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!

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.