Arbitrary Region Probabilities from Joint Representations

From a Region to a Probability

A joint distribution describes how probability is assigned to pairs of values. Many questions do not ask for a rectangle probability such as $P(a<X\le b,\ c<Y\le d)$. They instead describe a region: a collection of support points, a union of regions, or a set bounded by a line. The central task is to translate that description into the part of the joint representation that contributes probability.

Write $Z=(X,Y)$ for the random vector and $A$ for a region in the plane. The event of interest is $\{Z\in A\}$, and its probability is determined both by $A$ and by the way the joint law represents probability: through point masses, a density, or a combination of components. A geometric region alone does not determine its probability.

Definition: A region probability is $P((X,Y)\in A)$, where $A$ is a measurable subset of $\mathbb{R}^2$. For a joint PMF, calculate it by summing the PMF over support points in $A$. For a joint density $f$, calculate it by integrating $f$ over $A$. Probability outside the support contributes zero.

The General Calculation Rules

Suppose $(X,Y)$ has a joint PMF $p(x,y)=P(X=x,Y=y)$ on a finite or countable support $S$. Only support points in the region count. In particular, a point that satisfies the event description but is not in $S$ contributes nothing.

Discrete region rule: For any event region $A$, $$ P((X,Y)\in A)=\sum_{(x,y)\in S\cap A}p(x,y). $$

For a joint density $f$, the corresponding rule integrates over the part of the plane in the region:

Density region rule: If $(X,Y)$ has joint density $f$, then for every measurable region $A$, $$ P((X,Y)\in A)=\iint_A f(x,y)\,dx\,dy. $$

These rules follow directly from the defining representations of the joint law. In the discrete case, the event is the disjoint union of the point events $\{X=x,Y=y\}$ for $(x,y)\in S\cap A$. Countable additivity gives the stated sum. In the density case, the probability assigned to any measurable set is the integral of the density over that set; equivalently, it is the expectation of the indicator $1\{(X,Y)\in A\}$:

E[1\{(X,Y)\in A\}] =\iint_{\mathbb{R}^2}1\{(x,y)\in A\}f(x,y)\,dx\,dy =\iint_A f(x,y)\,dx\,dy.

For calculations with a density, it is often convenient to describe the region through vertical or horizontal slices. If $A=\{(x,y):a\le x\le b,\ g_1(x)\le y\le g_2(x)\}$, then, when the bounds describe a measurable region and $f$ is nonnegative and measurable, Tonelli's theorem gives

P((X,Y)\in A) =\int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx.

The same principle applies when the limits change across the region: split it into pieces on which simple bounds apply, then add the resulting probabilities. For a union, first check whether the pieces overlap. If they do, summing their probabilities without adjustment counts the overlap more than once.

Worked Examples

Worked Example: Selecting Support Points for a Discrete Event

Suppose a joint PMF is positive only at the following four points:

p(0,0)=\frac18,\qquad p(0,1)=\frac14,\qquad p(1,0)=\frac14,\qquad p(1,1)=\frac38.

Find $P(X+Y\ge 1)$. The event region consists of all points on or above the line $x+y=1$. Among the four support points, $(0,0)$ does not satisfy the inequality, while $(0,1)$, $(1,0)$, and $(1,1)$ do.

Sum the PMF values at those three points:

P(X+Y\ge 1) =p(0,1)+p(1,0)+p(1,1) =\frac14+\frac14+\frac38 =\frac28+\frac28+\frac38 =\frac78.

As a check, the four PMF values sum to $1/8+1/4+1/4+3/8=1$. The event excludes only $(0,0)$, so its probability is also $1-1/8=7/8$.

Worked Example: Integrating Over a Sloped Region

Let $(X,Y)$ have density $f(x,y)=1$ on the unit square $0\le x\le1,\ 0\le y\le1$, and density $0$ elsewhere. Find $P(Y\le X/2)$.

At each fixed $x$ between 0 and 1, the allowed values of $y$ run from 0 to $x/2$. This upper limit never exceeds 1, so the region can be integrated using $0\le x\le1$ and $0\le y\le x/2$:

\begin{aligned} P(Y\le X/2) &=\int_0^1\int_0^{x/2}1\,dy\,dx\\ &=\int_0^1\left[y\right]_0^{x/2}\,dx\\ &=\int_0^1\frac{x}{2}\,dx\\ &=\left[\frac{x^2}{4}\right]_0^1 =\frac14. \end{aligned}

The region is a triangle with base 1 and height $1/2$, so its area is also $1\cdot(1/2)/2=1/4$. This geometric check agrees with the integral. The boundary line itself has area zero, so using a strict rather than non-strict inequality would not change this probability.

Worked Example: A Union of Two Regions Under a Density

Suppose the joint density is $f(x,y)=e^{-x-y}$ for $x\ge0,\ y\ge0$, and zero elsewhere. Find $P(X\le1\text{ or }Y\le1)$.

The union has an overlapping part where both coordinates are at most 1. One way to avoid double-counting is to split the union into disjoint regions: first $0\le x\le1,\ y\ge0$, then $x>1,\ 0\le y\le1$. The first region includes the overlap; the second adds only the part of the event not already included.

\begin{aligned} P(X\le1\text{ or }Y\le1) &=\int_0^1\int_0^\infty e^{-x-y}\,dy\,dx +\int_1^\infty\int_0^1 e^{-x-y}\,dy\,dx\\ &=\int_0^1 e^{-x}\left(\int_0^\infty e^{-y}\,dy\right)dx +\int_1^\infty e^{-x}\left(\int_0^1 e^{-y}\,dy\right)dx\\ &=(1-e^{-1})+e^{-1}(1-e^{-1})\\ &=1-e^{-2}. \end{aligned}

For a check, the complement is $\{X>1,Y>1\}$. Its probability is

\int_1^\infty\int_1^\infty e^{-x-y}\,dy\,dx =e^{-1}e^{-1}=e^{-2},

so the event probability is $1-e^{-2}$, as calculated. Numerically, this is approximately $0.8647$.

Worked Example: A Region Under a Mixed Joint Law

Let the joint law assign probability $1/3$ to the point $(0,0)$. With the remaining probability $2/3$, let $(X,Y)=(U,1-U)$, where $U$ is uniform on $[0,1]$. Find $P(X<1/2)$.

The point-mass component contributes $1/3$, because its point has $x=0<1/2$. In the curve component, $X=U$, so the event is $U<1/2$, which has probability $1/2$ within that component. Weight this conditional component probability by its total mass $2/3$, then add the point contribution:

P(X<1/2) =\frac13+\frac23P(U<1/2) =\frac13+\frac23\cdot\frac12 =\frac23.

The line component lies on $x+y=1$, but it is not represented by an ordinary two-dimensional density. Integrating an area density over the region would miss this component. For a mixed law, calculate the probability contributed by each component and add those contributions.

Partitioning Regions Carefully

A useful general method is to partition a region into nonoverlapping pieces. If $A_1,\ldots,A_k$ are pairwise disjoint measurable regions whose union is $A$, then

Disjoint partition rule: For any joint law, $$ P((X,Y)\in A)=\sum_{j=1}^k P((X,Y)\in A_j). $$

Indeed, the events $\{(X,Y)\in A_j\}$ are disjoint because the regions are disjoint, and their union is $\{(X,Y)\in A\}$. Finite additivity of probability therefore gives the rule. Under a density, this is also the additivity of the integral over disjoint sets; under a PMF, it is the same regrouping of the point-mass sum.

Partitions are useful when a region has a boundary that changes form, when a union is easier to describe as separate nonoverlapping pieces, or when the support itself restricts which points can occur. For integrals, choose the order of integration that gives the clearest bounds. A vertical-slice description may require splitting at a boundary, while horizontal slices may avoid that split.

Common Mistakes and Why Region Calculations Matter

Integrating over a containing rectangle instead of the event. A bounding rectangle may include points that do not satisfy the event. Restrict the bounds to the actual region.
Summing every possible lattice point. A discrete calculation includes only points in the joint support and in the event region.
Adding probabilities of overlapping pieces directly. The overlap is then counted twice. Use disjoint pieces or subtract the intersection once.
Treating a mixed law as only a density. Point and curve components can carry positive probability even though they have zero area. Account for each component using its own representation.

Region probabilities connect the representation of a joint law to events stated in practical terms, such as a sum exceeding a threshold or a coordinate falling within a specified range. Reliable calculations start by identifying the support, writing the event as a set, and then choosing sums or integration bounds that cover that set exactly once.

Check Your Understanding

For each question, identify the relevant part of the joint representation before calculating.

A joint PMF has support at $(0,0),(0,1),(1,0),(1,1)$. Which support points satisfy $X+Y>1$?
A density is positive on $0\le x\le2,\ 0\le y\le1$. Describe vertical integration bounds for the region $0\le y\le x/2$.
Why can probabilities of two overlapping regions not usually be added without adjustment?
A joint law has a point mass and a density component. What separate contributions must be considered for a region probability?
For a continuous joint density, does including or excluding a straight-line boundary necessarily change the probability? State the relevant condition.