Multivariate Distribution Functions and Continuity Rectangles

Why Rectangles Matter in Several Dimensions

A distribution function in one dimension records probabilities of intervals extending to the left. In several dimensions, the analogous function records probabilities of lower orthants. Those probabilities contain enough information to recover probabilities of finite rectangles, and—when evaluated at suitable continuity points—enough information to characterize weak convergence.

The qualification about continuity points is essential. If a rectangle’s boundary carries positive probability under the limiting law, small movements of mass across that boundary can change the rectangle probability even when the distributions converge weakly. This tutorial develops the precise criterion and shows how continuity rectangles avoid that problem.

Multivariate Distribution Functions

For \(d=1\), this is the usual distribution function. In higher dimensions it records joint, rather than merely marginal, information. Knowing each coordinate’s distribution function separately does not in general determine the joint law.

A basic calculation recovers the probability of a half-open rectangle. Let \(a_i<b_i\) for each \(i\), and let \(R=\prod_{i=1}^d(a_i,b_i]\). For a subset \(S\) of \(\{1,\ldots,d\}\), define \(z^S_i=a_i\) if \(i\in S\), and \(z^S_i=b_i\) otherwise. Inclusion-exclusion gives

In two dimensions, this reads

For example, the first term counts the lower orthant ending at \((b_1,b_2)\). Subtracting the two terms with one lower endpoint removes the parts below \(a_1\) or \(a_2\); adding back the final term corrects for the overlap that was subtracted twice.

Continuity Rectangles

A Borel set \(A\subseteq\mathbb{R}^d\) is a continuity set for \(\mu\) if \(\mu(\partial A)=0\), where \(\partial A\) denotes its topological boundary. For a rectangle, a convenient sufficient condition is that every face lie in a coordinate hyperplane to which \(\mu\) assigns zero probability. More precisely, for \(R=\prod_i(a_i,b_i]\), it is sufficient that

The boundary of \(R\) is contained in the union of these finitely many hyperplanes, so this condition implies \(\mu(\partial R)=0\). It is sufficient, not necessary: parts of a hyperplane outside the rectangle may carry mass without contributing to the rectangle’s boundary.

The Continuity-Point Criterion

Weak convergence of probability measures on \(\mathbb{R}^d\), written \(\mu_n\Rightarrow\mu\), means that \(\int f\,d\mu_n\to\int f\,d\mu\) for every bounded continuous \(f:\mathbb{R}^d\to\mathbb{R}\). The criterion below translates this convergence into a statement about distribution functions.

Proof. First suppose \(\mu_n\Rightarrow\mu\), and fix a continuity point \(x=(x_1,\ldots,x_d)\) of \(F\). Let \(A_x=\prod_i(-\infty,x_i]\), and let \(x^{(m)}=x-(1/m,\ldots,1/m)\). The sets \(A_{x^{(m)}}\) increase to \(\prod_i(-\infty,x_i)\). Therefore

The last equality follows from continuity of \(F\) at \(x\). The boundary of \(A_x\) consists of points in \(A_x\) with at least one coordinate equal to its corresponding \(x_i\), exactly the part removed by the increasing union. Thus \(A_x\) is a continuity set. The Portmanteau theorem gives \(\mu_n(A_x)\to\mu(A_x)\), which is \(F_n(x)\to F(x)\).

Conversely, suppose \(F_n(x)\to F(x)\) at every continuity point of \(F\). For coordinate \(i\), let \(\mu_i\) be the \(i\)-th marginal of \(\mu\), and let \(D_i=\{t:\mu_i(\{t\})=0\}\). The complement of \(D_i\) is at most countable, since a probability measure has at most countably many atoms. Hence \(D_i\) is dense in \(\mathbb{R}\).

If \(x_i\in D_i\) for all \(i\), then \(F\) is continuous at \(x\). Indeed, when \(y\) is close to \(x\), the lower orthants \(A_y\) and \(A_x\) can differ only where at least one coordinate lies between \(x_i-\delta\) and \(x_i+\delta\), for a suitable \(\delta\) tending to zero with \(y\to x\). Thus

Consequently, \(F_n(x)\to F(x)\) at all points whose coordinates belong to the respective dense sets \(D_i\). For any rectangle \(Q=\prod_i(a_i,b_i]\) whose endpoints all belong to \(D_i\), every corner in the inclusion-exclusion formula has coordinates in the appropriate \(D_i\). The formula therefore implies \(\mu_n(Q)\to\mu(Q)\).

It remains to show convergence of integrals of bounded continuous functions. Choose \(M\) with \(M,-M\in D_i\) for every \(i\), and let \(K=(-M,M]^d\). Such choices can be made arbitrarily large because each \(D_i\) is dense. The rectangle formula gives \(\mu_n(K)\to\mu(K)\), and \(\mu(K)\) can be made arbitrarily close to \(1\) by taking \(M\) large. Therefore the probabilities of \(K^c\) under \(\mu\) and, for all sufficiently large \(n\), under \(\mu_n\), can both be made arbitrarily small.

Now fix bounded continuous \(f\), with \(|f|\le B\). On the compact set \([-M,M]^d\), \(f\) is uniformly continuous. Partition \(K\) into finitely many rectangles with endpoints in the respective dense sets \(D_i\), choosing the mesh small enough that the oscillation of \(f\) on each rectangle is at most \(\eta\). Choose a value of \(f\) at a point in each rectangle and define a step function \(h\) by that value on the rectangle and zero outside \(K\). Then \(|f-h|\le\eta\) on \(K\), while \(|f-h|\le B\) outside \(K\). Each rectangle in the partition has converging probability, so the finite sum defining \(\int h\,d\mu_n\) converges to \(\int h\,d\mu\). The remaining errors are bounded by \(\eta+B\mu_n(K^c)\) and \(\eta+B\mu(K^c)\), respectively. First choose \(M\) so the tail terms are small, then choose \(\eta\) small. It follows that \(\int f\,d\mu_n\to\int f\,d\mu\), proving \(\mu_n\Rightarrow\mu\). \(\square\)

Using the Criterion in Practice

The reverse direction of the theorem explains why distribution-function convergence is a strong test: values at continuity points yield rectangle probabilities, and fine partitions by continuity rectangles approximate bounded continuous functions. The forward direction explains why one should not demand convergence at every point. At a jump, probabilities can move across a threshold without violating weak convergence.

A useful practical sequence of checks is:

Why the Boundary Condition Matters

Rectangle probabilities are often the quantities of direct interest, but their convergence is not guaranteed for every rectangle under weak convergence. A threshold that passes through an atom is unstable: approximating distributions may place that mass just to one side or the other. Continuity rectangles rule out precisely this ambiguity.

In several dimensions, inspect every face, not just the corners or the marginal distributions in isolation. A joint law may put mass on a coordinate hyperplane, and that mass can make a rectangle boundary non-null. Conversely, a rectangle chosen with endpoints avoiding marginal atoms is a convenient sufficient choice, because its boundary is contained in finitely many null coordinate hyperplanes.