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: $\expect X = \dfrac {n + 1} 2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \paren {2 n^2 + 3 n + 1} - 3 \paren {n^2 + 2 n + 1} } {12}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^2 - 1} {12}\) |
$\blacksquare$