Exponential Distribution

Definition

Core Statement

The Exponential Distribution models the time between events in a Poisson process, where events occur continuously and independently at a constant average rate. It is the continuous analog of the geometric distribution.

Purpose

Model waiting times (time until next event).
Reliability analysis: Time until failure.
Queueing theory: Time between arrivals.
Foundation for survival analysis.

When to Use

Use Exponential Distribution When...

Modeling time until an event occurs.
Events occur at a constant rate ( $λ$ ).
Events are memoryless (past doesn't affect future).

Theoretical Background

Notation

X \sim Exponential (λ)

where $λ$ is the rate parameter (events per unit time).

Probability Density Function (PDF)

f (x | λ) = λ e^{- λ x}, x \geq 0

Cumulative Distribution Function (CDF)

F (x | λ) = 1 - e^{- λ x}

Memoryless Property

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

Meaning: If you've waited $s$ time units, the probability of waiting an additional $t$ units is the same as if you just started. The distribution "forgets" how long you've already waited.

Properties

Property	Formula
Mean	$μ = \frac{1}{λ}$
Variance	$σ^{2} = \frac{1}{λ^{2}}$
Median	$\frac{\ln (2)}{λ} \approx \frac{0.693}{λ}$
Mode	0 (peak at origin)

Relationship to Poisson

If the number of events in time $t$ follows $Poisson (λ t)$ , then the time between events follows $Exponential (λ)$ .

Worked Example: Call Center Waiting Time

Problem

A help desk receives calls at an average rate of 4 calls per hour ( $λ = 4$ ).

Questions:

What is the probability that the next call comes in less than 10 minutes?
What is the probability that you wait more than 30 minutes for a call?

Solution:

First, convert units to be consistent. Let's work in hours.

$λ = 4$ calls/hour.
10 minutes = $10 / 60 = 0.1667$ hours.
30 minutes = $30 / 60 = 0.5$ hours.

1. Probability wait < 10 mins ( $P (X \leq 0.1667)$ ):

F (x) = 1 - e^{- λ x}

P (X \leq 0.1667) = 1 - e^{- 4 \times 0.1667} = 1 - e^{- 0.667}

e^{- 0.667} \approx 0.513

P (X \leq 0.1667) = 1 - 0.513 = 0.487

Result: ~48.7% chance the next call arrives within 10 minutes.

2. Probability wait > 30 mins ( $P (X > 0.5)$ ):

P (X > x) = e^{- λ x}

P (X > 0.5) = e^{- 4 \times 0.5} = e^{- 2}

e^{- 2} \approx 0.135

Result: ~13.5% chance you wait more than 30 minutes.

Assumptions

Constant rate: Event rate $λ$ does not change over time.
Independence: Events occur independently.
Memorylessness: Past has no influence on future waiting times.

Limitations

Pitfalls

The "Real World Ages" Problem: Exponential implies components never wear out. A brand new bulb has the same failure probability as one that has run for 10 years. For mechanical wear, use Weibull.
Varying Rates: If calls peak at noon and drop at night, $λ$ is not constant. Use a Non-Homogeneous Poisson Process.
Clustering: If events trigger other events (e.g., earthquakes, stock trades), independence fails.

Python Implementation

from scipy.stats import expon
import numpy as np
import matplotlib.pyplot as plt

# Exponential with λ=0.5 (mean = 2)
lambda_param = 0.5
dist = expon(scale=1/lambda_param)  # scipy uses scale = 1/λ

# Mean
print(f"Mean: {dist.mean():.2f}")

# P(X > 3)
prob_gt_3 = 1 - dist.cdf(3)
print(f"P(X > 3): {prob_gt_3:.4f}")

# Visualize PDF
x = np.linspace(0, 10, 500)
plt.plot(x, dist.pdf(x), label=f'λ={lambda_param}')
plt.xlabel('Time')
plt.ylabel('Density')
plt.title('Exponential Distribution')
plt.legend()
plt.grid(alpha=0.3)
plt.show()

# Memoryless Property Verification
# P(X > 5 | X > 2) = P(X > 3)
cond_prob = (1 - dist.cdf(5)) / (1 - dist.cdf(2))
uncond_prob = 1 - dist.cdf(3)
print(f"Conditional P(X > 5 | X > 2): {cond_prob:.4f}")
print(f"Unconditional P(X > 3): {uncond_prob:.4f}")

R Implementation

# Exponential with λ=0.5
lambda_param <- 0.5

# Mean
1 / lambda_param

# P(X > 3)
pexp(3, rate = lambda_param, lower.tail = FALSE)

# Visualize
curve(dexp(x, rate = lambda_param), from = 0, to = 10,
      xlab = "Time", ylab = "Density",
      main = "Exponential Distribution", lwd = 2, col = "blue")

# Random sample
rexp(10, rate = lambda_param)

Interpretation Guide

Scenario	Interpretation
Scenario	Interpretation
----------	----------------
$λ = 0.2$	Mean wait = $1 / 0.2 = 5$ minutes.
High $λ$ (e.g., 100)	Events happen very frequently; waiting times are tiny.
Memorylessness	"It's been quiet for an hour" does not mean a call is more likely now.
Median < Mean	The distribution is right-skewed; most events happen early, but some take a long time.

Poisson Distribution - Number of events in fixed time.
Geometric Distribution - Discrete memoryless distribution.
Weibull Distribution - Generalizes exponential; allows aging.
Survival Analysis

Exponential Distribution

Definition

Purpose

When to Use

Theoretical Background

Notation

Probability Density Function (PDF)

Cumulative Distribution Function (CDF)

Properties

Relationship to Poisson

Worked Example: Call Center Waiting Time

Assumptions

Limitations

Python Implementation

R Implementation

Interpretation Guide

Related Concepts