The poisson approximation to binomial probabilities

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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.

$$P[x]={}_nC_x p^x q^{n-x}$$

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

$$P[x] = {}_nC_x \left[ \frac{\lambda}{n} \right]^x \left[ 1-\frac{\lambda}{n} \right]^{n-x}$$

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.

Poisson Approximation of Binomial Probabilities

This page need be used only for those binomial situations in which n is very large and p is very small. For example: The null hypothesis holds that a certain genetic characteristic will express itself in p=.001 of the population. In a sample of n=3000 subjects, k=7 are observed to display the characteristic, whereas only np=3 are expected. On the null hypothesis, how likely is it that a rate this great or greater could occur by mere chance? Your computer would not be able to perform the factorial and exponential operations required for direct calculation [Exact Binomial Probability Calculator], and np 20 and np < 5 OR nq < 5 then the Poisson is a good approximation.