Variance of Discrete Uniform Distribution

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Theorem

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


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:

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


So:

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


Then:

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

$\blacksquare$