Section 6.2 Discrete Uniform Distribution
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*}
is the probability function.
Theorem 6.2.1 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
- \(\mu = \frac{1+n}{2}\)
- \(\sigma^2 = \frac{n^2-1}{12}\)
- \(\gamma_1 = 0\)
- \(\gamma_2 = \frac{6}{5}\frac{1+n^2}{1-n^2}\)
- Distribution function F(x) = frac{x}{n} for \(x \in R\text{.}\)
Proof
- Trivially, by construction you get
\begin{equation*}
\sum_{k=1}^n \frac{1}{n} = 1
\end{equation*}
Also, 1/n is positive for all x values.
-
To determine the mean,
\begin{align*}
\mu & = \sum_{k=1}^n x \cdot \frac{1}{n}\\
& = \frac{1}{n}\sum_{k=1}^n x \\
& = \frac{1}{n} \frac{n(n+1)}{2}\\
& = \frac{1+n}{2}
\end{align*}
-
To determine the variance,
\begin{align*}
\sigma^2 & = \sum_{k=1}^n x^2 \cdot \frac{1}{n} - \mu^2\\
& = \frac{1}{n}\sum_{k=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,
\begin{align*}
\gamma_1 = & \sum_{k=1}^n x^3 \cdot \frac{1}{n} - 3 \mu \sum_{k=1}^n x^2 \cdot \frac{1}{n} + 2\mu^3\\
& = \frac{n^2(n+1)^2}{4n} - 3\frac{(n(n+1))}{2} \frac{1+n}{2} + 2 \left ( \frac{1+n}{2}\right )^3 \\
& =
\end{align*}
For Kurtosis, use the fourth moment and simplify...the algebra is performed using Sage in the active cell below this proof.
-
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 6.2.2 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.
Further, the probability of getting an outcome in A={2,3} would be f(2)+f(3) = 1/6 + 1/6 = 2/6.