The sample space for a discrete uniform distribution is the set of integers from \(a\) to \(b\), i.e., its parameters are \(a\) and \(b\). We denote it by \(\mathrm{Unif}(a,b)\). All elements of the sample space have equal probability. Since there are \(b-a+1\) elements in the sample space, the PMF for a discrete uniform distribution is
\[p_X(x)=\dfrac{1}{b-a+1}.\]
Expectation
The mean of a discrete uniform random variable is
\[\mathbb{E}[X]=\frac{a+b}{2}.\]
Proof
\[\begin{align}
\mathbb{E}[X]&=\frac{1}{b-a+1}\left(a+a+1+\ldots+b-1+b\right)\\
&=\frac{1}{b-a+1}\left(\frac{b-a+1}{2}\right)\left(a+b\right)\\
&=\frac{a+b}{2}
\end{align}\]
Variance
The variance of a discrete uniform random variable is
\[\mathrm{var}(X)=\tfrac{1}{12}(b-a)(b-a+2).\]
Proof
To calculate the variance, we have
$\(\begin{align}
\mathrm{var}(X)&=\mathbb{E}[X^2]-(\mathbb{E}[X])^2\\
&=\frac{1}{b-a+1}\left[(a)^2+(a+1)^2+\ldots+(b)^2\right]-\left(\frac{a+b}{2}\right)^2.
\end{align}\)$
Before we proceed, we have to figure out the sum \(a^2+\ldots+b^2.\)
In the following, we will calculate the sum \(a^2+(a+1)^2+\ldots+b^2.\)
Recall from high school algebra that
\[n^3-(n-1)^3=3n^2-3n+1.\]
Using this relation repeatedly, we have
\[\begin{align}
b^3-(b-1)^3&=3b^2-3b+1\\
(b-1)^3-(b-2)^3&=3(b-1)^2-3(b-1)+1\\
\vdots\qquad&=\qquad\vdots\\
a^3-(a-1)^3&=3a^2-3a+1.
\end{align}\]
Summing both sides, we get
\[b^3-(a-1)^3=3(a^2+\ldots+b^2)-3(a+\ldots+b)+(b-a+1).\]
Rearranging this equation gives
\[\begin{align}
a^2+\ldots+b^2&=\frac{1}{3}\left[b^3-(a-1)^3+3(a+\ldots+b)-(b-a+1)\right]\\
&=\frac{1}{3}\left[b^3-(a-1)^3+\frac{3}{2}(b-a+1)(a+b)-(b-a+1)\right].
\end{align}\]
Using the relation
\[x^3-y^3=(x-y)(x^2+xy+y^2),\]
we get
\[b^3-(a-1)^3=(b-a+1)\left[b^2+b(a-1)+(a-1)^2\right].\]
Consequently,
\[\begin{align}
a^2+\ldots+b^2&=\frac{b-a+1}{3}\left[b^2+b(a-1)+(a-1)^2+\frac{3}{2}(a+b)-1\right]\\
&=\frac{b-a+1}{3}\left[b^2+ab+a^2+\tfrac{1}{2}(b-a)\right].
\end{align}\]
Continuing our calculation of the variance, we have
\[\begin{align}
\mathrm{var}(X)&=\frac{1}{3}\left[b^2+ab+a^2+\tfrac{1}{2}(b-a)\right]-\frac{1}{4}(a+b)^2\\
&=\tfrac{1}{12}\left[4\!\left(b^2+ab+a^2\right)+2(b-a)-3(a+b)^2\right]\\
&=\tfrac{1}{12}\left[\left(b^2-2ab+a^2\right)+2(b-a)\right]\\
&=\tfrac{1}{12}(b-a)(b-a+2).
\end{align}\]