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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.

$$X = X_1+X_2+\dots+X_N$$

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Hi, welcome to Math.SE. Try to format your math by putting $$ around what you want formatted. –  Sujaan Kunalan Aug 25 '13 at 20:59
    
Well the permissible values are $\mathbb{Z}\cap[-N,N]$, to begin with. What is the probability that the sum evaluates to each such value? –  Jonathan Y. Aug 25 '13 at 21:18
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It's a discrete distribution, so "pmf" might be a more appropriate term. One can call it a pdf with respect to counting measure, but "discrete" is synonymous with saying it's characterized by a pmf, assigning probabilities to individual points rather than to more complicated sets of points. –  Michael Hardy Aug 25 '13 at 22:12

1 Answer 1

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.

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I think $k-l+m=n$ and in the summation limit it should be $k+l+m=N$,isn't it? –  hhsaffar Nov 30 '13 at 15:31

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