Poisson race distribution

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Definition

Let $X\sim$$\text{Poi}$$(\lambda_1)$ and $Y\sim\text{Poi}(\lambda_2)$ be given random variables (that are independent), then define a new random variable:

• $Z:\eq X-Y$

We^Alec: call this a "Poisson race" as in some sense they're racing, $Z>0$ then $X$ is winning, $Z\eq 0$, neck and neck and $Z<0$ then $Y$ is winning.

• Caveat:Remember that Poisson measures stuff per unit thing, meaning, for example, that say we are talking "defects per mile of track" with $\lambda_1$ being the defects per mile of the left rail (defined somehow) and $\lambda_2$ the defects per mile of the right rail.
  • If there are 5 on the left and 3 on the right $Z$ for that mile was $+2$ - the next mile is independent and doesn't start off with this bias, that is $Z\eq 0$ for the next mile it doesn't mean the right rail caught up with 2 more than the left for this mile.
  • So this doesn't model an ongoing race.

Descriptions

• for $k\ge 0$ and $k\in\mathbb{Z}$ we have:
  • Claim 1: $$\P{Z\eq k}\eq \lambda_1^k e^{-\lambda_1-\lambda_2} \sum_{i\in\mathbb{N}_0}\frac{(\lambda_1\lambda_2)^i}{i!(i+k)!}$$ , which may be written:
    • $$\P{Z\eq k}\eq \lambda_1^k e^{-\lambda_1-\lambda_2}\sum_{i\in\mathbb{N}_0}\left( \frac{(\lambda_1\lambda_2)^i}{(i!)^2}\cdot\frac{i!}{(i+k)!}\right)$$ - this has some terms that look a bit like a Poisson term (squared) - perhaps it might lead to a closed form.
• for $k\le 0$ and $k\in\mathbb{Z}$ we can re-use the above result with $X$ and $Y$ flipped:
  • Can't find where I've written it down, will not guess it

Proof of claims

Grade: B

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