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Weak Convergence on Euclidean Spaces · Tutorial 173 of 946

Why Marginal Convergence Does Not Give Joint Convergence

See why converging coordinate distributions do not, in general, determine the behavior of the joint distribution.

Intermediate 10 min read

What You'll Learn

  • Define marginal and joint weak convergence for pairs of random variables.
  • Prove that joint weak convergence implies weak convergence of each marginal.
  • Construct discrete and continuous examples with convergent marginals but nonconvergent joint laws.
  • Identify dependence information that marginal distributions leave unspecified.
  • Recognize the special case where both marginal limits are constants.

Why Marginals Are Not the Whole Joint Law

A random vector contains more information than the distributions of its individual coordinates. The marginals describe how each coordinate behaves separately; the joint distribution also describes how the coordinates behave together.

This distinction matters for weak convergence. If a sequence of random vectors converges weakly, then each coordinate converges weakly. The reverse implication is false in general: coordinate distributions can settle down while dependence keeps changing.

Definition: Let \((X_n,Y_n)\) be random vectors in \(\mathbb{R}^2\). Their joint laws converge weakly to a probability measure \(\mu\) if, for every bounded continuous function \(f:\mathbb{R}^2\to\mathbb{R}\), the expectations \(\mathbb{E}[f(X_n,Y_n)]\) converge to \(\int f\,d\mu\). The marginal laws are the laws of \(X_n\) and \(Y_n\) separately. Marginal weak convergence means that each of these one-dimensional laws converges weakly.

A bounded continuous test of the joint vector can detect a relationship between its coordinates. Tests of the marginals alone cannot, in general, detect whether the coordinates tend to move together, move oppositely, or become independent.

Joint Convergence Implies Marginal Convergence

The forward implication follows from coordinate projections. A projection is a continuous map from \(\mathbb{R}^2\) to \(\mathbb{R}\), so the continuous mapping theorem gives the same conclusion. Here is a direct proof using bounded continuous tests.

Theorem: If \((X_n,Y_n)\) converges weakly to a random vector \((X,Y)\) in \(\mathbb{R}^2\), then \(X_n\) converges weakly to \(X\), and \(Y_n\) converges weakly to \(Y\).
Proof. Let \(h:\mathbb{R}\to\mathbb{R}\) be any bounded continuous function. The function \(f(x,y)=h(x)\) is bounded and continuous on \(\mathbb{R}^2\). Joint weak convergence therefore gives \[ \mathbb{E}[h(X_n)] = \mathbb{E}[f(X_n,Y_n)] \longrightarrow \mathbb{E}[f(X,Y)] = \mathbb{E}[h(X)]. \] This holds for every bounded continuous \(h\), which is the test-function criterion for \(X_n\) to converge weakly to \(X\). Applying the same argument to \(f(x,y)=h(y)\) proves weak convergence of \(Y_n\) to \(Y\). \(\square\)

The proof uses only tests that look at one coordinate at a time. To establish the converse, one would need to control tests that look at both coordinates simultaneously. Marginal convergence supplies no such control by itself.

Worked Example: Bernoulli Marginals, Alternating Dependence

Worked Example: Bernoulli Marginals, Alternating Dependence

Let \(B\) have the Bernoulli distribution with probabilities \(1/2\) at both \(0\) and \(1\). Define \[ (X_n,Y_n)= \begin{cases} (B,B),& n\text{ odd},\\ (B,1-B),& n\text{ even}. \end{cases} \] For every \(n\), each coordinate has the same Bernoulli distribution. In particular, both marginal sequences converge weakly, and their limits are unchanged as \(n\) varies.

Now use the bounded continuous joint test \[ f(x,y)=g(x)g(y), \qquad g(t)=\frac{t^2}{1+t^2}. \] The function \(g\) is continuous and takes values in \([0,1)\), so \(f\) is a valid bounded continuous test on \(\mathbb{R}^2\). At the two possible Bernoulli values, \(g(0)=0\) and \(g(1)=1/2\).

For odd \(n\), the pair equals \((1,1)\) with probability \(1/2\) and \((0,0)\) with probability \(1/2\). Thus \[ \mathbb{E}[f(X_n,Y_n)] = \frac12\left(\frac12\cdot\frac12\right)+\frac12(0\cdot0) = \frac18. \] For even \(n\), the pair is either \((0,1)\) or \((1,0)\), and \(f\) vanishes at both points. Hence \(\mathbb{E}[f(X_n,Y_n)]=0\). These expectations alternate between \(1/8\) and \(0\), so they do not converge. The joint laws therefore do not converge weakly, despite the fixed marginal laws.

Worked Example: Continuous Marginals, Alternating Dependence

Worked Example: Continuous Marginals, Alternating Dependence

Let \(Z\) be standard normal, and set \[ (X_n,Y_n)= \begin{cases} (Z,Z),& n\text{ odd},\\ (Z,-Z),& n\text{ even}. \end{cases} \] The first coordinate is standard normal for every \(n\). The second is also standard normal for every \(n\), because the standard normal distribution is symmetric about zero. Thus both marginal sequences converge, and both limits are standard normal.

Consider the bounded continuous function \(f(x,y)=e^{-(x-y)^2}\). For odd \(n\), \(X_n-Y_n=0\), so \[ \mathbb{E}[f(X_n,Y_n)]=1. \] For even \(n\), \(X_n-Y_n=2Z\), and the standard normal density gives \[ \mathbb{E}[f(X_n,Y_n)] = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} e^{-4z^2}e^{-z^2/2}\,dz = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-9z^2/2}\,dz = \frac13. \] The last equality follows by rescaling the standard Gaussian integral. Since these test expectations alternate between \(1\) and \(1/3\), the joint laws do not converge weakly. Even smooth, fixed marginals do not prevent dependence from changing with \(n\).

Marginals Do Not Specify a Unique Joint Limit

The failure is not limited to sequences whose joint laws oscillate. Two different joint laws can have exactly the same marginals. The next example compares dependence that is perfect with dependence that is absent.

Worked Example: The Same Uniform Marginals, Different Joint Laws

Let \(U\) and \(V\) be independent random variables, each uniformly distributed on \([0,1]\). Compare the random vectors \((U,U)\) and \((U,V)\). In either vector, the first and second coordinates each have the uniform distribution on \([0,1]\).

Again test with \(f(x,y)=e^{-(x-y)^2}\), which is bounded and continuous on \(\mathbb{R}^2\). For the vector \((U,U)\), \(f(U,U)=1\) always, so its expectation is \(1\). For \((U,V)\), the event \(U\ne V\) has probability one: the diagonal \(\{(u,v):u=v\}\) has two-dimensional Lebesgue measure zero in the unit square. Consequently, \(e^{-(U-V)^2}<1\) with probability one, and therefore \[ \mathbb{E}[e^{-(U-V)^2}]<1. \] The two joint laws are distinct even though their marginals agree exactly. The difference is the dependence structure, not either coordinate distribution.

A Special Case: Both Marginal Limits Are Constants

There is an important exception to the general failure. If both marginal limits are point masses, the joint limit is forced: the random vector must concentrate near the corresponding constant pair. This does not require independence.

Theorem: Suppose \(X_n\) converges weakly to the point mass at \(a\in\mathbb{R}\), and \(Y_n\) converges weakly to the point mass at \(b\in\mathbb{R}\). Then \((X_n,Y_n)\) converges weakly to the point mass at \((a,b)\).
Proof. Weak convergence to a constant implies convergence in probability to that constant. In particular, for each \(\delta>0\), \[ \mathbb{P}(|X_n-a|\geq\delta)\longrightarrow 0, \qquad \mathbb{P}(|Y_n-b|\geq\delta)\longrightarrow 0. \] By the union bound, \[ \mathbb{P}\!\left(\max\{|X_n-a|,|Y_n-b|\}\geq\delta\right) \leq \mathbb{P}(|X_n-a|\geq\delta)+\mathbb{P}(|Y_n-b|\geq\delta) \longrightarrow 0. \] Thus \((X_n,Y_n)\) converges in probability to \((a,b)\), regardless of any dependence between \(X_n\) and \(Y_n\). To verify weak convergence directly, let \(f:\mathbb{R}^2\to\mathbb{R}\) be bounded and continuous, and write \(M=\sup_{(x,y)}|f(x,y)|<\infty\). Given \(\eta>0\), continuity at \((a,b)\) provides a \(\delta>0\) such that \(\max\{|x-a|,|y-b|\}<\delta\) implies \(|f(x,y)-f(a,b)|<\eta\). Therefore \[ \left|\mathbb{E}[f(X_n,Y_n)]-f(a,b)\right| \leq \eta+2M\,\mathbb{P}\!\left(\max\{|X_n-a|,|Y_n-b|\}\geq\delta\right). \] The probability term tends to zero, so the limit superior of the left side is at most \(\eta\). Since \(\eta>0\) was arbitrary, the expectations converge to \(f(a,b)\), proving weak convergence to the point mass at \((a,b)\). \(\square\)

What Marginal Convergence Leaves Open

For nonconstant limits, separate convergence of the coordinates does not say whether their dependence converges, or what the limiting dependence should be. It does not establish joint weak convergence and does not identify a joint limit if one exists. In particular, knowing two marginal laws does not determine the probability of joint events such as \(\{X_n\leq x,\ Y_n\leq y\}\).

A practical way to test a proposed joint convergence argument is to ask what happens to bounded continuous functions of both coordinates. Coordinatewise tests of the forms \(h(x)\) and \(k(y)\) address the marginals. A joint test, such as \(e^{-(x-y)^2}\), can detect whether the coordinates tend to coincide or separate. If the argument has no information about such joint behavior, marginal convergence alone cannot fill the gap.

Takeaway: Joint weak convergence always implies marginal weak convergence. The converse generally fails because marginal laws omit dependence information. When both marginal limits are constants, however, concentration of each coordinate forces joint convergence to the constant pair.

Check Your Understanding

Use bounded continuous tests to explain what the marginal laws do and do not determine.

  1. Why does joint weak convergence imply weak convergence of the first coordinate?
  2. In the Bernoulli example, why do the marginal laws stay fixed while the joint laws fail to converge?
  3. For the normal example, what are the two alternating values of the test expectation, and why do they rule out joint weak convergence?
  4. Can two different joint laws have the same two marginals? Give the construction from the uniform example.
  5. Why does convergence of both coordinates to constants imply joint convergence even when the coordinates may be dependent?

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