The Poisson distribution is a discrete probability distribution with identical mean and variance that expresses the probability of a given number of events occurring in a fixed interval of time. In this post, we derive the Poisson probability mass function beginning with the Binomial PDF.

The binomial distribution represents the number of successes, $$k$$ in $$n$$ independent Bernoulli trials, where the probability of success $$p$$ for any given trial is $$0 \leq p \leq 1$$. Formally, the binomial PDF is given by

$$f(k; p, n) = \binom{n}{k}p^{k}(1-p)^{n-k},$$

where $$\binom{n}{k}$$ represents the binomial coefficient, which equates to

$$\binom{n}{k} = \frac{n!}{n!(n-k)!}.$$

The expanded binomial PDF then appears as

$$f(k; p, n) = \frac{n!}{n!(n-k)!}p^{k}(1-p)^{n-k}.$$

The mean of the binomial distribution is

$$\mu = \lambda = np,$$

which can be rearranged in terms of $$p$$ to yield

$$p = \frac{\lambda}{n}.$$

Substituting this expression for $$p$$ into the binomial PDF results in

$$f(k; p, n) = \frac{n!}{n!(n-k)!}\Big(\frac{\lambda}{n}\Big)^{k}\Big(1-\frac{\lambda}{n}\Big)^{n-k}$$

If we let $$n$$ grow very large and $$p$$ grow very small, after a bit of re-expression we can reformulate the binomial distribution as

$$\lim_{n\to\infty , p \to 0} f(k; p, n) = \frac{n(n-1) \cdots (n-k+1)}{k!} \frac{\lambda^{k}}{n^{k}} \Big(1-\frac{\lambda}{n}\Big)^{n} \Big(1-\frac{\lambda}{n}\Big)^{-k}.$$

Swapping denominator terms yields

$$= \frac{n(n-1) \cdots (n-k+1)}{n^{k}} \frac{\lambda^{k}}{k!} \Big(1-\frac{\lambda}{n}\Big)^{n} \Big(1-\frac{\lambda}{n}\Big)^{-k}$$

Recall from calculus that $$e = \lim_{k\to\infty} (1 + 1/k)^{k}$$ and similarly $$e^{x} = \lim_{k\to\infty} (1 + x/k)^{k}$$. Leveraging these identities, $$(1 - \frac{\lambda}{n})^{n}$$ becomes $$e^{-\lambda}$$ in the limit. With very large $$n$$, the expression for the PDF reduces to

$$\lim_{n\to\infty} f(k; p, n) = (1) \Big( \frac{\lambda^{k}}{k!}\Big)(e^{-\lambda})(1).$$

After a little cleanup, we obtain the familiar expression for the Poisson density with mean $$\lambda$$:

$$f(k; \lambda) = \frac{\lambda^{k}}{k!}e^{-\lambda} \hspace{1cm} k \in 0, 1, 2, \cdots.$$

An interesting property of the Poisson distribution (as well as the Binomial and Negative Binomial distributions) is that successive probabilities can be found recursively. The recurrence relation is given by

$$\frac{p_{k}}{p_{k-1}} = a +\frac{b}{k}, \hspace{2mm} k = 2,3,4, \cdots,$$

where in the case of the Poisson distribution, $$a=0$$ and $$b=\lambda$$, resulting in

$$p_{k} = \frac{\lambda}{k}p_{k-1}, \hspace{2mm} k = 1,2,3, \cdots.$$

To demonstrate, for a system approximated by a Poisson distribution with $$\lambda = 3$$, the probabilities for $$0 \leq k \leq 6$$ are:

\begin{align*} p_{0} &= .049787068 \\ p_{1} &= .149361205 \\ p_{2} &= .224041808 \\ p_{3} &= .224041808 \\ p_{4} &= .168031356 \\ p_{5} &= .100818813 \\ p_{6} &= .050409407. \end{align*}

To use the recurrence relation, we first calculate $$p_0$$ as above. Then:

\begin{align*} p_{0} &=.049787068 \\ p_{1} &=p_{0}\frac{\lambda}{k} =.049787068*\frac{3}{1}=.149361205 \\ p_{2} &=p_{1}\frac{\lambda}{k} =.149361205*\frac{3}{2}=.224041808 \\ p_{3} &=p_{2}\frac{\lambda}{k} =.224041808*\frac{3}{3}=.224041808 \\ p_{4} &=p_{3}\frac{\lambda}{k} =.224041808*\frac{3}{4}=.168031356 \\ p_{5} &=p_{4}\frac{\lambda}{k} =.168031356*\frac{3}{5}=.100818813 \\ p_{6} &=p_{5}\frac{\lambda}{k} =.100818813*\frac{3}{6}=.050409407, \end{align*}

which are identical to the probabilities obtained from evaluating the Poisson PDF directly.