2.2b.9 Poisson Distribution

Poisson Distribution

A discrete random variable follows the Poisson distribution $X\sim \text{Pois}(\lambda )$ if it represents the number of events occurring in a fixed interval of time or space, given that these events occur independently and at a constant average rate $\lambda$. Formally, the Poisson distribution of a random variable is given by the probability mass function,
$p_X(x)=e^{-\lambda }\frac{{\lambda }^x}{x!},$
where if $\lambda <1$, the PMF is monotonically decreasing, and if $\lambda >1$, the PMF first increases then decreases as $x\to \infty$.

Intuition

The Poisson distribution is used in “rare” events such as in phone call disconnections, radioactive emissions, crossovers in chromosomes, and calculating the number of cases of a disease. This model is appropriate when some event can be assumed to happen randomly with a constant density in a continuous region (of time or space), such that occurrences in disjoint subregions are independent (each event is considered a separate occurrence).

Properties

(Mean) $\mathbb{E}[X\sim \text{Pois}(\lambda )]=\lambda$.

We calculate the expectation of a poisson random variable using the Taylor series expansion $e^x={\sum }_{n=0}^{\infty }\frac{x^n}{n!}$,

$$\begin{array}{c}\begin{array}{rl}\mathbb{E}[X]& =\sum _{i=0}^{\infty }i\times \mathbb{P}[X=i]\\ & =\sum _{i=1}^{\infty }i\frac{{\text{uplambda}}^i}{i!}{e}^{-\text{uplambda}}\\ & =\lambda e^{-\text{uplambda}}\sum _{i=1}^{\infty }\frac{{\text{uplambda}}^{i-1}}{(i-1)!}\\ & =\lambda e^{-\text{uplambda}}{e}^{\text{uplambda}}\\ & =\lambda \end{array}\end{array}$$

Note that we omit the 0th term due to it being equal to 0.

(Variance) $Var[X\sim \text{Pois}(\lambda )]=\lambda$.

We already know that $\mathbb{E}[X]=\text{uplambda}$. Therefore we need to calculate $\mathbb{E}[X(X-1)]$ (0 is again trivial in the summation, as well as 1 this time):

$$\begin{array}{c}\begin{array}{rl}\mathbb{E}[X(X-1)]& =\sum _{i=2}^{\infty }i(i-1)\frac{{\lambda }^i}{i!}{e}^{-\lambda }\\ & ={\lambda }^2{e}^{-\lambda }\sum _{i=2}^{\infty }\frac{{\lambda }^{i-2}}{(i-2)!}\\ & ={\lambda }^2\end{array}\end{array}$$

Knowing this, we can calculate the variance from the expansion $\mathbb{E}[X^2]=\mathbb{E}[X(X-1)]+\mathbb{E}[X]$

$$\begin{array}{c}\begin{array}{rl}Var(X)& =\mathbb{E}[X(X-1)]+\mathbb{E}[X]-\mathbb{E}[X{]}^2\\ & ={\text{uplambda}}^2+\text{uplambda}-{\text{uplambda}}^2\\ & =\text{uplambda}\end{array}\end{array}$$

(Sum) If $X\sim \mathrm{Pois}(\lambda )$ and $Y\sim \mathrm{Pois}(\mu )$, then $X+Y=\mathrm{Pois}(\lambda +\mu )$.

(Binomial Approximation)

Examples

Probability of Typos on a Page

Suppose we are writing an article and make on average 1 typo per page. We can then model a Poisson distribution with $\lambda =1$. So the probability that we make 5 typos on a page is
$\mathbb{P}[X=5]=\frac{1^5}{5!}{e}^{-1}=\frac{1}{120e}\approx \mathrm{0.003.}$
Now suppose the article has 200 pages. If the number of typos on each page is independent, then the probability that there is at least one page with exactly 5 typos is

\begin{equation}

\begin{aligned}

\mathbb{P}[\exists \text{a page with exactly 5 typos}] &= 1 - \mathbb{P}[\text{every page has } \neq 5 \text{ typos}] \
&= 1-\prod^{200}{i=1}\mathbb{P}[\text{page i has } \neq 5 \text{ typos}] \
&= 1-\prod^{200}{i=1}(1 - \mathbb{P}[X=5]) \
&= 1-\left(1 - \frac{1}{120e}\right)^{200} \approx 0.46
\end{aligned}
\end{equation}