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Continuous uniform distribution

A value equally likely to fall anywhere in an interval.

Parameters and support

$a$: lower endpoint — $a \in R$
$b$: upper endpoint — $b > a$
Support: $a \leq x \leq b$

The formulas

f(x): $f (x) = \frac{1}{b - a}$
F(x): $F (x) = \frac{x - a}{b - a}, a \leq x \leq b$
Mean: $E [X] = \frac{a + b}{2}$
Variance: $Var (X) = \frac{( b - a ) ^{2}}{12}$
MGF: $M (t) = \frac{e ^{t b} - e ^{t a}}{t ( b - a )}, t \neq = 0$

Memory hook. Variance is (range)²/12 — the continuous version uses the RANGE, the discrete version uses the COUNT. A U(0,1) has variance 1/12.

Where the moments come from

$E [X] = \int_{a}^{b} \frac{x}{b - a} d x = \frac{b ^{2} - a ^{2}}{2 ( b - a )} = \frac{a + b}{2}$ .
$E [X^{2}] = \int_{a}^{b} \frac{x ^{2}}{b - a} d x = \frac{b ^{3} - a ^{3}}{3 ( b - a )} = \frac{a ^{2} + ab + b ^{2}}{3}$ .
$Var (X) = \frac{a ^{2} + ab + b ^{2}}{3} - \frac{( a + b ) ^{2}}{4} = \frac{( b - a ) ^{2}}{12}$ .
Only the WIDTH matters for the variance, which is why shifting the interval changes nothing.

Worked example

A loss is uniformly distributed on [0, 500]. Find the expected payment under a policy with a 100 ordinary deductible.

The payment is $(X - 100)_{+}$ , which is 0 below 100 and $X - 100$ above it.
$E [(X - 100)_{+}] = \int_{100}^{500} (x - 100) \cdot \frac{1}{500} d x$ .
Substituting $u = x - 100$ gives $\frac{1}{500} \int_{0}^{400} u d u = \frac{1}{500} \cdot \frac{40 0 ^{2}}{2}$ .
$= \frac{80000}{500} = 160$ .

Answer: 160

The mean, variance, CDF and moment generating function above are re-derived numerically from this distribution’s own density on every test run — summed over the support for a discrete distribution, integrated by quadrature for a continuous one — and compared with the closed forms printed here. A typo on this page fails the build.

Traps

Using (b − a + 1)² − 1 over 12 — that is the DISCRETE uniform variance.
Computing E[X] − d for an expected payment with a deductible; the payment is truncated at 0, so the answer is not the mean minus the deductible.

Drill this: the Exam P question bank has original questions on this distribution, and the recall trainer builds its prompts from exactly the formulas above.

Continuous uniform distribution

Parameters and support

The formulas

Where the moments come from

Worked example

Traps

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