## The mean and variance of the discrete uniform distribution

### The discrete uniform distribution

The uniform discrete distribution has range of n points in [a, b], or
$\{a, a +1, a+2, .... b-1, b \}.$ Each point has a uniform pmf (probability mass function)

$Pr(x) = 1/n.$

where $n = b -a + 1.$

### Mean of the discrete uniform distribution

Let us compute the mean and variance. Intuitively, we feel that the mean is at the
midpoint $(a + b) / 2.$ We shall see that this intuition is correct.

$E(X) = \sum x \frac{1}{n} = \frac{1}{n} \sum {x}.$

But $\sum {x}$ is just a sum of an arithmetic progression where the start value $a$
and the last value is $b$ with common difference of 1. The sum is then

$S = n (a + b)/2$

from which the mean or expectation is computed to b be

.

### The variance of the uniform discrete distribution.

Let us first translate the random variable range from [a,b] to the range[1,n]. Then we know that the variance is not affected by the translation.

 $V(X)$ $= \overline{X^2} - (\overline{X})^2$ $= (n (n+1) (2n+1)/(6 n) - [(n+1)/2)]^2$ $= \frac{2n^2 + n + 2n + 1}{6}- \frac{(n^2 + 2n + 1)}{4}$ $= \frac{[4 n^2 + 2n + 4n + 2 - 3n^2 -6n -3]}{12}$ $= \frac{[n^2 -1] }{12}$ $= \frac{(n+1) (n -1) }{12}$

Here we used the fact that $\sum_{i=1}^n i^2 = \frac{(n (n+1) (2n+1)}{6}$ and $\sum_{i=1}^n i = \frac{n (n+1)}{2}$. These formulas are well known.

Substituting $n = b -a + 1$ in the above formula yields $V(X) = (b -a + 2) (b -a)/12 = (b -a +1)^2 -1.$

To recap: The mean and variance of a random variable X with domain $\{a, a+1, \cdots, b-1, b\}$
are given by

### 2 Responses to “The mean and variance of the discrete uniform distribution”

1. Rasoul Says:

I think the proof is not general because we cannot say that

\Sum_{x=0}^{n} x^2 = \Sum_{x=a}^{b} x^2

even if n = b - a + 1

From summation identities, we have

\Sum_{x=a}^{b} x^2 = \Sum_{x=0}^{b-a} (x+a)^2

2. ernie Says:

Thanks Rasoul, I will review this article when I have the time.