1

A follow-up to the following question: Sum of two independent Skellam distributed random variables

Is there an analytical result for the distribution corresponding to the sum of two independent random variables, where one is Poisson distributed and the other is Skellam distributed?

Wolpertinger
  • 131
  • 1
  • 12
  • There is a small mistake: $e^{-(\mu_1+\nu_1+\lambda)(...)}$. You can then conclude :) – Snoop Apr 06 '23 at 10:01
  • @Snoop oh boy, that's embarrassing... thanks! Also thanks again for the last answer, I did not know the method of characteristic functions. Very pretty :) – Wolpertinger Apr 06 '23 at 10:10

1 Answers1

1

After Snoops correction, I decided to turn my attempt into an answer:

As in Sum of two independent Skellam distributed random variables, one can proceed via characteristic functions as follows

$$\begin{aligned}E[e^{i\xi (X+Y)}]&=E_\mathrm{Skellam}[e^{i\xi X}]E_\mathrm{Poisson}[e^{i\xi Y}]=\\ &=e^{-(\mu_1+\mu_2)+\mu_1e^{i\xi }+\mu_2e^{-i\xi}} \,e^{\lambda(e^{i\xi}-1)}=\\ &=e^{-(\mu_1+\mu_1+\lambda)+(\mu_1+\lambda)e^{i\xi}+\mu_2e^{-i\xi}} \,.\end{aligned}$$

The result is then again a Skellam distribution with $\tilde{\mu_1}=\mu_1+\lambda$ and $\tilde{\mu_2} = \mu_2$.

I guess one can also understand this intuitively, since the Skellam distribution is obtained as the difference of two Poisson processes in the first place. So one really just adds two Poisson processes here and then subtracts another.

Wolpertinger
  • 131
  • 1
  • 12