I would like to prove the distribution of $S_n$ being the sum of $X_n$ Poisson distributed with mean $\mu$ is the sum of the means of the individual distribution, as stated by @Brian Tung in a comment, in this question it is calculated by the convolution of two independent Poisson distributions. So my try:

The pdf of a Poisson distribution is:

$$ \displaystyle P(X_n = k) = e^{-\mu} \frac{\mu^k}{k!}, \; \; k = 0,1,\dots $$

The Convolution of a function is defined as:

$$ \displaystyle (f \star g)(t) = \int_{-\infty}^{\infty} f(\tau)g(t-\tau) d \tau $$

Then applying this here get:

$$ \displaystyle P(S_n = k) = (P(S_n = \tau) \star P(S_n = \tau))(t) = \int_{-\infty}^{\infty} e^{-\mu} \frac{\mu^{\tau}}{\tau!} e^{-\mu} \frac{\mu^{\tau - t}}{(\tau-t)!} d \tau $$

$$ \displaystyle = \int_{-\infty}^{\infty} \frac{e^{-2 \mu} \mu^{-t}}{\Gamma (\tau+1) \Gamma (\tau-t+1} d\tau $$

And I'm totally stuck .. I think i haven't even started this correct?

Continuing from @echzhen's answer:

$$ \displaystyle = \sum_{i=0}^{k} e^{\lambda_2} \frac{\lambda_2^{k-i}}{(k-i)!} e^{\lambda_1} \frac{\lambda_1^{i}}{(i)!} $$

$$ \displaystyle = e^{\lambda_1 + \lambda_2} \sum_{i=0}^k \frac{1}{k!} \frac{k!}{(k-i)!(i)!} \lambda_1^i \lambda_2^{k-i} $$

Can we bring the $1/k!$ out infront of the sum? If so, then we apply the binomial theorem:

$$ \displaystyle = e^{\lambda_1 + \lambda_2} \frac{1}{k!} \sum_{i=0}^k {k \choose i} \lambda_1^i \lambda_2^{k-i} $$

$$ \displaystyle = e^{\lambda_1 + \lambda_2} \frac{1}{k!} (\lambda_1 + \lambda_2)^{k} $$

  • $\begingroup$ First notice, that if you proved that $X_1 + X_2$ is Poisson with $\lambda = \lambda_1 + \lambda_2$ you can extend this result to any finite sum. $\endgroup$ – echzhen Feb 24 '16 at 19:55
  • $\begingroup$ I added this proof, is it correct? .. although I only made this step because I know what the conclusion should be, how can I know before hand to plug in $\mu_1 + \mu_2$ into where the mean goes? $\endgroup$ – Sunhwa Feb 24 '16 at 20:02
  • $\begingroup$ On the one hand it is correct to put $\mu_1+\mu_2$, but on the other hand it is not a proof :) $\endgroup$ – echzhen Feb 24 '16 at 20:05
  • $\begingroup$ I am still interested in seeing the proof using convolutions .. but it must be beyond me, as I had trouble with this proof for $n=2$ >.< maybe in a couple years $\endgroup$ – Sunhwa Feb 24 '16 at 20:31
  • $\begingroup$ Don't you think that with convolution the proof is basically the same? Just compare the sum that you have obtained and the integral. $\endgroup$ – echzhen Feb 24 '16 at 20:37

Let $X_1, X_2$ be independent Poisson random variables with intensity $\lambda_1, \lambda_2$ respectfully. Let $S = X_1+X_2$, hence $$P\{X_1+X_2 = k\}=\sum\limits_{i=0}^{\infty}P\{X_1+X_2=k|X_1=i\}P\{X_1 = i\} =\sum\limits_{i=0}^{k}P\{X_2=k-i\}P\{X_1 = i\}$$ can you go from here?

  • $\begingroup$ How can you fix the sum finite in the last step here? .. because you first fix $X_1$ then $X_2$? $\endgroup$ – Sunhwa Feb 24 '16 at 20:32
  • 1
    $\begingroup$ Because $P\{X_2 < 0\} = 0$ $\endgroup$ – echzhen Feb 24 '16 at 20:35

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