Rectangle Tests for Joint Distributions

Why Test Independence with Distribution Functions?

Independence of random vectors is defined through factorization of probabilities for all Borel sets. That definition is exact, but it appears to require checking an enormous collection of sets. Distribution functions provide a more structured test: they record probabilities on lower-left rectangles, a family of sets that is simple to describe and still generates the Borel sets.

The key point is not that lower-left rectangles are the only sets that matter. It is that they form a sufficiently rich family: if the joint and product laws agree on them, then the laws agree everywhere. We will make that statement precise and prove it.

Joint Distribution Functions of Real Vectors

Let \(X=(X_1,\ldots,X_m)\) be a random vector with values in \(\mathbb{R}^m\). For \(x=(x_1,\ldots,x_m)\in\mathbb{R}^m\), write \(X\le x\) to mean \(X_i\le x_i\) for every coordinate \(i\). The distribution function of \(X\) is the function \(F_X:\mathbb{R}^m\to[0,1]\) defined by the probability of the lower-left rectangle ending at \(x\).

The event in this definition is the rectangle \[ (-\infty,x_1]\times\cdots\times(-\infty,x_m] \times(-\infty,y_1]\times\cdots\times(-\infty,y_n]. \] The inequalities are coordinatewise. In particular, when \(X\) is a vector, \(F_X(x)\) is not merely a list of its one-dimensional marginal distribution functions; it records the probability that all coordinates meet their respective bounds at once.

Recall that \(X\) and \(Y\) are independent if \(\mathbb{P}(X\in A,\ Y\in B)=\mathbb{P}(X\in A)\mathbb{P}(Y\in B)\) for every Borel set \(A\subseteq\mathbb{R}^m\) and every Borel set \(B\subseteq\mathbb{R}^n\). The rectangle test reduces this condition to the lower-left rectangles.

The Rectangle Criterion

Proof. First suppose \(X\) and \(Y\) are independent. For any fixed \(x\) and \(y\), the events \(\{X\le x\}\) and \(\{Y\le y\}\) belong to the respective sigma-algebras generated by the random vectors. Independence therefore gives

For the converse, let \(\mu\) be the law of \((X,Y)\) on \(\mathbb{R}^{m+n}\), and let \(\nu\) be the product of the marginal laws of \(X\) and \(Y\). For \(x\in\mathbb{R}^m\) and \(y\in\mathbb{R}^n\), set

The assumed factorization says exactly that \(\mu(C_x\times D_y)=\nu(C_x\times D_y)\) for all \(x,y\). The family of these product rectangles, together with the whole space, is closed under finite intersections: intersecting two such rectangles replaces each coordinate bound by the smaller of the two bounds. It is therefore a \(\pi\)-system.

This \(\pi\)-system generates the Borel sets of \(\mathbb{R}^{m+n}\). Indeed, lower-left orthants generate the Borel sets in each Euclidean factor, and products of generating sets generate the product Borel sigma-algebra, which is the Borel sigma-algebra of \(\mathbb{R}^{m+n}\). The two probability measures \(\mu\) and \(\nu\) agree on a generating \(\pi\)-system and both have total mass one. The \(\pi\)-\(\lambda\) theorem, equivalently the uniqueness theorem for finite measures on a generating \(\pi\)-system, gives \(\mu=\nu\). Thus the joint law is the product of its marginals, which is equivalent to independence of \(X\) and \(Y\).

For completeness, the measure-uniqueness step uses the following observation: the class of Borel sets on which two probability measures agree is a \(\lambda\)-system. It contains the whole space, is closed under complements, and is closed under countable disjoint unions by countable additivity. If it contains a generating \(\pi\)-system, the \(\pi\)-\(\lambda\) theorem says it contains the generated sigma-algebra. This proves the required equality of laws. \(\square\)

Worked Example: Independent Uniform Random Variables

Worked Example: A Factorization Failure Detects Dependence

Worked Example: A Vector-Valued Distribution

When a Countable Test Is Enough

The theorem asks for factorization at every real threshold. In practice, an equivalent countable test is often useful: it is enough to check thresholds whose coordinates are rational. No assumption of continuous distributions is needed.

Proof. Necessity follows from the rectangle criterion. For sufficiency, fix arbitrary \(x\in\mathbb{R}^m\) and \(y\in\mathbb{R}^n\). For each coordinate, choose rational sequences \(q_{i,k}\downarrow x_i\) and \(r_{j,k}\downarrow y_j\). Such sequences exist by the density of \(\mathbb{Q}\) in \(\mathbb{R}\). Write \(q_k=(q_{1,k},\ldots,q_{m,k})\) and \(r_k=(r_{1,k},\ldots,r_{n,k})\).

The events \(\{X\le q_k,\ Y\le r_k\}\) decrease as \(k\) increases, and their intersection is exactly \(\{X\le x,\ Y\le y\}\). Since probabilities are finite, continuity from above gives

The same argument for the marginal events gives \(F_X(x)=\lim_{k\to\infty}F_X(q_k)\) and \(F_Y(y)=\lim_{k\to\infty}F_Y(r_k)\). Applying the assumed rational-threshold factorization and taking limits,

This holds for all real \(x,y\), so the rectangle criterion proves independence. \(\square\)

What the Test Does—and Does Not—Require

A common mistake is to use only one-dimensional marginal distribution functions when the random elements are vectors. The relevant quantity is \(F_X(x)\), the probability that every coordinate of \(X\) lies below its own threshold, and likewise for \(Y\). The theorem compares the joint probability of the two vector events with the product of those two probabilities.

Another point is that factorization at a few selected thresholds does not establish independence. A failure at even one pair of thresholds disproves it, but to prove it one needs factorization on a determining family. All real thresholds work; all rational coordinate thresholds also work because approximation from above and continuity from above recover every real threshold. There is no need to assume a density, or that the distribution functions are continuous.

This criterion is useful because it translates a statement about all Borel events into an explicit identity for distribution functions. It also gives a practical route for calculations: derive the joint distribution function and compare it with the product of the marginals. The generating-family argument explains why this comparison is decisive rather than merely suggestive.