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The distribution of the sum of two (or more) independent Poisson distributed random variables is again a Poisson distribution (Poisson Distribution of sum of two random independent variables $X$, $Y$).

The distribution of the difference of two independent Poisson distributed random variables is also an analytically known distribution, the Skellam distribution.

Is there also an analytical result for the distribution corresponding to the sum of two (or more) independent Skellam distributed random variables?

Wolpertinger
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1 Answers1

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Let $X,Y$ be independent Skellam-distributed rvs with respective parameters $\mu_{1},\mu_2$ and $\nu_1,\nu_2$. Then by independence we have

$$\begin{aligned}E[e^{i\xi (X+Y)}]&=E[e^{i\xi X}]E[e^{i\xi Y}]=\\ &=(e^{-(\mu_1+\mu_2)+\mu_1e^{i\xi }+\mu_2e^{-i\xi}})(e^{-(\nu_1+\nu_2)+\nu_1e^{i\xi }+\nu_2e^{-i\xi}})=\\ &=e^{-(\mu_1+\nu_1+\mu_2+\nu_2)+(\mu_1+\nu_1)e^{i\xi}+(\mu_2+\nu_2)e^{-i\xi}}\end{aligned}$$ So $Z=X+Y$ is Skellam-distributed with parameters $\mu_1+\nu_1,\mu_2+\nu_2$. This generalizes to arbitrary finite sums of independent Skellam rvs. Also, the difference of two independent Skellam-distributed rvs is Skellam distributed with parameters $\mu_1+\nu_2,\mu_2+\nu_1$.

Snoop
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