Approximating a Binomial Random Variable Using A Poisson Random Variable

A random variable X, taking on values 0,1,2,… is said to be a Poisson random variable with parameter lambda if for some lambda >0,

$\begin{align} p(i) = P[X=i] = e^{-\lambda}(\frac{\lambda^i}{i!}) \quad\quad \text{ for i=0,1,2 ...} \end{align}$

A commonly known (or maybe not commonly known, depending on who you are) property of a Poisson random variable is that it can be used to approximate a Binomial random variable when n is large and p is small. The following is the derivation of why this is mathematical true:

First, lets set:

$\begin{align} \lambda = np \end{align}$ $\begin{align} P[X=i] = \frac{n!}{(n-i)!i!}p^i(1-p)^{n-i} \end{align}$ $\begin{align} P[X=i] = \frac{n!}{(n-i)!i!}\frac{\lambda}{n}^i(1-\frac{\lambda}{n})^{n-i} \end{align}$ $\begin{align} P[X=i] = \frac{n!}{(n-i)!i!}\frac{\lambda^i}{i!}\frac{ (1-\frac{\lambda}{n})^n }{ (1-\frac{\lambda}{n})^i } \end{align}$

Now, for a large n and small p

$\begin{align} \frac{n!}{(n-i)!i!} \quad\quad \text{ converges to 1} \end{align}$ $\begin{align} (1-\frac{\lambda}{n})^n \quad\quad \text{ converges to } \quad\quad e^{-\lambda} \end{align}$ $\begin{align} (1-\frac{\lambda}{n})^i \quad\quad \text{ converges to } \quad\quad 1 \end{align}$

Hence for large n and small p

$\begin{align} P[X=i] = e^{-\lambda}\frac{\lambda^i}{i!} \end{align}$