The exponential distribution is a continuous probability distribution that models the time between consecutive events in a poisson-process. It is the unique continuous distribution with the memoryless property.
A random variable \(X\) follows an exponential distribution with rate parameter \(\lambda > 0\) if
\[ X \sim \text{Exponential}(\lambda) \quad \text{or} \quad X \sim \mathcal{E}(\lambda). \]
An alternative parameterisation uses the scale parameter \(\theta = 1/\lambda\).
The probability density function is
\[ f(x) = \begin{cases} \lambda e^{-\lambda x} & x \geq 0, \\ 0 & x < 0. \end{cases} \]
Interpretation: The density decays exponentially, so short waiting times are the most likely and arbitrarily long waits become increasingly rare.
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
from scipy.stats import expon
sns.set_style('whitegrid')
sns.set_palette('Set2')
x = np.linspace(0, 5, 400)
fig, ax = plt.subplots(figsize=(7, 4))
for lam in (0.5, 1.0, 2.0):
ax.plot(x, expon.pdf(x, scale=1 / lam),
lw=2, label=rf'$\lambda = {lam}$')
ax.set_xlabel('$x$')
ax.set_ylabel('$f(x)$')
ax.set_title('Exponential density')
ax.legend()
plt.show()
The cumulative distribution function is
\[ F(x) = \begin{cases} 1 - e^{-\lambda x} & x \geq 0, \\ 0 & x < 0. \end{cases} \]
The expectation and variance of \(X \sim \text{Exponential}(\lambda)\) are
\[ \mathbb{E}[X] = \frac{1}{\lambda} \quad \& \quad \operatorname{Var}(X) = \frac{1}{\lambda^2}. \]
Proof. Integrating by parts,
\[ \mathbb{E}[X] = \int_0^\infty x \lambda e^{-\lambda x} \, dx = \frac{1}{\lambda}. \]
Similarly \(\mathbb{E}[X^2] = 2/\lambda^2\), so \(\operatorname{Var}(X) = 2/\lambda^2 - 1/\lambda^2 = 1/\lambda^2\).
The moment generating function (MGF) of \(X \sim \text{Exponential}(\lambda)\) is
\[ M_X(t) = \frac{\lambda}{\lambda - t}, \quad t < \lambda. \]
The characteristic function of \(X \sim \text{Exponential}(\lambda)\) is
\[ \varphi_X(t) = \frac{\lambda}{\lambda - it}. \]
The exponential distribution is the unique continuous distribution with the memoryless property: for any \(s, t > 0\),
\[ \mathbb{P}(X > s + t \mid X > s) = \mathbb{P}(X > t). \]
Proof. Since \(\mathbb{P}(X > u) = e^{-\lambda u}\),
\[ \mathbb{P}(X > s + t \mid X > s) = \frac{e^{-\lambda(s + t)}}{e^{-\lambda s}} = e^{-\lambda t} = \mathbb{P}(X > t). \]
Interpretation: Knowing that an event has not occurred for time \(s\) does not change the distribution of the remaining waiting time.
The time between customer arrivals at a service counter, or the service time in an M/M/1 queue, is commonly modelled as exponential.
The lifetime of an electronic component with a constant failure rate, or the time to radioactive decay, follows an exponential distribution.
The waiting time for the next earthquake at a location, or the gap between successive traffic accidents, is often modelled as exponential.