Poisson distribution

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My informal derivation feels too formal, but isn't formal enough to be a formal one! Work in progress!

Poisson distribution
[ilmath]X\sim\text{Poi}(\lambda)[/ilmath]
[ilmath]\lambda\in\mathbb{R}_{\ge 0} [/ilmath] ([ilmath]\lambda[/ilmath] - the average rate of events per unit)
file.png
Definition
Type	Discrete, over [ilmath]\mathbb{N}_{\ge 0} [/ilmath]
p.m.f	[math]\mathbb{P}[X\eq k]:\eq e^{-\lambda}\frac{\lambda^k}{k!} [/math]
c.d.f	[math]\mathbb{P}[X\le k]\eq e^{-\lambda}\sum^k_{i\eq 0}\frac{\lambda^i}{k!} [/math]
Characteristics
Expected value	[ilmath]\mathbb{E}[X]\eq\lambda[/ilmath]
Variance	[ilmath]\text{Var}(X)\eq\lambda[/ilmath]

Definition

• [ilmath]X\sim\text{Poisson}(\lambda)[/ilmath]
  • for [ilmath]k\in\mathbb{N}_{\ge 0} [/ilmath] we have: [math]\mathbb{P}[X\eq k]:\eq\frac{e^{-\lambda}\lambda^k}{k!} [/math]
    • the first 2 terms are easy to give: [ilmath]e^{\lambda} [/ilmath] and [ilmath]\lambda e^{-\lambda} [/ilmath] respectively, after that we have [ilmath]\frac{1}{2}\lambda^2 e^{-\lambda} [/ilmath] and so forth
  • for [ilmath]k\in\mathbb{N}_{\ge 0} [/ilmath] we have: [math]\mathbb{P}[X\le k]\eq e^{-\lambda}\sum^k_{j\eq 0}\frac{1}{j!}\lambda^j[/math]

As a formal random variable

[ilmath]\xymatrix{ & [0,1] \ar[r]^X & \mathbb{N}_0 \\ & \mathcal{B}([0,1]) \ar[dl]_-{\lambda} & \mathcal{P}(\mathbb{N}_0) \ar[l]_-{X^{-1} } \ar@{-->}@/^1em/[dll]^-{\mathbb{P}:\eq \lambda\circ X^{-1} } \\ \mathbb{R} & & } [/ilmath]
Situation for our RV

Caveat:[ilmath]\lambda[/ilmath] here is used to denote 2 things - the parameter to the Poisson distribution, and the restriction of the 1 dimensional Lebesgue measure to some region of interest.

There is no unique way to define a random variable, here is one way.

• Let [ilmath]\big([/ilmath][ilmath][0,1][/ilmath][ilmath],\ [/ilmath][ilmath]\mathcal{B}([0,1])[/ilmath][ilmath],\ [/ilmath][ilmath]\lambda[/ilmath][ilmath]\big)[/ilmath] be a probability space - which itself could be viewed as a rectangular distribution's random variable
  • Let [ilmath]\lambda\in\mathbb{R}_{>0} [/ilmath] be given, and let [ilmath]X\sim\text{Poi}(\lambda)[/ilmath]
    • Specifically consider [ilmath]\big(\mathbb{N}_0,\ [/ilmath][ilmath]\mathcal{P}(\mathbb{N}_0)[/ilmath][ilmath]\big)[/ilmath] as a sigma-algebra and [ilmath]X:[0,1]\rightarrow\mathbb{N}_0[/ilmath] by:
      • [ilmath]X:x\mapsto\left\{\begin{array}{lr}0&\text{if }x\in[0,p_1)\\1 & \text{if }x\in[p_1,p_2)\\ \vdots & \vdots \\ k & \text{if }x\in[p_k,p_{k+1})\\ \vdots & \vdots \end{array}\right.[/ilmath] for [math]p_1:\eq e^{-\lambda} \frac{\lambda^1}{1!} [/math] and [ilmath]p_k:\eq p_{k-1}+e^{-\lambda}\frac{\lambda^k}{k!} [/ilmath]

Giving the setup shown on the left.

To add to page

• The sum of two random variables with Poisson distributions is a Poisson distribution itself
  • If [ilmath]X\sim\text{Poi}(\lambda)[/ilmath] and [ilmath]Y\sim\text{Poi}(\lambda')[/ilmath] then [ilmath]X+Y\sim\text{Poi}(\lambda+\lambda') [/ilmath]
• Time between events of a Poisson distribution Alec (talk) 22:06, 11 November 2017 (UTC)

Mean

• [math]\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[/math][math]0+\left[ e^{-\lambda}\sum^\infty_{n\eq 1} \frac{\lambda^n}{(n-1)!}\right] [/math][math]\eq e^{-\lambda}\lambda\left[\sum^\infty_{n\eq 1}\frac{\lambda^{n-1} }{(n-1)!}\right][/math]

  [math]\eq \lambda e^{-\lambda}\left[\sum^{\infty}_{n\eq 0}\frac{\lambda^n}{n!}\right][/math][math]\eq \lambda e^{-\lambda}\left[\lim_{n\rightarrow\infty}\left(\sum^{n}_{k\eq 0}\frac{\lambda^k}{k!}\right)\right][/math]

  • But! [math]e^x\eq\lim_{n\rightarrow\infty}\left(\sum^n_{i\eq 0}\frac{x^i}{i!}\right)[/math]

  • So [math]\eq\lambda e^{-\lambda} e^\lambda[/math]
    • [ilmath]\eq\lambda[/ilmath]

Derivation

Standard Poisson distribution:

• Let [ilmath]S:\eq[0,1)\subseteq\mathbb{R} [/ilmath], recall that means [ilmath]S\eq\{x\in\mathbb{R}\ \vert\ 0\le x<1\} [/ilmath]
• Let [ilmath]\lambda[/ilmath] be the average count of some event that can occur [ilmath]0[/ilmath] or more times on [ilmath]S[/ilmath]

We will now divide [ilmath]S[/ilmath] up into [ilmath]N[/ilmath] equally sized chunks, for [ilmath]N\in\mathbb{N}_{\ge 1} [/ilmath]

• Let [ilmath]S_{i,N}:\eq\left[\frac{i-1}{N},\frac{i}{N}\right)[/ilmath] for [ilmath]i\in\{1,\ldots,N\}\subseteq\mathbb{N} [/ilmath]

We will now define a random variable that counts the occurrences of events per interval.

• Let [ilmath]C\big(S_{i,N}\big)[/ilmath] be the RV such that its value is the number of times the event occurred in the [ilmath]\left[\frac{i-1}{N},\frac{i}{N}\right)[/ilmath] interval

We now require:

• [math]\lim_{N\rightarrow\infty}\left(\mathbb{P}[C\big(S_{i,N}\big)\ge 2]\right)\eq 0[/math] - such that:
  • as the [ilmath]S_{i,N} [/ilmath] 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 [ilmath]S_{i,\infty} [/ilmath] so no [ilmath]\text{BORV}(\frac{\lambda}{N})[/ilmath] distribution then either

Note that:

• [math]\lim_{N\rightarrow\infty}\big(C(S_{i,N})\big)\eq\lim_{N\rightarrow\infty}\left(\text{BORV}\left(\frac{\lambda}{N}\right)\right) [/math]
  • This is supposed to convey that the distribution of [ilmath]C(S_{i,N})[/ilmath] as [ilmath]N[/ilmath] gets large gets arbitrarily close to [ilmath]\text{BORV}(\frac{\lambda}{N})[/ilmath]

So we may say for sufficiently large [ilmath]N[/ilmath] that:

• [math]C(S_{i,N})\mathop{\sim}_{\text{(approx)} } [/math][ilmath]\text{BORV}(\frac{\lambda}{N})[/ilmath], so that:
  • [ilmath]\mathbb{P}[C(S_{i,N})\eq 0]\approx(1-\frac{\lambda}{N}) [/ilmath]
  • [ilmath]\mathbb{P}[C(S_{i,N})\eq 1]\approx \frac{\lambda}{N} [/ilmath], and of course
  • [ilmath]\mathbb{P}[C(S_{i,N})\ge 2]\approx 0[/ilmath]

Assuming the [ilmath]C(S_{i,N})[/ilmath] are independent over [ilmath]i[/ilmath] (which surely we get from the [ilmath]\text{BORV} [/ilmath] distributions?) we see:

• [math]C(S)\mathop{\sim}_{\text{(approx)} } [/math][ilmath]\text{Bin} [/ilmath][math]\left(N,\frac{\lambda}{N}\right)[/math] or, more specifically: [math]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)[/math]

We see:

• [math]\mathbb{P}[C(S)\eq k]\eq\lim_{N\rightarrow\infty} [/math][ilmath]\Big(\mathbb{P}\big[\text{Bin}(N,\frac{\lambda}{N})\eq k\big]\Big)[/ilmath][math]\eq\lim_{N\rightarrow\infty}\left({}^N\!C_k\ \left(\frac{\lambda}{N}\right)^k\left(1-\frac{\lambda}{N}\right)^{N-k}\right)[/math]

We claim that:

• [math]\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} [/math]

We will tackle this in two parts:

• [math]\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)[/math] where [ilmath]B\rightarrow e^{-\lambda} [/ilmath] and [math]A\rightarrow \frac{\lambda^k}{k!} [/math]

Proof

Key notes:

A

Notice:

• [math]{}^N\!C_k\ \left(\frac{\lambda}{N}\right)^k \eq \frac{N!}{(N-k)!k!}\cdot\frac{1}{N^k}\cdot\lambda^k[/math]

  [math]\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[/math]

  • Notice that as [ilmath]N[/ilmath] gets bigger [ilmath]N-k+1[/ilmath] is "basically" [ilmath]N[/ilmath] so the [ilmath]N[/ilmath]s in the denominator cancel (in fact the value will be slightly less than 1, tending towards 1 as [ilmath]N\rightarrow\infty[/ilmath]) this giving:
    • [math]\frac{\lambda^k}{k!} [/math]

B

This comes from:

• [math]e^x:\eq\lim_{n\rightarrow\infty}\left(\left(1+\frac{x}{n}\right)^n\right)[/math], so we get the [ilmath]e^{-\lambda} [/ilmath] term.