The Poisson Distribution

The Poisson distribution counts how many rare events land in a fixed window of time or space: orders arriving in a second, defaults in a year, typos on a page, meteor strikes in a decade. The recipe is many opportunities for something to happen, each individually unlikely, with no fixed cap on the total. A single number, the average count $\lambda$, controls the whole shape.

From binomial to Poisson

Think of a minute split into thousands of tiny slivers, each of which might or might not contain an order. That is a The Binomial Distribution with a huge number of trials $n$ and a tiny success probability $p$. If you let $n$ grow without bound and $p$ shrink toward zero while keeping the average count $np$ pinned at a fixed value $\lambda$, the binomial formula converges to the Poisson. This is why it is nicknamed the law of rare events: it is what "many trials, each rarely a success" looks like in the limit.

The formula

$$P(X = k) = e^{-\lambda}\, \frac{\lambda^k}{k!} .$$

The symbols: $\lambda$ (lambda) is the average number of events expected in the window; $k$ is the specific count you are asking about ($0, 1, 2, \dots$, with no upper limit); $k!$ is $k$ factorial; and $e \approx 2.718$ is the base of the natural logarithm. The factor $e^{-\lambda}$ is a normalizer that makes all the probabilities sum to 1.

Slide the rate $\lambda$ below. At small $\lambda$ the distribution is lopsided and jammed against zero; as $\lambda$ grows the peak marches right, the mound fattens, and it starts to look like a bell — the same normal shape the The Central Limit Theorem promises.

Mean equals variance

The Poisson's defining fingerprint is that its mean and variance are the same number:

$$E[X] = \lambda, \qquad \operatorname{Var}(X) = \lambda .$$

This is a genuinely testable claim. If you collect count data and find the variance sitting far above the mean, the data is overdispersed and a Poisson model is wrong, a diagnostic you can run in your head before trusting any Poisson-based estimate.

Worked example

A trading desk receives large orders at an average rate of $\lambda = 3$ per minute. What is the probability that exactly 5 arrive in a given minute? Set $k = 5$:

$$P(X = 5) = e^{-3}\, \frac{3^5}{5!} = e^{-3}\, \frac{243}{120} .$$

Now $243/120 = 2.025$ and $e^{-3} \approx 0.0498$, so

$$P(X = 5) \approx 0.0498 \times 2.025 \approx 0.101 .$$

About a 10% chance of 5 orders in the minute. For contrast, the probability of a completely quiet minute is $P(X = 0) = e^{-3} \approx 0.0498$, just under 5%, and the expected number of orders is simply $\lambda = 3$, sitting right under the tallest bars.

Where it misleads

• Overdispersion is common. Real-world counts often clump, pushing the variance above the mean; when that happens the negative binomial, which adds a spread parameter, fits far better.
• Events must be independent and the rate steady. Poisson assumes arrivals don't trigger one another and $\lambda$ doesn't drift. Order flow that comes in bursts around news violates both, and the true tail is heavier than Poisson admits.
• The rate scales with the window. $\lambda$ is tied to the interval length: 3 per minute means 6 per two minutes and $0.05$ per second. And independent Poisson streams merge simply, their rates just add.
• Large $\lambda$ turns Gaussian. For big average counts a Poisson$(\lambda)$ is well approximated by a normal with mean and variance both $\lambda$, another face of the The Central Limit Theorem.

The Poisson is one node in the web of Common Distributions: it is the rare-event limit of the The Binomial Distribution on one side and the count that pairs with exponential waiting times on the other. Knowing its mean, its variance, and the fact that they coincide covers most of what interviews ask, and connects directly back to Expected Value.