View source for Poisson race distribution

{{ProbMacro}}{{M|\newcommand{\l}[1]{\lambda#1} }}
__TOC__
==Definition==
Let {{M|X\sim}}[[Poisson distribution|{{M|\text{Poi} }}]]{{M|(\lambda_1)}} and {{M|Y\sim\text{Poi}(\lambda_2)}} be given {{plural|random variable|s}} (that are [[independent random variables|independent]]), then define a new random variable:
* {{M|Z:\eq X-Y}}
We<sup><span style="color:#454545;">Alec:</span></sup><ref>I've heard this somewhere before, I can't remember where from - I've had a quick search, while not established it's in the right area</ref> call this a "''Poisson race''" as in some sense they're racing, {{M|Z>0}} then {{M|X}} is winning, {{M|Z\eq 0}}, neck and neck and {{M|Z<0}} then {{M|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 {{M|\lambda_1}} being the defects per mile of the left rail (defined somehow) and {{M|\lambda_2}} the defects per mile of the right rail.
** If there are 5 on the left and 3 on the right {{M|Z}} for that mile was {{M|+2}} - the next mile is independent and doesn't start off with this bias, that is {{M|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 {{M|k\ge 0}} and {{M|k\in\mathbb{Z} }} we have:
** '''Claim 1:''' <span style="font-size:1.0em;">{{MM|\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)!} }}</span><ref group="Note">Note that {{M|\sum_{i\in\mathbb{N}_0}a_i}} means {{M|\sum^\infty_{i\eq 0}a_i}} - see [[Notes:Infinity notation]]</ref>, which may be written:
*** {{MM|\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 {{M|k\le 0}} and {{M|k\in\mathbb{Z} }} we can re-use the above result with {{M|X}} and {{M|Y}} flipped:
** Can't find where I've written it down, will not guess it
==Proof of claims==
{{Requires proof|grade=B|msg=Page 384 document spans consecutive numbers, squared, A4, no holes
* Added proof, attention was all over the place so may be a bit disjoint [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 02:08, 7 January 2018 (UTC)
** Worth going through it again! [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 02:08, 7 January 2018 (UTC) 
STILL TODO:
* The {{M|k<0}} case! - Method outlined already. [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 02:08, 7 January 2018 (UTC)}}
<div style="float:right;margin-left:0.2em;margin-top:0.2em;margin-bottom:0.2em;">{{/CalcTable1}}</div>
===Finding {{M|\P{Z\eq k} }} for {{M|k\in\mathbb{N}_0}}===
{{Todo|Cover why we split the cases (from below: "''Whereas: if we have {{M|k<0}} then {{M|Y>X}} so if we have {{M|Y\eq 0}} then {{M|X< 0}} follows, which we can't do, so we must sum the other way around '') - good start}}
* Let {{M|k\in\mathbb{N}_0}} be given
** Suppose {{M|k\ge 0}}
*** Then {{M|X-Y\eq k}} giving {{M|X\eq Y+k}}
*** Specifically, {{M|k\ge 0}} means {{M|X-Y\ge 0}} so {{M|X\ge Y}} and so forth as shown for a handful of values in the '''''table on the right'''''
**** As [[Poisson distribution|Poisson]] is only defined for values in {{M|\mathbb{N}_0}} we can start with {{M|Y\eq 0}} and {{M|X}} will be {{M|\ge 0}} too
***** ''Whereas:'' if we have {{M|k<0}} then {{M|Y>X}} so if we have {{M|Y\eq 0}} then {{M|X< 0}} follows, which we can't do, so we must sum the other way around if {{M|k<0}}. '''''This is why we analyse {{M|\P{Z\eq k} }} as two cases!'''''
*** So we see that if {{M|Y}} takes any value in {{M|\mathbb{N}_0}} that {{M|X\eq Y+k}} means {{M|X\in\mathbb{N}_0}} too, the key point
*** Thus we want to sum over {{M|i\in\mathbb{N}_0}} of {{M|\P{Y\eq i\wedge X\eq i+k} }} as for these cases we see {{M|X-Y\eq i+k-i\eq k}} as required
**** So: {{MM|\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 {{M|X}} and {{M|Y}} are [[independent distributions]] we know {{M|\Pcond{X\eq u}{Y\eq v}\eq\P{X\eq u} }} we see:
****** {{MM|\P{Z\eq k}\eq\sum_{i\in\mathbb{N}_0} \P{Y\eq i}\cdot\P{X\eq i+k} }}
***** Then {{MM|\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 [[Poisson distribution|definition of the Poisson distribution]]
*****: {{MM|\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 {{M|(i+k)!\eq i!\big((i+1)(i+2)\cdots(i+k)\big)}}{{MM|\eq i!\frac{(i+k)!}{i!} }}, so {{MM|\frac{1}{(i+k)!}\eq \frac{1}{i!\frac{(i+k)!}{i!} } \eq \frac{i!}{i!(i+k)!} }}
***** So {{MM|\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:
******* {{MM|\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: ''''' <div style="display:inline-block;font-size:1.2em;border:1px solid #000;padding:0.25em;">{{MM|\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]} }}</div>

====NOTES for {{M|k<0}}====
Let {{M|k':\eq -k}} so it's positive and note that:
* {{M|X-Y\eq k}} means {{M|Y-X\eq -k\eq k'}}
** Importantly {{M|Y-X\eq k'}} for {{M|k' >0}} (as {{M|k<0}} we see {{M|k':\eq -k > 0}})
*** Let {{M|X':\eq Y}} and {{M|Y':\eq X}} then we have:
**** {{M|X'-Y'\eq k'}} for {{M|k'>0}} and both {{M|X'}} and {{M|Y'}} as [[Poisson distributions]] with given rates
***** We can use the {{M|k\ge 0}} formula for this as {{M|k'>0}} so is certainly {{M|\ge 0}}
==Notes==
<references group="Note"/>
==References==
<references/>
{{Fundamental probability distributions navbox|plain}}
{{Definition|Probability|Statistics|Elementary Probability}}
{{Probability Distribution|fund=true}}