Expectation 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:

$\expect X = \dfrac {a + b} 2$


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 the definition of the expected value of a continuous random variable:

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

So:

\(\ds \expect X\) \(=\) \(\ds \int_{-\infty}^a 0 x \rd x + \int_a^b \frac x {b - a} \rd x + \int_b^\infty 0 x \rd x\)
\(\ds \) \(=\) \(\ds \intlimits {\frac {x^2} {2 \paren {b - a} } } a b\) Primitive of Power, Fundamental Theorem of Calculus
\(\ds \) \(=\) \(\ds \frac {b^2 - a^2} {2 \paren {b - a} }\)
\(\ds \) \(=\) \(\ds \frac {\paren {b - a} \paren {b + a} } {2 \paren {b - a} }\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \frac {a + b} 2\)

$\blacksquare$


Also see


Sources

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