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
$X\sim\text{Poi}(\lambda)$
$\lambda\in\mathbb{R}_{\ge 0}$ ($\lambda$ - the average rate of events per unit)
file.png
Definition
Type	Discrete, over $\mathbb{N}_{\ge 0}$
p.m.f	$$\mathbb{P}[X\eq k]:\eq e^{-\lambda}\frac{\lambda^k}{k!}$$
c.d.f	$$\mathbb{P}[X\le k]\eq e^{-\lambda}\sum^k_{i\eq 0}\frac{\lambda^i}{k!}$$
Characteristics
Expected value	$\mathbb{E}[X]\eq\lambda$
Mdm	$$2\lambda e^{-\lambda}\frac{\lambda^{u} }{u!}$$ for: $u:\eq$$\text{Floor}$$(\lambda)$
Variance	$\text{Var}(X)\eq\lambda$

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: $$\mathbb{P}[X\le k]\eq e^{-\lambda}\sum^k_{j\eq 0}\frac{1}{j!}\lambda^j$$

As a formal random variable

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

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

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 $X\sim\text{Poi}(\lambda)$ and $Y\sim\text{Poi}(\lambda')$ then $X+Y\sim\text{Poi}(\lambda+\lambda')$
• Time between events of a Poisson distribution Alec (talk) 22:06, 11 November 2017 (UTC)
  • This follows an exponential distribution Alec (talk) 18:46, 2 April 2018 (UTC)
• Poisson race distribution - Alec (talk) 23:57, 26 November 2017 (UTC)
• Poisson mixture
  • Not to be confused with some other thing, see immediately below.

Other

• Notes:Poisson and Gamma distribution
• I have proved (ages ago) and found a source that if we have $X\sim\text{Poi}(\lambda)$ is the distribution of "observable events" and say that the probability of each event being observed is $p$, (that is observing an event is an i.i.d process, whether we observe or not being $\sim$$\text{BORV}$$(p)$, with $1$ indicating 'observed' and $0$ indicating 'missed') then the distribution of observations, say $Y$ follows $\sim\text{Poi}(p\lambda)$
  • As with the above and "Poisson+Poisson=Poisson" we'd see that the Poisson distributions form some sort of algebraic structure.
  • This is not to be confused with the definition of a Poisson mixture.

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

Mdm

See: Mdm of the Poisson distribution for proof.

For $X\sim\text{Poi}(\lambda)$ we have:

• $$\text{Mdm}(X)\eq 2\lambda e^{-\lambda}\frac{\lambda^u}{u!}$$ for
  • $u:\eq$$\text{Floor}$$(\lambda)$

Derivation of the Poisson distribution

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 $S_{i,N}:\eq\left[\frac{i-1}{N},\frac{i}{N}\right)$ 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.