Consider the following exercise taken from Probability-1 by Shiryaev:

Assuming that $\xi_1$ and $\xi_2$ are two independent Poisson random variables with parameters, respectively, $\lambda_1 > 0$ and $\lambda_2 > 0$, prove that $\xi_1 + \xi_2$ has Poisson distribution with parameter $\lambda_1 + \lambda_2$.

The exercise can be found on p. 294. Since the lecture I am attending (probability 1) is quite bad, I do self-teaching again. But I am a bit stuck on this exercise. I googled, but I think I need some more specific help understanding the concepts and not just the solution. The definition of a Poisson random variable is the following:

Let $X$ be a random variable taking values $k = 0,1,2,\dots$ with probabilities $p_k$. $X$ is a Poisson random variable if $$p_k = e^{-\lambda}\frac{\lambda^k}{k!} \qquad k = 0,1,2,\dots$$

On p. 291 I found the following (the question is from this chapter):

The distribution function of the sum of two independent random variables is the convolution of their distribution functions.

So I am asking myself: Is this useful here? How can I calculate the distribution function of the Poisson distribution? I know I am lacking knowledge, but I think it is hard to learn this by myself. Thanks for your time.

  • 2
    $\begingroup$ Why not just write it out? The probability that the sum will have value $k$ is $\sum_{i=0}^k e^{-\lambda_1}\frac {\lambda_1^i}{i!}\times e^{-\lambda_2}\frac {\lambda_2^{k-i}}{(k-i)!}$. Now just invoke the binomial theorem. $\endgroup$ – lulu Mar 10 '17 at 21:24
  • $\begingroup$ @lulu Thanks a lot! Yes, I confused myself a little bit. $\endgroup$ – TheGeekGreek Mar 10 '17 at 21:30

The simplest way I see is to use characteristic functions (if not for that particular problem, for many others that will arise when studying convergence of random variables). For a real-valued random variable $X$, the characteristic function $\phi_X\colon\mathbb{R}\mapsto \mathbb{C}$ is defined by $$ \phi_X(t) = \mathbb{E}[e^{itX}] $$ and fully characterizes the distribution of $X$.

Now, given the characteristic functions of a Poisson random variable, we can write, for any $t\in\mathbb{R}$, $$\begin{align} \phi_{\xi_1+\xi_2}(t) &= \mathbb{E}[e^{it(\xi_1+\xi_2)}] = \mathbb{E}[e^{it\xi_1}e^{it\xi_2}] = \mathbb{E}[e^{it\xi_1}]\mathbb{E}[e^{it\xi_2}] \tag{Independence} \\ &= \phi_{\xi_1}(t)\phi_{\xi_2}(t) = e^{\lambda_1(e^{it}-1)}e^{\lambda_2(e^{it}-1)} \tag{Known expression}\\ &= e^{\lambda_1(e^{it}-1)+\lambda_2(e^{it}-1)} = e^{(\lambda_1+\lambda_2)(e^{it}-1)} = e^{\lambda'(e^{it}-1)} \end{align}$$ for $\lambda'\stackrel{\rm def}{=} \lambda_1+\lambda_2$. Since this last expression, $e^{\lambda'(e^{it}-1)}$, is the expression of the characteristic function of a Poisson random variable with parameter $\lambda'$, we can conclude that $\xi_1+\xi_2$ is Poisson with parameter $\lambda'$.

Additional: if you want to prove that $t\mapsto e^{\lambda(e^{it}-1)}$ is the expression of the characteristic function a Poisson distributed r.v. $X$ with parameter $\lambda$ (instead of saying "it is standard"), here is the derivation: for $t\in\mathbb{R}$ $$ \phi_X(t) = \mathbb{E}[e^{itX}] = \sum_{n=0}^\infty e^{itn} \frac{e^{-\lambda} \lambda^n}{n!} = e^{-\lambda} \sum_{n=0}^\infty \frac{(e^{it} \lambda)^n}{n!} = e^{-\lambda} e^{e^{it} \lambda} = e^{\lambda(e^{it} -1)} $$ using the fact that $e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$ for $z\in\mathbb{C}$.

  • $\begingroup$ How do we solve with moment generating function?@Clement C. $\endgroup$ – Pole_Star Oct 17 '20 at 14:11
  • $\begingroup$ @Pole_Star Try to mimic the proof which uses the characteristic function. This should go through (the MGF and CF have very similar properties). $\endgroup$ – Clement C. Oct 17 '20 at 16:19

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