Waiting for the next event

A Poisson process counts how many events happen in a given interval. But flip the question: how long until the next event?

Buses arriving at a stop. Radioactive atoms decaying. Customers entering a shop. If the events follow a Poisson process, the waiting time between them follows a specific distribution.

What do you think?

If earthquakes hit a region at an average rate of 2 per year (Poisson), what's the average waiting time between consecutive earthquakes?

Definition

Exponential Distribution

A random variable $X$ has an Exponential distribution with rate $\lambda > 0$ if its PDF is: $f(x) = \lambda e^{-\lambda x}, \quad x \geq 0$ and $f(x) = 0$ for $x < 0$ . We write $X \sim \text{Expo}(\lambda)$ .

The PDF starts at height $\lambda$ and decays exponentially. Higher $\lambda$ means faster decay — shorter waits are more likely.

The CDF has a clean closed form:

$F(x) = P(X \leq x) = 1 - e^{-\lambda x}, \quad x \geq 0$

Exponential Distribution

Rate (λ): 1.00

0.253

Query time (t): 1.0

P(X ≤ t)

63.21%

P(X > t)

36.79%

E[X]

1.00

Var(X)

1.00

Adjust the rate $\lambda$ and the query time $t$ . Switch between PDF and CDF views. Then toggle the memoryless property to see the core idea.

Key properties

Exponential Moments

If $X \sim \text{Expo}(\lambda)$ : $E[X] = \Large\frac{1}{\lambda}, \qquad \text{Var}(X) = \frac{1}{\lambda^2}$ The standard deviation equals the mean: $\text{SD}(X) = 1/\lambda$ .

The rate and the expected wait are reciprocals.

If calls arrive at rate λ = 4 per hour, what's the average wait between calls (in minutes)? (whole number, in minutes)

For Expo(2), what is P(X > 1)? (Hint: use the survival function) (decimal to 3 places, e.g. 0.456)

The memoryless property

The Geometric distribution was memoryless in discrete time. The Exponential is its continuous counterpart.

Memoryless Property

If $X \sim \text{Expo}(\lambda)$ , then for any $s, t > 0$ : $P(X > s + t \mid X > s) = P(X > t)$

Given that you've already waited $s$ minutes with no event, the remaining wait has the same distribution as if you just started. The past is irrelevant.

Proof

$P(X > s + t \mid X > s) = \Large\frac{P(X > s + t)}{P(X > s)} = \frac{e^{-\lambda(s+t)}}{e^{-\lambda s}} \normalsize= e^{-\lambda t} = P(X > t)$

The process has no memory. If a lightbulb has an exponentially distributed lifetime and has been running for 1000 hours, the chance it lasts another 100 hours is the same as a brand-new bulb lasting 100 hours. Real lightbulbs age, but atoms don't.

See the memoryless property in action — pick how long you've already waited and watch the conditional survival curve land right on top of the original:

Memoryless Property

Distribution

Success probability p: 0.30

0.10.8

Already waited s = 3 trials: 3

110

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

The curves overlap!

P(X > 3+2 | X > 3) = 0.4900 = P(X > 2) = 0.4900

Simulate: given survival past 3, what’s the remaining wait?

Samples

Avg remaining

0.000

E[X] (unconditional)

3.33

What do you think?

A radioactive atom has an exponential lifetime with mean 10 years. It's already survived 50 years. What's its expected remaining lifetime?

Enter a whole number

The Exponential is the only continuous distribution with the memoryless property, just as the Geometric is the only discrete one.

Connection to Poisson

The Exponential and Poisson are two sides of the same coin:

Poisson	Exponential
Counts events in fixed time	Measures time between events
$X \sim \text{Pois}(\lambda t)$	$T \sim \text{Expo}(\lambda)$
Discrete (0, 1, 2, ...)	Continuous ( $t \geq 0$ )
"How many in this interval?"	"How long until the next one?"

If events arrive at rate $\lambda$ per unit time:

The count in $t$ units is $\text{Pois}(\lambda t)$
The wait for the next event is $\text{Expo}(\lambda)$

Applications

Scenario	Rate $\lambda$	Mean wait $1/\lambda$
Server requests	100/sec	10 ms
Customer arrivals	30/hour	2 min
Radioactive decay	0.01/year	100 years
Machine failures	0.5/month	2 months

P(Expo(3) ≤ 0.5)? (decimal to 3 places, e.g. 0.456)

If Expo(λ) has E[X] = 5, what is λ? (decimal, e.g. 0.42)

Var(Expo(4)) = ? (decimal, e.g. 0.42)

Summary

Property	Formula
PDF	$f(x) = \lambda e^{-\lambda x}$ , $x \geq 0$
CDF	$F(x) = 1 - e^{-\lambda x}$
Mean	$E[X] = 1/\lambda$
Variance	$\text{Var}(X) = 1/\lambda^2$
Memoryless	$P(X > s+t \mid X > s) = P(X > t)$
Poisson link	Expo wait ↔ Poisson count

The Exponential distribution models waiting times in memoryless processes. Whenever events arrive randomly at a constant rate, the gaps between them are Exponential.

What's next

That covers the main continuous distributions. Coming up: deeper connections between distributions and tools for working with functions of random variables.