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Vector Slutsky theorem

Statement

Let $d, k, m \geq 1$ be fixed. For each $n$ , let $X_{n} \in R^{d}$ and $Y_{n} \in R^{k}$ be random vectors on one probability space; the probability space may vary with $n$ . Let $X \in R^{d}$ be a random vector on any probability space, and let $c \in R^{k}$ be constant. If $X_{n} \Rightarrow X$ and $Y_{n} - c = o_{P} (1)$ , then $(X_{n}, Y_{n}) \Rightarrow (X, c) .$ Consequently, for every continuous $h : R^{d + k} \to R^{m}$ , $h (X_{n}, Y_{n}) \Rightarrow h (X, c)$ . No independence between $X_{n}$ and $Y_{n}$ is required.

Facts & Assumptions

Given: The random vectors and convergence hypotheses in the statement.

[F1]

Finite-dimensional stochastic-order calculus and tightness: If $X_{n} \Rightarrow X$ in a finite-dimensional Euclidean space, then $X_{n} = O_{P} (1)$ .

[F2]

Stochastic order and convergence notation for finite-dimensional random vectors: $Y_{n} - c = o_{P} (1)$ means $P_{n} (∥ Y_{n} - c ∥_{2} > ε) \to 0$ for each $ε > 0$ .

[F3]

Weak convergence of borel probability measures: For Borel probability measures $μ_{n}, μ$ on a metric space $S$ , weak convergence means convergence of integrals for every bounded continuous real-valued function.

[F4]

Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line: A subset of finite-dimensional Euclidean space is compact if and only if it is closed and bounded.

[F5]

Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous: A continuous map from a compact metric space to a metric space is uniformly continuous.

[F6]

Continuous mapping theorem: If $Z_{n} \Rightarrow Z$ and $g$ is measurable with a $Z$ -null discontinuity set, then $g (Z_{n}) \Rightarrow g (Z)$ .

[F7]

Convergence in distribution of random elements: Random elements converge in distribution when their laws converge weakly; they need not share a probability space.

[F8]

Euclidean spheres and closed balls as subspaces of $R^{n}$ : The Euclidean closed ball is the set of vectors whose Euclidean distance from its centre is at most its radius.

[F9]

$R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it: On $R^{N}$ , the Euclidean metric is $d_{2} (x, y) = \sqrt{\sum_{i < N} (x_{i} - y_{i})^{2}}$ .

[F10]

Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric: A metric satisfies symmetry and the triangle inequality.

[F11]

Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed: Closed balls are closed subsets of a metric space.

Proof

Given: The objects and hypotheses in the statement. Write $P_{n}$ for the probability measure carrying $(X_{n}, Y_{n})$ .

1.1givenF1

By [F1], $X_{n} = O_{P} (1)$ . Fix a bounded continuous $f : R^{d + k} \to R$ , and let $η > 0$ . The definition of $O_{P} (1)$ gives $R > 0$ with ${lim sup}_{n} P_{n} (∥ X_{n} ∥_{2} > R) \leq η$ .

1.2F2F4F5F8F9F10F11givenchoosealgebra

Let $K = {\overline{B}}_{2} (0, R) \times {\overline{B}}_{2} (c, 1) \subseteq R^{d + k}$ . The factor balls are closed by [F11]. The coordinate projections are continuous because [F9] gives $d_{2} (x, x^{'}) \leq d_{2} ((x, y), (x^{'}, y^{'}))$ and the analogous inequality for $y$ , so $K$ is closed. For $(x, y) \in K$ , [F9] and [F10] give $d_{2} ((x, y), (0, 0))^{2} = ∥ x ∥_{2}^{2} + ∥ y ∥_{2}^{2} \leq R^{2} + (∥ c ∥_{2} + 1)^{2}$ , so $K$ is bounded. By [F4], $K$ is compact. By [F5], $f ∣_{K}$ is uniformly continuous. For any $γ > 0$ , choose $0 < δ \leq 1$ such that points of $K$ at distance less than $δ$ have $f$ -values differing by less than $γ$ . By [F8], on ${∥ X_{n} ∥_{2} \leq R, ∥ Y_{n} - c ∥_{2} \leq δ / 2}$ both pair values lie in $K$ , and their Euclidean distance is at most $δ / 2$ . Therefore $∣ E_{n} f (X_{n}, Y_{n}) - E_{n} f (X_{n}, c) ∣ \leq γ + 2 ∥ f ∥_{\infty} (P_{n} (∥ X_{n} ∥_{2} > R) + P_{n} (∥ Y_{n} - c ∥_{2} > δ / 2)) .$ By [F2], the second tail tends to zero. Taking the limiting upper bound, then letting $γ$ and $η$ decrease to zero, proves that the expectation difference tends to zero.

2.1F3F7givenstep 1.2

By [F7], $X_{n} \Rightarrow X$ means weak convergence of its laws. The function $x \mapsto f (x, c)$ is bounded and continuous, so [F3] gives $E_{n} f (X_{n}, c) \to E f (X, c)$ . Together with step 1.2, bounded-continuous-test convergence gives $(X_{n}, Y_{n}) \Rightarrow (X, c)$ .

3.1F6step 2.1∎

The map $h$ is continuous everywhere, hence measurable with empty discontinuity set. Apply [F6] to step 2.1 to obtain $h (X_{n}, Y_{n}) \Rightarrow h (X, c)$ .

Depends on

Used by

Dependency tree · two levels

63 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Peter Bartlett, Theoretical Statistics, Stat 210B Lecture 16, PDF pp. 9–11 (standard reference, not scraped)
John Duchi, Theory of Statistics, Stats 300B Lecture 2, Winter 2018, PDF p. 1 (standard reference, not scraped)