Joint Densities and Area Integrals

From Joint Probabilities to Area Integrals

For a pair of random variables, probabilities need not be concentrated at isolated points. They can instead be spread continuously across the plane. A joint density describes that spread: the probability assigned to a region is the integral of the density over that region. The area of the region matters as well as the density values within it.

This is the continuous counterpart of adding joint probabilities over points in a countable support. Here the sum is replaced by a double integral, and the probability of an event is obtained by integrating over all coordinate pairs that satisfy the event.

Definition: A pair $(X,Y)$ has a joint density $f$ if $f:\mathbb{R}^2\to[0,\infty]$ is measurable and, for every Borel set $A\subseteq\mathbb{R}^2$, $$ P\bigl((X,Y)\in A\bigr)=\iint_A f(x,y)\,dx\,dy. $$ The integrals are with respect to two-dimensional Lebesgue area. In particular, the density must satisfy $$ \iint_{\mathbb{R}^2} f(x,y)\,dx\,dy=1. $$

A density is not itself a probability at a point. It measures probability per unit area near that point. For a small region where $f$ is nearly constant, the probability is approximately the density value multiplied by the region's area. The exact probability is the integral, which accounts for any changes in density across the region.

The joint density is determined only up to changes on sets of area zero: changing its values there does not change any of the integrals in the definition. In particular, a pair with a joint density assigns probability zero to any single point, since a point has area zero.

Normalization and the Probability Measure

The total integral is not a technical afterthought. It ensures that the density assigns total probability one. More generally, any nonnegative measurable function with integral one defines a probability law by integrating it over events.

Theorem: Suppose $f:\mathbb{R}^2\to[0,\infty]$ is measurable and $$ \iint_{\mathbb{R}^2} f(x,y)\,dx\,dy=1. $$ For each Borel set $A$, define $P_f(A)=\iint_A f(x,y)\,dx\,dy$. Then $P_f$ is a probability measure on the Borel subsets of $\mathbb{R}^2$.

Proof. Since $f$ is nonnegative, every integral $P_f(A)$ is nonnegative. Also, $P_f(\varnothing)=0$, and the assumed total integral gives $P_f(\mathbb{R}^2)=1$. It remains to check countable additivity. Let $A_1,A_2,\ldots$ be pairwise disjoint Borel sets. The indicator of their union is the pointwise sum of their indicators: $$ \mathbf{1}_{\cup_{i=1}^{\infty} A_i}(x,y) =\sum_{i=1}^{\infty}\mathbf{1}_{A_i}(x,y). $$ Multiplying by $f$ preserves this equality because $f\geq 0$. The partial sums on the right increase to the full sum. By the monotone convergence theorem, $$ P_f\left(\bigcup_{i=1}^{\infty}A_i\right) =\iint_{\mathbb{R}^2}f\sum_{i=1}^{\infty}\mathbf{1}_{A_i}\,dx\,dy =\sum_{i=1}^{\infty}\iint_{\mathbb{R}^2}f\mathbf{1}_{A_i}\,dx\,dy =\sum_{i=1}^{\infty}P_f(A_i). $$ Thus $P_f$ is nonnegative, has total mass one, and is countably additive, so it is a probability measure. $\square$

Worked Example: Checking a Constant Density and Computing a Rectangle Probability

Consider the candidate function $$ f(x,y)= \begin{cases} \frac{1}{2}, & 0\leq x\leq 2,\ 0\leq y\leq 1,\\ 0, & \text{otherwise}. \end{cases} $$ First verify that it is normalized, then find $P(1/2\leq X\leq 3/2,\ 1/4\leq Y\leq 3/4)$.

The function is nonnegative and measurable. Its total integral is its constant value multiplied by the area of the rectangle on which it is nonzero: $$ \iint_{\mathbb{R}^2} f(x,y)\,dx\,dy =\int_0^2\int_0^1 \frac{1}{2}\,dy\,dx =\int_0^2 \frac{1}{2}\,dx =1. $$ So it is a valid joint density.

The event rectangle has width $3/2-1/2=1$ and height $3/4-1/4=1/2$. Its density is $1/2$, so direct integration gives $$ P\left(\frac12\leq X\leq\frac32,\ \frac14\leq Y\leq\frac34\right) =\int_{1/2}^{3/2}\int_{1/4}^{3/4}\frac12\,dy\,dx =\int_{1/2}^{3/2}\frac14\,dx =\frac14. $$ Equivalently, the probability is density times area: $\frac12\times(1\times\frac12)=\frac14$.

Iterated Integrals Over Rectangles

For a bounded rectangle $R=[a,b]\times[c,d]$, the probability is the double integral of $f$ over $R$. Because $f$ is nonnegative and measurable, Tonelli's theorem allows this integral to be evaluated as an iterated integral in either order. The result is

Theorem: If $(X,Y)$ has joint density $f$, then for finite $a\leq b$ and $c\leq d$, $$ P(a\leq X\leq b,\ c\leq Y\leq d) =\int_a^b\int_c^d f(x,y)\,dy\,dx =\int_c^d\int_a^b f(x,y)\,dx\,dy. $$

Proof. The event is that $(X,Y)$ belongs to the rectangle $R=[a,b]\times[c,d]$. By the definition of a joint density, its probability is $\iint_R f(x,y)\,dx\,dy$. Write this as an integral over the plane by multiplying $f$ by the indicator of $R$: $$ P\bigl((X,Y)\in R\bigr) =\iint_{\mathbb{R}^2} f(x,y)\mathbf{1}_{[a,b]}(x)\mathbf{1}_{[c,d]}(y)\,dx\,dy. $$ The integrand is nonnegative and measurable, so Tonelli's theorem permits either order of integration. Integrating first in $y$ restricts that integral to $[c,d]$, and then integrating in $x$ restricts the outer integral to $[a,b]$. Reversing the order gives the other expression. The probability is finite because it is at most the total integral, which is one. $\square$

Worked Example: A Density That Changes Across the Unit Square

Let $f(x,y)=4xy$ for $0\leq x\leq1$ and $0\leq y\leq1$, and let $f(x,y)=0$ elsewhere. Verify the total integral and calculate the probability that both coordinates lie between zero and one half.

The total integral factors into two one-dimensional integrals: $$ \int_0^1\int_0^1 4xy\,dy\,dx =4\left(\int_0^1x\,dx\right)\left(\int_0^1y\,dy\right) =4\left(\frac12\right)\left(\frac12\right) =1. $$ Thus this is a normalized joint density.

For the requested rectangle, integrate over $0\leq x\leq1/2$ and $0\leq y\leq1/2$: $$ P\left(0\leq X\leq\frac12,\ 0\leq Y\leq\frac12\right) =\int_0^{1/2}\int_0^{1/2}4xy\,dy\,dx. $$ The inner integral is $2xy^2$ evaluated from $0$ to $1/2$, which equals $x/2$. Therefore $$ \int_0^{1/2}\frac{x}{2}\,dx =\left[\frac{x^2}{4}\right]_0^{1/2} =\frac{1}{16}. $$ The density is smaller near the origin, so this probability is not the density at a representative point multiplied by the rectangle's area; the integral accounts for its variation.

Integrating Over a Simple Bounded Region

Not every event describes a rectangle. A region may have a boundary whose location depends on one coordinate. The same principle still applies: integrate the density over the points in the event, expressing the bounds so that each point in the region is counted once.

Worked Example: Probability Inside a Bounded Triangle

Consider $$ f(x,y)= \begin{cases} 2, & 0\leq x\leq1,\ 0\leq y\leq1-x,\\ 0, & \text{otherwise}. \end{cases} $$ Verify the total integral, then find the probability of the bounded rectangle $0\leq X\leq1/2,\ 0\leq Y\leq1/2$.

For each $x$ between zero and one, the allowed $y$-values run from zero to $1-x$. Thus $$ \iint_{\mathbb{R}^2} f(x,y)\,dx\,dy =\int_0^1\int_0^{1-x}2\,dy\,dx =\int_0^1 2(1-x)\,dx =2\left[x-\frac{x^2}{2}\right]_0^1 =1. $$ The function is nonnegative and has total integral one.

On the requested rectangle, every point satisfies the triangular restriction: when $0\leq x\leq1/2$, $1-x\geq1/2$, so all $y$ between zero and $1/2$ are allowed. Therefore $$ P\left(0\leq X\leq\frac12,\ 0\leq Y\leq\frac12\right) =\int_0^{1/2}\int_0^{1/2}2\,dy\,dx =\int_0^{1/2}1\,dx =\frac12. $$

Common Mistakes and Why Area Matters

A probability is an integral, not a density value. For a constant density over a rectangle, the calculation reduces to the constant multiplied by the rectangle's area. For a varying density, the integral adds up the changing contributions across the region. In either case, leaving out the area factor can change the answer.

Using the density value as the event probability. A density can exceed one, and it is not a point probability. Integrate it over the event region.
Checking only that a function is nonnegative. A candidate must also have total integral one. For example, the constant function $1/2$ on the unit square integrates to $1/2$, not one, so it is not a joint density.
Forgetting one of the region's dimensions. On the unit square, the density $1$ assigns probability $1\times(1/2)\times1=1/2$ to a strip of width $1/2$ and height $1$, not $1/4$.
Integrating outside the event. For a region with a sloping boundary, use limits that describe the region itself, as in $0\leq y\leq1-x$, rather than integrating over a larger rectangle.

Joint densities give a unified way to calculate probabilities for continuous pairs: identify the event as a region in the plane, integrate over that region, and check that the density's integral over the whole plane is one. The geometry determines the limits of integration; the density determines how probability is distributed within those limits.

Check Your Understanding

For each question, identify the relevant area integral before evaluating it.

A density is constant at $1/3$ on $[0,3]\times[0,1]$ and zero elsewhere. What is its total integral?
For density $1$ on the unit square, what is the probability of $0\leq X\leq1/4$ and $0\leq Y\leq1$?
Why does a nonnegative measurable function with total integral $1/2$ fail to be a joint density?
For $f(x,y)=4xy$ on the unit square, write the iterated integral for $P(X\leq1/2,\ Y\leq1)$.
For the density equal to $2$ on $0\leq x\leq1,\ 0\leq y\leq1-x$, explain why the upper limit for $y$ depends on $x$.