The Exponential Distribution

The exponential distribution models the waiting time until the next event in a process where events occur at a constant average rate: the time between new infections, the length of an infectious period, the interval between radioactive decays, or the time until a component fails. It is the continuous companion of the Poisson distribution.

Figure 1. The exponential density; the mean 1/lambda marks the average waiting time.

Definition#

Let $X\sim\mathrm{Exponential}(\lambda)$ be the waiting time with rate $\lambda$ . Its probability density function is $f(x)=\lambda e^{-\lambda x},\qquad x\ge 0,$ with cumulative distribution function $F(x)=1-e^{-\lambda x}$ .

Support: $x\in[0,\infty)$ .
Parameter: rate $\lambda>0$ (events per unit time). The reciprocal $1/\lambda$ is the mean waiting time, sometimes called the scale.
Mean: $\mathbb{E}[X]=\dfrac{1}{\lambda}$ .
Variance: $\mathrm{Var}(X)=\dfrac{1}{\lambda^2}$ .

Show the mean, variance, and memoryless property

Mean and variance. Integrate by parts with the density $f(x) = \lambda e^{-\lambda x}$ on $[0,\infty)$ :

$\mathbb{E}[X] = \int_0^{\infty} x\,\lambda e^{-\lambda x}\,dx = \Big[-x e^{-\lambda x}\Big]_0^{\infty} + \int_0^{\infty} e^{-\lambda x}\,dx = 0 + \frac{1}{\lambda} = \frac{1}{\lambda} .$

A second integration by parts gives $\mathbb{E}[X^2] = \int_0^\infty x^2 \lambda e^{-\lambda x}\,dx = 2/\lambda^2$ , so

$\operatorname{Var}(X) = \mathbb{E}[X^2] - \big(\mathbb{E}[X]\big)^2 = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2} .$

Memorylessness. The survival function is $P(X > t) = \int_t^\infty \lambda e^{-\lambda x}\,dx = e^{-\lambda t}$ , so

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

Having already waited a time $s$ tells you nothing about the remaining wait — the defining property of the exponential.

The memoryless property#

The exponential is the only continuous distribution that is memoryless: $P(X>s+t\mid X>s)=P(X>t).$ Having already waited $s$ units tells you nothing about how much longer you must wait — the process “forgets” the past. This is why it models the time to the next event in a constant-rate process, and why it is a common (if simplistic) model for infectious periods that end at a constant hazard.

Link to the Poisson#

If events occur according to a Poisson process with rate $\lambda$ — so the count in any window is Poisson( $\lambda\cdot\text{length}$ ) — then the waiting times between consecutive events are independent Exponential( $\lambda$ ) variables. Counting events and timing events are two views of the same underlying process.

When it arises#

The exponential arises for inter-event times at a constant rate: time between infections in early epidemic growth, duration of an infectious or latent period, survival times with constant hazard, and service or arrival times in queues.

In code#

Watch the parametrization carefully. R uses the rate $\lambda$ , SciPy uses a scale $=1/\lambda$ , and Julia’s Exponential(θ) also uses the scale $\theta=1/\lambda$ .

R#

# R parametrizes by rate = lambda
dexp(1, rate = 0.5)    # density at x = 1
pexp(1, rate = 0.5)    # P(X <= 1) = 1 - exp(-0.5)
qexp(0.95, rate = 0.5) # 95% quantile

set.seed(123)
x <- rexp(10000, rate = 0.5)  # random sample, mean = 1/0.5 = 2
hist(x, breaks = 40, freq = FALSE)     # histogram
curve(dexp(x, rate = 0.5), add = TRUE) # overlay the density

Python#

Python

import numpy as np
from scipy import stats

# scipy.stats.expon uses SCALE = 1/lambda (not the rate!)
lam = 0.5
stats.expon.pdf(1, scale=1/lam)    # density at 1
stats.expon.cdf(1, scale=1/lam)    # P(X <= 1)
stats.expon.ppf(0.95, scale=1/lam) # 95% quantile

rng = np.random.default_rng(123)
x = rng.exponential(scale=1/lam, size=10000)  # numpy also uses scale = 1/lambda
# plt.hist(x, bins=40, density=True); overlay stats.expon.pdf on a grid

Julia#

Julia

using Distributions, Random

# Julia's Exponential(theta) uses the SCALE theta = 1/lambda, NOT the rate.
lam = 0.5
d = Exponential(1 / lam)  # scale theta = 1/0.5 = 2  => rate lambda = 0.5
pdf(d, 1)                 # density at 1
cdf(d, 1)                 # P(X <= 1)
quantile(d, 0.95)         # 95% quantile

Random.seed!(123)
x = rand(d, 10_000)       # random sample, mean = theta = 2
# histogram(x, normalize=:pdf); plot!(t -> pdf(d, t)) to overlay the density

Simulation#

Many draws with rate $\lambda=0.5$ have empirical mean near $1/\lambda=2$ and variance near $1/\lambda^2=4$ .

set.seed(7)
x <- rexp(1e6, rate = 0.5)
mean(x)  # ~ 2.0  (theoretical mean = 1/lambda)
var(x)   # ~ 4.0  (theoretical variance = 1/lambda^2)

Why it matters for statistics#

The exponential is the building block of survival analysis and reliability, the simplest hazard model, and the inter-arrival law of Poisson processes used throughout epidemiology and queueing theory. Understanding its rate-vs-scale parametrization prevents a common and costly software bug when moving between R, Python, and Julia.