The Poisson Distribution

STA 211 · Lecture 09

Introduction

The Poisson distribution models counts of events occurring randomly in a fixed interval of time, space, area or volume. It is appropriate when events occur independently at a stable average rate and simultaneous events in a very small interval are negligible.

Learning Objectives

You should be able to:

  • recognise a Poisson count model;
  • calculate Poisson probabilities;
  • derive its mean, variance and MGF;
  • explain its process assumptions and additivity; and
  • use it as an approximation to the binomial distribution.

Poisson Probability Function

A random variable XX has a Poisson distribution with parameter λ>0\lambda > 0, written X∼Pois(λ)X \sim Pois(\lambda), if

P(X=x)=e−λλxx!,x=0,1,2,….P(X = x) = e^{- \lambda}\frac{\lambda^{x}}{x!},\quad\quad x = 0,1,2,\ldots.

The parameter λ\lambda is both the mean and variance:

E(X)=λ,E(X) = \lambda,

Var(X)=λ.Var(X) = \lambda.

The MGF is

MX(t)=exp{λ(et−1)}.M_{X}(t) = exp\{\lambda\left( e^{t} - 1 \right)\}.

Poisson probability mass function

Poisson probability mass function

Poisson Process Interpretation

For a homogeneous Poisson process with rate ρ\rho per unit interval, the count in an interval of length tt has parameter

λ=ρt.\lambda = \rho t.

The model assumes:

  1. counts in disjoint intervals are independent;
  2. the event rate is constant;
  3. the probability of one event in a very short interval is approximately proportional to its length; and
  4. the probability of two or more events in a very short interval is negligible relative to its length.

Example 9.1: Calls

A help desk receives an average of 4 calls per 10 minutes. The probability of exactly 6 calls in 10 minutes is

e−4466!≈0.1042.e^{- 4}\frac{4^{6}}{6!} \approx 0.1042.

The probability of no calls is

P(X=0)=e−4≈0.0183.P(X = 0) = e^{- 4} \approx 0.0183.

Recurrence Relation

Successive probabilities satisfy

P(X=x+1)=λx+1P(X=x).P(X = x + 1) = \frac{\lambda}{x + 1}P(X = x).

Additivity

If Xi∼Pois(λi)X_{i} \sim Pois\left( \lambda_{i} \right) are independent, then

∑i=1nXi∼Pois(∑i=1nλi).\sum_{i = 1}^{n}X_{i} \sim Pois\left( \sum_{i = 1}^{n}\lambda_{i} \right).

This follows immediately by multiplying their MGFs.

Poisson Limit of the Binomial

If Xn∼Bin(n,pn)X_{n} \sim Bin\left( n,p_{n} \right), n→∞n \rightarrow \infty, pn→0p_{n} \rightarrow 0, and npn→λnp_{n} \rightarrow \lambda, then

P(Xn=x)→e−λλxx!.P\left( X_{n} = x \right) \rightarrow e^{- \lambda}\frac{\lambda^{x}}{x!}.

This justifies the approximation

Bin(n,p)≈Pois(np)Bin(n,p) \approx Pois(np)

when nn is large and pp is small. A common practical guide is p≤0.05p \leq 0.05 and npnp not too large, though accuracy should be assessed in context.

Example 9.2: Rare defects

In 200 independent items, each has defect probability 0.01. With λ=np=2\lambda = np = 2,

P(X≤1)≈e−2(1+2)=3e−2≈0.4060.P(X \leq 1) \approx e^{- 2}(1 + 2) = 3e^{- 2} \approx 0.4060.

Post-Test

  1. State the Poisson PMF and support.
  2. If X∼Pois(3)X \sim Pois(3), find P(X=0)P(X = 0) and P(X≥1)P(X \geq 1).
  3. Calls arrive at 6 per hour. Find the probability of 2 calls in 20 minutes.
  4. If X∼Pois(2)X \sim Pois(2) and Y∼Pois(5)Y \sim Pois(5) are independent, state the distribution of X+YX + Y.
  5. Approximate P(X=0)P(X = 0) for X∼Bin(500,0.004)X \sim Bin(500,0.004) using a Poisson model.

Selected answers: 2. e−3e^{- 3} and 1−e−31 - e^{- 3}; 3. with λ=2\lambda = 2, P(X=2)=2e−2P(X = 2) = 2e^{- 2}; 4. Pois(7)Pois(7); 5. e−2e^{- 2}.