Discrete Uniform Distribution

Section 6.2 Discrete Uniform Distribution

In this section, you will investigate distributions that begin with individual outcomes that are equally likely and expand into more general settings.

Theorem 6.2.1. Discrete Uniform Distribution.

Assume outcomes in \(R\) = {1, 2, 3, ..., n} are equally likely. Then, the probability function for the discrete uniform variable \(X\) is

\begin{equation*} f(x) = \frac{1}{n} \end{equation*}

for \(x \in R\text{.}\)

Proof.

Assume that you have a variable with space \(R\) = {1, 2, 3, ..., n} so that the likelihood of each value is equally likely. Then, the probability function satisfies \(f(x) = c\) for any \(x \in R\text{.}\) As before, since \(\sum_{x \in R} f(x) = 1\text{,}\) then

\begin{equation*} f(x) = \frac{1}{n}. \end{equation*}

Theorem 6.2.2. Properties of the Discrete Uniform Probability Function.

\(f(x) = \frac{1}{n}\) over \(R\) = {1, 2, 3, ..., n} satisfies the properties of a discrete probability function and

\(\displaystyle \mu = \frac{1+n}{2}\)
\(\displaystyle \sigma^2 = \frac{n^2-1}{12}\)
\(\displaystyle \gamma_1 = 0\)
\(\displaystyle \gamma_2 = 3 - \frac{6}{5}\frac{n^2+1}{n^2-1}\)
Distribution function \(F(x) = \frac{x}{n}\) for \(x \in R\text{.}\)

Proof.

Trivially, by construction you get by summing over \(R = \{1, 2, ... , n \}\)

\begin{equation*} \sum_{x=1}^n \frac{1}{n} = 1 \end{equation*}

Also, 1/n is positive for all x values.

To determine the mean 1,

\begin{align*} \mu & = \sum_{x=1}^n x \cdot \frac{1}{n}\\ & = \frac{1}{n}\sum_{x=1}^n x \\ & = \frac{1}{n} \frac{n(n+1)}{2}\\ & = \frac{1+n}{2} \end{align*}

To determine the variance 2,

\begin{align*} \sigma^2 & = \sum_{x=1}^n x^2 \cdot \frac{1}{n} - \mu^2\\ & = \frac{1}{n}\sum_{x=1}^n x^2 - \left ( \frac{1+n}{2}\right )^2 \\ & = \frac{1}{n} \frac{n(n+1)(2n+1)}{6} - \frac{1+2n+n^2}{4}\\ & = \frac{(2n^2+3n+1)}{6} - \frac{1+2n+n^2}{4}\\ & = \frac{(4n^2+6n+2)}{12} - \frac{3+6n+3n^2}{12}\\ & = \frac{(n^2-1)}{12} \end{align*}

For skewness 3,

\begin{align*} \gamma_1 = & \sum_{x=1}^n x^3 \cdot \frac{1}{n} - 3 \mu \sum_{x=1}^n x^2 \cdot \frac{1}{n} + 2\mu^3\\ & = \frac{n^2(n+1)^2}{4n} - 3\frac{(n(n+1)(2n+1))}{2n} \frac{1+n}{2} + 2 \left ( \frac{1+n}{2}\right )^3 \\ & = \frac{n^2(n+1)^2}{4n} - \frac{(n+1)^2 (n(2n+1)}{4n} + \frac{(n+1)^3}{4}\\ & = \frac{(n+1)^2}{4} \left [ n - 2n -1 + (n+1) \right ]\\ & = 0 \end{align*}

which should be obvious since the histogram for this distribution is constantly flat.

For kurtosis 4, use the fourth moment and simplify. This this is tedious, the algebra is performed using Sage in the active cell below this proof. However, you might want to supply the remainder of this proof using the fact that

\begin{equation*} \sum_{x=1}^n x^4 = \frac{6n^5 + 15n^4 + 10n^3 - n}{30}. \end{equation*}

Sage can also do the algebra for you to determine each of these measures. Notice, as n increases the Kurtosis approaches \(\frac{6}{5}\) which indicates that there is (obviously) no tend toward central tendency over time.

Example 6.2.3. Rolling one die.

When you consider rolling a regular, fair, single 6-sided die, each side is equally likely. The sample space consists of the 6 sides, each with a unique number of physical dots. Let the random variable \(X\) correspond each side with the number corresponding to the number of dots. Then, \(R\) = {1, 2, 3, 4, 5, 6}. Since each side is equally likely then \(f(x) = 1/6\text{.}\)

Further, the probability of getting an outcome in A = {2,3} would be \(f(2)+f(3) = 1/6 + 1/6 = 2/6\text{.}\)