What's the PMF of the Difference between 2 Independent Poisson RV?

Question

I searched around and found that the difference between 2 independent Poisson RV $X_1$ (mean $\mu_1$) and $X_2$ (mean $\mu_2$) follows the Skellam distribution such that its PMF is: $$f(k; \mu_1, \mu_2) = e^{-(\mu_1 + \mu_2)}(\frac{\mu_1}{\mu_2})^{\frac{k}{2}}I_{|k|}(2\sqrt{\mu_1\mu_2})$$

where $I_{|k|}$ is a Bessel function...

and it gets quite complicated...

so erm... is there a simpler way to calculate something like e.g. $$P(X_1 - X_2 = 1)$$ or $$P(X_1 - X_2 < 2)$$ ?

Answer 1

If you want to directly compute, then \begin{split} P(X_1 - X_2 = 1) &= \sum_{k=1}^{\infty} P(X_1-X_2 = 1|X_1 = k)P(X_1=k) \\ &= \sum_{k=1}^{\infty} P(X_2 = k-1)P(X_1=k) \\ &= \sum_{k=1}^{\infty} e^{-\mu_2}\frac{\mu_2^{(k-1)}}{(k-1)!}e^{-\mu_1}\frac{\mu_1^{k}}{k!} \\ &= \frac{e^{-(\mu_1+\mu_2)}}{\mu_2}\sum_{k=1}^{\infty} \frac{(\mu_1\mu_2)^{k}}{k!(k-1)!} \\ \end{split} If we use the formula based on Bessel's function, then the infinite sum calculation will be subsumed into Bessel function computation.

Answer 2

$\begin{align} \mathsf P(X_1-X_2 = 1) &= \sum_{k=0}^\infty \mathsf P(X_2=k, X_1=1+k) & \text{by total probability, partitioning over $X_2=k$} \\ &= \sum_{k=0}^\infty \mathsf P(X_2=k)\mathsf P(X_1=1+k) & \text{by independence of the random variables} \\ & =\sum_{k=0}^\infty \frac{\mu_2^k e^{-\mu_2}}{k!}\frac{\mu_1^{k+1}e^{-\mu_1}}{(k+1)!} & \text{because }X_1\sim\mathcal P(\mu_1), X_2\sim\mathcal P(\mu_2) \\ & = e^{-\mu_1-\mu_2}\sum_{k=0}^\infty \frac{\mu_1^{k+1}\mu_2^k}{k!(k+1)!} \\ & = e^{-\mu_1-\mu_2}\sqrt{\frac{\mu_1}{\mu_2}}I_1(2\sqrt{\mu_1\mu_2}) & \text{$I_1(z)$ is a Modified Bessel Function of the first kind} \\[5ex] \mathsf P(X_1-X_2 \leq z) & = \sum_{k=1}^\infty \mathsf P(X_2=k, X_1\leq z+k) \\ & = \sum_{k=1}^\infty \mathsf P(X_2=k) \sum_{h=0}^{z+k}\mathsf P(X_1=h) \end{align}$et cetera