Uniform distribution

The story

Pick a point at random on a ruler between $a$ and $b$ , with no preference for any part. The chance of landing in any sub-interval depends only on its length: $P(c \le X \le d) = (d - c)/(b - a)$ . The CDF is therefore a straight ramp rising from 0 at $a$ to 1 at $b$ .

Its mean is the midpoint $(a + b)/2$ and its variance is $(b - a)^2/12$ . The Uniform(0, 1) is also the raw material of simulation: computers generate uniform random numbers and transform them into everything else, for instance by feeding them through an inverse CDF.

Formulas

Support

x \in [a, b]

PDF

f(x) = \frac{1}{b - a}

CDF

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

Expectation

\mathbb{E}[X] = \frac{a + b}{2}

Variance

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

Parameters

Parameter	What it controls
$a$	The lower end of the interval. Keep it below $b$ .
$b$	The upper end of the interval. Wider intervals give a lower, flatter rectangle, since the total area must stay 1.

Where it shows up

The output of a random number generator such as Math.random(), on $[0, 1)$ .
The rounding error when a measurement is rounded to the nearest unit, roughly uniform on $[-0.5, 0.5]$ .
The arrival time of a bus that comes once every ten minutes, if you turn up at a random moment.
The angle at which a spun bottle comes to rest, uniform on $[0^\circ, 360^\circ)$ .

Try this

Set $a = 0$ and $b = 1$ , then toggle to the CDF view: it is a straight line, so $P(X \le x)$ is simply $x$ . Shade from 0.2 to 0.5 and confirm the probability is 0.3.
With $a = 0$ and $b = 1$ , switch to Sample means with $n = 2$ . The averages form a triangle, not a rectangle. Try $n = 3$ and then $n = 12$ and watch the Central Limit Theorem take over.
Press Draw 1000 on Uniform(0, 1) and compare the sample variance with $1/12 \approx 0.0833$ . Then widen $b$ to 4 and check that the variance grows to $16/12 \approx 1.33$ , because it scales with the square of the width.

Related distributions

It is the continuous counterpart of the Discrete Uniform.
Uniform(0, 1) is the Beta with $\alpha = \beta = 1$ .
If $U$ is Uniform(0, 1), then $-\ln(U)/\lambda$ is Exponential with rate $\lambda$ ; this is the inverse-CDF trick in action.
The mean of $n$ independent Uniforms tends to a Normal as $n$ grows; even $n = 12$ already looks convincingly bell-shaped.