September 7, 2026
Today we are going to talk about the Poisson distribution, or rather, the intuition behind it and what makes a process truly random, and how its seemingly cryptic probability mass function (PMF) emerges from this random process.
Imagine a unit period of time. Define the random variable \(X\) as the number of times an event occurs in this period of time, assuming the expected number of occurrences is \(\lambda\). That is, the rate of occurrence is \(\lambda\) events per unit time. Obviously, \(X\) is a discrete random variable and its possible values are all non-negative integers (\(0, 1, 2, 3, \dots\)).
What is its pmf? I.e. what is \(P(X = k)\)?
To start, we need to make some assumptions about the nature of how those "events" arrive. The assumption is that those events arrive truly randomly, not influenced by other arrivals nor the exact timings of the arrivals within this unit time period.
This sounds super hand-wavy, so let's formalize it into two conditions:
Let's look at two counterexamples showing just how those two conditions uphold the notion of total "randomness" in the arrival process of corresponding events. Suppose we are to tally the number of goals that a soccer player is to score, case 1: in a match, and case 2: in their lifetime. Now let's see if each case satisfies those conditions by mapping onto real-world assumptions.
Will the number of events, i.e. the number of goals scored in a single match, satisfy the independent intervals constraint? No. Think about it, if say in the first half, the team scored a deluge of goals, and the player in question is a striker, then it is likely that the coach would sub the player off, or at least adopt a more defensive strategy, and we would expect fewer goals to be scored in the second half, and vice versa. This means that the number of events, or goals scored, in the first half affects that in the second half, so this does NOT realistically justify the independent interval condition if modeled as such.
For case 2, if we were to model a soccer player's goals scored within that player's entire lifetime, then the stationarity condition (condition 2) would not be a realistic assumption. Why? Because a player's performance is affected by age and body conditions, and those change throughout one's lifetime, so a match played in that player's prime would not have the same distribution as that played, say, in that player's youth, or later ages.
The above examples illustrate how the two conditions model an intuitively "random" arrival process. Something that has no memory of its past behavior nor of the present time, and that has the same event distribution over every equal-length interval, is considered, for all practical purposes, random, and can be reasonably modeled by the Poisson distribution.
So consider a unit of time where the arriving events are truly random, as they satisfy both the independent interval and the stationarity conditions, and therefore the random variable \(X\) denoting the number of events occurring in that unit of time is considered Poisson. It is also expected that \(\lambda\) events on average occur during that unit period of time. Now what?
Answer: divide and conquer. Or more specifically, let's divide the unit time period into \(N\) equal time slots. Now ask ourselves, can we analyze what happens in each of the time slots independently and identically? The answer is yes! Why? The independent interval condition means that in each time slot, events occur independently of other time slots. Stationarity says that since the time slots are of equal length and only the length of an interval affects event arrivals, each slot has the same event distribution. I.e. the number of events occurring in each of the \(N\) time slots is independent and identically distributed (i.i.d.)!
Now it remains to figure out what that distribution is for each individual time slot. Remember, we will let \(N \to \infty\), hence the duration of each time slot goes to zero. In each time slot, there are three scenarios: the number of events being 0, 1, or greater than 1. Another key assumption for the "random" events occurring in our premise is that the probability that 2 or more events occur in an ever-shrinking time slot goes to 0 (as the probability that even one event occurrence gets rarer and rarer as \(N \to \infty\)). This can be considered a "third" condition, if you will. Thus, the random variable representing the number of events occurring in each time slot actually approaches a Bernoulli random variable as \(N \to \infty\): either the event occurred in that time slot or it didn't. Since we know the rate of occurrence for the entire unit time period is \(\lambda\) events per unit time period, then for each sub-interval time slot, the probability that an event occurred is simply \(\frac{\lambda}{N}\). Thus, the probability that the event didn't occur in that time slot is \(1 - \frac{\lambda}{N}\), and this is i.i.d. across all time slots.
Now, \(X\) is simply the sum of \(N\) i.i.d. Bernoulli random variables with probability of success being \(\frac{\lambda}{N}\), and the sum of i.i.d. Bernoulli random variables yields a binomial random variable. \(X\), being Poisson, is simply the limit of that binomial random variable as \(N \to \infty\). So we have:
\begin{align} P(X = k) &= \lim\limits_{N \to \infty} \binom{N}{k} \left(\frac{\lambda}{N}\right)^k \left(1 - \frac{\lambda}{N}\right)^{N - k} \\ \\ &= \lim\limits_{N \to \infty} \frac{N!}{k!(N - k)!} \left(\frac{\lambda}{N}\right)^k \left(1 - \frac{\lambda}{N}\right)^{N - k} \\ \\ &= \lim\limits_{N \to \infty} \frac{\lambda^k}{k!} \left(\frac{N^k + \text{lower degree terms of } N}{N^k}\right) \left(1 + \frac{(-\lambda)}{N}\right)^{N} \left(1 - \frac{\lambda}{N}\right)^{-k} \\ \\ &= \frac{\lambda^k}{k!}e^{-\lambda} \end{align}
After some simplifications, we see that we can separate out the \(\frac{\lambda^k}{k!}\) part as it is independent of \(N\). Note that \(\frac{N!}{(N-k)!} = N(N-1)\cdots(N-k+1)\), which is a degree-\(k\) polynomial in \(N\), so the second product term is a ratio of polynomials in terms of \(N\) and hence its limit is the ratio of the dominant terms, which is \(1\). The last product term \(\left(1 - \frac{\lambda}{N}\right)^{-k}\) also goes to \(1\), since \(k\) is held fixed while the base goes to \(1\). Finally, the \(\left(1 + \frac{(-\lambda)}{N}\right)^{N}\) product term as \(N \to \infty\) is the famous exponential function limit equalling \(e^{-\lambda}\).
Therefore, the pmf of the Poisson distribution with rate \(\lambda\) is:
\begin{align} P(X = k) &= \frac{\lambda^k e^{-\lambda}}{k!} \end{align}
First, let's do a sanity check. From the pmf, we can see that for \(k = 0, 1, 2, 3, \dots\), \(k! > 0\). Also, since \(\lambda\) denotes the rate of events occurring, we can reasonably assume that \(\lambda > 0\). Therefore both \(\lambda^k\) and \(e^{-\lambda}\) are greater than 0. This means \(P(X = k) > 0\) for all accepted values of \(k\).
Next, we ensure that the total pmf over acceptable values of \(k\) adds up to 1:
\begin{align} \sum_{k = 0}^{\infty}P(X = k) &= \sum_{k = 0}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} \\ \\ &= e^{-\lambda} \sum_{k = 0}^{\infty} \frac{\lambda^k}{k!} \\ \\ &= e^{-\lambda} e^{\lambda} \\ \\ &= 1 \end{align}
The second-to-last step uses the famous Taylor series expansion of the exponential function.
Now, \(\lambda\) is given as the rate of events occurring, that is, number of events occurring per unit period of time. Therefore, in that unit period of time, the expected number of events occurring is \(\lambda\). Now let's use the official definition of expectation to see if the result corresponds to the given.
\begin{align} E[X] &= \sum_{k = 0}^{\infty} kP(X = k) \\ \\ &= \sum_{k = 0}^{\infty} k \frac{\lambda^k e^{-\lambda}}{k!} \\ \\ &= \sum_{k = 1}^{\infty} k \frac{\lambda^k e^{-\lambda}}{k!} \\ \\ &= \sum_{k = 1}^{\infty} \frac{\lambda^k e^{-\lambda}}{(k-1)!} \\ \\ &= \sum_{m = 0}^{\infty} \frac{\lambda^{m + 1} e^{-\lambda}}{m!} \quad \text{(letting } m = k - 1 \text{)}\\ \\ &= \lambda \sum_{m = 0}^{\infty} \frac{\lambda^{m} e^{-\lambda}}{m!} \\ \\ &= \lambda \end{align}
As expected.
Now let's look at the variance of \(X\):
\begin{align} \operatorname{Var}[X] &= E[X^2] - \left(E[X]\right)^2 \\ \\ &= \sum_{k = 0}^{\infty} k^2 \frac{\lambda^k e^{-\lambda}}{k!} - \lambda^2 \\ \\ &= \sum_{k = 1}^{\infty} k \frac{\lambda^k e^{-\lambda}}{(k-1)!} - \lambda^2 \\ \\ &= \sum_{m = 0}^{\infty} (m + 1) \frac{\lambda^{m + 1} e^{-\lambda}}{m!} - \lambda^2 \quad \text{(letting } m = k - 1 \text{)}\\ \\ &= \lambda \left(\sum_{m = 0}^{\infty} m\frac{\lambda^{m} e^{-\lambda}}{m!} + \sum_{m = 0}^{\infty} \frac{\lambda^{m} e^{-\lambda}}{m!}\right) - \lambda^2 \\ \\ &= \lambda \left(\lambda + 1\right) - \lambda^2 \\ \\ &= \lambda^2 + \lambda - \lambda^2 \\ \\ &= \boxed{\lambda} \end{align}
So it turns out that the variance for a Poisson distribution is also \(\lambda\). What does it mean? This means that as the rate increases, the variance in the number of arrivals also increases. This is why sometimes we get "bursts" (or "droughts") of events occurring in a period that deviates greatly from the expected average.