Continuous Random Variable¶

Definition¶

Under the course's absolutely continuous convention, a real-valued random variable $X$ is continuous if it has a nonnegative integrable density $f_X$ such that

\mathbb{P}(X\in A)=\int_A f_X(x)\,\mathrm{d}x

for every interval $A\subseteq\mathbb{R}$ $R$ . Necessarily $\int_{\mathbb{R}}f_X(x)\,\mathrm{d}x=1$ $1$ . A continuous CDF by itself does not guarantee the existence of a density.

Interpretation¶

A continuous random variable has its probability spread over intervals rather than concentrated on single values. Probabilities are areas under its density $f_X$ , and every single value has probability zero.

A density curve with the area between a and b shaded as the probability that X lies between them, and a vertical line at a point c, which has no area and so probability 0.

In the figure the shaded region between $a$ and $b$ has area $\mathbb{P}(a\le X\le b)$ $b)$ , the integral of $f_X$ from $a$ to $b$ . At the point $c$ the vertical line under the curve has no width and therefore no area: $\mathbb{P}(X=c)=0$ $0$ , although $f_X(c)$ is positive.

Two consequences follow. Endpoints do not matter, so $\mathbb{P}(a<X<b)=\mathbb{P}(a\le X\le b)$ $b)$ . And $f_X(x)$ is not a probability: it is probability per unit length, and it may exceed $1$ ; the uniform distribution on $(0,\tfrac{1}{2})$ has density $2$ there. Where $f_X$ is continuous, a short interval of length $h$ starting at $x$ has probability approximately $f_X(x)\,h$ .

Compared with a discrete random variable, sums of masses become definite integrals of a density: the PMF value $p_X(x)$ is a probability, while $f_X(x)$ is not. The normal distribution is another standard example. Probability zero does not mean impossible: in each experiment $X$ takes some value, although every single value had probability zero.

Key Results¶

$\mathbb{P}(X=x)=0$ for every real $x$ .
Its CDF is absolutely continuous and equals the integral of its density.

Examples¶

A uniform variable on $(0,1)$ has density $1$ there and no atom at $1/2$ .
An exponential waiting time has $\mathbb{P}(a<X<b)=\int_a^b\lambda \mathrm{e}^{-\lambda x}\,\mathrm{d}x$ for $0\le a<b$ $b$ .

Prerequisites¶

Sources¶

Probability Theory for EOR · Week 5
Nicky van Foreest · Probability Distributions · Week 2

Aliases: continuous r.v., absolutely continuous random variable, Continuous rv, continuously distributed random variable Symbols: element of, integral, lambda, real numbers, subset

Last updated: 20 October 2026 03:02