Relation between Binomial and Poisson distributions

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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.

$$P(x)=\left(\genfrac{}{}{0ex}{}{n}{x}\right)p^x{q}^{n-x}$$

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

$$P(x)=\left(\genfrac{}{}{0ex}{}{n}{x}\right){\left(\frac{\lambda }{n}\right)}^x{\left(1-\frac{\lambda }{n}\right)}^{n-x}$$

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,

$$\underset{n\to \infty }{\lim }P(x)=\frac{e^{-\lambda }{\lambda }^x}{x!}$$

Which is what we wished to show.