When a Joint Law Lives on a Curve
A pair of random variables need not spread probability across a two-dimensional region. For example, if \(Y=X\) with probability one, every possible pair \((X,Y)\) lies on the diagonal line \(y=x\). The marginal distributions of \(X\) and \(Y\) might each be continuous, but the joint law is confined to a set with zero area. This distinction matters: a joint density describes probabilities by integrating over area, whereas a curve-concentrated law assigns all its probability to a set of area zero.
A useful way to describe this kind of dependence is to generate both coordinates from one random variable. The latent variable determines a point on a curve, and probabilities for the pair can then be calculated by finding which latent-variable values place that point in the event of interest.
The curve representation is a pushforward description: each value of \(T\) is mapped to a point in the plane. More precisely, for every Borel set \(A\subseteq\mathbb R^2\),
Here \(F_T\) is the distribution of \(T\), and \(\mathbf{1}_A\) is one when its argument belongs to \(A\) and zero otherwise. This probability formula applies whether \(T\) is discrete, continuous, or has some other distribution. When \(T\) has a density, the integral can be written as an ordinary integral against that density.
Why a Curve Law Has No Joint Density
A joint density \(f\) with respect to two-dimensional Lebesgue measure would have to satisfy \(P((X,Y)\in A)=\int_A f(x,y)\,dx\,dy\) for every Borel set \(A\). In particular, it would assign probability zero to every set of area zero. Therefore, if the pair lies with probability one on a curve of area zero, it cannot have such a density.
For a common class of curves, the zero-area property follows directly from the sections of the set. Suppose \(g\) is continuous on a compact interval \(I\), and consider its graph \(\Gamma=\{(x,g(x)):x\in I\}\). This graph is Borel measurable because it is compact. For each fixed \(x\), its vertical section contains either one point, \(g(x)\), or no points. In either case, that section has one-dimensional Lebesgue measure zero. Tonelli's theorem applied to the nonnegative indicator of \(\Gamma\) gives the area of the graph as the integral of these section lengths, which is zero.
Proof. Suppose, to the contrary, that the joint law has a density \(f\). The defining density formula applied to \(\Gamma\) gives
because \(\Gamma\) has area zero. This contradicts \(P((X,Y)\in\Gamma)=1\). Thus no such density exists. \(\square\)
The theorem is about a joint density with respect to area, not about whether the coordinates have marginal densities. A curve can be thin in the plane while each coordinate varies continuously along it.
Worked Probabilities from the Support
To calculate an event probability, translate its conditions on \(X\) and \(Y\) into conditions on the single parameter \(T\). This reduces a two-coordinate event to a one-dimensional event.
Worked Example: A Uniform Pair on the Diagonal
Let \(U\sim\operatorname{Uniform}(0,1)\), and set \(X=U\), \(Y=U\). Find \(P(X\leq 0.7,\ Y>0.2)\), and identify the marginal distributions.
The support is the diagonal segment \(\{(u,u):0\leq u\leq 1\}\). Both event conditions concern the same value \(U\):
Since \(U\) is uniform on an interval of length one, the probability is the length of this interval:
Also, \(X=U\) and \(Y=U\), so each marginal is \(\operatorname{Uniform}(0,1)\). The joint law nevertheless has no two-dimensional density: it assigns probability one to the diagonal, which has area zero. The coordinates are not independent; in fact, \(Y=X\) with probability one.
Worked Example: A Parabolic Support
Let \(U\sim\operatorname{Uniform}(0,1)\), and define \(X=U\), \(Y=U^2\). Calculate \(P(X\leq 0.6,\ 0.16\leq Y\leq 0.49)\), and find the marginal distribution of \(Y\).
Because \(U\) is nonnegative, the restrictions on \(Y=U^2\) are equivalent to \(0.4\leq U\leq 0.7\). The additional condition \(X\leq 0.6\) is \(U\leq 0.6\). Their intersection is \(0.4\leq U\leq 0.6\). Endpoints do not affect this probability because \(U\) is continuous, so
For the marginal distribution of \(Y\), its support is \([0,1]\). If \(y<0\), then \(P(Y\leq y)=0\). For \(0\leq y\leq 1\), the event \(Y\leq y\) is \(U^2\leq y\), or equivalently \(U\leq\sqrt{y}\). Thus
Differentiating on \(0<y<1\) gives the marginal density \(f_Y(y)=1/(2\sqrt{y})\); this integrates to one since \(\int_0^1(2\sqrt{y})^{-1}\,dy=1\). The pair is concentrated on the parabola \(y=x^2\), even though both marginal distributions have densities. The joint distribution has no density with respect to area.
Worked Example: An Exponential Pair on a Ray
Let \(T\) have the exponential distribution with rate \(1\), so its density is \(f_T(t)=e^{-t}\) for \(t\geq 0\). Set \(X=T\), \(Y=2T\). Find \(P(1\leq X\leq 2,\ 2.5\leq Y\leq 3.5)\).
The first condition restricts \(T\) to \([1,2]\). The second gives \(2.5\leq 2T\leq 3.5\), or \(1.25\leq T\leq 1.75\). Both conditions hold exactly when \(1.25\leq T\leq 1.75\). Integrating the exponential density over that interval,
Numerically this is approximately \(0.2865-0.1738=0.1127\). The support is the ray \(y=2x\), \(x\geq0\), and all probability lies on that area-zero set.
Marginals from the Curve Representation
For \(X=g(T)\) and \(Y=h(T)\), the marginal distribution functions follow by imposing a condition on one coordinate at a time:
Equivalently, the probability of a Borel set \(B\subseteq\mathbb R\) under the \(X\)-marginal is \(P(g(T)\in B)\), and the corresponding statement for \(Y\) is \(P(h(T)\in B)\). These formulas do not require a one-to-one parameterization. If several values of \(T\) map to the same coordinate, all of them must be included in the relevant event.
Justification. For any Borel set \(A\subseteq\mathbb R^2\), the events \(\{(X,Y)\in A\}\) and \(\{(g(T),h(T))\in A\}\) are equal by the definition of \(X\) and \(Y\). Taking probabilities gives the joint-event formula. For a Borel set \(B\subseteq\mathbb R\), the equality \(X=g(T)\) gives \(\{X\in B\}=\{g(T)\in B\}\), and likewise for \(Y=h(T)\). Taking probabilities proves the marginal statements. \(\square\)
Common Pitfall and Why the Distinction Matters
A frequent mistake is to try to describe a curve-concentrated law by assigning an ordinary joint density along the curve and then integrating that expression over two-dimensional regions. This cannot work with an area density: every area-zero curve has integral zero under any density, while the curve may carry probability one. A formula that resembles a density in a parameter is instead a distribution for the latent variable, and probabilities must be calculated through the parameter map.
For example, in the parabolic model above, \(f_X\) and \(f_Y\) are valid one-dimensional marginal densities, but multiplying them would describe a different joint law—one that spreads probability over a two-dimensional region and treats the coordinates as independent. The actual model enforces \(Y=X^2\) exactly. The support representation preserves that dependence and makes joint-event calculations transparent.
A curve representation also does not by itself say that every point of the displayed curve is part of the support in the strict topological sense. The latent variable may put no probability near some parameter values, or the parameterization may trace the same point more than once. What the representation guarantees is that the pair lies in the image of the map with probability one; probability calculations still depend on the distribution of \(T\), not just on the geometric curve.
Check Your Understanding
For each question, use the curve representation to identify the relevant latent-variable event before calculating.
- Let \(U\sim\operatorname{Uniform}(0,1)\), \(X=U\), and \(Y=1-U\). Find \(P(X\leq 0.4,\ Y\leq 0.7)\).
- If \(X=U\) and \(Y=U^2\), where \(U\sim\operatorname{Uniform}(0,1)\), write \(F_Y(y)\) for \(0\leq y\leq1\).
- Explain why a joint law that assigns probability one to the diagonal \(y=x\) cannot have a density with respect to area.
- Let \(T\) be exponential with rate \(1\), \(X=T\), and \(Y=3T\). Express \(P(X\leq 1,\ Y\geq 1.5)\) as a probability involving \(T\).
- Can both marginals have densities when the joint law has no two-dimensional density? Explain using one of the examples.