Joint Laws With Point, Curve, and Area Components

Why One Joint Law Can Have Different Components

A bivariate random vector need not place all its probability in the same geometric way. Some outcomes may occur with positive probability at particular points. Other probability may be spread continuously along a curve, while the remaining probability is spread over a two-dimensional region. A single ordinary joint density cannot describe all three contributions: a point and a curve have zero area, even when they carry positive probability.

The useful approach is to describe the law as a sum of measures. To calculate the probability of an event, find how much probability each component assigns to it, then add those amounts. This tutorial develops that representation and applies it to a concrete model.

Definition of a Three-Component Joint Law

Let $(X,Y)$ be a random vector taking values in $\mathbb{R}^2$. A Borel set $A\subseteq\mathbb{R}^2$ is an event for this vector when $\{(X,Y)\in A\}$ is measurable. The joint law assigns the probability $P((X,Y)\in A)$ to each such set.

Definition: A joint law with point, curve, and area components has the form $$ P((X,Y)\in A) = \sum_{j=1}^{\infty} p_j\,\mathbf{1}_{\{z_j\in A\}} + c\int \mathbf{1}_{\{\gamma(t)\in A\}}\,Q(dt) + \int_A h(x,y)\,dx\,dy $$ for every Borel set $A$. Here $z_j\in\mathbb{R}^2$ are point locations, $p_j\geq 0$ are their probability masses, and $\mathbf{1}$ is an indicator. The map $\gamma$ is a measurable parameterization of a curve, $Q$ is a probability distribution on its parameter space, and $c\geq 0$ is the total mass assigned to that curve component. The function $h$ is a nonnegative measurable area contribution, integrated with respect to two-dimensional Lebesgue measure.

For this to be a probability distribution, the total mass must satisfy $$ \sum_{j=1}^{\infty}p_j+c+\int_{\mathbb{R}^2}h(x,y)\,dx\,dy=1. $$ The area contribution $h$ need not integrate to one: its integral is the probability assigned to the area component. Likewise, $Q$ describes the distribution of the curve parameter conditional on using that component; the curve’s total probability is $c$.

The curve term is a probability measure carried by the image of $\gamma$. In particular, it is not an area density evaluated “on the curve.” The parameterization and $Q$ determine how probability is distributed along the curve. Different parameterizations can describe the same geometric curve but induce different probability distributions if the parameter law is not adjusted.

Why the Representation Is a Probability Law

Theorem: Suppose the point weights are nonnegative and sum to a finite value, $Q$ is a probability distribution, $\gamma$ is measurable, $c\geq 0$, and $h$ is nonnegative and measurable with finite integral. If $$ \sum_{j=1}^{\infty}p_j+c+\int_{\mathbb{R}^2}h(x,y)\,dx\,dy=1, $$ then the displayed formula defines a probability distribution on the Borel sets of $\mathbb{R}^2$.

Proof. Each point contribution assigns mass $p_j$ to the location $z_j$, and hence defines a measure. The curve contribution is also a measure: for disjoint Borel sets $A_1,A_2,\ldots$, their inverse images under $\gamma$ are disjoint, so countable additivity of $Q$ gives countable additivity of $cQ(\gamma^{-1}(A))$. The area contribution is a measure because it is the integral of the nonnegative function $h$ over each set.

A sum of nonnegative measures is countably additive. In detail, for pairwise disjoint Borel sets $A_k$, countable additivity holds for each component, and the nonnegative series can be rearranged: $$ \sum_j p_j\mathbf{1}_{\{z_j\in\cup_k A_k\}} +cQ\!\left(\gamma^{-1}\!\left(\cup_k A_k\right)\right) +\int_{\cup_k A_k}h = \sum_k P((X,Y)\in A_k). $$ Finally, applying the formula to the whole plane gives total mass one by the assumed normalization. Thus it is a probability measure. $\square$

A Concrete Mixed Joint Distribution

Consider the following model. With probability $1/4$, the vector is exactly $(2,2)$. With probability $1/3$, draw $T$ uniformly on $[0,1]$ and set $(X,Y)=(T,1-T)$. With the remaining probability $5/12$, draw a point uniformly over the unit square. The area component therefore has contribution $5/12$ per unit area on that square.

Model: For every Borel set $A$, $$ P((X,Y)\in A) = \frac14\mathbf{1}_{\{(2,2)\in A\}} + \frac13\int_0^1\mathbf{1}_{\{(t,1-t)\in A\}}\,dt + \frac{5}{12}\int_0^1\int_0^1\mathbf{1}_{\{(x,y)\in A\}}\,dy\,dx. $$

The point mass is $1/4$. The curve contribution has mass $1/3$, since the uniform distribution of $T$ integrates to one. The area contribution has mass $$ \frac{5}{12}\int_0^1\int_0^1 1\,dy\,dx = \frac{5}{12}. $$ The total is $1/4+1/3+5/12=3/12+4/12+5/12=1$. The curve is the line segment $x+y=1$ inside the unit square. The point $(2,2)$ lies outside both continuous components’ supports.

Worked Example: Probability Above the Diagonal

Find $P(X+Y>1)$ under the model.

Point contribution. At $(2,2)$, the sum is $4$, so this point belongs to the event. Its contribution is $1/4$.

Curve contribution. On the curve, $X+Y=T+(1-T)=1$ for every $T\in[0,1]$. The strict inequality is never satisfied, so the curve contributes zero.

Area contribution. Within the unit square, $x+y>1$ is the triangle above the line $y=1-x$. Its area is $$ \int_0^1\int_{1-x}^1 dy\,dx = \int_0^1 x\,dx = \frac12. $$ Multiplying by the area contribution per unit area gives $(5/12)(1/2)=5/24$. Therefore $$ P(X+Y>1)=\frac14+0+\frac{5}{24}=\frac{11}{24}. $$

Worked Example: A Rectangle That Touches the Curve at One Point

Find the probability of $A=\{(x,y):x\leq 1/2,\ y\leq 1/2\}$.

The atom $(2,2)$ is not in $A$, so it contributes zero. For the curve component, both inequalities require $T\leq 1/2$ and $1-T\leq 1/2$. Together these hold only when $T=1/2$. Since a uniform random variable has probability zero of equaling a specified point, the curve contribution is zero.

The part of the unit square in $A$ is the square $[0,1/2]\times[0,1/2]$, with area $1/4$. The area contribution is thus $(5/12)(1/4)=5/48$. Adding all three contributions gives $$ P((X,Y)\in A)=0+0+\frac{5}{48}=\frac{5}{48}. $$

Worked Example: A Vertical Strip

Find $P(X\leq 1/2)$.

The point $(2,2)$ does not satisfy the inequality, so its contribution is zero. On the curve, $X=T$, and therefore $$ \frac13P(T\leq 1/2)=\frac13\cdot\frac12=\frac16. $$ The area portion of the unit square with $x\leq 1/2$ has area $1/2$, so its contribution is $(5/12)(1/2)=5/24$. The total probability is $$ P(X\leq 1/2)=0+\frac16+\frac{5}{24} =\frac{4}{24}+\frac{5}{24} =\frac38. $$

Reading the Components Correctly

The event formula is a sum of contributions, not a choice of only one representation. If a set includes an atom, a portion of the curve, and an area region, all three terms may be positive. The event’s probability is their sum. This remains true if the geometric supports meet or overlap: each term is a measure contributed by a component of the model.

A curve can carry positive probability even though it has zero area. In the example, the line segment $x+y=1$ has two-dimensional area zero, but it receives total probability $1/3$. Consequently, integrating an ordinary two-dimensional density over that line would not recover the curve probability. Similarly, the point $(2,2)$ has probability $1/4$, even though a single point has area zero.

Key takeaway: For a joint law with point, curve, and area components, evaluate the event separately under each component and add the results. The point terms use membership, the curve term uses the parameter values mapped into the event, and the area term uses a two-dimensional integral.

Common Mistake

Wrong reasoning: “The curve has area zero, so it has probability zero; only the area density matters.”

Correction: Zero area implies zero probability only for a law that is absolutely continuous with respect to area. The model explicitly assigns probability $1/3$ to the curve component, independently of its area. Likewise, a point mass is not obtained by integrating the area contribution over that point. Always inspect every component in the joint representation before calculating an event probability.

Check Your Understanding

For each question, identify the relevant component contributions before combining them.

In the model, what is the total probability assigned to the line $x+y=1$? Why does its zero area not determine that probability?
For the event $X+Y=1$, calculate the point, curve, and area contributions separately.
Suppose a Borel event contains the atom but none of the curve points. What term in the event formula gives the atom’s contribution?
Why must the area function $h$ integrate to the mass of the area component, rather than necessarily to one?
In the vertical-strip example, which component contributes $1/6$, and which contributes $5/24$?