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

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

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

Then, the probability density function of $X$ is

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

Proof: A continuous uniform variable is defined as having a constant probability density between minimum $a$ and maximum $b$. Therefore,

$\label{eq:cuni-pdf-s1} \begin{split} f_X(x) &\propto 1 \quad \text{for all} \quad x \in [a,b] \quad \text{and} \\ f_X(x) &= 0, \quad\!\! \text{if} \quad x < a \quad \text{or} \quad x > b \; . \end{split}$

To ensure that $f_X(x)$ is a proper probability density function, the integral over all non-zero probabilities has to sum to $1$. Therefore,

$\label{eq:cuni-pdf-s2} f_X(x) = \frac{1}{c(a,b)} \quad \text{for all} \quad x \in [a,b]$

where the normalization factor $c(a,b)$ is specified, such that

$\label{eq:cuni-pdf-s3} \frac{1}{c(a,b)} \int_{a}^{b} 1 \, \mathrm{d}x = 1 \; .$

Solving this for $c(a,b)$, we obtain:

$\label{eq:cuni-pdf-s4} \begin{split} \int_{a}^{b} 1 \, \mathrm{d}x &= c(a,b) \\ [x]_a^b &= c(a,b) \\ c(a,b) &= b-a \; . \end{split}$
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Metadata: ID: P37 | shortcut: cuni-pdf | author: JoramSoch | date: 2020-01-31, 15:41.