\begin{align}
\end{align}
Motivation
Suppose \( X \) follows a binomial distribution \( X \equiv bin(n,p) \) with n large say, 10,000 , and its mean a reasonable value of around 3,4.
\begin{align}
P(X = k) &= \binom{n}{p} p^k (1-p)^{n-k} \\
& = \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}
\end{align}
Let \( \frac{\lambda}{n} = p \)
Making the following two approximations, for large n
$$ \binom{n}{k} k! \approx n^k $$ and $$ e^{-\lambda} \approx (1-\frac{\lambda}{n})^n$$
\begin{align}
P(X=k) = \frac{\lambda^k}{k!}e^{-\lambda}
\end{align}
Let’s check the probability sums to 1.
\begin{align}
&\sum_{k=0}^{\infty} P(X=k) = \sum_{k=0}^{\infty} \frac{\lambda^k}{k!}e^{-\lambda} \\
&= e^{-\lambda}\sum_{k=0}^{\infty} \frac{\lambda^k}{k!} = e^{-\lambda} .e^{\lambda} = 1
\end{align}
A random variable which satisfied \( P(X=k) = \frac{\lambda^k}{k!}e^{-\lambda}\) is known as a Poisson random variable. \begin{align} \mathbb{E}[X] &= \sum_{k=0}^{\infty} k \frac{\lambda^k e^{-\lambda}}{k!} = \sum_{k=1}^{\infty} \frac{\lambda^k e^{-\lambda}}{(k-1)!}\\ &\lambda \sum_{k=1}^{\infty} \frac{\lambda^{k-1} e^{-\lambda}}{(k-1)!} = \lambda \end{align} \begin{align} \mathbb{E}[X^2]& = \sum_{k=0}^{\infty} k^2 \frac{\lambda^k e^{-\lambda}} {k!} \\ &=\sum_{k=1}^{\infty} (k-1)\frac{\lambda^k e^{-\lambda}} {(k-1)!} + \sum_{k=1}^{\infty} \frac{\lambda^k e^{-\lambda}} {(k-1)!} \\ &= \lambda^2 \sum_{k=2}^{\infty} \frac{\lambda^{k-2} e^{-\lambda}} {(k-2)!} + \lambda \sum_{k=1}^{\infty} \frac{\lambda^{k-1} e^{-\lambda}} {(k-1)!} \\ = \lambda^2 + \lambda \end{align} so \( \text{Var}[X] = \lambda^2 + \lambda -\lambda^2 = \lambda \)
Some Examples of modelling with a Poisson distribution.
1. Number of busses arriving in one hour.
2. Number of defective products produced in a factory in a day.