Exponential & Poisson

Back to the bus stop the last lesson left us at. Buses come at random — sometimes two within a minute, sometimes a long empty stretch. Two natural questions hang in the air. How long until the next bus? That answer is a continuous waiting time. How many buses arrive in the next hour? That answer is a whole-number count. Remarkably, one single dial governs both — a rate of events per unit time, written λ — so we learn the two together.

The waiting time

The time you wait for the next event is the exponential distribution. Its density slopes down from the start — short waits are common, long ones rare, with a thin tail trailing off — exactly the shape the last lesson predicted for a waiting time.

• Density: f(x) = λ·e^(−λx) for x ≥ 0.
• CDF: F(x) = 1 − e^(−λx), so the chance of still waiting past x is P(X > x) = e^(−λx).
• Mean: E(X) = 1/λ — a faster rate λ means a shorter average wait, just as you would expect.
• Variance: Var(X) = 1/λ².

The strange property the bus has

Here is the “no more due now” idea the last lesson dangled. The exponential forgets the past. If you have already waited s minutes, the chance of waiting at least t more is exactly the chance a fresh arrival waits t — your ten minutes of standing there bought you nothing:

P(X > s + t | X > s) = P(X > t)

A bus that has not come in ten minutes is no more “due” than one you just started waiting for. The exponential is the only continuous distribution with this memorylessness, and GATE leans on it constantly. Open the Exponential tab below, drag the rate λ, and watch the curve steepen as faster events shorten the wait.

The count

Now the other question — how many events land in one interval. That count is the Poisson distribution, the exponential’s discrete partner, sharing the same rate λ:

P(X = k) = e^(−λ) · λ^k / k! ,   k = 0, 1, 2, …

Its signature fact, and a GATE favourite, is that its two summary numbers are equal:

For a Poisson, both the mean and the variance equal λ — a true fact, tested again in GATE DA 2024.

A worked example — a real 2024 question

Let X be Exponential(λ). If 5·E(X) = Var(X), find λ.

Substitute the two formulas and the rest is algebra:

E(X) = 1/λ,   Var(X) = 1/λ²

5 · E(X) = Var(X)
5 · (1/λ) = 1/λ²
5/λ      = 1/λ²

multiply both sides by λ²:
5λ = 1   ⟹   λ = 0.2

So λ = 0.2. The whole question was really “do you know E(X) = 1/λ and Var(X) = 1/λ²?” — answer that and it falls out in two lines.

A question to carry forward

The exponential and Poisson are the shapes of random arrivals. But the most famous continuous shape of all is none of these — it appears whenever many small, independent effects pile up: heights, measurement errors, the average of a big sample. Here is the thread onward: what is that bell-shaped distribution, and why does looking things up in it always start by “standardising” a value first?

In one breath

• One rate λ governs both: exponential = continuous waiting time to the next event; Poisson = discrete count of events in an interval.
• Exponential(λ): f = λe^(−λx), F = 1 − e^(−λx), P(X>x) = e^(−λx); mean 1/λ, variance 1/λ²; the unique memoryless distribution.
• Poisson(λ): P(X=k) = e^(−λ)·λ^k/k!; mean = variance = λ (the signature fact).
• The reliable NAT: solve for λ from a mean/variance relation — 2024: 5·E(X)=Var(X) → 5/λ = 1/λ² → λ = 0.2.
• Do not swap: exponential variance 1/λ² (square of the mean); Poisson variance equals the mean.

Practice