Why Test Functions Characterize a Distributional Limit
Convergence in distribution is often introduced by tracking cumulative distribution functions at their continuity points. The bounded-continuous-function criterion gives another way to describe the same limiting behavior: instead of checking probabilities of half-lines, we check expectations of a broad class of functions of the random variable.
This formulation is useful because many quantities in probability and statistics are expectations. It also clarifies an important limitation: distributional convergence controls expectations of bounded continuous functions, but it does not by itself control expectations of unbounded functions such as \(g(x)=x\).
The characterization below says that, for real-valued random variables, it is enough to compare the expectations of all such test functions. Write \(\|g\|_\infty=\sup_x|g(x)|\). Each expectation exists because \(g\) is bounded.
Worked Examples Using Test Functions
Worked Example: A Two-Point Distribution Converging to Zero
For each positive integer \(n\), suppose \(X_n\) takes the values \(-1/n\) and \(1/n\), each with probability \(1/2\). Verify convergence in distribution to the constant zero using bounded continuous test functions.
Let \(g:\mathbb{R}\to\mathbb{R}\) be any bounded continuous function. The expectation is $$ E[g(X_n)]=\frac{1}{2}g(-1/n)+\frac{1}{2}g(1/n). $$
Since \(-1/n\to0\) and \(1/n\to0\), continuity of \(g\) at zero gives \(g(-1/n)\to g(0)\) and \(g(1/n)\to g(0)\). Therefore $$ E[g(X_n)]\longrightarrow \frac{1}{2}g(0)+\frac{1}{2}g(0)=g(0)=E[g(0)]. $$ This holds for every bounded continuous \(g\), so the characterization yields \(X_n\xrightarrow{d}0\).
Worked Example: Rare Mass Escaping to Infinity
Let \(X_n=0\) with probability \(1-1/n\), and \(X_n=n\) with probability \(1/n\). Verify the distributional limit using test functions, and compare it with the behavior of the mean.
For any bounded continuous \(g\), $$ E[g(X_n)]=\left(1-\frac{1}{n}\right)g(0)+\frac{1}{n}g(n). $$ Thus $$ \left|E[g(X_n)]-g(0)\right| =\frac{1}{n}|g(n)-g(0)| \leq \frac{2\|g\|_\infty}{n}\longrightarrow 0. $$ Because \(E[g(0)]=g(0)\), the test-function criterion proves \(X_n\xrightarrow{d}0\).
The mean behaves differently: \(E[X_n]=(1-1/n)0+(1/n)n=1\) for every \(n\), whereas the limiting constant has mean zero. There is no contradiction. The identity function \(g(x)=x\) is continuous but unbounded, so it is not among the test functions controlled by this criterion. The probability \(1/n\) at the increasingly large value \(n\) disappears from bounded tests, but its contribution to the mean remains one.
Worked Example: A Discrete Grid Converging to a Uniform Law
Let \(X_n\) be uniform on the \(n\) points \(1/n,2/n,\ldots,n/n\). Show that \(X_n\) converges in distribution to \(U\sim\operatorname{Uniform}(0,1)\).
For any bounded continuous \(g:\mathbb{R}\to\mathbb{R}\), $$ E[g(X_n)]=\frac{1}{n}\sum_{k=1}^{n}g(k/n). $$ The restriction of \(g\) to \([0,1]\) is continuous, so its right-endpoint Riemann sums converge to its integral: $$ \frac{1}{n}\sum_{k=1}^{n}g(k/n)\longrightarrow\int_0^1 g(x)\,dx. $$ A uniform random variable on \((0,1)\) has density one on that interval, and hence \(E[g(U)]=\int_0^1g(x)\,dx\). Therefore \(E[g(X_n)]\to E[g(U)]\) for every bounded continuous \(g\), which proves \(X_n\xrightarrow{d}U\).
Proof of the Characterization
The two directions use different ideas. From convergence of distribution functions, we first restrict attention to a large bounded interval and approximate a continuous function there by finitely many constants. Conversely, continuous functions that rise or fall across a short interval can approximate the indicator of a half-line.
Control the tails, partition a bounded interval using continuity points of the limiting distribution function, and approximate the function by a step function.
Bracket each half-line indicator by bounded continuous functions and shrink the interval over which the brackets differ.
First direction. Suppose \(X_n\xrightarrow{d}X\), and fix a bounded continuous \(g\). Set \(M=\|g\|_\infty\). Given \(\varepsilon>0\), choose continuity points \(A<B\) of \(F\) so far into the tails that $$ P(X\leq A)+P(X>B)<\varepsilon. $$ Such continuity points can be chosen because a distribution function has at most countably many discontinuities and its tail probabilities tend to zero.
Because \(A\) and \(B\) are continuity points, \(F_n(A)\to F(A)\) and \(F_n(B)\to F(B)\). Consequently, $$ P(X_n\leq A)+P(X_n>B)=F_n(A)+1-F_n(B) \longrightarrow P(X\leq A)+P(X>B)<\varepsilon. $$ The probabilities outside \((A,B]\) are therefore small for \(X\) and, for all sufficiently large \(n\), for \(X_n\) as well.
The function \(g\) is uniformly continuous on the compact interval \([A,B]\). Choose a partition \(A=t_0<t_1<\cdots<t_m=B\), with all \(t_j\) continuity points of \(F\), fine enough that the oscillation of \(g\) on each subinterval is less than \(\varepsilon\). This is possible because continuity points are dense. Choose \(\xi_j\in[t_{j-1},t_j]\) and define \(s(x)=g(\xi_j)\) for \(t_{j-1}<x\leq t_j\), with \(s(x)=0\) outside \((A,B]\).
On \((A,B]\), \(|g(x)-s(x)|<\varepsilon\); outside this interval, \(|g(x)-s(x)|\leq M\). Hence $$ |E[g(X_n)]-E[s(X_n)]| \leq \varepsilon+M P(X_n\leq A\text{ or }X_n>B), $$ and the same bound holds with \(X\) in place of \(X_n\). Meanwhile, the step-function expectation is a finite sum: $$ E[s(X_n)]=\sum_{j=1}^{m}g(\xi_j)\{F_n(t_j)-F_n(t_{j-1})\}. $$ Every \(t_j\) is a continuity point, so this sum converges to the corresponding expression for \(X\), namely \(E[s(X)]\). The two approximation bounds and the small tail probabilities imply that \(|E[g(X_n)]-E[g(X)]|\) can be made arbitrarily small. Thus \(E[g(X_n)]\to E[g(X)]\).
Second direction. Suppose \(E[g(X_n)]\to E[g(X)]\) for every bounded continuous \(g\). Fix a continuity point \(x\) of \(F\). For each \(\delta>0\), define two bounded continuous functions. Let \(g^-_\delta\) equal 1 on \((-\infty,x-\delta]\), decrease linearly from 1 to 0 on \([x-\delta,x]\), and equal 0 on \([x,\infty)\). Let \(g^+_\delta\) equal 1 on \((-\infty,x]\), decrease linearly from 1 to 0 on \([x,x+\delta]\), and equal 0 on \([x+\delta,\infty)\). They satisfy, for every real \(y\), $$ g^-_\delta(y)\leq \mathbf{1}_{(-\infty,x]}(y)\leq g^+_\delta(y). $$ Taking expectations at \(X_n\) gives $$ E[g^-_\delta(X_n)]\leq F_n(x)\leq E[g^+_\delta(X_n)]. $$ By the assumed test-function convergence, the left and right sides converge respectively to \(E[g^-_\delta(X)]\) and \(E[g^+_\delta(X)]\). As \(\delta\downarrow0\), \(g^-_\delta(y)\) converges to \(\mathbf{1}_{(-\infty,x)}(y)\), while \(g^+_\delta(y)\) converges to \(\mathbf{1}_{(-\infty,x]}(y)\). Both functions are bounded by 1, so the bounded convergence theorem applies. Since \(x\) is a continuity point, \(P(X=x)=0\), and both limiting expectations equal \(F(x)\). The inequalities therefore imply $$ \liminf_{n\to\infty}F_n(x)\geq F(x) \quad\text{and}\quad \limsup_{n\to\infty}F_n(x)\leq F(x). $$ Thus \(F_n(x)\to F(x)\) at every continuity point of \(F\), proving \(X_n\xrightarrow{d}X\). \(\square\)
What the Criterion Does—and Does Not—Control
A bounded continuous test function averages the values of a random variable after passing them through a continuous, bounded transformation. The theorem says that all these transformed expectations converge precisely when the distributions converge. In applications, a carefully chosen test function can make a convergence argument short: the grid example reduces to a Riemann sum, and the rare-mass example reduces to a bound proportional to \(1/n\).
The restriction to bounded functions is essential. In the rare-mass example, the identity function detects a contribution from a very small probability at a very large value. Distributional convergence alone does not prevent that contribution. To infer convergence of expectations for an unbounded function, additional assumptions are needed; the bounded-continuous criterion makes no such claim.
Check Your Understanding
Use the test-function criterion and its proof to answer the following questions.
- For \(X_n\) uniform on \(\{-1/n,1/n\}\), what property of \(g\) ensures that \(E[g(X_n)]\to g(0)\)?
- In the rare-mass example, calculate \(E[X_n]\). Why does this not contradict convergence in distribution to zero?
- Why are continuity points of the limiting distribution function used as partition endpoints in the first direction of the proof?
- What happens to the two continuous bounds on the half-line indicator as \(\delta\downarrow0\), and why do their expectations have the same limit at a continuity point?
- For the discrete grid example, identify the Riemann sum and its limiting integral for a bounded continuous test function \(g\).