Poisson Distribution

Poisson Distribution

A random variable X has a poisson distribution with parameter μ>0 if the Probability Distribution, PMF of X is:

p(x;μ)=e−μμxx!:x=0,1,2,…

Where did this come from? Unlike the Binomial Experiment that came from doing trials, we'll see that it can be obtained from the binomial distribution by certain limiting operations.

We'll see that μ is the expected value of this distribution. Since μ>0 then we have p(x;μ)>0. The fact that ∑x=0∞p(x;μ)=1 is a consequence of the Maclaurin Series of eμ:

eμ=1+μ+μ22!+⋯=∑x=0∞μxx!

Multiply both sides by e−μ:

1=∑x=0∞e−μμxx!=∑x=0∞p(x;μ)
Proposition

Suppose that in the binomial Probability Distribution, PMF b(x;n,p) we let n→∞ and p→0 in such a way that np→μ>0. Then b(x;n,p)→p(x;μ) (see Binomial Experiment), the Poisson pmf.

Proof
Begin with the binomial pmf:

b(x;n,p)=(nx)px(1−p)n−x=n!x!(n−x)!px(1−p)n−x=n⋅(n−1)⋅⋯⋅(n−x+1)x!px(1−p)n−x=nn⋅n−1n⋅⋯⋅n−x+1n⋅(np)xx!⋅(1−p)n(1−p)−xMult. by nx

By taking n→∞ and p→0 with np→μ. Then:

b(x;n,p)→1⋅1⋅⋯⋅1⋅μxx!⋅limn→∞,np→μ(1−npn)n⋅(1−0)−x

From calculus (1−an)n→e−a as n→∞. It's also the same that (1−ann)n→e−a as n→∞ if an→a. Here since we have an=np→μ then:

b(x;n,p)→μxx!⋅e−μ=p(x;μ)

☐
The idea here is that if:

  1. Our population is large: n→∞
  2. Our probability is small: p→0
    Then b(x;n,p)≈p(x;μ) where μ=np.

The Mean, Variance, and Moment Generating Function

Proposition

If X has a Poisson distribution with parameter μ, then E(X)=Var=μ.

These are derived directly from the Expected Value and Variance and Standard Deviation of X definitions.

The Poisson Process

Any process that has a distribution where the "events" are just arrivals of some object is called a Poisson Process. Here we have μ=λt where λ is the rate of the process. It's good when we need to count these events over a period of time.