Multivariate Densities on Irregular Supports

Why the Shape of the Support Matters

A joint density describes how probability is distributed across the plane, but its formula alone does not determine where probability can occur. The support region also matters: if the random vector can only fall inside a triangle, disk, or curved region, probabilities must be calculated over the part of that region satisfying the event. Integrating over a convenient rectangle that includes points outside the support can give the wrong answer.

For a bivariate random vector $(X,Y)$, the relevant notion of area is two-dimensional Lebesgue measure. In ordinary examples, it agrees with geometric area. The same ideas extend to a vector in $d$ dimensions by replacing area with $d$-dimensional volume.

Definition: Let $S$ be a measurable subset of $\mathbb{R}^2$. A joint density of $(X,Y)$ supported on $S$ is a nonnegative measurable function $f_{X,Y}$ on $\mathbb{R}^2$ such that $f_{X,Y}(x,y)=0$ outside $S$ (apart from sets of area zero) and $$ \iint_{\mathbb{R}^2} f_{X,Y}(x,y)\,dx\,dy=1. $$ For every measurable event region $A\subseteq\mathbb{R}^2$, $$ P\big((X,Y)\in A\big)=\iint_A f_{X,Y}(x,y)\,dx\,dy. $$

The density value at a point is not itself a probability. Probability is the integral of the density over a region. A density can therefore exceed one, especially when probability is concentrated in a small area, without violating any probability rule.

Density Normalization and Region Probabilities

The basic test for a proposed density has two parts: it must be nonnegative, and its total integral over the plane must be one. When a formula is given only on a support $S$, the total integral reduces to an integral over $S$, since the density is zero outside it.

Theorem: Suppose $f$ is a nonnegative measurable function on $\mathbb{R}^2$ with $\iint_{\mathbb{R}^2} f(x,y)\,dx\,dy=1$. Defining the probability of each measurable region $A$ by $P((X,Y)\in A)=\iint_A f(x,y)\,dx\,dy$ gives a probability law. In particular, the probability of the whole plane is one, and probabilities of disjoint regions add.

Proof. The integral of a nonnegative measurable function over a measurable set is nonnegative, so every assigned probability is nonnegative. For the whole plane the assigned probability is the assumed integral, which is one. If $A_1,A_2,\ldots$ are pairwise disjoint measurable regions, countable additivity of the integral for nonnegative functions gives $$ \iint_{\bigcup_{j=1}^{\infty}A_j} f(x,y)\,dx\,dy =\sum_{j=1}^{\infty}\iint_{A_j} f(x,y)\,dx\,dy. $$ Thus the assignment satisfies the probability axioms. $\square$

A useful special case is a uniform density over a finite-area region. Here “uniform” means the density has the same value at every point of the region and is zero elsewhere.

Result: If $S$ has finite, positive area $|S|$, then the density that is constant on $S$ and zero outside it must equal $1/|S|$ on $S$. Under this density, for every measurable $A$, $$ P\big((X,Y)\in A\big)=\frac{|A\cap S|}{|S|}. $$

Derivation. Write the constant density as $c$. Its total integral is $c|S|$, so normalization requires $c|S|=1$, hence $c=1/|S|$. Integrating this constant over the event’s part of the support gives $$ \iint_A f(x,y)\,dx\,dy =\iint_{A\cap S}\frac{1}{|S|}\,dx\,dy =\frac{|A\cap S|}{|S|}. $$ This area-ratio shortcut applies only when the density is constant on the support.

Worked Examples on Nonrectangular Supports

Worked Example: A Uniform Density on a Triangle

Let $S=\{(x,y):x\geq 0,\ y\geq 0,\ x+y\leq 1\}$, and suppose $f_{X,Y}(x,y)=c$ on $S$, with density zero elsewhere. Find $c$, then calculate $P(X+Y\leq 1/2)$.

Normalize. At a fixed $x$ between 0 and 1, $y$ ranges from 0 to $1-x$. Therefore the area of $S$ is $$ \int_0^1\int_0^{1-x}1\,dy\,dx =\int_0^1(1-x)\,dx =\left[x-\frac{x^2}{2}\right]_0^1 =\frac12. $$ The total probability is $c/2$, so $c/2=1$ and $c=2$.

Calculate the event probability. The event region within $S$ has $0\leq x\leq 1/2$ and $0\leq y\leq 1/2-x$. Thus $$ P(X+Y\leq 1/2) =2\int_0^{1/2}\int_0^{1/2-x}1\,dy\,dx =2\int_0^{1/2}(1/2-x)\,dx =2\left[\frac{x}{2}-\frac{x^2}{2}\right]_0^{1/2} =\frac14. $$ Equivalently, this smaller triangle has one quarter the area of $S$, so its probability is $1/4$.

Worked Example: A Uniform Density on a Disk

Suppose $(X,Y)$ is uniform on the disk $S=\{(x,y):x^2+y^2\leq 4\}$. Find the density on $S$ and the probability that the vector lies within distance 1 of the origin.

The support is a disk of radius 2, so its area is $\pi(2)^2=4\pi$. The uniform density is consequently $1/(4\pi)$ on $S$. The event $X^2+Y^2\leq 1$ is the disk of radius 1, with area $\pi$. Its probability is $$ P(X^2+Y^2\leq 1) =\frac{\pi}{4\pi} =\frac14. $$ The event disk lies entirely inside the support disk, so no part of its area needs to be removed. This calculation uses geometric areas directly; it does not require changing coordinates.

Worked Example: A Density Under a Curved Boundary

Let the support be $S=\{(x,y):0\leq x\leq 1,\ 0\leq y\leq x^2\}$, and let the density be the unknown constant $c$ on $S$. Find $c$ and calculate $P(X>1/2)$.

For each $x$, the vertical extent of the support is from $y=0$ to $y=x^2$. Its area, and hence the total integral of the constant $c$, is $$ \int_0^1\int_0^{x^2}c\,dy\,dx =c\int_0^1 x^2\,dx =\frac{c}{3}. $$ Setting this equal to one gives $c=3$.

The event $X>1/2$ restricts the outer integration range to $1/2<x\leq1$; the curved upper bound for $y$ is still $x^2$. Therefore $$ P(X>1/2) =\int_{1/2}^1\int_0^{x^2}3\,dy\,dx =3\int_{1/2}^1 x^2\,dx =\left[x^3\right]_{1/2}^1 =1-\frac18 =\frac78. $$ The result is large because most of the support’s area lies to the right of $x=1/2$.

Setting Up an Integral for an Event

For an irregular support, first identify the points that are both in the support and in the event. Then describe that intersection with integration bounds. If the density is given by a formula rather than a constant, integrate that formula—not merely the area—over the intersection. The order of integration is a matter of convenience, but the bounds must describe exactly the same region in either order.

Write the support.
Translate its inequalities into a precise region $S$, including any curved or slanted boundaries.

Check normalization.
Verify nonnegativity, integrate the proposed density over $S$, and confirm that the result is one. If a constant is unknown, solve for it.

Intersect with the event.
Restrict the region to points satisfying the event, then choose bounds that cover this intersection without adding points outside it.

Integrate and check.
Integrate the density over those bounds. Confirm that the answer is between zero and one and is consistent with the event’s size and location.

Boundaries are sometimes written with strict inequalities and sometimes with non-strict inequalities. If the boundary has area zero, this choice does not change a density-based probability: the integral of an integrable density over a set of area zero is zero. Straight-line edges and circle boundaries have area zero, so the examples above are unaffected by their boundary conventions.

Common Mistake: Using the Enclosing Rectangle

A frequent error is to integrate over the smallest convenient rectangle containing the support, as though every point in that rectangle could occur. For the triangular support in the first example, the enclosing rectangle is $0\leq x\leq1,\ 0\leq y\leq1$, which has area 1, while the triangle has area $1/2$. If the density is 2 on the triangle and zero outside, integrating the value 2 over the entire rectangle gives 2, not 1. That calculation silently assigns positive density to points outside the support.

The correction is to keep the support in view throughout. Integrate only over the triangle, or over the event’s intersection with the triangle. A rectangle may be used as an outer range for setting up an integral only when the integrand is also restricted to the support—for example, by using appropriate inner bounds or an indicator that is zero outside the support.

Key takeaway: A joint density is normalized with respect to area over its actual support. For any event, its probability is the integral of the density over the event’s intersection with that support—not over a larger region chosen for convenience.

Check Your Understanding

For each question, identify the region that should be integrated over before calculating.

A density is constant on a triangle with area $3$. What constant makes it a valid density?
A vector is uniform on a disk of radius $5$. What is the probability it lies in the concentric disk of radius $2$?
For support $0\leq x\leq1,\ 0\leq y\leq x$, describe the bounds for the event $X>1/2$.
Why does integrating a density over a rectangle containing its support sometimes give an incorrect normalization?
If an event boundary is a circle, under what condition can including or excluding that boundary leave its probability unchanged?