Geometric Supports and Integration Limits

Why Support Geometry Matters

A joint density is not integrated over every possible pair of values of its random variables. It is integrated over the region where the density is positive, or over the part of that region corresponding to the event of interest. The limits must describe that region precisely: bounds that are too wide include points outside it, while bounds that are too narrow leave out possible outcomes.

The practical task is to translate between two descriptions of the same set: inequalities such as $0<y<x<1$, and iterated integral limits such as $0<x<1$ with $0<y<x$. Once the limits are correct, evaluating the integral is usually routine. This skill is aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: The support of a continuous joint density $f_{X,Y}(x,y)$ is the region of the plane where the density can be positive. A point outside the support has density zero. To find the probability of an event $A$, integrate the density over the points in the support that also satisfy the conditions defining $A$: $$ P\big((X,Y)\in A\big)=\iint_{A\cap R} f_{X,Y}(x,y)\,dA, $$ where $R$ is the support and $dA$ denotes area in the plane.

For continuous distributions, changing whether a boundary line is included does not change an integral: a line has area zero. Thus $0<y<x<1$ and $0\leq y\leq x\leq1$ describe the same probabilities. Boundaries still help identify the region, but the important issue is which two-dimensional points the inequalities include.

From Inequalities to Integral Limits

An iterated integral describes a region by choosing one variable as the outer variable and giving an interval for the inner variable at each outer value. For example, limits of the form $a<x<b$ and $g_1(x)<y<g_2(x)$ say that $x$ ranges from $a$ to $b$, and, for each such $x$, $y$ lies between the two boundary curves. The order of integration matters to the written limits, so each bound must agree with the inequalities defining the region.

A reliable setup follows three steps: write the support inequalities, add the event inequalities, and identify the resulting range of the outer variable before writing the inner limits. If the density is specified as zero outside its support, integrating over a larger rectangle can be mathematically possible only if the zero-density portions are handled correctly; it is generally safer to use limits that describe the actual region.

State the support.
Write every inequality that defines where the joint density is positive.

Intersect with the event.
Add the event’s inequalities. The probability region must satisfy both the support and event conditions.

Choose an outer range.
Find the full range of the outer variable among points in the intersection.

Set the inner bounds and check.
For each outer value, use the inequalities to give the full interval of inner values in the region, with no extra points.

Worked Examples

Worked Example: A Rectangle Intersected With an Event

Suppose $(X,Y)$ has joint density $f_{X,Y}(x,y)=1/6$ on the rectangle $0<x<2,\ 0<y<3$, and density zero elsewhere. Find $P(X+Y<2)$. The stated density is a valid model: the support has area $2\cdot3=6$, so its total probability is $6(1/6)=1$.

Method: intersect the support and event, then integrate. The support requires $0<x<2$ and $0<y<3$. The event adds $y<2-x$. Since $x\geq0$, the event’s upper bound $2-x$ is at most 2, so it is already below the support’s upper bound 3. It is positive only when $x<2$, which matches the support’s $x$-range. Therefore, for each $x$ from 0 to 2, $y$ runs from 0 to $2-x$.

\begin{aligned} P(X+Y<2) &=\int_0^2\int_0^{2-x}\frac{1}{6}\,dy\,dx\\ &=\int_0^2\frac{2-x}{6}\,dx\\ &=\frac{1}{6}\left[2x-\frac{x^2}{2}\right]_0^2\\ &=\frac{1}{6}(4-2) =\frac{1}{3} =0.3333. \end{aligned}

The result is consistent with the geometry: the event occupies a right triangle of area 2 inside a support where density is $1/6$, giving probability $2/6$. The common mistake is to use $0<y<3$ for the event at every $x$. That would integrate over the entire support and give 1, including points such as $(1,2)$ that fail $x+y<2$.

Worked Example: A Triangular Support

Let $f_{X,Y}(x,y)=2$ when $0<y<x<1$, and zero elsewhere. Find $P(Y<X/2)$. First verify that the proposed density integrates to 1, so the model is a valid joint density.

Method: use the support to determine the outer range, then impose the event on the inner variable. The support gives $0<x<1$ and $0<y<x$. The event gives $y<x/2$. For positive $x$, $x/2<x$, so the event’s upper bound is stricter than the support’s upper bound. The integration region is therefore $0<x<1$ and $0<y<x/2$.

\begin{aligned} \int_0^1\int_0^x 2\,dy\,dx &=\int_0^1 2x\,dx =\left[x^2\right]_0^1=1,\\ P(Y<X/2) &=\int_0^1\int_0^{x/2}2\,dy\,dx\\ &=\int_0^1 x\,dx =\left[\frac{x^2}{2}\right]_0^1 =\frac{1}{2} =0.5000. \end{aligned}

The normalization check confirms the stated density, and the event probability is half of the total mass. The common mistake is to use $0<y<x$ for the event integral as well as for the support check. Those are the support limits, not the event limits; using them would return the total probability 1 rather than the probability of the smaller event.

Worked Example: Reading a Slanted Boundary

Suppose the joint density is $f_{X,Y}(x,y)=1$ on $0<y<2,\ y/2<x<1$, and zero elsewhere. Find $P(Y>1)$. Before calculating, check that the density is normalized: for each $y$ from 0 to 2, the permitted $x$-interval has length $1-y/2$.

Method: keep $y$ as the outer variable. The support already gives $0<y<2$, with $y/2<x<1$. The event $Y>1$ restricts the outer range to $1<y<2$ but does not further restrict $x$ at a fixed $y$. Thus the inner limits remain $y/2$ and 1.

\begin{aligned} \int_0^2\int_{y/2}^1 1\,dx\,dy &=\int_0^2\left(1-\frac{y}{2}\right)dy\\ &=\left[y-\frac{y^2}{4}\right]_0^2 =2-1=1,\\ P(Y>1) &=\int_1^2\int_{y/2}^1 1\,dx\,dy\\ &=\left[y-\frac{y^2}{4}\right]_1^2\\ &=1-\left(1-\frac{1}{4}\right) =\frac{1}{4} =0.2500. \end{aligned}

This verifies both the normalization and the requested probability. The common mistake is to set the inner lower bound to 0. The support requires $x>y/2$; dropping that slanted boundary adds points outside the support and changes the integral.

Why the Integral Gives the Probability

For a continuous pair with joint density $f_{X,Y}$, probability over a small planar patch is approximately the density at that patch multiplied by its area. Adding these contributions over a region gives the integral of the density over that region. In particular, if the region can be described by outer and inner intervals, the area integral becomes an iterated integral.

This also explains why the support and the event must both be used. The density contributes probability only where it is positive; the event selects which of those possible outcomes count. The desired region is their intersection, not the support alone and not the event considered without regard to where the density is positive.

Key takeaway: Write the support and event as inequalities before writing an integral. Then verify that the outer limits cover the full region and that, at every outer value, the inner limits include exactly the permitted values. Boundaries may be included or excluded without changing a continuous probability.

Common Exam Traps

Using support limits as event limits. The support describes all possible values; an event usually selects only part of it.
Forgetting to intersect inequalities. A bound supplied by the event may be tighter than the corresponding support bound, as $y<x/2$ is tighter than $y<x$ for positive $x$.
Ignoring a slanted boundary. If the support says $x>y/2$, the inner $x$-integral must start at $y/2$, not at zero.
Choosing an outer range from only one condition. The outer range must contain only values for which the inner interval is nonempty and must include every such value in the region.
Assuming boundaries carry probability. For a joint density, a boundary line has area zero and contributes no separate probability mass.

Check Your Understanding

For each problem, identify the support and event region before evaluating the probability.

A pair has density $1/4$ on $0<x<2,\ 0<y<2$, and zero elsewhere. Find $P(Y>X)$. (Answer: $0.5000$)
A pair has density 1 on $0<x<1,\ 0<y<2x$, and zero elsewhere. Find $P(Y<X)$. (Answer: $0.5000$)
A pair has density $2/9$ on $0<x<3,\ 0<y<3-x$, and zero elsewhere. Find $P(X<1)$. (Answer: $0.5556$)
A pair has density $2/3$ on $0<y<1,\ y<x<2$, and zero elsewhere. Find $P(X>3/2)$. (Answer: $0.3333$)
A pair has density 2 on $0<x<1,\ x<y<2x$, and zero elsewhere. Find $P(Y>1)$. (Answer: $0.5000$)

What You'll Learn