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  • $\begingroup$ Very good answer. What I wanted to propose are probability generating functions. In this case for $X_i$: $g_i(s) = E[s^{X_i}] = (1-p_i) + p_is$ and the generating function of the sum is $g(s) = \prod_{i=1}^n g_i(s)$. Taking the nth derivative wrt $s$ evaluated at $0$ gives all the point probabilites. Finally sum them up for the cdf. I just can not offer code for this, so your answer can be readily applied. $\endgroup$ Commented Feb 7, 2013 at 8:50
  • $\begingroup$ @Richard The pgf approach works just fine--in fact, I used that method (in Mathematica, where it's a one-liner) to verify my example here. But under the hood, the product is carried out as a convolution, so it amounts to the same solution. BTW, there's no need to take a derivative: you only need to expand the product as a polynomial in $s$ and pick out the probabilities from the coefficients. $\endgroup$ Commented Feb 7, 2013 at 16:38
  • $\begingroup$ yes, of course the derivatives are just there to extract the coefficients. And yes - in R, it has to be the way you presented it. I just thought I write down the pgf approach for its mathematical beauty ;) $\endgroup$ Commented Feb 7, 2013 at 17:05