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Discrete uniform distribution

A finite set of equally likely integer outcomes, such as one fair die.

Parameters and support

$a$

smallest value in the support — $a\in \mathbb{Z}$

$b$

largest value in the support — $b\ge a$

Support

$x\in \{a,a+1,\dots ,b\}$

The formulas

p(x)

$$p(x)=\frac{1}{b-a+1}$$

F(x)

$$F(x)=\frac{\lfloor x\rfloor -a+1}{b-a+1},a\le x\le b$$

Mean

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

Variance

$$\mathrm{Var}(X)=\frac{(b-a+1{)}^2-1}{12}$$

MGF

$$M(t)=\frac{e^{at}\left(e^{nt}-1\right)}{n\left(e^t-1\right)},n=b-a+1$$

Where the moments come from

1. Shift the support to $\{1,2,\dots ,n\}$ with $n=b-a+1$; shifting changes the mean by a constant and leaves the variance alone.
2. $E[X]=\frac{1}{n}{\sum }_{k=1}^{n}k=\frac{n+1}{2}$ using the arithmetic-series sum.
3. $E[X^2]=\frac{1}{n}{\sum }_{k=1}^n{k}^2=\frac{(n+1)(2n+1)}{6}$ using the sum of squares.
4. $\mathrm{Var}(X)=\frac{(n+1)(2n+1)}{6}-{\left(\frac{n+1}{2}\right)}^2=\frac{n^2-1}{12}$.
5. Shifting back adds $a-1$ to the mean and leaves the variance unchanged.

Worked example

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 the range b − a instead of the count b − a + 1 in the variance.
• Assuming the variance formula (n² − 1)/12 applies to a CONTINUOUS uniform — there it is (b − a)²/12.

Related

• Continuous uniform distribution
• Binomial distribution

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.