A spinner can land anywhere on a circle, a measurement instrument can report a real number, and a waiting time can take values in an interval rather than in a countable list. In these situations the most tempting discrete question, "what is the probability of this exact value?", gives the least useful answer. For an idealized uniform measurement on $[0,1]$, every point has probability $0$, yet the interval $[0,1/2]$ has probability $1/2$. The theory of continuous random variables begins with this failure: probabilities are no longer recovered by adding masses at individual points, but by measuring regions and integrating a density over them.

This page is a child of the general notion of a [random variable](/page/Random%20Variable). The parent concept says that a random variable is a measurable map out of a [probability space](/page/Probability%20Space). The continuous child concept adds a structural assumption on its distribution: the law is absolutely continuous with respect to [Lebesgue measure](/page/Lebesgue%20Measure), so probabilities can be computed by integration against a density. That extra assumption is powerful, but it is also restrictive; many random variables are neither discrete nor continuous in this sense.

The first warning is worth making concrete. A continuous model can assign probability $0$ to every point without assigning probability $0$ to every set. We write $\mathbb 1_E$ for the indicator function of a set $E$, so $\mathbb 1_E(x)=1$ when $x\in E$ and $\mathbb 1_E(x)=0$ otherwise. We also write $\mathcal L^1$ for one-dimensional Lebesgue measure, which assigns intervals their usual lengths, and $\mathcal B(\mathbb R)$ for the Borel $\sigma$-algebra, the standard collection of measurable subsets of the real line.

[example: A Point Has Probability Zero but an Interval Does Not] Let $X \sim \operatorname{Unif}(0,1)$, so its density is $f_X(x)=\mathbb 1_{(0,1)}(x)$. For any $a\in[0,1]$, the event $\{X=a\}$ is the same as $\{X\in\{a\}\}$, and the density formula gives \begin{align*} \mathbb P(X=a)=\int_{\{a\}}\mathbb 1_{(0,1)}(x)\,d\mathcal L^1(x). \end{align*} Since $0\le \mathbb 1_{(0,1)}(x)\le 1$ for all $x$ and $\mathcal L^1(\{a\})=0$, \begin{align*} 0\le \int_{\{a\}}\mathbb 1_{(0,1)}(x)\,d\mathcal L^1(x)\le \int_{\{a\}}1\,d\mathcal L^1(x)=\mathcal L^1(\{a\})=0. \end{align*} Therefore $\mathbb P(X=a)=0$. For the interval event, \begin{align*} \mathbb P(0\le X\le 1/2)=\int_{[0,1/2]}\mathbb 1_{(0,1)}(x)\,d\mathcal L^1(x). \end{align*} Changing the integrand at the single point $0$ does not change the integral, because $\mathcal L^1(\{0\})=0$, so \begin{align*} \int_{[0,1/2]}\mathbb 1_{(0,1)}(x)\,d\mathcal L^1(x)=\int_{(0,1/2]}1\,d\mathcal L^1(x). \end{align*} The last integral is the length of $(0,1/2]$: \begin{align*} \int_{(0,1/2]}1\,d\mathcal L^1(x)=\mathcal L^1((0,1/2])=\frac{1}{2}. \end{align*} Thus every individual point has probability $0$, while the interval $[0,1/2]$ has probability $1/2$. This is why a continuous random variable is described by probabilities of intervals, or more generally Borel sets, rather than by point probabilities. [/example]

The lesson is not that individual outcomes are impossible in ordinary language. The mathematical point is sharper: the event $\{X=a\}$ is too small for the density model to detect. Continuous probability lives at the scale of sets with positive Lebesgue measure.

## Definition

The parent definition of a random variable gives measurability, but it does not say what the distribution looks like. To isolate the class for which probabilities are integrals over subsets of Euclidean space, we ask whether the law of the random variable is controlled by Lebesgue measure.

[definition: Continuous Random Variable] Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. A real-valued random variable $X: (\Omega, \mathcal F) \to (\mathbb R, \mathcal B(\mathbb R))$ is a continuous random variable if there exists a [measurable function](/page/Measurable%20Function) $f_X: \mathbb R \to [0,\infty)$ such that for every Borel set $A \in \mathcal B(\mathbb R)$, \begin{align*} \mathbb P(X \in A) = \int_A f_X(x)\,d\mathcal L^1(x). \end{align*} [/definition]

This definition identifies the class. It says that the law $\mu_X=\mathbb P\circ X^{-1}$ is absolutely continuous with respect to $\mathcal L^1$, and the function $f_X$ is the object that carries the probability information. It is called a probability density function for $X$.

## Densities and Distribution Functions

### Densities as Probability per Unit Length

The density deserves its own name because it is the computational replacement for a [probability mass function](/page/Probability%20Mass%20Function). A probability mass function assigns probability directly to a value; a density assigns probability per unit length, so only integrals over sets have probabilistic meaning.

[definition: Probability Density Function] Let $X: (\Omega, \mathcal F) \to (\mathbb R, \mathcal B(\mathbb R))$ be a continuous random variable. A probability density function for $X$ is a measurable function $f_X: \mathbb R \to [0,\infty)$ satisfying \begin{align*} \mathbb P(X \in A) = \int_A f_X(x)\,d\mathcal L^1(x) \end{align*} for every Borel set $A \in \mathcal B(\mathbb R)$. [/definition]

A density is determined only up to equality $\mathcal L^1$-a.e. Changing $f_X$ at a single point, or on any Lebesgue null set, does not change any probability. This raises a necessary validation question: when does a nonnegative function actually define a probability model?

[quotetheorem:10097]

The normalization condition is the test that separates a candidate density from a merely nonnegative function. Necessity comes from applying the density formula to the whole real line: the event $X\in\mathbb R$ has probability $1$, so the total integral of $f_X$ must be $1$. Sufficiency says that this same condition is enough to build a probability law by setting the probability of each Borel set to the integral over that set. If the total integral is less than $1$, some probability mass is missing; if it is greater than $1$, the proposed rule assigns too much total mass. Thus normalization is the first check before using a displayed formula as a density.

A one-dimensional density assigns mass by integrating over intervals and Borel sets in $\mathbb R$. For several coordinates, point probabilities are still zero, but the relevant events are regions in $\mathbb R^n$ rather than intervals. The same normalization issue remains, now with volume measure in place of length. The obstruction is to distinguish an arbitrary nonnegative function of several variables from one that controls the whole joint law through volume integration.

[definition: Continuous Random Vector] Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. A random vector $X: (\Omega, \mathcal F) \to (\mathbb R^n, \mathcal B(\mathbb R^n))$ is a continuous random vector if there exists a measurable function $f_X: \mathbb R^n \to [0,\infty)$ such that for every Borel set $A \in \mathcal B(\mathbb R^n)$, \begin{align*} \mathbb P(X \in A) = \int_A f_X(x)\,d\mathcal L^n(x). \end{align*} [/definition]

This vector version is essential for multivariate statistics and stochastic processes. A pair of measurements, a location in the plane, or a collection of errors in a regression model is usually studied through a joint density.

### Cumulative Probability