Coordinate Projections and Vector-Valued Weak Convergence

From a Joint Limit to Marginal and Transformed Limits

A random vector records several coordinates together, and its law is a probability measure on a Euclidean space. Once the vector laws converge weakly, continuous functions of those vectors inherit a weak limit. Coordinate projections are an especially useful case: selecting one or several coordinates is a continuous map, so convergence of the vector laws implies convergence of the corresponding marginal laws.

The central method is to test the transformed vectors with bounded continuous functions. Composing such a test function with a continuous transformation gives another bounded continuous test function on the original space. This simple observation is the mechanism behind both marginal convergence and convergence after a continuous vector-valued transformation.

Weak Convergence of Vector Laws

Definition: Let $\mu_n$ and $\mu$ be probability measures on $\mathbb{R}^d$. The measures $\mu_n$ converge weakly to $\mu$ if, for every bounded continuous function $f:\mathbb{R}^d\to\mathbb{R}$, the integrals of $f$ with respect to $\mu_n$ converge to the integral with respect to $\mu$. In symbols, $$ \mu_n \Rightarrow \mu \quad\Longleftrightarrow\quad \int_{\mathbb{R}^d} f\,d\mu_n \longrightarrow \int_{\mathbb{R}^d} f\,d\mu \quad\text{for every bounded continuous } f:\mathbb{R}^d\to\mathbb{R}. $$

For a random vector $X$ taking values in $\mathbb{R}^d$, write $\mathcal{L}(X)$ for its law. We write $X_n\Rightarrow X$ when $\mathcal{L}(X_n)\Rightarrow\mathcal{L}(X)$. The random vectors need not be defined on the same probability space: weak convergence is a statement about their laws.

A map $g:\mathbb{R}^d\to\mathbb{R}^k$ is continuous if it is continuous at every point of its domain. For a random vector $X$, the law of $g(X)$ is the image of the law of $X$ under $g$: for a Borel set $B\subseteq\mathbb{R}^k$, its probability is $\mathbb{P}(g(X)\in B)$. The next result explains why continuous transformations preserve weak limits.

The Continuous Mapping Principle for Vector-Valued Maps

Theorem: Suppose $X_n\Rightarrow X$ in $\mathbb{R}^d$, and $g:\mathbb{R}^d\to\mathbb{R}^k$ is continuous. Then $g(X_n)\Rightarrow g(X)$ in $\mathbb{R}^k$.

Proof. Let $h:\mathbb{R}^k\to\mathbb{R}$ be any bounded continuous function. Since $g$ and $h$ are continuous, their composition $h\circ g:\mathbb{R}^d\to\mathbb{R}$ is continuous. It is bounded because $h$ is bounded. The definition of $X_n\Rightarrow X$, applied to $h\circ g$, gives $$ \mathbb{E}[h(g(X_n))] = \mathbb{E}[(h\circ g)(X_n)] \longrightarrow \mathbb{E}[(h\circ g)(X)] = \mathbb{E}[h(g(X))]. $$ This holds for every bounded continuous $h$ on $\mathbb{R}^k$, which is precisely $g(X_n)\Rightarrow g(X)$. $\square$

The argument uses the full vector-valued definition of weak convergence: the test function $h$ may depend on all $k$ output coordinates at once. It does not require a separate limit calculation for each output coordinate.

Coordinate Projections Give Marginal Limits

For $1\leq j\leq d$, define the $j$-th coordinate projection by $\pi_j(x_1,\ldots,x_d)=x_j$. It is continuous, so the theorem immediately gives convergence of each coordinate marginal. More generally, one can retain any chosen group of coordinates.

Corollary: Suppose $X_n\Rightarrow X$ in $\mathbb{R}^d$. For indices $1\leq i_1<\cdots<i_r\leq d$, define $$ \pi_I(x_1,\ldots,x_d)=(x_{i_1},\ldots,x_{i_r}), \qquad I=\{i_1,\ldots,i_r\}. $$ Then $\pi_I(X_n)\Rightarrow\pi_I(X)$ in $\mathbb{R}^r$. In particular, each coordinate of $X_n$ converges weakly to the corresponding coordinate of $X$.

Proof. Each component of $\pi_I$ is a coordinate function and is continuous; hence $\pi_I:\mathbb{R}^d\to\mathbb{R}^r$ is continuous. Apply the theorem with $g=\pi_I$. $\square$

This result concerns projecting a vector whose joint law already has a weak limit. It identifies the limits of its coordinate marginals, and even of any selected subvector, by applying a continuous map to that joint limit.

Worked Example: Selecting Coordinates from a Convergent Vector

Let $U$ be uniformly distributed on $[0,1]$, and define $X_n=(U,U+1/n,2U)$ and $X=(U,U,2U)$. For every outcome, $X_n\to X$ in $\mathbb{R}^3$. If $f:\mathbb{R}^3\to\mathbb{R}$ is bounded and continuous, then $f(X_n)\to f(X)$ pointwise and $|f(X_n)|\leq \sup|f|$. The dominated convergence theorem therefore gives $\mathbb{E}[f(X_n)]\to\mathbb{E}[f(X)]$, so $X_n\Rightarrow X$.

Projecting onto the first and third coordinates gives $(X_{n,1},X_{n,3})=(U,2U)$ for every $n$, and the limiting projection is also $(U,2U)$. Thus the selected-coordinate laws converge (in fact, they are identical for every $n$). The third marginal is the law of $2U$, which is uniform on $[0,2]$: for $0\leq t\leq 2$, $\mathbb{P}(2U\leq t)=\mathbb{P}(U\leq t/2)=t/2$. The second coordinate, $U+1/n$, likewise converges weakly to $U$, as follows either from the projection result or directly by bounded convergence applied to any bounded continuous test function.

Continuous Transformations Can Produce New Vector Limits

Coordinate selection is only one use of the theorem. A continuous transformation may combine coordinates, apply nonlinear formulas, or change the dimension. The output is a vector-valued random variable, and its weak limit is obtained by applying the same map to the limiting vector.

Worked Example: A Nonlinear Transformation of Two Coordinates

Let $U$ and $V$ be independent uniform random variables on $[0,1]$, and put $X_n=(U,V+1/n)$ and $X=(U,V)$. For every outcome, $X_n\to X$. For each bounded continuous $f:\mathbb{R}^2\to\mathbb{R}$, continuity gives $f(X_n)\to f(X)$, and bounded convergence gives convergence of expectations. Hence $X_n\Rightarrow X$.

Consider the continuous map $g:\mathbb{R}^2\to\mathbb{R}^3$ given by $g(x,y)=(x+y,x-y,xy)$. Applying it explicitly, $$ g(X_n) = \left(U+V+\frac1n,\ U-V-\frac1n,\ UV+\frac{U}{n}\right), \qquad g(X)=(U+V,\ U-V,\ UV). $$ The map is continuous because each component is a polynomial. The theorem therefore gives $(U+V+1/n,\ U-V-1/n,\ UV+U/n)\Rightarrow(U+V,\ U-V,\ UV)$. This is convergence of the entire three-dimensional output vector, tested against every bounded continuous function on $\mathbb{R}^3$.

Affine Maps and Matrix Transformations

A common continuous transformation is affine: $g(x)=Ax+b$, where $A$ is a fixed matrix and $b$ is a fixed vector of the appropriate dimension. Every affine map between finite-dimensional Euclidean spaces is continuous. Thus if $X_n\Rightarrow X$, then $AX_n+b\Rightarrow AX+b$. This statement covers linear combinations, coordinate rearrangements, and fixed shifts of vector-valued limits.

Worked Example: Transforming a Gaussian Vector

Let $Z$ and $W$ be independent standard normal random variables, and define $X_n=(Z+1/n,W-2/n)$ and $X=(Z,W)$. The vectors converge pointwise, so bounded convergence applied to bounded continuous tests shows $X_n\Rightarrow X$. Use the linear map $g(x,y)=(x+y,x-y)$. Then $$ g(X_n)=(Z+W-1/n,\ Z-W+3/n), \qquad g(X)=(Z+W,\ Z-W). $$ The continuous mapping theorem gives convergence of these transformed vectors.

The limit is a centered bivariate normal vector. Each component has variance $\operatorname{Var}(Z+W)=\operatorname{Var}(Z)+\operatorname{Var}(W)=2$ and $\operatorname{Var}(Z-W)=\operatorname{Var}(Z)+\operatorname{Var}(W)=2$, using independence. Their covariance is $\operatorname{Cov}(Z+W,Z-W)=\operatorname{Var}(Z)-\operatorname{Var}(W)=1-1=0$. Because the vector is jointly normal, its zero covariance implies that its two components are independent. Thus the transformed limit has independent $N(0,2)$ coordinates. The theorem gives the vector limit directly; identifying its distribution uses the usual facts about linear transformations of Gaussian vectors.

What the Test-Function Argument Does—and Does Not—Require

For a proposed transformation, the essential check is continuity on the domain where the limiting vector takes values. When the map is continuous everywhere on $\mathbb{R}^d$, no extra condition on the distribution of $X$ is needed. The proof then consists of composing a bounded continuous output test with the map and invoking the definition of weak convergence.

The output dimension need not equal the input dimension. A projection can reduce dimension, while a continuous map can increase it. In either case, the limiting law is the law of the transformed limit. One should distinguish this direction of reasoning from trying to reconstruct a joint limit using only separate coordinate information: here the starting assumption is convergence of the vector laws themselves.

Key takeaway: If $X_n\Rightarrow X$ in $\mathbb{R}^d$, then every continuous map $g:\mathbb{R}^d\to\mathbb{R}^k$ satisfies $g(X_n)\Rightarrow g(X)$. Coordinate marginals are the special case in which $g$ selects coordinates.

Check Your Understanding

Answer each question by using the bounded-continuous-test definition or the continuous mapping theorem.

Why is the coordinate projection $\pi_j:\mathbb{R}^d\to\mathbb{R}$ continuous?
In the proof of the theorem, why is $h\circ g$ a valid test function for the convergence $X_n\Rightarrow X$?
If $X_n\Rightarrow X$ in $\mathbb{R}^d$, what is the weak limit of $AX_n+b$ for a fixed matrix $A$ and vector $b$?
For $g(x,y)=(x+y,xy)$, state the limit of $g(X_n)$ when $X_n\Rightarrow X$ in $\mathbb{R}^2$.
What assumption about the vector laws is the starting point for deducing convergence of coordinate marginals?