Is there a probabilistic (not analytical) argument for why the sum of independent Poissons is Poisson? I was just wondering if there was a simple probabilistic argument for why a Poisson random variable with parameter $\lambda_1$ plus a Poisson random variable with parameter $\lambda_2$ (both independent) is a Poisson random variable with the sum of the parameters as a parameter.
I know one exists for binomial but can't think of one for Poisson.
 A: As Glen_b noted in comments, think of a Poisson variable as counting the number of events for a Poisson process of intensity $\alpha$ in a fixed window of size $w$: the Poisson parameter for this variable is $\alpha w$.
Suppose now there are two independent Poisson processes separately running in the same window, one of intensity $\alpha_1$ with $\alpha_1 w = \lambda_1$ and the other with intensity $\alpha_2$ with $\alpha_2 w = \lambda_2.$  

This figure shows realizations of two independent Poisson processes on the line within the window displayed as a gray line segment: one is shown as the orange points and the other as blue points.  Their counts are realizations of two Poisson variables.  By ignoring the colors, as shown at the bottom, we obtain a realization of a Poisson process whose intensity is the sum of the original intensities.  The count is the sum of the counts.
The sum of these counts clearly is a Poisson process of intensity $\alpha_1 + \alpha_2$ because it trivially satisfies all the requirements of Poisson process (the count is proportional to the size of the window, events in non-overlapping windows are independent, and the chance of a count exceeding $1$ grows vanishingly small as the window size shrinks to zero).  Therefore the sum of counts is a Poisson random variable with parameter $(\alpha_1+\alpha_2)w = \lambda_1+\lambda_2,$ QED.
A: If you're satisfied with the fact that binomials sum to binomials when their parameter $p$ is the same, then you can view Poisson distributions as binomials with rare success that scales like $p\approx c/n$. This correspondence is exact in the limit so two Poissons sum to another Poisson immediately follows from the binomial case. In the case where the binomials have different probabilities $p_1,p_2$, they sum to a binomial with an adjusted parameter, whose adjustment approaches $p_1+p_2$ as both $p$'s get very small.
