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It is well known that if $X \sim B(m, p)$ and $Y \sim B(n, p)$ are independent then $X+Y \sim B(m+n, p)$ but what is the distribution of $X-Y$?

Here is what I have tried. $\Pr[X-Y = c] = \sum_{i=0}^n \Pr[X = c+i] \Pr[Y = i]$ which can be simplified to $\left(\frac{p}{1-p}\right)^c (1-p)^{m+n} \sum_{i=0}^n \binom{m}{c+i} \binom{n}{i} \left(\frac{p}{1-p}\right)^{2i}$. Letting $q = \left(\frac{p}{1-p}\right)^2$, simplifying this expression reduces to simplifying $\sum_{i=0}^n \binom{m}{c+i} \binom{n}{i} q^i$. But I am stuck here and perhaps there is a better approach.

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marked as duplicate by Surb, Joel Reyes Noche, kingW3, quid, Namaste Mar 2 '15 at 13:01

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