Variance of Discrete Uniform Distribution

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Theorem

Let $X$ be a discrete random variable with the discrete uniform distribution with parameter $n$.


Then the variance of $X$ is given by:

$\var X = \dfrac {n^2 - 1} {12}$


Proof

From the definition of Variance as Expectation of Square minus Square of Expectation:

$\var X = \expect {X^2} - \paren {\expect X}^2$

From Expectation of Function of Discrete Random Variable:

$\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \map \Pr {X = x}$


So:

\(\ds \expect {X^2}\) \(=\) \(\ds \sum_{k \mathop = 1}^n k^2 \paren {\frac 1 n}\) Definition of Discrete Uniform Distribution
\(\ds \) \(=\) \(\ds \frac 1 n \sum_{k \mathop = 1}^n k^2\)
\(\ds \) \(=\) \(\ds \frac 1 n \frac {n \paren {n + 1} \paren {2 n + 1} } 6\) Sum of Sequence of Squares
\(\ds \) \(=\) \(\ds \frac {\paren {n + 1} \paren {2 n + 1} } 6\)


Then:

\(\ds \var X\) \(=\) \(\ds \expect {X^2} - \paren {\expect X}^2\)
\(\ds \) \(=\) \(\ds \frac {\paren {n + 1} \paren {2 n + 1} } 6 - \frac {\paren {n + 1}^2} 4\) Expectation of Discrete Uniform Distribution: $\expect X = \dfrac {n + 1} 2$
\(\ds \) \(=\) \(\ds \frac {2 \paren {2 n^2 + 3 n + 1} - 3 \paren {n^2 + 2 n + 1} } {12}\)
\(\ds \) \(=\) \(\ds \frac {n^2 - 1} {12}\)

$\blacksquare$