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Continuous uniform distribution

A value equally likely to fall anywhere in an interval.

Parameters and support

aa
lower endpoint - a∈Ra \in \mathbb{R}
bb
upper endpoint - b>ab > a
Support
a≤x≤ba \le x \le b

The formulas

f(x)
f(x)=1b−af(x) = \dfrac{1}{b-a}
F(x)
F(x)=x−ab−a,a≤x≤bF(x) = \dfrac{x-a}{b-a}, \quad a \le x \le b
Mean
E[X]=a+b2E[X] = \dfrac{a+b}{2}
Variance
Var⁡(X)=(b−a)212\operatorname{Var}(X) = \dfrac{(b-a)^{2}}{12}
MGF
M(t)=etb−etat(b−a),t≠0M(t) = \dfrac{e^{tb} - e^{ta}}{t(b-a)}, \quad t \ne 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

  1. E[X]=∫abxb−a dx=b2−a22(b−a)=a+b2E[X] = \int_a^b \frac{x}{b-a}\,dx = \frac{b^2-a^2}{2(b-a)} = \frac{a+b}{2}.
  2. E[X2]=∫abx2b−a dx=b3−a33(b−a)=a2+ab+b23E[X^2] = \int_a^b \frac{x^2}{b-a}\,dx = \frac{b^3-a^3}{3(b-a)} = \frac{a^2+ab+b^2}{3}.
  3. Var⁡(X)=a2+ab+b23−(a+b)24=(b−a)212\operatorname{Var}(X) = \frac{a^2+ab+b^2}{3} - \frac{(a+b)^2}{4} = \frac{(b-a)^2}{12}.
  4. 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.

  1. The payment is (X−100)+(X - 100)_+, which is 0 below 100 and X−100X - 100 above it.
  2. E[(X−100)+]=∫100500(x−100)⋅1500 dxE[(X-100)_+] = \int_{100}^{500} (x-100)\cdot\dfrac{1}{500}\,dx.
  3. Substituting u=x−100u = x-100 gives 1500∫0400u du=1500⋅40022\dfrac{1}{500}\int_0^{400} u\,du = \dfrac{1}{500}\cdot\dfrac{400^2}{2}.
  4. =80000500=160= \dfrac{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 that test run.

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.

Related

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.