## Motivation

Often used as a model for data we’re *counting*. This implies non-negative values and integers in a sequence, like $0,1,2,3,…$

A r.v. $X∼Pois$ models the total number of “successes” (in this case, something that we count) as a sequence of Bernoulli trials, where we have a large amount of trials and the probability of “success” (being counted) is small. Some examples:

- Number of text messages during this
*class period* - Number of car accidents
*within the next hour* - Number of raisins in a loaf of raisin bread: here, a “trial” is a
*very small*surface area of the bread, where we can either have 0 or 1 raisin (count, or no count).

Notice the similarities with the Binomial distribution. Often, when the number of trials is exceedingly large, calculating the PMF for $X∼Bin$ takes more compute time (a lot of multiplications), and accuracy issues (multiplying very large numbers). So the Poisson is often used as a *good approximation* in this case.

## PMF

Start with the PMF of the Binomial distribution, but take the number of trials to infinity, and $p$ to zero. Keep $np$ constant though (otherwise this would make no sense), so let $λ=np$. This is called the **rate**.

Calculate $lim_{n→∞}lim_{p→0}P(X=0)=(1−p)_{n}=(1−nλ )_{n}$. Taking the limit, $P(X=0)→e_{−λ}$.

For $k=1,2,3,…$, we calculate the ratio $P(X=k−1)P(X=k) $, beginning with $k=1$, because we have the denominator $P(X=k−1)=P(X=0)$.

Taking the limits, we have $P(X=k−1)P(X=k) →kλ $. From this, we can see $P(X=1)=e_{−λ}λ$, and generalizing this we have

$P(X=k)=k!e_{−λ}λ_{k} $There is no upper bound on the support $k$ for a r.v. $X$ that has the Poisson distribution. It takes one parameter, $λ$, so we can say $X∼Pois(λ)$.

## Properties of the distribution

If $X∼Bin(n,p)$, then as we take the limits of $n$ and $p$ to infinity and 0 respectively, then the PMF of $X$ converges to the PMF of the Poisson with $λ=np$.

The sum of independent Poissons is also a Poisson: given $X∼Pois(λ_{1})$ and $Y∼Pois(λ_{2})$, if $X$ and $Y$ are independent, then $X+Y∼Pois(λ_{1}+λ_{2})$.

If $X∼Pois(λ)$, then both Expectation $E(X)=λ$ and Variance $Var(X)=λ$.

### Poisson paradigm (informal)

This distribution is a good approximation when we have $n$ Bernoulli trials that each have different success probabilities, and aren’t perfectly independent.

Let $A_{1},A_{2},…,A_{n}$ be events where $p_{i}=P(A_{i})$, where $n≫p_{i}$, and $A_{i}$ is independent or weakly dependent. Let $I_{i}$ be an Indicator variable for $A_{i}$, and $X=∑_{i=1}I_{i}$. Then $X$ is approximately $Pois(λ)$, where $λ=∑_{i=1}p_{i}$.