The formula for Poisson Distribution is as follows:
$$P(X=x) = \frac{e^{-\lambda} \cdot \lambda^x }{x!}$$
Where,
- $\lambda =$ average (or, mean) rate of an event to take place. In other words, mean number of events in the interval.
- $x = $ actual number of events in the interval.
My question is,
What is the explanation behind each of these components and their operators?
For instance,
- $x$ has a factorial
- $\lambda$ is raised to the power of $x$
- exponent $e$ is multiplied
- $e$ is raised to the power of $-\lambda$
What do all these things imply individually?
The PMF for a binomially distributed R.V. with parameters $n,p$ is given by: $$ P(X = k) = {{n} \choose {k}}p^{k}(1-p)^{n-k} $$ Set $\lambda = np$, so that $p = \frac{\lambda}{n}$. Our goal is to show that:
$$ \lim_{n \to \infty} P(X = k) = \lim_{n\to\infty}{{n} \choose {k}}p^{k}(1-p)^{n-k} = \frac{e^{-\lambda}\lambda^k}{k!} $$ Substituting for $p$, we have: $$ P(X = k) = {{n} \choose {k}}\left(\frac{\lambda}{n}\right)^{k}\left(1-\frac{\lambda}{n}\right)^{n-k} $$ Expanding, we get: $$ P(X = k) = \frac{n!}{k! (n-k)!}\left(\frac{\lambda}{n}\right)^{k}\left(1-\frac{\lambda}{n}\right)^{n-k} =\frac{\lambda^k}{k!} \frac{n!}{(n-k)!}\frac{1}{n^k}\left(1-\frac{\lambda}{n}\right)^{n-k} $$ Notice now that: $$ \frac{n!}{(n-k)!n^{k}} = \frac{n(n-1)(n-2) \cdots (n-k)}{n^k} $$ Which in the limit becomes $1$ as there are $k$ powers of $n$ in the numerator and $k$ in the denominator (the constants don't matter as $n$ grows large). We are thus left to evaluate:
$$ \lim_{n\to \infty} \left(1-\frac{\lambda}{n}\right)^{n-k} = \lim_{n\to \infty} \left(1-\frac{\lambda}{n}\right)^{n} \left(1-\frac{\lambda}{n}\right)^{-k} = \lim_{n\to \infty} \left(1-\frac{\lambda}{n}\right)^{n} = e^{-\lambda} $$ as we wanted. Putting things together, we get from the first and last steps: $$ \lim_{n \to \infty} P(X = k) = \frac{\lambda^ke^{-\lambda}}{k!} $$