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Finite-dimensional stochastic-order calculus and tightness

Statement

Let all random vectors with index $n$ be defined on a common probability space for that $n$ ; the spaces may vary with $n$ . In any fixed finite-dimensional Euclidean space, the following rules hold:

If $U_{n} = o_{P} (1)$ and $V_{n} = o_{P} (1)$ , then $U_{n} + V_{n} = o_{P} (1)$ . If $U_{n} = O_{P} (1)$ and $V_{n} = O_{P} (1)$ , then $U_{n} + V_{n} = O_{P} (1)$ .
For positive integers $p, q, r$ , let $B : R^{p} \times R^{q} \to R^{r}$ be a fixed map for which there is a constant $C \geq 0$ with $∥ B (u, v) ∥_{2} \leq C ∥ u ∥_{2} ∥ v ∥_{2}$ for all $u, v$ . Assume in each application that $B (U_{n}, V_{n})$ is a random vector on the same probability space as $U_{n}, V_{n}$ , for every $n$ . If $U_{n} = o_{P} (1)$ and $V_{n} = O_{P} (1)$ , or conversely, then $B (U_{n}, V_{n}) = o_{P} (1)$ . If both are $O_{P} (1)$ , then $B (U_{n}, V_{n}) = O_{P} (1)$ .
If $X_{n} \Rightarrow X$ in $R^{d}$ , then $X_{n} = O_{P} (1)$ .

The last conclusion is finite-dimensional boundedness in probability. No independence or Axiom of Choice is assumed.

Facts & Assumptions

Given: Random vectors on probability spaces as in the statement; in part 3, $X_{n} \Rightarrow X$ .

[F1]

Stochastic order and convergence notation for finite-dimensional random vectors: $X_{n} = o_{P} (1)$ means $P_{n} (∥ X_{n} ∥_{2} > ε) \to 0$ for every $ε > 0$ ; $X_{n} = O_{P} (1)$ means that for every $η > 0$ there is $M > 0$ with ${lim sup}_{n} P_{n} (∥ X_{n} ∥_{2} > M) \leq η$ .

[F2]

Convergence in distribution of random elements: Random elements $X_{n}, X$ with values in the same metric space converge in distribution if their laws converge weakly. They may be defined on different probability spaces.

[F3]

Portmanteau theorem: For Borel probabilities $μ_{n}, μ$ on a metric space $S$ , the closed-set bound is ${lim sup}_{n} μ_{n} (F) \leq μ (F)$ for every closed $F$ .

[F4]

Continuity from above when one set has finite measure: For a decreasing sequence of measurable sets, if one set has finite measure then the measure of the intersection is the infimum of the measures.

[F5]

Euclidean spheres and closed balls as subspaces of $R^{n}$ : On $R^{d}$ , $d_{2} (x, y) = ∥ x - y ∥_{2}$ is the Euclidean metric, and every vector has finite Euclidean norm.

[F6]

Open ball, closed ball and sphere in a metric space: The open ball is $B (c, r) = {x : d (c, x) < r}$ for radius $r > 0$ .

[F7]

Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed: Every open metric ball is open.

[F8]

The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement: A set is closed when its complement is open.

Proof

Given: The random vectors and hypotheses in the statement.

1.1F1F2F3F4F5F6F7F8givenchoose

Suppose $X_{n} \Rightarrow X$ in $R^{d}$ . For positive integers $m$ , set $F_{m} = R^{d} ∖ B (0, m) = {x : ∥ x ∥_{2} \geq m}$ , where [F5] identifies the Euclidean distance with the norm and [F6] gives the metric-ball definition. By [F7] each ball is open, so [F8] makes $F_{m}$ closed. The sets decrease to the empty set because every vector has finite norm, and their probabilities are at most one. By [F4], $P_{X} (F_{m}) \to 0$ . For each $m$ , [F2] and [F3] give ${lim sup}_{n} P_{n} (∥ X_{n} ∥_{2} \geq m) \leq P_{X} (F_{m})$ . Choosing $m$ with right side below any prescribed $η > 0$ proves $X_{n} = O_{P} (1)$ by [F1].

1.2F1givenalgebra

If $U_{n} = o_{P} (1)$ and $V_{n} = o_{P} (1)$ in one Euclidean space, the triangle inequality gives ${∥ U_{n} + V_{n} ∥_{2} > ε} \subseteq {∥ U_{n} ∥_{2} > ε / 2} \cup {∥ V_{n} ∥_{2} > ε / 2}$ . The union bound and [F1] show that the left event has probability tending to zero.

1.3F1givenchoosealgebra

If $U_{n} = O_{P} (1)$ and $V_{n} = O_{P} (1)$ , choose $M_{U}, M_{V} > 0$ using [F1]. The event ${∥ U_{n} + V_{n} ∥_{2} > M_{U} + M_{V}}$ is contained in ${∥ U_{n} ∥_{2} > M_{U}} \cup {∥ V_{n} ∥_{2} > M_{V}}$ , so its limiting upper probability is at most the sum of the two chosen tail bounds. Since the bounds can be arbitrarily small, $U_{n} + V_{n} = O_{P} (1)$ .

1.4F1givenalgebra

If $C = 0$ , then $B (U_{n}, V_{n}) = 0$ for all $n$ . Suppose $C > 0$ , $U_{n} = o_{P} (1)$ , and $V_{n} = O_{P} (1)$ . For any $η > 0$ , choose $M > 0$ with ${lim sup}_{n} P_{n} (∥ V_{n} ∥_{2} > M) \leq η$ . The norm bound for $B$ gives ${∥ B (U_{n}, V_{n}) ∥_{2} > ε} \subseteq {∥ V_{n} ∥_{2} > M} \cup {∥ U_{n} ∥_{2} > ε / (C M)}$ . By [F1], the second event has probability tending to zero, so the limiting upper probability of the left event is at most $η$ ; arbitrariness of $η$ proves $B (U_{n}, V_{n}) = o_{P} (1)$ . The same argument applies when the small factor is in the second coordinate.

2.1F1givenchoosealgebra∎

If both inputs to $B$ are $O_{P} (1)$ , choose $M_{U}, M_{V} > 0$ so that the limiting upper probability of either input exceeding its bound is at most an arbitrary $η > 0$ . Outside the union of these two tail events, $∥ B (U_{n}, V_{n}) ∥_{2} \leq C M_{U} M_{V}$ . The union bound makes the limiting upper probability above this fixed bound at most $2 η$ . Letting $η$ decrease to zero proves $B (U_{n}, V_{n}) = O_{P} (1)$ .

Remarks

The composite random-vector hypothesis in part 2 is necessary. A norm bound on an arbitrary map does not establish measurability of its composites. For an explicit counterexample to the original claim without that premise, suppose a non-Borel set $A \subset (0, 1)$ is supplied, and let $Z (ω) = ω$ on the supplied uniform probability space $(0, 1)$ with its Borel sigma-algebra. Set $B (u, v) = ∣ u ∣ ∣ v ∣ 1_{A} (v)$ on $R^{2}$ , where $1_{A}$ is zero outside $A$ . This map satisfies $∣ B (u, v) ∣ \leq ∣ u ∣ ∣ v ∣$ for every pair. Taking $U_{n} \equiv 1$ and $V_{n} = Z$ gives bounded-in-probability inputs, but ${B (U_{n}, V_{n}) > 0} = A$ is not a measurable event, so the output is not a random variable and has no $O_{P} (1)$ classification under the definition. Taking instead $U_{n} \equiv 1 / n$ gives an $o_{P} (1)$ input and the same nonmeasurable positive-set inverse image, despite the pointwise bound by $1 / n$ . This argument assumes the stated non-Borel set and uniform Borel probability space; it does not supply an existence proof for that set. Requiring only the actually used composites to be random vectors retains arbitrary norm-bounded maps, without imposing Borel measurability, bilinearity, or continuity on $B$ itself.

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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 pp. 1–3 (standard reference, not scraped)