The Poisson Process

The Poisson process counts independent, exponentially-spaced arrivals in a fixed interval.

Previously, we modeled the time between independent events using the exponential distribution. We gave the example of customers arriving at a coffee shop at a rate of $\lambda = 3$ per hour (where the tick is one hour). The time between arrivals follows $\text{Exp}(3)$. This gives us an expected gap of 20 minutes between customers.

Note: If the tick were 1 minute, instead of 1 hour, $\lambda = 0.05$. The tick is important!

Now we flip the question. Instead of asking, “How long until the next customer?”, we ask: “How many customers arrive in a given window of time?” If we watch the shop for an entire 8-hour shift, how many customers should we expect?

The expected value is obvious. An average of 3 customers per hour. 8 hours. We should expect 24 customers in 8 hours. But what about every other possibility?

The answer is the Poisson distribution.

From Exponential to Poisson

The derivation starts with the same setup. Events arrive at a constant average rate $\lambda$. We want to count how many events occur in a fixed interval of length $t$.

Step 1: Partition the interval. Slice $[0, t]$ into $n$ equal subintervals, each of length $\frac{t}{n}$.

Step 2: Probability per subinterval. For large $n$, each subinterval is so short that at most one event can fall in it. The probability of exactly one event in a subinterval of length $\frac{t}{n}$ is approximately:

$P(\text{one event}) \approx \lambda \cdot \frac{t}{n}$

Step 3: Recognize the binomial. Each subinterval either contains an event or it doesn’t, independently of the others. That’s a classic binomial setup: $n$ independent trials, each with success probability $p = \frac{\lambda t}{n}$. The total number of events across all $n$ subintervals follows:

$X \sim \text{Binomial}\!\left(n,\, p\right) = \text{Binomial}\!\left(n,\, \frac{\lambda t}{n}\right)$

Here $n$ is the number of subintervals (trials) and $p = \frac{\lambda t}{n}$ is the probability of exactly one event in each subinterval. Notice that as $n$ grows, $p$ shrinks proportionally. We continue to slice the interval ever finer while keeping the total expected count $np = \lambda t$ fixed.

The probability of exactly $k$ events is:

$P(X = k) = \binom{n}{k} \left(p\right)^k \left(1 - p\right)^{n-k}, \quad \text{where} \space p = \frac{\lambda t}{n}$

$P(X = k) = \binom{n}{k} \left(\frac{\lambda t}{n}\right)^k \left(1 - \frac{\lambda t}{n}\right)^{n-k}$

Step 4: Take the limit. Let $n \to \infty$ — infinitely many, infinitely thin subintervals. Each factor converges separately:

$\binom{n}{k} \cdot \frac{1}{n^k} \;\to\; \frac{1}{k!} \qquad \left(\frac{\lambda t}{n}\right)^k \cdot n^k = (\lambda t)^k \qquad \left(1 - \frac{\lambda t}{n}\right)^n \;\to\; e^{-\lambda t}$

Multiplying the pieces together:

$P(X = k) = \frac{(\lambda t)^k \, e^{-\lambda t}}{k!}, \quad k \in \mathbb{N}_0$

This is the Poisson distribution.

The Poisson Distribution

The Poisson probability mass function (PMF) has one parameter: $\mu = \lambda t$, the expected number of events in the interval.

Note: We say PMF rather than PDF because $k$ is discrete. You can have 0, 1, 2, or 24 customers, but you can’t have 3.7 customers. The exponential distribution used a PDF because time is continuous and can take any non-negative real value.

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

Its key properties:

The mean and variance are equal. This is the fingerprint of the Poisson distribution — if you fit a Poisson model to real data and find the sample mean and variance are very different, that’s a signal the Poisson isn’t the right model.

Back to the coffee shop. With $\lambda = 3$ and $t = 8$, we get $\mu = \lambda t = 24$. Over a full shift, the expected number of customers is 24. But how does the probability spread?

$P(X = 20) = \dfrac{24^{20} \, e^{-24}}{20!} \approx 0.0624$

$P(X = 24) = \dfrac{24^{24} \, e^{-24}}{24!} \approx 0.0812$

$P(X = 30) = \dfrac{24^{30} \, e^{-24}}{30!} \approx 0.0363$

Even the most likely outcome ($k = 24$) only occurs about 8% of the time. The probability spreads across a wide range. That’s typical for a Poisson distribution with a large mean — it looks roughly bell-shaped, which is no accident (the central limit theorem explains why).

Interactive PMF

Use the slider to see how $\mu = \lambda t$ shapes the distribution.

A few things to notice:

• Small $\lambda t$: the distribution is right-skewed, concentrated near zero.
• Large $\lambda t$: the distribution spreads and approaches a normal distribution.
• The peak always sits near $k = \lambda t$, confirming that $\lambda t$ is both the mean and the most likely value (approximately).

Two Sides of the Same Coin

The exponential distribution and the Poisson distribution describe the same underlying process from two different perspectives. Both stem from the same assumption: events arrive independently at a constant average rate $\lambda$.

If you know $\lambda$ for one, you know it for the other. The two distributions are mathematically inseparable. They are two ways of viewing the same process.

Real-World Examples

Anywhere events arrive at a roughly constant, independent rate, you can count them with the Poisson distribution:

• Customer arrivals. A coffee shop sees $\lambda = 3$ arrivals per hour. In an 8-hour shift, the customer count follows $\text{Poisson}(24)$.

• Server requests. A web server handles $\lambda = 500$ requests per minute. The number of requests in a 10-second window follows $\text{Poisson}(83.3)$.

• Manufacturing defects. A factory produces defects at a rate of $\lambda = 0.02$ per item. In a batch of 1,000 items, the defect count follows $\text{Poisson}(20)$.

• Natural events. A seismograph records minor tremors at $\lambda = 2$ per day. The weekly count follows $\text{Poisson}(14)$.

The same caveats apply as with the exponential: the Poisson is the right model when events are independent and the rate is roughly constant. If customers tend to arrive in groups, or if the rush-hour rate is three times the off-peak rate, a more nuanced model is needed.