Supports, Zero-Density Regions, and Normalization

Why Support Matters

A joint-density formula does not describe probability uniformly over the whole plane. Its constraints identify where the density may be positive, and its values determine how probability is distributed within that region. To check or normalize a proposed density, the integral must cover the region where the formula contributes—not merely a convenient rectangle containing it.

This distinction matters especially when a formula includes an indicator function or when the variables satisfy an inequality such as $0<y<x<1$. The latter describes a triangle, not the entire square $0<x<1,\ 0<y<1$. Integrating over the square would count points where the density is zero.

Density, Positivity Region, and Support

Definition: A joint density of a random vector $(X,Y)$ is a nonnegative measurable function $f$ on $\mathbb{R}^2$ such that, for every measurable region $A\subseteq\mathbb{R}^2$, $$ P\bigl((X,Y)\in A\bigr)=\iint_A f(x,y)\,dx\,dy, $$ and $\iint_{\mathbb{R}^2}f(x,y)\,dx\,dy=1$. The positivity region of a specified version of $f$ is the set $\{(x,y):f(x,y)>0\}$. The density is zero outside this region.

In formulas, constraints are often written using an indicator: $f(x,y)=c\,g(x,y)\mathbf{1}_{S}(x,y)$, where $S$ is a specified region and $\mathbf{1}_{S}$ equals 1 on $S$ and 0 elsewhere. The indicator makes the density zero outside $S$. If $g$ is itself zero at some points of $S$, the positivity region can be smaller than $S$.

A density is defined only up to changes on sets of area zero. For example, assigning a different value to a density along a line does not change any probabilities, because a line has two-dimensional Lebesgue measure zero. Thus the exact inclusion or exclusion of ordinary boundaries usually does not affect normalization. It is still useful to state the constraints clearly: they determine the region used in the integral.

The support of the probability distribution is a related but slightly different idea. It is the set of points for which every open neighborhood has positive probability. For a density, this support is the closure of the region where probability can occur, with the understanding that modifying the density on an area-zero set does not change the distribution. In calculations, “effective support” will mean the region on which the given formula contributes to the integral, considered up to area-zero differences.

Key distinction: A bounding rectangle may contain the effective support without being the effective support. Integrate the density over its positivity region, or over all of $\mathbb{R}^2$ while correctly treating it as zero elsewhere.

Normalizing a Nonnegative Formula

Often a problem gives a nonnegative kernel $g(x,y)$ and asks for a constant $c$ such that $f(x,y)=c\,g(x,y)$ is a joint density. The kernel specifies the shape of the density, while $c$ sets its total integral equal to one.

Theorem: Let $g:\mathbb{R}^2\to[0,\infty]$ be measurable and let $$ I=\iint_{\mathbb{R}^2}g(x,y)\,dx\,dy. $$ If $0<I<\infty$, then $f=g/I$ is a joint density. It is the unique density of the form $c\,g$ with a finite constant $c$. If $I=0$, no finite constant can make $c\,g$ integrate to one; if $I=\infty$, no positive finite constant can do so.

Proof. Since $g\geq 0$, the function $f=g/I$ is nonnegative and measurable. Its integral is $$ \iint_{\mathbb{R}^2} f(x,y)\,dx\,dy =\frac{1}{I}\iint_{\mathbb{R}^2}g(x,y)\,dx\,dy =\frac{I}{I}=1. $$ Thus $f$ is a density. If $c\,g$ is a density, then $1=\iint c\,g=cI$, so $c=1/I$. If $I=0$, every finite multiple has integral zero. If $I=\infty$, every positive finite multiple has infinite integral, while $c=0$ gives integral zero. None of these integrals equals one. $\square$

For a nonnegative measurable kernel, Tonelli’s theorem permits the total integral to be evaluated as an iterated integral, including when the integral might be infinite. In the examples below, the constraints give finite, explicitly bounded regions, and the displayed iterated integrals calculate the total mass directly.

Worked Examples

Worked Example: Normalizing a Density on a Rectangle

Suppose a proposed joint density has the form $$ f(x,y)=c(x+2y)\mathbf{1}_{\{0<x<1,\ 0<y<2\}}(x,y). $$ Find $c$ and state the effective support.

The indicator restricts the calculation to the rectangle $0<x<1,\ 0<y<2$. On this rectangle, $x+2y>0$, so the kernel is positive throughout its interior. The total integral before normalization is $$ \int_0^1\int_0^2(x+2y)\,dy\,dx =\int_0^1\left[xy+y^2\right]_{y=0}^{y=2}\,dx =\int_0^1(2x+4)\,dx =\left[x^2+4x\right]_{0}^{1}=5. $$ Therefore $c=1/5$, and $$ f(x,y)=\frac{x+2y}{5}\mathbf{1}_{\{0<x<1,\ 0<y<2\}}(x,y). $$ This function is nonnegative and its integral is $(1/5)5=1$. Its effective support is the stated rectangle, up to its boundary.

Worked Example: A Disconnected Effective Support

Let $g(x,y)$ equal 1 on the union of the two rectangles $$ R_1=\{0<x<1,\ 0<y<1/2\},\qquad R_2=\{2<x<3,\ 0<y<1\}, $$ and equal 0 elsewhere. Find the constant $c$ for which $f=cg$ is a density.

The rectangles are disjoint, so the integral of $g$ is the sum of their areas. The first area is $1\cdot(1/2)=1/2$; the second is $1\cdot1=1$. Hence $$ \iint_{\mathbb{R}^2}g(x,y)\,dx\,dy =\iint_{R_1}1\,dx\,dy+\iint_{R_2}1\,dx\,dy =\frac12+1=\frac32. $$ It follows that $c=2/3$. Indeed, $$ \iint_{\mathbb{R}^2}f(x,y)\,dx\,dy =\frac23\left(\frac12+1\right)=1. $$ The effective support has two separated pieces. Replacing them with the large rectangle $0<x<3,\ 0<y<1$ would include regions where $g=0$, and would give the wrong normalizing constant.

Worked Example: Normalizing on a Triangular Region

Suppose $$ f(x,y)=c(x+y)\mathbf{1}_{\{0<y<x<1\}}(x,y). $$ Find $c$, showing how the inequalities determine the integration limits.

The constraint $0<y<x<1$ means that $x$ ranges from 0 to 1 and, for each fixed $x$, $y$ ranges from 0 to $x$. Thus the unnormalized integral is $$ \int_0^1\int_0^x(x+y)\,dy\,dx =\int_0^1\left[xy+\frac{y^2}{2}\right]_{y=0}^{y=x}\,dx =\int_0^1\frac{3x^2}{2}\,dx =\frac32\left[\frac{x^3}{3}\right]_0^1 =\frac12. $$ Therefore $c=2$. The proposed density is $$ f(x,y)=2(x+y)\mathbf{1}_{\{0<y<x<1\}}(x,y), $$ and its total integral is $2(1/2)=1$. Integrating over the whole unit square would be incorrect: the region $0<x<y<1$ is outside the specified support.

Zero-Density Regions and Integration Limits

A reliable way to read a piecewise or indicator-based formula is to separate two tasks. First identify the constraints that restrict where the function can be nonzero. Then check whether its nonnegative kernel vanishes at any additional points within those constraints. The integral only receives a contribution from points where the formula is positive, although changing values on an area-zero set cannot affect its value.

For a region described by inequalities, integration limits should reflect those inequalities. Rectangles often permit independent limits for $x$ and $y$, as in the first example. For a triangular region such as $0<y<x<1$, the permissible range of $y$ depends on $x$. Reversing the order of integration is possible, but the bounds must still describe the same triangle: $0<x<1$ and $0<y<x$ are equivalent to $0<y<1$ and $y<x<1$.

Result: If two nonnegative measurable functions agree except on a set of area zero, their integrals over every measurable region are equal. In particular, changing a density on an area-zero set leaves all probabilities unchanged. If $N$ has area zero, then a random vector with density $f$ satisfies $$ P\bigl((X,Y)\in N\bigr)=\iint_N f(x,y)\,dx\,dy=0. $$

Justification. A nonnegative measurable function has integral zero over a set of measure zero. Therefore, if $f$ and $h$ agree outside such a set $N$, their integrals over any region $A$ can differ only by their integrals over $A\cap N$. Both of those integrals are zero, so $\iint_A f=\iint_A h$. Applying the density definition to $N$ gives the displayed probability statement. $\square$

This result explains why open versus closed boundaries normally makes no difference when normalizing a two-dimensional density: line segments have area zero. It does not mean every zero-density region can be ignored when setting up bounds. A region of positive area on which the formula is zero contributes zero to the integral, but accidentally treating the density as positive there changes the calculation.

Common Pitfall: Normalizing Over the Bounding Box

A frequent error is to find a rectangle enclosing the constraints, integrate the kernel over that rectangle, and treat the result as the total mass. In the triangular example, the bounding box is the unit square, but the formula is supported only where $y<x$. The other half of the square must not contribute. A second error is to assume that a formula is already a density because it is nonnegative; nonnegativity is necessary, but its total integral must also equal one.

Before accepting a proposed joint density, check three things: the formula is measurable, it is nonnegative, and its integral over its actual effective support is one. If it is a kernel with unknown constant, calculate its integral first and verify that the result is finite and strictly positive. The constant is then its reciprocal.

Check Your Understanding

Use the constraints in each formula to identify where the density can contribute before setting up an integral.

For $f(x,y)=c\,\mathbf{1}_{\{1<x<3,\ 0<y<2\}}(x,y)$, what is the effective support, and what value of $c$ makes $f$ a density?
Why would integrating $c(x+y)$ over the entire unit square give the wrong normalization for the density in the triangular worked example?
A nonnegative kernel $g$ has integral $I=0$. Can any finite constant $c$ make $cg$ a density? Explain.
Does changing a density along the boundary of a rectangle change its total probability? State the condition that makes the answer valid.
A kernel is equal to 2 on each of two disjoint rectangles of areas $1/4$ and $3/4$, and zero elsewhere. What constant normalizes it?