Why a Joint Support Restriction Matters
Independence means that learning one variable does not change the probabilities of events involving the other. A particularly direct way to disprove it is to find two events that each have positive probability but cannot happen together. This often follows from a restriction on the joint support: the marginal laws allow each event separately, while the joint law forbids their combination.
For example, if a joint law only places probability on pairs with \(x<y\), then the event \(X>3/4\) cannot occur together with \(Y<1/4\). If both events have positive marginal probabilities, independence would require their intersection to have positive probability, so the restriction proves dependence.
Support and a Rectangle Test
For a continuous distribution, the support is not simply the set of points where a chosen density is positive. It is the set of points every neighborhood of which has positive probability. Thus, an isolated point where a density is zero may still belong to the support. The support describes which combinations are locally possible under the probability law.
The first statement is the defining event-factorization property of independence. For the consequence, independence would imply that the joint probability equals the product of the two marginal probabilities. That product is strictly positive, contradicting the stated zero joint probability. This test needs positive probabilities for both marginal events; an impossible intersection alone is not enough.
Proof. First suppose \((x,y)\) is outside \(\operatorname{supp}(X)\times\operatorname{supp}(Y)\). Then either \(x\notin\operatorname{supp}(X)\) or \(y\notin\operatorname{supp}(Y)\). In the first case, some open interval \(I\) containing \(x\) has \(P(X\in I)=0\). The open rectangle \(I\times\mathbb{R}\) is a neighborhood of \((x,y)\) with joint probability zero. In the second case, an open interval \(J\) containing \(y\) has \(P(Y\in J)=0\), and \(\mathbb{R}\times J\) is a zero-probability neighborhood. Either way, \((x,y)\) is outside the joint support.
For the reverse inclusion, take \(x\in\operatorname{supp}(X)\) and \(y\in\operatorname{supp}(Y)\). Every open neighborhood \(G\) of \((x,y)\) contains a product \(I\times J\) of open intervals with \(x\in I\) and \(y\in J\). By the support definitions, \(P(X\in I)>0\) and \(P(Y\in J)>0\). Independence gives \[ P((X,Y)\in I\times J)=P(X\in I)P(Y\in J)>0. \] Since \(I\times J\subseteq G\), the neighborhood \(G\) has positive joint probability. Thus \((x,y)\) belongs to the joint support, proving the equality. Its contrapositive gives the dependence conclusion. \(\square\)
Worked Examples
Worked Example: A Continuous Law Restricted to an Ordered Region
Let \((X,Y)\) have joint density \(f(x,y)=2\) when \(0<x<y<1\), and \(f(x,y)=0\) otherwise. Show that \(X\) and \(Y\) are dependent using events with positive marginal probabilities.
The density integrates to one: the region has area \(1/2\), so its integral is \(2(1/2)=1\). Integrating out \(Y\), for \(0<x<1\), \[ f_X(x)=\int_x^1 2\,dy=2(1-x). \] Integrating out \(X\), for \(0<y<1\), \[ f_Y(y)=\int_0^y 2\,dx=2y. \]
Take \(A=(3/4,1)\) and \(B=(0,1/4)\). Their marginal probabilities are \[ P(X\in A)=\int_{3/4}^1 2(1-x)\,dx =[2x-x^2]_{3/4}^1 =1-\frac{15}{16} =\frac{1}{16}, \] and \[ P(Y\in B)=\int_0^{1/4}2y\,dy =[y^2]_0^{1/4} =\frac{1}{16}. \] But \(x>3/4\) and \(y<1/4\) cannot both hold when the joint law is restricted to \(x<y\). Therefore \(P(X\in A,Y\in B)=0\), whereas independence would require this probability to be \((1/16)(1/16)=1/256\). The variables are dependent.
Worked Example: A Forbidden Pair in a Discrete Law
Suppose \((X,Y)\) takes the values \((0,0),(0,1),(1,1)\), each with probability \(1/3\). Use the joint support to check independence.
The marginal probabilities are \(P(X=1)=1/3\) and \(P(Y=0)=1/3\). Each event is possible, but the pair \((1,0)\) is absent from the joint support, so \(P(X=1,Y=0)=0\). If \(X\) and \(Y\) were independent, this probability would have to equal \[ P(X=1)P(Y=0)=\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}. \] Since \(0\ne 1/9\), the variables are not independent. Both marginal supports are \(\{0,1\}\), so their product includes the forbidden pair \((1,0)\).
Worked Example: A Mixed Law Concentrated on the Axes
Let \(C\) equal 0 or 1 with probability \(1/2\) each. Let \(U,V\) be independent uniform random variables on \((0,1)\), independent of \(C\), and define \((X,Y)=(U,0)\) if \(C=0\), and \((X,Y)=(0,V)\) if \(C=1\). Show that the marginal supports do not imply independence.
If \(C=0\), then \(X>0\); if \(C=1\), then \(X=0\). Hence \(P(X>0)=1/2\). Similarly, \(P(Y>0)=1/2\). The events \(X>0\) and \(Y>0\) cannot occur together, because the construction places the pair on one axis or the other. Thus \(P(X>0,Y>0)=0\), while independence would require \(1/2\cdot1/2=1/4\). The variables are dependent.
Each marginal support is \([0,1]\): every neighborhood of every point in that interval has positive probability under the corresponding marginal, including neighborhoods of zero. But the joint support consists only of the two axes segments \([0,1]\times\{0\}\) and \(\{0\}\times[0,1]\). For instance, \((1/2,1/2)\) is in the product of marginal supports but not in the joint support.
How to Use the Support Test Carefully
The support theorem gives a necessary condition for independence: the joint support must include every combination of points from the marginal supports. A structural restriction that removes one such combination rules out independence. The rectangle test is often the most efficient way to turn that geometric observation into a probability calculation: find marginal events of positive probability whose intersection is forbidden.
Support equality is not, by itself, a sufficient condition for independence. Two dependent variables can have a joint support equal to the product of their marginal supports; the support records where probability can occur locally, not how probability is distributed across that region. A full independence check requires factorization of probabilities (or an equivalent distributional criterion), not just a comparison of support sets.
- Do not rely on one zero-density point. A density can vanish at a point that remains in the support because every neighborhood of the point has positive probability.
- Do not confuse separate possibility with joint possibility. Positive marginal probabilities for two events do not guarantee that their intersection has positive probability.
- Do not infer independence from matching supports. The product-support condition is necessary for independence, but it does not describe the probability allocation within that support.
This method is valuable because it can establish dependence without calculating a full joint distribution or covariance. It is especially useful when the model has a visible geometric, ordering, or structural constraint: translate that constraint into events, verify their marginal probabilities are positive, and compare the forbidden joint probability with the positive product independence requires.
Check Your Understanding
Answer each question using the definitions and tests above.
- If \(X\) and \(Y\) are independent and \(x,y\) belong to their respective marginal supports, what must be true of \((x,y)\) in the joint support?
- A joint law is restricted to \(x+y<1\). Which events involving \(X>3/4\) and \(Y>3/4\) provide a possible rectangle test, and what marginal probabilities must be checked?
- Why does a single point where a joint density is zero not necessarily lie outside the joint support?
- In the discrete example, calculate the probability independence would require for the pair \((1,0)\). Why does its actual probability disprove independence?
- Does equality of the joint support and the product of the marginal supports guarantee independence? Explain briefly.