Portmanteau Calculations for Balls and Rectangles

From Weak Convergence to Event Probabilities

Weak convergence describes how probability distributions behave when tested against bounded continuous functions. In applications, however, the quantity of interest is often an event probability: the chance that a random vector falls inside a ball, within a specified range of coordinates, or on one side of a threshold.

For an event defined by a fixed set, weak convergence does not by itself guarantee convergence of its probabilities. The practical test is the mass that the limiting distribution assigns to the set’s boundary. This tutorial develops that test for Euclidean balls and rectangles, and works through cases where the probabilities do and do not converge.

Let \(X_n\) and \(X\) be random vectors taking values in \(\mathbb{R}^d\), and suppose \(X_n\Rightarrow X\). Write \(\mu_n\) and \(\mu\) for their respective distributions. For a Borel set \(A\), the notation \(\mu_n(A)\) means \(\mathbb{P}(X_n\in A)\).

Portmanteau bounds: For every open set \(G\) and every closed set \(F\) in \(\mathbb{R}^d\), \(\mu(G)\leq\liminf_{n\to\infty}\mu_n(G)\) and \(\limsup_{n\to\infty}\mu_n(F)\leq\mu(F)\). In particular, if \(\mu(\partial A)=0\), then \(\mu_n(A)\to\mu(A)\). Such a set \(A\) is called a continuity set for \(\mu\).

The two inequalities point in different directions: the open-set bound is a lower bound, while the closed-set bound is an upper bound. When the boundary has zero limiting probability, these bounds squeeze the event probabilities to the desired limit. The calculations below are ways to check that boundary condition.

Balls: The Boundary Is a Sphere

For \(x\in\mathbb{R}^d\) and \(r>0\), define the open and closed Euclidean balls by \(B(x,r)=\{y:\|y-x\|<r\}\) and \(\overline B(x,r)=\{y:\|y-x\|\leq r\}\). Their common boundary is the sphere \(S(x,r)=\{y:\|y-x\|=r\}\).

Thus, for either the open ball or the closed ball, the relevant question is whether \(\mathbb{P}(\|X-x\|=r)=0\). If it is zero, weak convergence implies convergence of the probabilities of both balls. The shape of the sphere alone does not settle this question: a distribution may place all its mass on that sphere.

Result: If \(X_n\Rightarrow X\) in \(\mathbb{R}^d\) and \(\mathbb{P}(\|X-x\|=r)=0\), then \[ \mathbb{P}(X_n\in B(x,r))\to\mathbb{P}(X\in B(x,r)) \quad\text{and}\quad \mathbb{P}(X_n\in\overline B(x,r))\to\mathbb{P}(X\in\overline B(x,r)). \]

This follows directly from the continuity-set criterion: the boundary of either ball is \(S(x,r)\), and the stated condition says precisely that this boundary has zero \(\mu\)-mass.

Worked Example: A Ball Under a Bivariate Standard Normal Limit

Suppose \(X_n\Rightarrow X\), where \(X\) has the standard bivariate normal density \(f(u,v)=(2\pi)^{-1}\exp(-(u^2+v^2)/2)\). Consider the open ball of radius \(r>0\) centered at the origin.

The limiting density is with respect to two-dimensional Lebesgue measure. The sphere \(u^2+v^2=r^2\) has Lebesgue measure zero, so it has probability zero under this density. The continuity-set criterion therefore gives convergence of the probabilities of the open balls.

The limiting probability can also be calculated. In polar coordinates, the density times the area element is \((2\pi)^{-1}e^{-s^2/2}s\,ds\,d\theta\). Hence

\mathbb{P}(\|X\|<r) =\int_0^{2\pi}\int_0^r \frac{1}{2\pi}e^{-s^2/2}s\,ds\,d\theta =1-e^{-r^2/2}.

Consequently, \(\mathbb{P}(\|X_n\|<r)\to 1-e^{-r^2/2}\). The same limit holds for the closed balls, since the boundary circle also has probability zero.

Why Densities Give Zero Mass to These Boundaries

The preceding example used a geometric fact: spheres have zero \(d\)-dimensional Lebesgue measure. Here is a proof that works in every positive integer dimension.

Proposition: For \(d\geq 1\), every sphere \(S(x,r)\) in \(\mathbb{R}^d\), with \(r\geq 0\), has \(d\)-dimensional Lebesgue measure zero.

Proof: For \(r=0\), the sphere is the singleton \(\{x\}\), which has Lebesgue measure zero. Now let \(r>0\). For \(0<\varepsilon<r\), the sphere \(S(x,r)\) is contained in the annulus \(\{y:r-\varepsilon<\|y-x\|<r+\varepsilon\}\), up to its two bounding spheres; equivalently, it is contained in the closed annulus with those radii. The volume of that closed annulus is \[ v_d\big((r+\varepsilon)^d-(r-\varepsilon)^d\big), \] where \(v_d\) is the finite volume of the unit ball in \(\mathbb{R}^d\). As \(\varepsilon\downarrow0\), this expression tends to zero, by continuity of the function \(t\mapsto t^d\). The sphere’s Lebesgue measure is bounded above by this volume for every such \(\varepsilon\); it must therefore be zero. \(\square\)

If the law of \(X\) has a density with respect to Lebesgue measure, every Lebesgue-null set has probability zero. The proposition therefore ensures that every sphere, at every center and radius, has zero probability under such a law. The same conclusion does not hold for a law without a density: for example, a vector supported on a circle assigns probability one to that circle.

Rectangles: Check the Coordinate Faces

A (bounded) rectangle in \(\mathbb{R}^d\) has the form \(R=\prod_{j=1}^d [a_j,b_j]\), where \(a_j<b_j\). Its boundary is contained in the union of the coordinate faces where at least one coordinate equals an endpoint. More precisely, \(\partial R\subseteq\bigcup_{j=1}^d(\{y:y_j=a_j\}\cup\{y:y_j=b_j\})\). The same boundary describes the corresponding open rectangle \(\prod_{j=1}^d(a_j,b_j)\).

A coordinate face can have positive probability even when the distribution is otherwise spread across several dimensions. The relevant quantities are the endpoint atom probabilities of the coordinate marginals.

Proposition: Let \(X=(X_1,\ldots,X_d)\) have any probability distribution on \(\mathbb{R}^d\). For \(R=\prod_{j=1}^d[a_j,b_j]\), \[ \mathbb{P}(X\in\partial R) \leq \sum_{j=1}^d\big(\mathbb{P}(X_j=a_j)+\mathbb{P}(X_j=b_j)\big). \] In particular, if every coordinate marginal assigns zero probability to each rectangle endpoint, then \(R\) is a continuity set.

Proof: If \(X\in\partial R\), then at least one coordinate of \(X\) equals one of the endpoints of its interval. Thus \[ \{X\in\partial R\}\subseteq \bigcup_{j=1}^d\big(\{X_j=a_j\}\cup\{X_j=b_j\}\big). \] Taking probabilities and applying the union bound gives the stated inequality. If all the endpoint probabilities on the right are zero, then \(\mathbb{P}(X\in\partial R)=0\), which is the continuity-set condition. \(\square\)

This test only uses the coordinate marginals to control the boundary; it does not require the coordinates to be independent. When the limiting vector has a joint density, each face is Lebesgue-null, so the boundary has probability zero. The marginal test is often more convenient when the limiting distribution is known through its one-dimensional marginals.

Worked Example: A Rectangle Under Independent Normal Coordinates

Suppose \(X_n\Rightarrow X=(X_1,X_2)\), where \(X_1\) and \(X_2\) are independent standard normal random variables. Consider the rectangle \(R=(a_1,b_1]\times(a_2,b_2]\), with finite \(a_1<b_1\) and \(a_2<b_2\).

Each normal marginal has a density, so it assigns probability zero to each of the four endpoints. The boundary criterion shows that the probabilities of \(R\) under \(X_n\) converge to the probability under \(X\). Independence lets us calculate that limit. Writing \(\Phi\) for the standard normal distribution function,

\mathbb{P}(X\in R) =\big(\Phi(b_1)-\Phi(a_1)\big) \big(\Phi(b_2)-\Phi(a_2)\big).

For example, if \(R=(0,1]\times(-1,1]\), the limit is \((\Phi(1)-\Phi(0))(\Phi(1)-\Phi(-1))\). Both factors are positive, and no boundary term is needed: each coordinate has zero probability of equalling any of the four endpoints.

When Boundary Mass Prevents Convergence

A nonzero boundary probability means that the continuity-set criterion cannot guarantee convergence. It does not, by itself, prove that the probabilities fail to converge. A direct example shows why the distinction matters.

Worked Example: An Open Rectangle Whose Probability Jumps

In \(\mathbb{R}^2\), let \(X_n=(1/n,0)\) deterministically and let \(X=(0,0)\) deterministically. Then \(X_n\to X\), so \(X_n\Rightarrow X\). Take the open rectangle \(R=(0,1)\times(-1,1)\).

For every \(n\geq2\), \(0<1/n<1\) and \(-1<0<1\), so \(X_n\in R\) and \(\mathbb{P}(X_n\in R)=1\). But \(X\notin R\), because its first coordinate equals the excluded endpoint \(0\), so \(\mathbb{P}(X\in R)=0\). The probabilities do not converge to the limiting probability.

The limiting vector lies on the boundary of \(R\) with probability one. This is exactly the situation in which the continuity-set criterion does not apply. It is not the fact that the rectangle is open that causes a universal failure; the decisive feature here is the positive boundary mass.

A Practical Boundary Check

For a fixed ball or rectangle, the calculation can be organized in a short sequence. First identify the boundary, then evaluate its probability under the limiting law, and only then invoke the event-probability conclusion. This avoids assuming that weak convergence automatically transfers to every event.

Identify the event set.
Write the ball or rectangle precisely, including whether its endpoints or sphere are included.

Find its boundary.
For a ball, this is the sphere at the specified radius. For a rectangle, it is contained in the union of its coordinate faces.

Check limiting boundary mass.
Use the sphere probability, endpoint atoms of the coordinate marginals, or a density argument.

Apply the appropriate conclusion.
If the boundary mass is zero, the event probabilities converge. If it is positive, the Portmanteau bounds still hold, but convergence may fail.

One common pitfall is to confuse a convenient sufficient condition with a necessary one. A density guarantees zero mass on spheres and rectangle faces, but a density is not required: a discrete or singular distribution may also assign zero mass to the particular boundary under consideration. Conversely, even a distribution with a density can give positive probability to a set with nonempty interior; what matters here is its mass on the boundary, not the probability of the set itself.

The useful conclusion is local to the chosen set. A limiting law may give zero mass to the boundary of one ball and positive mass to the boundary of another, or have no atoms at the endpoints of one rectangle but atoms at those of another. Check the actual boundary in the problem rather than relying on a broad label such as “continuous distribution.”

Check Your Understanding

For each question, identify the boundary-mass calculation that controls the conclusion.

If \(X_n\Rightarrow X\) in \(\mathbb{R}^3\), what probability must be zero to conclude convergence for the open ball \(B(x,r)\)?
Why does a density with respect to three-dimensional Lebesgue measure imply that a sphere has probability zero?
For a rectangle \([a_1,b_1]\times[a_2,b_2]\), which four coordinate endpoint probabilities can be used to bound its boundary probability?
Does positive limiting probability on a rectangle’s boundary prove that the event probabilities fail to converge? Explain.
Can a rectangle be a continuity set when the limiting distribution has no joint density? Give the boundary condition that answers the question.