Exponential Distribution

Motivation

The exponential distribution models the time until an event occurs when events happen:

• Continuously over time
• At a constant average rate
• Independently of past history

Examples:

• Time until radioactive decay
• Time until customer arrives
• Lifetime of a component
• Time between earthquakes (if Poisson process)

Definition

$X \sim \text{Exp}(\lambda)$ has PDF:

$f(x) = \lambda e^{-\lambda x}, \quad x \geq 0$

where $\lambda > 0$ is the rate parameter.

Range

$X \in [0, \infty)$

CDF

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

Survival Function

$P(X > x) = e^{-\lambda x}$

Expectation and Variance

$E[X] = \frac{1}{\lambda}$

Interpretation: If events occur at rate $\lambda$ per unit time, average wait time is $1/\lambda$.

$\text{Var}(X) = \frac{1}{\lambda^2}$

$\sigma_X = \frac{1}{\lambda} = E[X]$

For exponential, mean equals standard deviation.

Memoryless Property

The exponential is the only continuous memoryless distribution:

$P(X > s + t \mid X > s) = P(X > t)$

Interpretation: If a component has survived $s$ time units, its remaining lifetime distribution is the same as a new component.

Proof

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

Practical Meaning

A lightbulb with exponential lifetime has no “aging.” A 10-year-old bulb (that still works) is as good as new. This is reasonable for some random failures but not for wear-out failures.

Quantiles

For $X \sim \text{Exp}(\lambda)$:

$P(X \leq x_p) = 1 - e^{-\lambda x_p} = p$

$x_p = -\frac{\ln(1-p)}{\lambda}$

Median:

$x_{0.5} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}$

Note: median $<$ mean since the distribution is right-skewed.

Relation to Poisson

If events occur as a Poisson process with rate $\lambda$:

• Number of events in time $t$ ~ $\text{Poisson}(\lambda t)$
• Time until first event ~ $\text{Exp}(\lambda)$
• Time until $k$th event ~ $\text{Gamma}(k, \lambda)$

Example

Customers arrive at rate 2 per minute. Time until first customer:

$X \sim \text{Exp}(\lambda = 2)$

$E[X] = \frac{1}{2} = 0.5 \text{ minutes}$

$P(X > 1) = e^{-2 \cdot 1} = e^{-2} \approx 0.135$

Minimum of Exponentials

If $X_1, \ldots, X_n$ are independent with $X_i \sim \text{Exp}(\lambda_i)$, then:

$\min(X_1, \ldots, X_n) \sim \text{Exp}\left(\sum_{i=1}^n \lambda_i\right)$

Intuition: If $n$ independent processes are running, and we wait for the first to “fire,” the minimum follows exponential with combined rate.

Example: Competing Risks

A system has two failure modes, with failure rates $\lambda_1 = 0.01$/hour and $\lambda_2 = 0.02$/hour. Time to failure:

$\min(X_1, X_2) \sim \text{Exp}(0.03)$

$E[\text{failure time}] = \frac{1}{0.03} \approx 33.3 \text{ hours}$

Visualisation

The exponential PDF starts at $\lambda$ and decays toward 0. It is always decreasing (no mode except at 0).

Summary

Property	Value
PDF	$f(x) = \lambda e^{-\lambda x}$, $x \geq 0$
CDF	$F(x) = 1 - e^{-\lambda x}$
Mean	$\frac{1}{\lambda}$
Variance	$\frac{1}{\lambda^2}$
Median	$\frac{\ln 2}{\lambda}$
Memoryless	$P(X > s+t \mid X > s) = P(X > t)$