Difference between revisions of "Poisson race distribution"
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Revision as of 19:11, 31 December 2017
[ilmath]\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} }[/ilmath]
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 X-Y[/ilmath]
WeAlec:[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 re-use 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
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:
The message provided is:
Page 384 document spans consecutive numbers, squared, A4, no holes
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
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