Conditional Probability Density Function¶

Definition¶

Let $X$ and $Y$ have joint density $f_{X,Y}$ , and let $y$ satisfy $0<f_Y(y)<\infty$ $\infty$ . The conditional PDF of $X$ given $Y=y$ is

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

The denominator is the density $f_Y(y)$ , not $\mathbb{P}(Y=y)=0$ $0$ . On a set of conditioning values $y$ of probability zero, for example finitely many points, conditional laws may be chosen arbitrarily.

The Marginal Probability Density Function in the denominator is obtained by integrating the joint density over the other coordinate. Dividing by it makes the conditional density integrate to one.

Interpretation¶

A conditional density describes one variable after a continuous value of another is observed. It uses a ratio of densities, rather than dividing by the probability of a zero-probability event: the slice of the joint density at $Y=y$ is rescaled by $f_Y(y)$ so that it has total area $1$ .

The triangle 0 less than x less than y less than 1 with joint density 2, and a horizontal slice at Y = y; rescaled, the slice gives the conditional density 1/y, uniform on 0 to y.

The joint density equals $2$ on the triangle $0<x<y<1$ $1$ . Along the horizontal slice at height $y$ it equals $2$ on $0<x<y$ $y$ , a segment of length $y$ , so the marginal density is $f_Y(y)=2y$ $2 y$ . Dividing gives $f_{X\mid Y}(x\mid y)=2/(2y)=1/y$ on $(0,y)$ : given $Y=y$ $y$ , $X$ is uniform on $(0,y)$ .

The formula mirrors the conditional PMF $p_{X,Y}(x,y)/p_Y(y)$ , with densities in place of masses. Heuristically, conditioning on a thin strip $y<Y\le y+h$ $h$ , which has positive probability, produces a factor $h$ in both the joint and the marginal probability; it cancels, leaving the ratio of densities. Like any density, $f_{X\mid Y}(x\mid y)$ is not a probability and can exceed $1$ : in the triangle it equals $4$ when $y=\tfrac{1}{4}$ $\frac{1}{4}$ . Changing the conditional density for a set of $y$ -values of probability zero changes no probability, which is why the definition allows arbitrary choices there.

Multiplying back, $f_{X,Y}=f_{X\mid Y}f_Y$ builds a joint model in two stages, and integrating over $y$ gives the continuous law of total probability. For independent $X$ and $Y$ the conditional density is simply $f_X$ , and integrating $f_{X\mid Y}$ up to $x$ gives the conditional CDF.

Key Results¶

For every $y$ with $0<f_Y(y)<\infty$ $\infty$ , $\int f_{X\mid Y}(x\mid y)\,\mathrm{d}x=1$ $1$ .
$f_{X,Y}(x,y)=f_{X\mid Y}(x\mid y)f_Y(y)$ except possibly on a set of zero area.

Examples¶

For joint density $2$ on $0<x<y<1$ $1$ , $f_{X\mid Y}(x\mid y)=1/y$ on $0<x<y$ $y$ : a uniform conditional law.
For independent $X\sim\operatorname{Exp}(2)$ and $Y\sim\operatorname{Exp}(3)$ $Exp (3)$ , $f_{X\mid Y}(x\mid y)=2\mathrm{e}^{-2x}$ for $x>0$ $0$ .

Prerequisites¶

Sources¶

Nicky van Foreest · Probability Distributions · Week 2

Aliases: conditional PDF, conditional density, conditional density function Symbols: infinity, integral

Last updated: 20 October 2026 03:02