Continuous Distributions: Uniform and Exponential
Continuous Distributions: Uniform and Exponential
Beyond the normal distribution from core maths, Further Statistics introduces other continuous probability distributions. The continuous uniform and exponential distributions are key models with distinct applications.
Continuous Random Variables: Recap
A continuous random variable X has a probability density function (pdf) f(x) satisfying:
- f(x) ≥ 0 for all x
- ∫₋∞^∞ f(x) dx = 1
- P(a ≤ X ≤ b) = ∫_a^b f(x) dx
The cumulative distribution function (cdf): F(x) = P(X ≤ x) = ∫₋∞^x f(t) dt
Key relationships:
- F'(x) = f(x) (differentiate the cdf to get the pdf)
- P(X > x) = 1 − F(x)
- E(X) = ∫ x f(x) dx (mean/expected value)
- Var(X) = E(X²) − [E(X)]² where E(X²) = ∫ x² f(x) dx
The Continuous Uniform Distribution U(a, b)
The simplest continuous distribution: every value in the interval [a, b] is equally likely.
PDF:
f(x) = 1/(b − a) for a ≤ x ≤ b, and 0 otherwise
CDF:
F(x) = 0 for x < a, (x − a)/(b − a) for a ≤ x ≤ b, 1 for x > b
Properties:
| Property | Formula |
|---|---|
| Mean | E(X) = (a + b)/2 |
| Variance | Var(X) = (b − a)²/12 |
| Standard deviation | σ = (b − a)/(2√3) |
Derivation of variance:
E(X²) = ∫_a^b x²/(b−a) dx = (b³ − a³)/(3(b−a)) = (a² + ab + b²)/3
Var(X) = (a² + ab + b²)/3 − (a+b)²/4 = (a² + ab + b²)/3 − (a² + 2ab + b²)/4
= (4a² + 4ab + 4b² − 3a² − 6ab − 3b²)/12 = (a² − 2ab + b²)/12 = (b−a)²/12
Applications: Random number generators, modelling arrival times within a known interval, rounding errors.
Example: X ~ U(2, 8). Find P(3 ≤ X ≤ 5) and E(X).
P(3 ≤ X ≤ 5) = (5 − 3)/(8 − 2) = 2/6 = 1/3
E(X) = (2 + 8)/2 = 5
Var(X) = (8 − 2)²/12 = 36/12 = 3
The Exponential Distribution Exp(λ)
Models the time between events in a Poisson process (where events occur randomly at a constant average rate).
If events occur at rate λ per unit time, the time T between consecutive events follows T ~ Exp(λ).
PDF:
f(t) = λe^(−λt) for t ≥ 0, and 0 for t < 0
CDF:
F(t) = 1 − e^(−λt) for t ≥ 0
Properties:
| Property | Formula |
|---|---|
| Mean | E(T) = 1/λ |
| Variance | Var(T) = 1/λ² |
| Standard deviation | σ = 1/λ |
| Median | (ln 2)/λ ≈ 0.693/λ |
| Mode | 0 |
Key probabilities:
- P(T > t) = e^(−λt) — the survival function
- P(T ≤ t) = 1 − e^(−λt)
The Memoryless Property
The exponential distribution is the only continuous distribution with the memoryless property:
P(T > s + t | T > s) = P(T > t)
In words: given that you have already waited s units, the probability of waiting at least t more units is the same as the original probability of waiting at least t units. The past waiting time gives no information about the future.
Proof: P(T > s + t | T > s) = P(T > s + t)/P(T > s) = e^(−λ(s+t))/e^(−λs) = e^(−λt) = P(T > t) ✓
This makes the exponential distribution appropriate for modelling lifetimes of components that do not age (constant failure rate), but inappropriate for modelling human lifetimes (which have an increasing failure rate).
Connection to the Poisson Distribution
If events follow a Poisson process with rate λ:
- Number of events in time t: N(t) ~ Poisson(λt)
- Time between events: T ~ Exp(λ)
- Time until the nth event: T₁ + T₂ + ... + Tₙ ~ Gamma(n, λ)
Derivation: P(T > t) = P(no events in time t) = P(N(t) = 0) = e^(−λt)
Worked Examples
Example 1: Buses arrive at a stop according to a Poisson process with rate 4 per hour. Find the probability of waiting more than 20 minutes.
T ~ Exp(4) (rate per hour). Convert 20 min = 1/3 hour.
P(T > 1/3) = e^(−4 × 1/3) = e^(−4/3) ≈ 0.264
Example 2: The lifetime of a light bulb (in thousands of hours) follows Exp(0.5). Find the probability it lasts between 1000 and 3000 hours.
P(1 < T < 3) = F(3) − F(1) = (1 − e^(−1.5)) − (1 − e^(−0.5)) = e^(−0.5) − e^(−1.5) ≈ 0.6065 − 0.2231 = 0.383
Example 3: Using the memoryless property. Given a bulb has lasted 2000 hours, find the probability it lasts at least 4000 hours total.
P(T > 4 | T > 2) = P(T > 2) = e^(−0.5 × 2) = e^(−1) ≈ 0.368
Fitting Distributions to Data
To check if data follows a particular distribution:
1. Estimate parameters from the data (e.g., λ = 1/x̄ for exponential, a = min, b = max for uniform)
2. Compute expected frequencies
3. Perform a χ² goodness-of-fit test
Remember to reduce degrees of freedom by 1 for each parameter estimated from the data.
Exam Tips
- For U(a, b): the pdf is a constant — probabilities are just proportions of the interval length
- For Exp(λ): P(T > t) = e^(−λt) is often quicker than using the cdf 1 − e^(−λt) and subtracting from 1
- The mean of Exp(λ) is 1/λ, not λ — a common confusion (if average wait is 5 min, λ = 1/5 per min)
- Always check your units: if λ is per hour, t must be in hours
- The memoryless property is unique to the exponential distribution — state it clearly when using it
- For multi-part questions, drawing the pdf helps visualise which area you need
- In exam mark schemes, stating H₀, computing E, applying the formula, finding ν, comparing with the critical value, and writing a conclusion in context each earn separate marks