Point Masses and Finite-Support Weak Convergence

Weak Convergence for Atomic Measures

Weak convergence is defined by what bounded continuous functions can detect. For measures concentrated on finitely many points, those integrals reduce to finite sums. This makes point masses and finite mixtures useful settings in which to see exactly how weak convergence works—and what it does not require.

Throughout, measures are Borel probability measures on \(\mathbb{R}^d\), with its usual Euclidean norm \(\|\cdot\|\). We use the bounded-continuous-test definition of weak convergence: \(\mu_n\) converges weakly to \(\mu\), written \(\mu_n\Rightarrow\mu\), when the integrals of every bounded continuous real-valued function converge.

Definition: For \(x\in\mathbb{R}^d\), the Dirac probability measure at \(x\), denoted \(\delta_x\), assigns probability one to \(x\). Equivalently, for every Borel set \(A\), \(\delta_x(A)=1\) if \(x\in A\), and \(\delta_x(A)=0\) otherwise. For any bounded Borel function \(f\), \(\int f\,d\delta_x=f(x)\).

Definition: A finite mixture of point masses is a probability measure of the form \(\mu=\sum_{i=1}^m p_i\delta_{x_i}\), where \(m\in\mathbb{N}\), \(p_i\geq 0\), \(\sum_{i=1}^m p_i=1\), and \(x_i\in\mathbb{R}^d\). Its integral against a bounded Borel function is \(\int f\,d\mu=\sum_{i=1}^m p_i f(x_i)\).

The Exact Criterion for Point Masses

For a point mass, weak convergence is equivalent to ordinary convergence of its location. The forward direction follows immediately by testing a continuous function at nearby points. For the reverse direction, one bounded continuous test function can detect whether the location remains away from the proposed limit.

Theorem: Let \(x_n,x\in\mathbb{R}^d\). Then \(\delta_{x_n}\Rightarrow\delta_x\) if and only if \(x_n\to x\) in Euclidean distance.

Proof. Suppose first that \(x_n\to x\). For every bounded continuous \(f:\mathbb{R}^d\to\mathbb{R}\), continuity at \(x\) gives \(f(x_n)\to f(x)\). Since \(\int f\,d\delta_{x_n}=f(x_n)\) and \(\int f\,d\delta_x=f(x)\), the defining test-integral condition gives \(\delta_{x_n}\Rightarrow\delta_x\).

Conversely, suppose \(\delta_{x_n}\Rightarrow\delta_x\). Choose the bounded continuous function \(g(y)=\min\{1,\|y-x\|\}\). Weak convergence implies

\min\{1,\|x_n-x\|\} =\int g\,d\delta_{x_n} \longrightarrow \int g\,d\delta_x =0.

For any \(\varepsilon>0\), eventually \(\min\{1,\|x_n-x\|\}<\min\{1,\varepsilon\}\). If \(\varepsilon\leq 1\), this implies \(\|x_n-x\|<\varepsilon\); if \(\varepsilon>1\), it implies \(\|x_n-x\|<1<\varepsilon\). Thus \(x_n\to x\). \strong>□

Worked Example: A Moving Point in the Plane

Let \(x_n=(1/n,2/n)\) and \(x=(0,0)\). The Euclidean distance is

\|x_n-x\|=\sqrt{\frac{1}{n^2}+\frac{4}{n^2}}=\frac{\sqrt{5}}{n}\longrightarrow 0.

Therefore \(\delta_{x_n}\Rightarrow\delta_{(0,0)}\). Directly, for any bounded continuous \(f\), the integrals are \(f(1/n,2/n)\) and \(f(0,0)\), and continuity at \((0,0)\) makes their difference tend to zero.

Worked Example: A Point Mass That Does Not Converge Weakly

In \(\mathbb{R}\), let \(x_n=(-1)^n\). This sequence does not converge to \(0\). The test function \(g(y)=\min\{1,|y|\}\) is bounded and continuous, and

\int g\,d\delta_{x_n}=g((-1)^n)=1 \qquad\text{while}\qquad \int g\,d\delta_0=g(0)=0.

The test integrals do not converge to the integral under \(\delta_0\), so \(\delta_{(-1)^n}\not\Rightarrow\delta_0\). The theorem says more generally that no other point mass can be its weak limit, since the locations themselves have no limit.

Finite Mixtures with Moving Atoms

For mixtures, the integral of a test function is a weighted sum of its values at the atoms. This gives a direct sufficient condition for convergence: each labelled atom approaches a limit location, and each corresponding weight approaches a limit weight.

Theorem: Fix \(m\in\mathbb{N}\). For each \(n\), let \(p_{n,i}\geq0\) with \(\sum_{i=1}^m p_{n,i}=1\), and let \(x_{n,i}\in\mathbb{R}^d\), for \(1\leq i\leq m\). Suppose \(p_{n,i}\to p_i\) and \(x_{n,i}\to x_i\) for every \(i\). Then \(\sum_{i=1}^m p_{n,i}\delta_{x_{n,i}}\Rightarrow\sum_{i=1}^m p_i\delta_{x_i}\). In particular, \(\sum_i p_i=1\).

Proof. Since there are finitely many indices and each \(p_{n,i}\to p_i\), taking limits in \(\sum_i p_{n,i}=1\) gives \(\sum_i p_i=1\); also \(p_i\geq0\). Let \(f\) be any bounded continuous function. For each \(i\), continuity gives \(f(x_{n,i})\to f(x_i)\), and hence \(p_{n,i}f(x_{n,i})\to p_i f(x_i)\). Taking limits in a finite sum yields

\int f\,d\left(\sum_{i=1}^m p_{n,i}\delta_{x_{n,i}}\right) =\sum_{i=1}^m p_{n,i}f(x_{n,i}) \longrightarrow \sum_{i=1}^m p_i f(x_i) =\int f\,d\left(\sum_{i=1}^m p_i\delta_{x_i}\right).

This holds for every bounded continuous \(f\), which is precisely weak convergence. □

Worked Example: Two Moving Atoms and Changing Weights

In \(\mathbb{R}\), consider \(\mu_n=(1-1/n)\delta_{1/n}+(1/n)\delta_{2+1/n}\), for \(n\in\mathbb{N}\). The weights converge to \(1\) and \(0\), and the atom locations converge to \(0\) and \(2\), respectively. Thus the theorem gives \(\mu_n\Rightarrow\delta_0\).

Indeed, for any bounded continuous \(f\),

\int f\,d\mu_n =\left(1-\frac1n\right)f(1/n)+\frac1n f(2+1/n) \longrightarrow f(0).

The first term tends to \(f(0)\) by continuity and \(1-1/n\to1\). If \(|f|\leq M\), the second term has absolute value at most \(M/n\), so it tends to zero. The limiting atom at \(2\) has zero weight; it contributes nothing to the limit.

Mixtures on a Fixed Finite Set

When the possible atom locations are fixed and distinct, weak convergence is equivalent to convergence of each weight. Distinctness matters: if two listed locations coincide, their weights cannot be distinguished by any test function; only their sum can be observed.

Theorem: Let \(x_1,\ldots,x_m\) be distinct points in \(\mathbb{R}^d\). For each \(n\), let \(p_{n,i}\geq0\) with \(\sum_i p_{n,i}=1\), and let \(p_i\geq0\) with \(\sum_i p_i=1\). Then \(\sum_i p_{n,i}\delta_{x_i}\Rightarrow\sum_i p_i\delta_{x_i}\) if and only if \(p_{n,i}\to p_i\) for every \(i\).

Proof. If every weight converges, then for any bounded continuous \(f\), the finite sums \(\sum_i p_{n,i}f(x_i)\) converge to \(\sum_i p_i f(x_i)\). This proves weak convergence.

For the converse, fix \(i\). If \(m=1\), both weights equal \(1\) and the conclusion holds. If \(m\geq2\), set \(r_i=\min_{j\ne i}\|x_i-x_j\|\), which is positive because the locations are distinct and there are finitely many of them. Define

f_i(y)=\max\left\{0,1-\frac{\|y-x_i\|}{r_i}\right\}.

This function is bounded and continuous. It equals \(1\) at \(x_i\). For \(j\ne i\), \(\|x_j-x_i\|\geq r_i\), so \(f_i(x_j)=0\). Therefore the two integrals of \(f_i\) are exactly \(p_{n,i}\) and \(p_i\). Weak convergence forces \(p_{n,i}\to p_i\). This works for every \(i\). □

Worked Example: Reading Mixture Weights with Test Functions

Take the distinct locations \(0\) and \(1\), and let \(\mu_n=(1/2+(-1)^n/4)\delta_0+(1/2-(-1)^n/4)\delta_1\). The weight at \(0\) alternates between \(3/4\) and \(1/4\), so it does not converge. The function \(f(y)=\max\{0,1-|y|\}\) is bounded and continuous, with \(f(0)=1\) and \(f(1)=0\). Consequently,

\int f\,d\mu_n=\frac12+\frac{(-1)^n}{4},

which does not converge. Hence the measures have no weak limit of the form \(p\delta_0+(1-p)\delta_1\). The calculation also shows why distinct fixed locations allow tests to isolate their weights.

A Small Amount of Mass May Escape

Weak convergence does not require every atom in a representation to converge to a finite location. An atom whose weight tends to zero can move arbitrarily far away without affecting the weak limit, because the test functions used in the definition are bounded.

Worked Example: Vanishing Weight at a Distant Atom

On \(\mathbb{R}\), let \(\nu_n=(1-1/n)\delta_0+(1/n)\delta_n\). For any bounded continuous \(f\), with \(|f|\leq M\),

\left|\int f\,d\nu_n-f(0)\right| =\left|-\frac1n f(0)+\frac1n f(n)\right| \leq \frac{2M}{n} \longrightarrow 0.

Thus \(\nu_n\Rightarrow\delta_0\), even though the second location \(n\) diverges. This is consistent with the moving-atom theorem: that theorem assumes each labelled atom has a finite limiting location, so it is sufficient but not necessary for mixture convergence.

What the Criteria Do—and Do Not—Say

The point-mass criterion is an exact equivalence: there is only one location to track, and weak convergence of the Dirac measures forces that location to converge. For mixtures, convergence of labelled locations and weights is a convenient sufficient criterion. It is not a necessary criterion for every possible representation, because labels can be rearranged, atoms can merge, or some weights can vanish.

For a fixed set of distinct locations, the weight criterion is exact. If locations are repeated, first combine the weights at each distinct location: for example, \(a\delta_x+b\delta_x=(a+b)\delta_x\). Weak convergence can detect the combined mass at \(x\), not the arbitrary division of that mass among duplicate labels.

Boundedness of the test functions also explains the distant-atom example. Weak convergence controls integrals against bounded continuous functions; it does not, by itself, control unbounded quantities such as moments. A tiny probability at a very large value may have little effect on bounded tests while having a substantial effect on an expectation. The convergence question must therefore be matched to the class of functions being tested.

Key takeaway: For point masses, weak convergence is exactly convergence of their locations. For finite mixtures, convergence of corresponding weights and locations guarantees weak convergence; on a fixed finite set of distinct locations, convergence of the weights is also necessary.

Check Your Understanding

Use bounded continuous test functions and the finite-mixture formulas to answer the following.

Why does \(\delta_{x_n}\Rightarrow\delta_x\) imply \(x_n\to x\)? Identify a bounded continuous function that proves the implication.
Suppose two mixture weights converge and their corresponding atom locations converge. What is the limiting measure?
Why can a bounded continuous function isolate one weight when the support consists of finitely many distinct fixed points?
Can \((1-1/n)\delta_0+(1/n)\delta_n\) converge weakly to \(\delta_0\) even though one location diverges? Explain using a bound on test integrals.
If two labels in a finite mixture refer to the same point, which quantity can weak convergence detect: each labelled weight separately, or their sum?