Integrating Over Irregular Joint Supports

Why the Support Determines the Calculation

For a pair of continuous random variables, a joint density describes probability across a region in the plane. When that region is triangular, curved, or otherwise constrained, the main challenge is often not finding an antiderivative. It is identifying which values of one variable are possible after the other variable has been fixed. Correct bounds follow from the support—the region where the joint density is positive.

A rectangular integration region can often be described with constant limits. On an irregular support, a slice through the region may begin or end at a boundary that depends on the other coordinate. Choosing the direction of integration carefully can make a multi-step probability calculation shorter and help prevent missing or adding part of the region.

Definition: A joint density \(f_{X,Y}(x,y)\) is nonnegative, equals zero outside the support, and satisfies \(\iint f_{X,Y}(x,y)\,dx\,dy=1\) over the entire support. For a region \(A\) in the plane, \(P((X,Y)\in A)=\iint_A f_{X,Y}(x,y)\,dx\,dy\). A marginal density is obtained by integrating the joint density over all feasible values of the other variable: \(f_X(x)=\int f_{X,Y}(x,y)\,dy\) and \(f_Y(y)=\int f_{X,Y}(x,y)\,dx\), with bounds determined by the support.

Reading a Support and Choosing Bounds

Suppose the support is specified by inequalities. To integrate with respect to \(y\) first, fix \(x\) and determine the lowest and highest feasible \(y\)-values. This describes vertical slices and gives an integral of the form \(\int \int f_{X,Y}(x,y)\,dy\,dx\). To integrate with respect to \(x\) first, fix \(y\), determine the feasible \(x\)-values, and use horizontal slices. The two orders describe the same region, but their bounds can look quite different.

A useful procedure is to sketch the boundary curves, mark their intersections, and then test a point in the proposed region. For a probability over only part of the support, also draw that event on the sketch. An event boundary can split the support into pieces, so the bounds may need separate integrals.

Write the support as inequalities.
Record the full range of both variables and any constraint linking them.

Choose a slicing direction.
Fix one variable and solve the support inequalities for the other. Prefer bounds that require fewer pieces.

Integrate over the target region.
For a marginal, include every feasible value of the variable being integrated out. For a probability, restrict the integral to the event as well.

Check the result.
Marginal densities must integrate to 1, and probabilities must lie between 0 and 1. The chosen bounds must stay inside the support.

Worked Examples with Triangular Supports

Worked Example: Marginals and a Probability on a Unit Triangle

Let \((X,Y)\) have constant joint density on the triangular support \(0<y<x<1\). Find the density, both marginal densities, and \(P(Y<1/2)\).

Method and normalization. The support has vertices \((0,0)\), \((1,0)\), and \((1,1)\), so its area is \(1/2\). If the constant density is \(c\), total probability requires \(c(1/2)=1\), giving \(c=2\). Thus \(f_{X,Y}(x,y)=2\) on \(0<y<x<1\), and zero elsewhere.

Marginal of \(X\). For a fixed \(x\), the support allows \(0<y<x\). Therefore,

f_X(x)=\int_0^x 2\,dy=2x,\qquad 0<x<1.

Outside this range \(f_X(x)=0\). The check is \(\int_0^1 2x\,dx=1\).

Marginal of \(Y\). For fixed \(y\), the support condition \(y<x<1\) gives \(y<x<1\), so

f_Y(y)=\int_y^1 2\,dx=2(1-y),\qquad 0<y<1.

This also integrates to 1: \(\int_0^1 2(1-y)\,dy=1\).

Probability. Using the marginal, \(P(Y<1/2)=\int_0^{1/2}2(1-y)\,dy\). Evaluating gives \([2y-y^2]_0^{1/2}=1-1/4=3/4=0.7500\). As a check using vertical slices, for \(0<x<1/2\), all feasible \(y\) satisfy \(y<1/2\); for \(1/2<x<1\), only \(0<y<1/2\) qualifies. The probability is therefore \(\int_0^{1/2}2x\,dx+\int_{1/2}^1 1\,dx=1/4+1/2=3/4\).

Common mistake avoided: At fixed \(y\), the feasible \(x\)-values run from \(y\) to 1, not from 0 to \(y\). Reversing those limits describes points outside the triangle and gives the wrong marginal.

Worked Example: A Weighted Density on a Sloping Triangle

Suppose a hypothetical model for two nonnegative quantities has support \(0<x<2\), \(0<y<2-x\), and joint density \(f_{X,Y}(x,y)=c(x+y)\) there. Find \(c\), the marginal density of \(X\), and \(P(X<1,\ Y<1)\).

Method and normalization. For each \(x\) from 0 to 2, \(y\) ranges from 0 to \(2-x\). Thus,

1=c\int_0^2\int_0^{2-x}(x+y)\,dy\,dx.

The inner integral is \(x(2-x)+(2-x)^2/2=2-x^2/2\). Integrating over \(x\) gives \([2x-x^3/6]_0^2=4-8/6=8/3\). Hence \(c=3/8\).

Marginal of \(X\). Integrate across the full feasible range of \(y\) at each fixed \(x\):

f_X(x)=\frac{3}{8}\left(2-\frac{x^2}{2}\right) =\frac{3}{4}-\frac{3x^2}{16},\qquad 0<x<2.

It is zero outside \(0<x<2\). As a check, \(\int_0^2(3/4-3x^2/16)\,dx=3/2-1/2=1\).

Joint probability. The event \(X<1,\ Y<1\) is the unit square within the support: every point in it has \(x+y<2\), apart from boundary points that do not affect a continuous probability. Integrate over that square:

P(X<1,Y<1)=\frac{3}{8}\int_0^1\int_0^1(x+y)\,dy\,dx =\frac{3}{8}\left(\frac12+\frac12\right) =\frac{3}{8}=0.3750.

The two inner contributions are \(\int_0^1\int_0^1 x\,dy\,dx=1/2\) and \(\int_0^1\int_0^1 y\,dy\,dx=1/2\). The answer is a valid probability, and the marginal check confirms the normalization. The common mistake is to integrate the entire support when the event restricts both coordinates; here the event requires the unit-square bounds.

Curved Boundaries and Reversing the Order

Curved supports work by the same principle: solve the support inequalities for whichever variable is being integrated out. A boundary that is simple when written as \(y\) in terms of \(x\) may yield different bounds when written as \(x\) in terms of \(y\). Inverting a boundary requires attention to its domain and, when necessary, to multiple branches.

Justification: If the joint density is nonnegative and integrable, integrating it over a region can be done using vertical slices or horizontal slices, provided the bounds in each description cover exactly that same region. This is the change in order of integration for a nonnegative function. For a marginal, fixing \(x\) and integrating across every feasible \(y\) adds the joint probability over all possible \(Y\)-values associated with that \(x\). The resulting function \(f_X(x)\) is therefore the density of \(X\); integrating it over an interval gives the probability that \(X\) lies in that interval.

Worked Example: A Curved Support with a Horizontal Slice

Let \(f_{X,Y}(x,y)=6\) on the curved support \(0<x<1,\ x^2<y<x\), and zero elsewhere. Find the marginal density of \(Y\), then calculate \(P(X<1/2)\).

Method and support check. The support area is \(\int_0^1(x-x^2)\,dx=1/2-1/3=1/6\), so the constant density 6 is normalized. For fixed \(y\), the constraints \(x^2<y\) and \(y<x\), with \(x>0\), imply \(x<\sqrt y\) and \(x>y\). Thus \(y<x<\sqrt y\), for \(0<y<1\).

Marginal of \(Y\). Integrating over the horizontal slice gives

f_Y(y)=\int_y^{\sqrt y}6\,dx=6(\sqrt y-y),\qquad 0<y<1.

The marginal integrates to \(6(2/3-1/2)=1\), confirming the density.

Probability. For each fixed \(x\), the support gives \(x^2<y<x\). The event \(X<1/2\) therefore has probability

\int_0^{1/2}\int_{x^2}^{x}6\,dy\,dx =6\int_0^{1/2}(x-x^2)\,dx =6\left(\frac18-\frac1{24}\right) =\frac12=0.5000.

The result is plausible because the density puts more vertical width near the middle of the \(x\)-range than near either endpoint, but the integral—not visual intuition—determines the probability. A common mistake is to use \(x\) from \(y\) to \(\sqrt y\) when computing \(f_Y\); those are the bounds for \(x\) at fixed \(y\), while the probability calculation here uses \(y\) from \(x^2\) to \(x\) at fixed \(x\).

Common Exam Traps and Checks

On an irregular support, algebraic bounds can be internally consistent yet describe the wrong region. The most reliable check is to test a simple point known to lie in the support. For example, if \(0<y<x<1\), the point \((x,y)=(3/4,1/4)\) is allowed. A proposed vertical slice at \(x=3/4\) should include \(y=1/4\); bounds that exclude it cannot be correct.

Do not use the full coordinate range automatically. At a fixed coordinate, the other variable may have a restricted range.
Do not confuse the two slicing directions. Bounds for \(y\) at fixed \(x\) differ from bounds for \(x\) at fixed \(y\).
Split the integral when a boundary changes. An event can intersect the support so that one pair of limits does not describe the whole target region.
Keep normalization separate from the event calculation. First ensure the density integrates to 1 on its full support; then integrate only over the event of interest.

These checks are useful in integrated probability problems because a correct density formula does not compensate for incorrect limits. Sketching the support and describing each slice in words are often faster than trying to repair a mistaken integral afterward.

Key takeaway: Write the support as inequalities, solve those inequalities for the inner variable, and make sure the resulting slices cover exactly the region required. Marginals integrate across the entire support; joint probabilities integrate only across the event within that support.

Check Your Understanding

For each problem, identify the support bounds before integrating. Probabilities are given to four decimal places as well as in exact form when convenient.

A joint density is constant on \(0<y<x<1\). Find \(P(X<1/2)\). (Answer: \(1/4=0.2500\))
For the same density and support, find \(P(Y<1/2)\). (Answer: \(3/4=0.7500\))
A joint density is constant on \(0<x<2,\ 0<y<2-x\). Find \(P(X>1)\). (Answer: \(1/4=0.2500\))
Let \(f_{X,Y}(x,y)=6\) on \(0<x<1,\ x^2<y<x\). Find \(P(X<1/2)\). (Answer: \(1/2=0.5000\))
Let \(f_{X,Y}(x,y)=\frac{3}{8}(x+y)\) on \(0<x<2,\ 0<y<2-x\). Find \(P(Y<1)\). (Answer: \(11/16=0.6875\))

What You'll Learn