1
$\begingroup$

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.

$\endgroup$
  • 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
2
$\begingroup$

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 $z\in\mathbb{C}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.