Difference between revisions of "Poisson race distribution"
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Let {{MX\sim}}[[Poisson distribution{{M\text{Poi} }}]]{{M(\lambda_1)}} and {{MY\sim\text{Poi}(\lambda_2)}} be given {{pluralrandom variables}} (that are [[independent random variablesindependent]]), then define a new random variable:  Let {{MX\sim}}[[Poisson distribution{{M\text{Poi} }}]]{{M(\lambda_1)}} and {{MY\sim\text{Poi}(\lambda_2)}} be given {{pluralrandom variables}} (that are [[independent random variablesindependent]]), then define a new random variable:  
* {{MZ:\eq XY}}  * {{MZ:\eq XY}}  
−  We  +  We<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, {{MZ>0}} then {{MX}} is winning, {{MZ\eq 0}}, neck and neck and {{MZ<0}} then {{MY}} is winning. 
* {{CaveatRemember 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.  * {{CaveatRemember 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 {{MZ}} for that mile was {{M+2}}  the next mile is independent and doesn't start off with this bias, that is {{MZ\eq 0}} for the next mile it doesn't mean the right rail caught up with 2 more than the left for this mile.  ** If there are 5 on the left and 3 on the right {{MZ}} for that mile was {{M+2}}  the next mile is independent and doesn't start off with this bias, that is {{MZ\eq 0}} for the next mile it doesn't mean the right rail caught up with 2 more than the left for this mile. 
Latest revision as of 10:33, 24 December 2018
Contents
Definition
Let [ilmath]X\sim[/ilmath][ilmath]\text{Poi} [/ilmath][ilmath](\lambda_1)[/ilmath] and [ilmath]Y\sim\text{Poi}(\lambda_2)[/ilmath] be given random variables (that are independent), then define a new random variable:
 [ilmath]Z:\eq XY[/ilmath]
We^{[1]} call this a "Poisson race" as in some sense they're racing, [ilmath]Z>0[/ilmath] then [ilmath]X[/ilmath] is winning, [ilmath]Z\eq 0[/ilmath], neck and neck and [ilmath]Z<0[/ilmath] then [ilmath]Y[/ilmath] 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 [ilmath]\lambda_1[/ilmath] being the defects per mile of the left rail (defined somehow) and [ilmath]\lambda_2[/ilmath] the defects per mile of the right rail.
 If there are 5 on the left and 3 on the right [ilmath]Z[/ilmath] for that mile was [ilmath]+2[/ilmath]  the next mile is independent and doesn't start off with this bias, that is [ilmath]Z\eq 0[/ilmath] 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 [ilmath]k\ge 0[/ilmath] and [ilmath]k\in\mathbb{Z} [/ilmath] we have:
 Claim 1: [math]\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)!} [/math]^{[Note 1]}, which may be written:
 [math]\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) [/math]  this has some terms that look a bit like a Poisson term (squared)  perhaps it might lead to a closed form.
 Claim 1: [math]\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)!} [/math]^{[Note 1]}, which may be written:
 for [ilmath]k\le 0[/ilmath] and [ilmath]k\in\mathbb{Z} [/ilmath] we can reuse the above result with [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] flipped:
 Can't find where I've written it down, will not guess it
Proof of claims
The message provided is:
[ilmath]k[/ilmath]  [ilmath]X[/ilmath]  [ilmath]Y[/ilmath] 

[ilmath]0[/ilmath]  [ilmath]X\eq 0[/ilmath]  [ilmath]Y\eq 0[/ilmath] 
[ilmath]X\eq 1[/ilmath]  [ilmath]Y\eq 1[/ilmath]  
[ilmath]X\eq 2[/ilmath]  [ilmath]Y\eq 2[/ilmath]  
[ilmath]\vdots[/ilmath]  [ilmath]\vdots[/ilmath]  
[ilmath]X\eq i[/ilmath]  [ilmath]Y\eq i[/ilmath]  
[ilmath]1[/ilmath]  [ilmath]X\eq 1[/ilmath]  [ilmath]Y\eq 0[/ilmath] 
[ilmath]X\eq 2[/ilmath]  [ilmath]Y\eq 1[/ilmath]  
[ilmath]X\eq 3[/ilmath]  [ilmath]Y\eq 2[/ilmath]  
[ilmath]\vdots[/ilmath]  [ilmath]\vdots[/ilmath]  
[ilmath]X\eq i+1[/ilmath]  [ilmath]Y\eq i[/ilmath]  
[ilmath]2[/ilmath]  [ilmath]X\eq 2[/ilmath]  [ilmath]Y\eq 0[/ilmath] 
[ilmath]X\eq 3[/ilmath]  [ilmath]Y\eq 1[/ilmath]  
[ilmath]X\eq 4[/ilmath]  [ilmath]Y\eq 2[/ilmath]  
[ilmath]\vdots[/ilmath]  [ilmath]\vdots[/ilmath]  
[ilmath]X\eq i+2[/ilmath]  [ilmath]Y\eq i[/ilmath]  
[ilmath]\cdots[/ilmath]  
[ilmath]j[/ilmath]  [ilmath]X\eq j[/ilmath]  [ilmath]Y\eq 0[/ilmath] 
[ilmath]X\eq j+1[/ilmath]  [ilmath]Y\eq 1[/ilmath]  
[ilmath]X\eq j+2[/ilmath]  [ilmath]Y\eq 2[/ilmath]  
[ilmath]\vdots[/ilmath]  [ilmath]\vdots[/ilmath]  
[ilmath]X\eq i+j[/ilmath]  [ilmath]Y\eq i[/ilmath] 
Finding [ilmath]\P{Z\eq k} [/ilmath] for [ilmath]k\in\mathbb{N}_0[/ilmath]
TODO: Cover why we split the cases (from below: "Whereas: if we have [ilmath]k<0[/ilmath] then [ilmath]Y>X[/ilmath] so if we have [ilmath]Y\eq 0[/ilmath] then [ilmath]X< 0[/ilmath] follows, which we can't do, so we must sum the other way around )  good start
 Let [ilmath]k\in\mathbb{N}_0[/ilmath] be given
 Suppose [ilmath]k\ge 0[/ilmath]
 Then [ilmath]XY\eq k[/ilmath] giving [ilmath]X\eq Y+k[/ilmath]
 Specifically, [ilmath]k\ge 0[/ilmath] means [ilmath]XY\ge 0[/ilmath] so [ilmath]X\ge Y[/ilmath] and so forth as shown for a handful of values in the table on the right
 As Poisson is only defined for values in [ilmath]\mathbb{N}_0[/ilmath] we can start with [ilmath]Y\eq 0[/ilmath] and [ilmath]X[/ilmath] will be [ilmath]\ge 0[/ilmath] too
 Whereas: if we have [ilmath]k<0[/ilmath] then [ilmath]Y>X[/ilmath] so if we have [ilmath]Y\eq 0[/ilmath] then [ilmath]X< 0[/ilmath] follows, which we can't do, so we must sum the other way around if [ilmath]k<0[/ilmath]. This is why we analyse [ilmath]\P{Z\eq k} [/ilmath] as two cases!
 As Poisson is only defined for values in [ilmath]\mathbb{N}_0[/ilmath] we can start with [ilmath]Y\eq 0[/ilmath] and [ilmath]X[/ilmath] will be [ilmath]\ge 0[/ilmath] too
 So we see that if [ilmath]Y[/ilmath] takes any value in [ilmath]\mathbb{N}_0[/ilmath] that [ilmath]X\eq Y+k[/ilmath] means [ilmath]X\in\mathbb{N}_0[/ilmath] too, the key point
 Thus we want to sum over [ilmath]i\in\mathbb{N}_0[/ilmath] of [ilmath]\P{Y\eq i\wedge X\eq i+k} [/ilmath] as for these cases we see [ilmath]XY\eq i+ki\eq k[/ilmath] as required
 So: [math]\P{Z\eq k}:\eq\sum_{i\in\mathbb{N}_0} \P{Y\eq i}\cdot\Pcond{X\eq i+k}{Y\eq i} [/math]
 As [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] are independent distributions we know [ilmath]\Pcond{X\eq u}{Y\eq v}\eq\P{X\eq u} [/ilmath] we see:
 [math]\P{Z\eq k}\eq\sum_{i\in\mathbb{N}_0} \P{Y\eq i}\cdot\P{X\eq i+k} [/math]
 Then [math]\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]} [/math]  by definition of the Poisson distribution
 [math]\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)!} [/math]
 Notice that [ilmath](i+k)!\eq i!\big((i+1)(i+2)\cdots(i+k)\big)[/ilmath][math]\eq i!\frac{(i+k)!}{i!} [/math], so [math]\frac{1}{(i+k)!}\eq \frac{1}{i!\frac{(i+k)!}{i!} } \eq \frac{i!}{i!(i+k)!} [/math]
 So [math]\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)!} [/math]
 Finally giving:
 [math]\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)!} [/math] or
 is is perhaps better written as: [math]\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]} [/math]
 Finally giving:
 As [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] are independent distributions we know [ilmath]\Pcond{X\eq u}{Y\eq v}\eq\P{X\eq u} [/ilmath] we see:
 So: [math]\P{Z\eq k}:\eq\sum_{i\in\mathbb{N}_0} \P{Y\eq i}\cdot\Pcond{X\eq i+k}{Y\eq i} [/math]
 Suppose [ilmath]k\ge 0[/ilmath]
NOTES for [ilmath]k<0[/ilmath]
Let [ilmath]k':\eq k[/ilmath] so it's positive and note that:
 [ilmath]XY\eq k[/ilmath] means [ilmath]YX\eq k\eq k'[/ilmath]
 Importantly [ilmath]YX\eq k'[/ilmath] for [ilmath]k' >0[/ilmath] (as [ilmath]k<0[/ilmath] we see [ilmath]k':\eq k > 0[/ilmath])
 Let [ilmath]X':\eq Y[/ilmath] and [ilmath]Y':\eq X[/ilmath] then we have:
 [ilmath]X'Y'\eq k'[/ilmath] for [ilmath]k'>0[/ilmath] and both [ilmath]X'[/ilmath] and [ilmath]Y'[/ilmath] as Poisson distributions with given rates
 We can use the [ilmath]k\ge 0[/ilmath] formula for this as [ilmath]k'>0[/ilmath] so is certainly [ilmath]\ge 0[/ilmath]
 [ilmath]X'Y'\eq k'[/ilmath] for [ilmath]k'>0[/ilmath] and both [ilmath]X'[/ilmath] and [ilmath]Y'[/ilmath] as Poisson distributions with given rates
 Let [ilmath]X':\eq Y[/ilmath] and [ilmath]Y':\eq X[/ilmath] then we have:
 Importantly [ilmath]YX\eq k'[/ilmath] for [ilmath]k' >0[/ilmath] (as [ilmath]k<0[/ilmath] we see [ilmath]k':\eq k > 0[/ilmath])
Notes
 ↑ Note that [ilmath]\sum_{i\in\mathbb{N}_0}a_i[/ilmath] means [ilmath]\sum^\infty_{i\eq 0}a_i[/ilmath]  see Notes:Infinity notation
References
 ↑ 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
