Conditional Cumulative Distribution Function¶

Definition¶

Let $X$ and $Y$ be continuous random variables with joint density $f_{X,Y}$ , and let $y$ satisfy $0<f_Y(y)<\infty$ $\infty$ . The conditional CDF of $X$ given $Y=y$ is

\begin{aligned} F_{X\mid Y}(x\mid y) =\int_{-\infty}^x f_{X\mid Y}(t\mid y)\,\mathrm{d}t. \end{aligned}

Here

\begin{aligned} f_{X\mid Y}(t\mid y)=\frac{f_{X,Y}(t,y)}{f_Y(y)}. \end{aligned}

We write $F_{X\mid Y}(x\mid y)=\mathbb{P}(X\leq x\mid Y=y)$ $y)$ . Note that for continuous $Y$ , this cannot be defined by the elementary formula $\mathbb{P}(A\mid B)=\mathbb{P}(A\cap B)/\mathbb{P}(B)$ $B) / P (B)$ , because $\mathbb{P}(Y=y)=0$ $0$ .

The integrand is the Conditional Probability Density Function. Integrating it up to $x$ gives the conditional threshold probability in exactly the same way that an unconditional PDF gives a CDF.

Interpretation¶

A conditional CDF gives threshold probabilities within a specified conditional distribution: $F_{X\mid Y}(x\mid y)$ is the probability that $X\le x$ once $Y=y$ is known. For each fixed $y$ it is an ordinary CDF in $x$ , obtained from the conditional density just as a CDF is obtained from a PDF.

Two conditional CDFs of X given Y = y, straight lines from 0 to 1 over the interval from 0 to y, for y = 1/2 and y = 1; each reaches 1/2 at x = y/2.

The diagram shows the conditional CDFs of $X\mid Y=y\sim\operatorname{Unif}(0,y)$ for $y=\tfrac{1}{2}$ (blue) and $y=1$ (red). Each graph rises in a straight line from $0$ at $x=0$ to $1$ at $x=y$ $y$ , following $F_{X\mid Y}(x\mid y)=x/y$ $x / y$ , and passes $\tfrac{1}{2}$ at $x=y/2$ $y /2$ : at $x=\tfrac{1}{4}$ for $y=\tfrac{1}{2}$ and at $x=\tfrac{1}{2}$ for $y=1$ $1$ . Changing $y$ changes the whole conditional distribution, not just one value.

The notation $\mathbb{P}(X\le x\mid Y=y)$ cannot be read through the elementary formula $\mathbb{P}(A\mid B)=\mathbb{P}(A\cap B)/\mathbb{P}(B)$ $B) / P (B)$ , because $\mathbb{P}(Y=y)=0$ for continuous $Y$ . It is defined through the conditional density instead, and the condition $0<f_Y(y)<\infty$ makes the ratio $f_{X,Y}(t,y)/f_Y(y)$ meaningful. For independent $X$ and $Y$ , the conditional CDF equals the unconditional $F_X$ for $Y$ -almost every conditioning value $y$ . One may choose a version equal to $F_X$ for every $y$ ; other versions can differ on conditioning values of probability zero.

Averaging the conditional CDFs over $Y$ gives back $F_X$ : integrating the continuous law of total probability in $x$ gives $F_X(x)=\int F_{X\mid Y}(x\mid y)f_Y(y)\,\mathrm{d}y$ $y) f_{Y} (y) d y$ . In the triangle model of the conditional density page, $f_Y(y)=2y$ on $(0,1)$ and $F_{X\mid Y}(x\mid y)=1$ when $y\le x$ $x$ , so for $0<x<1$

F_X(x)=\int_0^x 2y\,\mathrm{d}y+\int_x^1\frac{x}{y}\,2y\,\mathrm{d}y =x^2+2x(1-x)=2x-x^2.

Key Results¶

For every $y$ with $0<f_Y(y)<\infty$ $\infty$ , it is a nondecreasing, right-continuous CDF in $x$ .
Where differentiation is justified, $\partial F_{X\mid Y}(x\mid y)/\partial x=f_{X\mid Y}(x\mid y)$ except possibly on a set of zero length.

Examples¶

For the uniform conditional law on $(0,y)$ , $F_{X\mid Y}(x\mid y)=x/y$ for $0<x<y$ $y$ , so $F_{X\mid Y}(y/2\mid y)=1/2$ $1/2$ .
If $X\sim\operatorname{Exp}(2)$ is independent of $Y$ , then $F_{X\mid Y}(x\mid y)=1-\mathrm{e}^{-2x}$ for $x\ge0$ $0$ .

Prerequisites¶

Sources¶

Nicky van Foreest · Probability Distributions · Week 2

Aliases: conditional CDF, conditional distribution function Symbols: infinity, integral, intersection, partial

Last updated: 27 October 2026 11:52