Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate continuous distributions ▷ Continuous uniform distribution ▷ Cumulative distribution function

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

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

Then, the cumulative distribution function of $X$ is

\[\label{eq:cuni-cdf} F_X(x) = \left\{ \begin{array}{rl} 0 \; , & \text{if} \; x < a \\ \frac{x-a}{b-a} \; , & \text{if} \; a \leq x \leq b \\ 1 \; , & \text{if} \; x > b \; . \end{array} \right.\]

Proof: The probability density function of the continuous uniform distribution is:

\[\label{eq:cuni-pdf} \mathcal{U}(x; a, b) = \left\{ \begin{array}{rl} \frac{1}{b-a} \; , & \text{if} \; a \leq x \leq b \\ 0 \; , & \text{otherwise} \; . \end{array} \right.\]

Thus, the cumulative distribution function is:

\[\label{eq:cuni-cdf-s1} F_X(x) = \int_{-\infty}^{x} \mathcal{U}(z; a, b) \, \mathrm{d}z\]

First of all, if $x < a$, we have

\[\label{eq:cuni-cdf-s2a} F_X(x) = \int_{-\infty}^{x} 0 \, \mathrm{d}z = 0 \; .\]

Moreover, if $a \leq x \leq b$, we have using \eqref{eq:cuni-pdf}

\[\label{eq:cuni-cdf-s2b} \begin{split} F_X(x) &= \int_{-\infty}^{a} \mathcal{U}(z; a, b) \, \mathrm{d}z + \int_{a}^{x} \mathcal{U}(z; a, b) \, \mathrm{d}z \\ &= \int_{-\infty}^{a} 0 \, \mathrm{d}z + \int_{a}^{x} \frac{1}{b-a} \, \mathrm{d}z \\ &= 0 + \frac{1}{b-a} [z]_a^x \\ &= \frac{x-a}{b-a} \; . \end{split}\]

Finally, if $x > b$, we have

\[\label{eq:cuni-cdf-s2c} \begin{split} F_X(x) &= \int_{-\infty}^{b} \mathcal{U}(z; a, b) \, \mathrm{d}z + \int_{b}^{x} \mathcal{U}(z; a, b) \, \mathrm{d}z \\ &= F_X(b) + \int_{b}^{x} 0 \, \mathrm{d}z \\ &= \frac{b-a}{b-a} + 0 \\ &= 1 \; . \end{split}\]

This completes the proof.

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Metadata: ID: P38 | shortcut: cuni-cdf | author: JoramSoch | date: 2020-01-02, 18:05.