Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Continuous uniform distribution ▷ Mean

Theorem: Let $X$ be a random variable following a continuous uniform distribution:

$\label{eq:cuni} X \sim \mathcal{U}(a, b) \; .$

Then, the mean or expected value of $X$ is

$\label{eq:cuni-mean} \mathrm{E}(X) = \frac{1}{2} (a+b) \; .$

Proof: The expected value is the probability-weighted average over all possible values:

$\label{eq:mean} \mathrm{E}(X) = \int_{\mathcal{X}} x \cdot f_X(x) \, \mathrm{d}x \; .$

With the probability density function of the continuous uniform distribution, this becomes:

$\label{eq:cuni-mean-qed} \begin{split} \mathrm{E}(X) &= \int_a^b x \cdot \frac{1}{b-a} \, \mathrm{d}x \\ &= \left[ \frac{1}{2} \, \frac{x^2}{b-a} \right]_a^b \\ &= \frac{1}{2} \, \frac{b^2 - a^2}{b-a} \\ &= \frac{1}{2} \, \frac{(b+a)(b-a)}{b-a} \\ &= \frac{1}{2} (a+b) \; . \end{split}$
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Metadata: ID: P82 | shortcut: cuni-mean | author: JoramSoch | date: 2020-03-16, 16:12.