TITLE: Sum of N i.i.d. random variables
QUESTION [6 upvotes]: Suppose I have a random variable $X_i$ with pdf
$$X_i = \begin{cases}1 & P(X_i=1)=p\\-1 & P(X_i=-1)=q\\0 & P(X_i=0)=1-p-q\end{cases}$$
What is the pdf of sum of $N$ such i.i.d. random variables, i.e.
$$X = X_1+X_2+\dots+X_N$$

REPLY [2 votes]: For $n\in\{-N,-N+1,-N+2,\ldots,-2,-1,0,1,2,\ldots, N-2,N-1,N\}$, we have
$$
\Pr(X=n) = \sum_{k,\ell,m\,:\,k+\ell+m=n} \frac{n!}{k!\ell!m!} p^k q^\ell (1-p-q)^m.
$$
In one sense, that's an answer.  One could wonder about a combinatorial problem: How many triples $(k,\ell,m)$ are there for which $k+\ell+m=n$?  And does the sum admit useful simplifications?  But I'm going to leave this answer possibly less than complete for now.