X_k) \sim Mult_k(n , \vec{p})$ find $Cov(X_i,X_j)$ for all $i,j$. 
\begin{aligned}
    & If \ i = j, Cov(X_i, X_i) = Var(X_i) = np_i(1 - p_i)
    \\
    \\
    & If \ i \neq j, Cov(X_i, X_j) = C \ \ \text{ i.e. what we are trying to find}
    \\
    \\
 & Var(X_i + X_j) = Var(X_i) + Var(X_j) + 2Cov(X_i, X_j)
 \\
    \\
 & Var(X_i + X_j) = np_i (1 - p_i) + np_j (1 - p_j)+ 2C
 \\
 \\
 &  \text{By the lumping property }  X_i + X_j \sim Bin(n, p_i + p_j)
 \\
 \\
 &  n(p_i + p_j)(1 - (p_i + p_j))  = np_i(1 - p_i) + np_j(1 - p_j) + 2C
 \\
 &  (p_i + p_j)(1 - (p_i + p_j))  = p_i(1 - p_i) + p_j(1 - p_j) + \frac{2C}{n}
 \\
 & C = - n p_i p_j
\end{aligned}
