Difference between revisions of "Poisson distribution"

Revision as of 06:53, 20 September 2017

Definition

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

Mean

• $$\sum^\infty_{n\eq 0} n\times\mathbb{P}[X\eq n]\eq\sum^\infty_{n\eq 0}\left[ n\times e^{-\lambda}\frac{\lambda^n}{n!}\right]\eq$$ $$0+\left[ e^{-\lambda}\sum^\infty_{n\eq 1} \frac{\lambda^n}{(n-1)!}\right]$$ $$\eq e^{-\lambda}\lambda\left[\sum^\infty_{n\eq 1}\frac{\lambda^{n-1} }{(n-1)!}\right]$$

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

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

  • So $$\eq\lambda e^{-\lambda} e^\lambda$$
    • $\eq\lambda$

Derivation

Standard Poisson distribution:

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

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

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

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

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

We now require:

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

Note that:

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

So we may say for sufficiently large $N$ that:

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

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

• $$C(S)\mathop{\sim}_{\text{(approx)} }$$ $\text{Bin}$ $$\left(N,\frac{\lambda}{N}\right)$$ or, more specifically: $$C(S)\eq\lim_{N\rightarrow\infty}\Big(\sum^N_{i\eq 1}C(S_{i,N})\Big)\eq\lim_{N\rightarrow\infty}\left(\text{Bin}\left(N,\frac{\lambda}{N}\right)\right)$$

We see:

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

We claim that:

• $$\lim_{N\rightarrow\infty}\left({}^N\!C_k\ \left(\frac{\lambda}{N}\right)^k\left(1-\frac{\lambda}{N}\right)^{N-k}\right)\eq \frac{\lambda^k}{k!}e^{-\lambda}$$

We will tackle this in two parts:

• $$\lim_{N\rightarrow\infty}\Bigg(\underbrace{ {}^N\!C_k\ \left(\frac{\lambda}{N}\right)^k}_{A}\ \underbrace{\left(1-\frac{\lambda}{N}\right)^{N-k} }_{B}\Bigg)$$ where $B\rightarrow e^{-\lambda}$ and $$A\rightarrow \frac{\lambda^k}{k!}$$

Proof

Key notes:

A

Notice:

• $${}^N\!C_k\ \left(\frac{\lambda}{N}\right)^k \eq \frac{N!}{(N-k)!k!}\cdot\frac{1}{N^k}\cdot\lambda^k$$

  $$\eq\frac{1}{k!}\cdot\frac{\overbrace{N(N-1)\cdots(N-k+2)(N-k+1)}^{k\text{ terms} } }{\underbrace{N\cdot N\cdots N}_{k\text{ times} } } \cdot\lambda^k$$

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

B

This comes from:

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

Notes

1. Jump up ↑ Recall again that means $\{x\in\mathbb{R}\ \vert\ \frac{i-1}{N}\le x < \frac{i}{N} \}$