Variance of Continuous Uniform Distribution

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Theorem

Let $a, b \in \R$ such that $a < b$.

Let $X \sim \ContinuousUniform a b$ be the continuous uniform distribution over $\closedint a b$.

Then:

$\var X = \dfrac {\paren {b - a}^2} {12}$


Proof

From the definition of the continuous uniform distribution, $X$ has probability density function:

$\map {f_X} x = \begin{cases} \dfrac 1 {b - a} & a \le x \le b \\ 0 & \text{otherwise} \end{cases}$

From Variance as Expectation of Square minus Square of Expectation:

$\ds \var X = \int_{-\infty}^\infty x^2 \map {f_X} x \rd x - \paren {\expect X}^2$

So:

\(\ds \var X\) \(=\) \(\ds \int_{-\infty}^a 0 x^2 \rd x + \int_a^b \frac {x^2} {b - a} \rd x + \int_b^\infty 0 x^2 \rd x - \frac {\paren {a + b}^2} 4\) Expectation of Continuous Uniform Distribution
\(\ds \) \(=\) \(\ds \intlimits {\frac {x^3} {3 \paren {b - a} } } a b - \frac {\paren {a + b}^2} 4\) Primitive of Power, Fundamental Theorem of Calculus
\(\ds \) \(=\) \(\ds \frac {b^3 - a^3} {3 \paren {b - a} } - \frac {\paren {a + b}^2} 4\)
\(\ds \) \(=\) \(\ds \frac {4 \paren {b - a} \paren {a^2 + a b + b^2} } {12 \paren {b - a} } - \frac {3 \paren {a + b}^2} {12}\) Difference of Two Cubes
\(\ds \) \(=\) \(\ds \frac {4 a^2 + 4 a b + 4 b^2 - 3 a^2 - 6 a b - 3 b^2} {12}\) Square of Sum
\(\ds \) \(=\) \(\ds \frac {b^2 - 2 a b + a^2} {12}\)
\(\ds \) \(=\) \(\ds \frac {\paren {b - a}^2} {12}\) Square of Difference

$\blacksquare$


Also see


Sources

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