The continuous uniform distribution

What you’ll learn

• Use a rectangular density to find interval probabilities.
• Find the mean and variance from the endpoints.
• Relate uniform moments to integration.

Understand the idea

For $X\sim U(a,b)$ the density is constant between $a$ and $b$ and zero elsewhere. Since the rectangle must have area one, its height is $1/(b-a)$. Equal-length subintervals inside the support have equal probabilities, regardless of their positions.

Integrating the constant density gives a linear CDF within the support. Integrating $xf(x)$ gives the midpoint as the mean, and integration of $x^2f(x)$ leads to variance $(b-a)^2/12$. Wider intervals have greater variance even when the centre is unchanged.

The uniform model is appropriate only when there is a reason to treat all locations in the interval as equally likely. A bounded waiting time is not automatically uniform; a timetable or a non-random arrival process can change its shape.

Key formulae

$$f(x)=\frac1{b-a}\quad(a<x<b)$$

$$F(x)=\frac{x-a}{b-a}\quad(a\leq x\leq b)$$

$$E(X)=\frac{a+b}{2}$$

$$\operatorname{Var}(X)=\frac{(b-a)^2}{12}$$

Worked example

A waiting time follows $U(4,10)$ minutes. Find $P(6<X<9)$, its mean and its variance.

1. The whole interval has length $10-4=6$. The target interval has length $9-6=3$.
2. Its probability is $3/6=1/2$.
3. The mean is $(4+10)/2=7$ minutes.
4. The variance is $(10-4)^2/12=3$ minutes squared.

Answer: Probability $0.5$; mean $7$ minutes; variance $3$ minutes squared.