View source for Poisson distribution

{{Stub page|grade=A*|msg=My informal derivation feels too formal, but isn't formal enough to be a formal one! Work in progress!}}
{{infobox
|title=Poisson distribution
|above=<span style="font-size:1.5em;">{{M|X\sim\text{Poi}(\lambda)}}</span>
|subheader=<span style="font-size:1.2em;">{{M|\lambda\in\mathbb{R}_{\ge 0} }}</span><br/>''({{M|\lambda}} - the average rate of events per unit)''
|image=file.png
|header1=Definition
|label1=Type
|data1=[[Discrete probability distribution|Discrete]], over {{M|\mathbb{N}_{\ge 0} }}
|label2=[[Probability mass function|p.m.f]]
|data2={{MM|\mathbb{P}[X\eq k]:\eq e^{-\lambda}\frac{\lambda^k}{k!} }}
|label3=[[Cumulative distribution function|c.d.f]]
|data3={{MM|\mathbb{P}[X\le k]\eq e^{-\lambda}\sum^k_{i\eq 0}\frac{\lambda^i}{k!} }}
|header10=Characteristics
|label12=[[Expected value]]
|data12={{M|\mathbb{E}[X]\eq\lambda}}
|label13=[[Variance]]
|data13={{M|\text{Var}(X)\eq\lambda}}
}}
==Definition==
* {{M|X\sim\text{Poisson}(\lambda)}}
** for {{M|k\in\mathbb{N}_{\ge 0} }} we have: {{MM|\mathbb{P}[X\eq k]:\eq\frac{e^{-\lambda}\lambda^k}{k!} }}
*** the first 2 terms are easy to give: {{M|e^{\lambda} }} and {{M|\lambda e^{-\lambda} }} respectively, after that we have {{M|\frac{1}{2}\lambda^2 e^{-\lambda} }} and so forth
** for {{M|k\in\mathbb{N}_{\ge 0} }} we have: {{MM|\mathbb{P}[X\le k]\eq e^{-\lambda}\sum^k_{j\eq 0}\frac{1}{j!}\lambda^j}}
===[[/RV|As a formal random variable]]===
{{/RV|home=true}}
==To add to page==
* [[The sum of two random variables with Poisson distributions is a Poisson distribution itself]]
** If {{M|X\sim\text{Poi}(\lambda)}} and {{M|Y\sim\text{Poi}(\lambda')}} then  {{M|X+Y\sim\text{Poi}(\lambda+\lambda') }}
* [[Time between events of a Poisson distribution]] [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 22:06, 11 November 2017 (UTC)
* [[Poisson race distribution]] - [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 23:57, 26 November 2017 (UTC)
==Mean==
* {{MM|\sum^\infty_{n\eq 0} n\times\mathbb{P}[X\eq n]\eq\sum^\infty_{n\eq 0}\left[ n\times e^{-\lambda}\frac{\lambda^n}{n!}\right]\eq}}{{MM|0+\left[ e^{-\lambda}\sum^\infty_{n\eq 1} \frac{\lambda^n}{(n-1)!}\right] }}{{MM|\eq e^{-\lambda}\lambda\left[\sum^\infty_{n\eq 1}\frac{\lambda^{n-1} }{(n-1)!}\right]}}
*: {{MM|\eq \lambda e^{-\lambda}\left[\sum^{\infty}_{n\eq 0}\frac{\lambda^n}{n!}\right]}}{{MM|\eq \lambda e^{-\lambda}\left[\lim_{n\rightarrow\infty}\left(\sum^{n}_{k\eq 0}\frac{\lambda^k}{k!}\right)\right]}}
*:* But! {{MM|e^x\eq\lim_{n\rightarrow\infty}\left(\sum^n_{i\eq 0}\frac{x^i}{i!}\right)}}
** So {{MM|\eq\lambda e^{-\lambda} e^\lambda}}
*** {{M|\eq\lambda}}
==Derivation==
Standard Poisson distribution:
* Let {{M|S:\eq[0,1)\subseteq\mathbb{R} }}, recall that means {{M|S\eq\{x\in\mathbb{R}\ \vert\ 0\le x<1\} }}
* Let {{M|\lambda}} be the average count of some event that can occur {{M|0}} or more times on {{M|S}}

We will now divide {{M|S}} up into {{M|N}} equally sized chunks, for {{M|N\in\mathbb{N}_{\ge 1} }} 
* Let {{M|S_{i,N}:\eq\left[\frac{i-1}{N},\frac{i}{N}\right)}}<ref group="Note">Recall again that means {{M|\{x\in\mathbb{R}\ \vert\ \frac{i-1}{N}\le x < \frac{i}{N} \} }}</ref> for {{M|i\in\{1,\ldots,N\}\subseteq\mathbb{N} }}

We will now define a random variable that counts the occurrences of events per interval.
* Let {{M|C\big(S_{i,N}\big)}} be the RV such that its value is the number of times the event occurred in the {{M|\left[\frac{i-1}{N},\frac{i}{N}\right)}} interval

We now require:
* <span style="font-size:1.2em;">{{MM|\lim_{N\rightarrow\infty}\left(\mathbb{P}[C\big(S_{i,N}\big)\ge 2]\right)\eq 0}}</span> - such that:
** as the {{M|S_{i,N} }} get smaller the chance of 2 or more events occurring in the space reaches zero.
** {{Warning|This is phrased as a limit, I'm not sure it should be as we don't have any {{M|S_{i,\infty} }} so no {{M|\text{BORV}(\frac{\lambda}{N})}} distribution then either}}

Note that:
* {{MM|\lim_{N\rightarrow\infty}\big(C(S_{i,N})\big)\eq\lim_{N\rightarrow\infty}\left(\text{BORV}\left(\frac{\lambda}{N}\right)\right) }}
** This is supposed to convey that the distribution of {{M|C(S_{i,N})}} as {{M|N}} gets large gets arbitrarily close to {{M|\text{BORV}(\frac{\lambda}{N})}}

So we may say for ''sufficiently large'' {{M|N}} that:
* {{MM|C(S_{i,N})\mathop{\sim}_{\text{(approx)} } }}{{M|\text{BORV}(\frac{\lambda}{N})}}, so that:
** {{M|\mathbb{P}[C(S_{i,N})\eq 0]\approx(1-\frac{\lambda}{N}) }}
** {{M|\mathbb{P}[C(S_{i,N})\eq 1]\approx \frac{\lambda}{N} }}, and of course
** {{M|\mathbb{P}[C(S_{i,N})\ge 2]\approx 0}}

Assuming the {{M|C(S_{i,N})}} are independent over {{M|i}} (which surely we get from the {{M|\text{BORV} }} distributions?) we see:
* {{MM|C(S)\mathop{\sim}_{\text{(approx)} } }}[[Binomial distribution|{{M|\text{Bin} }}]]{{MM|\left(N,\frac{\lambda}{N}\right)}} or, more specifically:<!--

--> {{MM|C(S)\eq\lim_{N\rightarrow\infty}\Big(\sum^N_{i\eq 1}C(S_{i,N})\Big)\eq\lim_{N\rightarrow\infty}\left(\text{Bin}\left(N,\frac{\lambda}{N}\right)\right)}}


We see:
* {{MM|\mathbb{P}[C(S)\eq k]\eq\lim_{N\rightarrow\infty} }}{{M|\Big(\mathbb{P}\big[\text{Bin}(N,\frac{\lambda}{N})\eq k\big]\Big)}}{{MM|\eq\lim_{N\rightarrow\infty}\left({}^N\!C_k\ \left(\frac{\lambda}{N}\right)^k\left(1-\frac{\lambda}{N}\right)^{N-k}\right)}}

We claim that: 
* {{MM|\lim_{N\rightarrow\infty}\left({}^N\!C_k\ \left(\frac{\lambda}{N}\right)^k\left(1-\frac{\lambda}{N}\right)^{N-k}\right)\eq \frac{\lambda^k}{k!}e^{-\lambda} }}
We will tackle this in two parts:
* {{MM|\lim_{N\rightarrow\infty}\Bigg(\underbrace{ {}^N\!C_k\ \left(\frac{\lambda}{N}\right)^k}_{A}\ \underbrace{\left(1-\frac{\lambda}{N}\right)^{N-k} }_{B}\Bigg)}} where {{M|B\rightarrow e^{-\lambda} }} and {{MM|A\rightarrow \frac{\lambda^k}{k!} }}

===Proof===
Key notes:
====A====
Notice:
* {{MM|{}^N\!C_k\ \left(\frac{\lambda}{N}\right)^k \eq \frac{N!}{(N-k)!k!}\cdot\frac{1}{N^k}\cdot\lambda^k}}
*: {{MM|\eq\frac{1}{k!}\cdot\frac{\overbrace{N(N-1)\cdots(N-k+2)(N-k+1)}^{k\text{ terms} } }{\underbrace{N\cdot N\cdots N}_{k\text{ times} } } \cdot\lambda^k}}
** Notice that as {{M|N}} gets bigger {{M|N-k+1}} is "basically" {{M|N}} so the {{M|N}}s in the denominator cancel (in fact the value will be slightly less than 1, tending towards 1 as {{M|N\rightarrow\infty}}) this giving:
*** {{MM|\frac{\lambda^k}{k!} }}

====B====
This comes from:
* {{MM|e^x:\eq\lim_{n\rightarrow\infty}\left(\left(1+\frac{x}{n}\right)^n\right)}}, so we get the {{M|e^{-\lambda} }} term.
==Notes==
<references group="Note"/>
{{Definition|Statistics|Probability|Elementary Probability}}