Poisson race distribution

$\newcommand{\P}[2][]{\mathbb{P}#1{\left[{#2}\right]} } \newcommand{\Pcond}[3][]{\mathbb{P}#1{\left[{#2}\!\ \middle\vert\!\ {#3}\right]} } \newcommand{\Plcond}[3][]{\Pcond[#1]{#2}{#3} } \newcommand{\Prcond}[3][]{\Pcond[#1]{#2}{#3} }$

$\newcommand{\l}[1]{\lambda#1}$

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 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

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.

Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:

Page 384 document spans consecutive numbers, squared, A4, no holes

• Added proof, attention was all over the place so may be a bit disjoint Alec (talk) 02:08, 7 January 2018 (UTC)
  • Worth going through it again! Alec (talk) 02:08, 7 January 2018 (UTC)

STILL TODO:

• The $k<0$ case! - Method outlined already. Alec (talk) 02:08, 7 January 2018 (UTC)

$k$	$X$	$Y$
$0$	$X\eq 0$	$Y\eq 0$
	$X\eq 1$	$Y\eq 1$
	$X\eq 2$	$Y\eq 2$
	$\vdots$	$\vdots$
	$X\eq i$	$Y\eq i$
$1$	$X\eq 1$	$Y\eq 0$
	$X\eq 2$	$Y\eq 1$
	$X\eq 3$	$Y\eq 2$
	$\vdots$	$\vdots$
	$X\eq i+1$	$Y\eq i$
$2$	$X\eq 2$	$Y\eq 0$
	$X\eq 3$	$Y\eq 1$
	$X\eq 4$	$Y\eq 2$
	$\vdots$	$\vdots$
	$X\eq i+2$	$Y\eq i$
$\cdots$
$j$	$X\eq j$	$Y\eq 0$
	$X\eq j+1$	$Y\eq 1$
	$X\eq j+2$	$Y\eq 2$
	$\vdots$	$\vdots$
	$X\eq i+j$	$Y\eq i$

Finding $\P{Z\eq k}$ for $k\in\mathbb{N}_0$

---

TODO: Cover why we split the cases (from below: "Whereas: if we have $k<0$ then $Y>X$ so if we have $Y\eq 0$ then $X< 0$ follows, which we can't do, so we must sum the other way around ) - good start

---

• Let $k\in\mathbb{N}_0$ be given
  • Suppose $k\ge 0$
    • Then $X-Y\eq k$ giving $X\eq Y+k$
    • Specifically, $k\ge 0$ means $X-Y\ge 0$ so $X\ge Y$ and so forth as shown for a handful of values in the table on the right
      • As Poisson is only defined for values in $\mathbb{N}_0$ we can start with $Y\eq 0$ and $X$ will be $\ge 0$ too
        • Whereas: if we have $k<0$ then $Y>X$ so if we have $Y\eq 0$ then $X< 0$ follows, which we can't do, so we must sum the other way around if $k<0$. This is why we analyse $\P{Z\eq k}$ as two cases!
    • So we see that if $Y$ takes any value in $\mathbb{N}_0$ that $X\eq Y+k$ means $X\in\mathbb{N}_0$ too, the key point
    • Thus we want to sum over $i\in\mathbb{N}_0$ of $\P{Y\eq i\wedge X\eq i+k}$ as for these cases we see $X-Y\eq i+k-i\eq k$ as required
      • So: $$\P{Z\eq k}:\eq\sum_{i\in\mathbb{N}_0} \P{Y\eq i}\cdot\Pcond{X\eq i+k}{Y\eq i}$$
        • As $X$ and $Y$ are independent distributions we know $\Pcond{X\eq u}{Y\eq v}\eq\P{X\eq u}$ we see:
          • $$\P{Z\eq k}\eq\sum_{i\in\mathbb{N}_0} \P{Y\eq i}\cdot\P{X\eq i+k}$$
        • Then $$\P{Z\eq k}\eq\sum_{i\in\mathbb{N}_0}{\Bigg[ \underbrace{e^{-\l{_2} }\cdot\frac{\l{_2}^i}{i!} }_{\P{Y\eq i} }\cdot\underbrace{e^{-\l{_1} }\cdot\frac{\l{_1}^{i+k} }{(i+k)!} }_{\P{X\eq i+k} }\cdot\Bigg]}$$ - by definition of the Poisson distribution

          $$\eq e^{-\lambda_1-\l{_2} }\sum_{i\in\mathbb{N}_0}\frac{(\lambda_1\cdot\lambda_2)^i}{i!}\cdot\frac{\lambda_1^k}{(i+k)!}$$

          • Notice that $(i+k)!\eq i!\big((i+1)(i+2)\cdots(i+k)\big)$ $$\eq i!\frac{(i+k)!}{i!}$$ , so $$\frac{1}{(i+k)!}\eq \frac{1}{i!\frac{(i+k)!}{i!} } \eq \frac{i!}{i!(i+k)!}$$
        • So $$\P{Z\eq k} \eq e^{-\lambda_1-\l{_2} }\sum_{i\in\mathbb{N}_0}\frac{(\lambda_1\cdot\lambda_2)^i\lambda_1^k}{i!}\cdot\frac{i!}{i!(i+k)!}$$
          • Finally giving:
            • $$\P{Z\eq k} \eq e^{-\lambda_1-\l{_2} }\sum_{i\in\mathbb{N}_0}\lambda_1^k\cdot\frac{(\lambda_1\cdot\lambda_2)^i}{i!i!}\cdot\frac{i!}{(i+k)!}$$ or
            • is is perhaps better written as:
              $$\P{Z\eq k} \eq e^{-\lambda_1-\l{_2} }\sum_{i\in\mathbb{N}_0}{\left[\lambda_1^k\cdot\frac{(\lambda_1\cdot\lambda_2)^i}{(i!)^2}\cdot\frac{1}{\prod_{j\eq 1}^{k}(i+j)}\right]}$$

NOTES for $k<0$

Let $k':\eq -k$ so it's positive and note that:

• $X-Y\eq k$ means $Y-X\eq -k\eq k'$
  • Importantly $Y-X\eq k'$ for $k' >0$ (as $k<0$ we see $k':\eq -k > 0$)
    • Let $X':\eq Y$ and $Y':\eq X$ then we have:
      • $X'-Y'\eq k'$ for $k'>0$ and both $X'$ and $Y'$ as Poisson distributions with given rates
        • We can use the $k\ge 0$ formula for this as $k'>0$ so is certainly $\ge 0$