Absolutely Continuous Distribution¶

Definition¶

The distribution of a real-valued random variable $X$ is absolutely continuous if there is a nonnegative integrable function $f_X$ such that

F_X(x)=\int_{-\infty}^{x} f_X(t)\,\mathrm{d}t,\qquad \int_{-\infty}^{\infty}f_X(t)\,\mathrm{d}t=1,

where $F_X$ is its cumulative distribution function. The function $f_X$ is a probability density. “Law” and “distribution” mean the same thing here.

Interpretation¶

An absolutely continuous distribution spreads probability along the real line instead of placing it at single points. All probabilities come from areas under a density $f_X$ ; for an interval,

\mathbb{P}(a<X\le b)=\int_a^b f_X(t)\,\mathrm{d}t.

The two conditions in the Definition have separate roles. The first says that the CDF at $x$ is the area under $f_X$ to the left of $x$ ; the definite integral then turns $F_X(b)-F_X(a)$ into the area between $a$ and $b$ . The second, total area $1$ , says that $X$ takes some real value with probability one. At points where $f_X$ is continuous, the fundamental theorem of calculus gives $F_X'(x)=f_X(x)$ $f_{X} (x)$ : the density is the slope of the CDF.

Because a single point has length zero, $\mathbb{P}(X=a)=0$ for every $a$ , and it does not matter whether the endpoints of an interval are included. For the same reason the density is not unique: changing it at finitely many points changes no integral and therefore no probability.

The continuous random variable page uses exactly this convention. The property concerns the distribution, rather than the particular experiment used to construct $X$ . A continuous CDF alone is not enough, as the Key Results note: some continuous CDFs have no density. A distribution with a positive point mass is not absolutely continuous, and a mixture of a point mass and a density is neither discrete nor absolutely continuous.

Key Results¶

Every single value has probability zero: $\mathbb{P}(X=a)=0$ $0$ .
Its CDF is continuous. Continuity of a CDF alone does not imply the existence of a density.
Changing a density on a set of zero length does not change the distribution; density identities are understood in this sense.

Examples¶

For the uniform distribution on $(0,1)$ , take $f_X(t)=1$ on $(0,1)$ and zero elsewhere. Then $\mathbb{P}(1/4<X\le3/4)=1/2$ $1/2$ . A distribution with a positive point mass is not absolutely continuous.

Prerequisites¶

Sources¶

Wiki terminology supplement to the PT density convention; placement alongside Week 5 is provisional pending course review.
Marco Taboga, Continuous random variable, definition and interval-probability interpretation.

Aliases: absolutely continuous law, absolutely continuous probability distribution Symbols: infinity, integral

Last updated: 22 October 2026 19:04