Relation between Binomial and Poisson distributions

The Poisson distribution is actually a limiting case of a Binomial distribution when the number of trials, $n$ , gets very large and $p$ , the probability of success, is small. As a rule of thumb, if $n \ge 100$ and $np \le 10$ , the Poisson distribution (taking $\lambda = np$ ) can provide a very good approximation to the binomial distribution.

This is particularly useful as calculating the combinations inherent in the probability formula associated with the binomial distribution can become difficult when $n$ is large.

To better see the connection between these two distributions, consider the binomial probability of seeing x successes in $n$ trials, with the aforementioned probability of success, $p$ , as shown below.

P(x) = \binom{n}{x} p^x q^{n-x}

Let us denote the expected value of the binomial distribution, $np$ , by $\lambda$ . Note, this means that

p=\frac{\lambda}{n}

and since $q=1-p$ ,

q=1-\frac{\lambda}{n}

Now, if we use this to rewrite $P(x)$ in terms of $\lambda$ , $n$ , and $x$ , we obtain

P(x) = \binom{n}{x} \left( \frac{\lambda}{n} \right)^x \left( 1-\frac{\lambda}{n} \right)^{n-x}

Using the standard formula for the combinations of $n$ things taken $x$ at a time and some simple properties of exponents, we can further expand things to

P(x) = \frac{n(n-1)(n-2) \cdots (n-x+1)}{x!} \cdot \frac{\lambda^x}{n^x} \left( 1 - \frac{\lambda}{n} \right)^{n-x}

Notice that there are exactly $x$ factors in the numerator of the first fraction. Let us swap denominators between the first and second fractions, splitting the $n^x$ across all of the factors of the first fraction’s numerator.

P(x) = \frac{n}{n} \cdot \frac{n-1}{n} \cdots \frac{n-x+1}{n} \cdot \frac{\lambda^x}{x!}\left( 1 - \frac{\lambda}{n} \right)^{n-x}

Finally, let us split the last factor into two pieces, noting (for those familiar with Calculus) that one has a limit of $e^{-\lambda}$ .

P(x) = \frac{n}{n} \cdot \frac{n-1}{n} \cdots \frac{n-x+1}{n} \cdot \frac{\lambda^x}{x!}\left( 1 - \frac{\lambda}{n} \right)^n \left( 1 - \frac{\lambda}{n} \right)^{-x}

It should now be relatively easy to see that if we took the limit as n approaches infinity, keeping $x$ and $\lambda$ fixed, the first $x$ fractions in this expression would tend towards $1$ , as would the last factor in the expression. The second to last factor, as was mentioned before, tends towards $e^{-\lambda}$ , and the remaining factor stays unchanged as it does not depend on $n$ . As such,

\lim_{n \rightarrow \infty} P(x) = \frac{e^{-\lambda} \lambda^x}{x!}

Which is what we wished to show.