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✓ 4 results · all verified · 3 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 1 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Consistency and Limiting Distributions

1 · Prerequisites

2 · Summary

This page fixes the stochastic-order notation OP(1) and oP(1) using the Euclidean norm for fixed finite-dimensional random vectors. It proves the finite-dimensional sum and bilinear-product rules, tightness of weakly convergent sequences, and vector Slutsky convergence with a constant probability limit. The vector result is proved locally; the scalar Slutsky theorem is not treated as a vector theorem.

For an IID statistical experiment, an estimator sequence is a measurable map on each finite product sample space. Consistency here is pointwise at a fixed parameter, and an asymptotic distribution is a weak limit of laws; neither term implies uniformity or moment convergence. If two parameter values induce the same observation law but have distinct target values, no estimator sequence can be consistent at both. For a real-valued observation with finite first absolute moment, Khinchin's weak law gives consistency of the sample mean at each fixed parameter. The law's integrability hypothesis is retained exactly.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Stochastic order and convergence notation for finite-dimensional random vectors

Definition

For each positive integer n, let (Ωn,An,Pn) be a probability space and let Xn:Ωn→Rd be measurable into the Borel sigma-algebra generated by the Euclidean open sets, where d≥1 is fixed and Rd has the Euclidean norm ∥⋅∥2 (Euclidean spheres and closed balls as subspaces of Rn, Random elements and real random variables, Probability measures and probability spaces). The spaces and measures may depend on n; no common probability space is required.

Write Xn=oP(1) if for every ε>0, Pn(∥Xn∥2>ε)⟶0. Write Xn=OP(1), or say that (Xn) is bounded in probability, if for every η>0 there is M>0 such that lim sup⁡n→∞Pn(∥Xn∥2>M)≤η. For real-valued variables on one common probability space, oP(1) is convergence in probability to zero as defined in Convergence in probability.

The norms and tail probabilities are part of the definition. Independence is not implicit in either notation.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Asymptotic distribution

Definition

Let d≥1 be fixed, and equip Rd with the Borel sigma-algebra generated by its Euclidean open sets. Let Zn be measurable random elements of this Borel space, not necessarily defined on a common probability space, and let ν be a Borel probability measure on Rd. The sequence (Zn) has asymptotic distribution ν if its laws converge weakly to ν; equivalently, Zn⇒Z for any random vector Z with law ν (Convergence in distribution of random elements, Weak convergence of borel probability measures).

For an estimator, any centering or scaling, such as rn(Tn−ψ(θ)), is part of the sequence whose asymptotic distribution is being specified; it is not implicit in the term. Weak convergence alone does not assert convergence of moments.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

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:

  1. If Un=oP(1) and Vn=oP(1), then Un+Vn=oP(1). If Un=OP(1) and Vn=OP(1), then Un+Vn=OP(1).
  2. For positive integers p,q,r, let B:Rp×Rq→Rr be a fixed map for which there is a constant C≥0 with ∥B(u,v)∥2≤C∥u∥2∥v∥2 for all u,v. Assume in each application that B(Un,Vn) is a random vector on the same probability space as Un,Vn, for every n. If Un=oP(1) and Vn=OP(1), or conversely, then B(Un,Vn)=oP(1). If both are OP(1), then B(Un,Vn)=OP(1).
  3. If Xn⇒X in Rd, then Xn=OP(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, Xn⇒X.

[F1]

Stochastic order and convergence notation for finite-dimensional random vectors: Xn=oP(1) means Pn(∥Xn∥2>ε)→0 for every ε>0; Xn=OP(1) means that for every η>0 there is M>0 with lim sup⁡nPn(∥Xn∥2>M)≤η.

[F2]

Convergence in distribution of random elements: Random elements Xn,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 μn,μ on a metric space S, the closed-set bound is lim sup⁡nμn(F)≤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 Rn: On Rd, d2(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.

Proof

Given: The random vectors and hypotheses in the statement.

1.1F1F2F3F4F5F6F7F8givenchoose

Suppose Xn⇒X in Rd. For positive integers m, set Fm=Rd∖B(0,m)={x:∥x∥2≥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 Fm closed. The sets decrease to the empty set because every vector has finite norm, and their probabilities are at most one. By [F4], PX(Fm)→0. For each m, [F2] and [F3] give lim sup⁡nPn(∥Xn∥2≥m)≤PX(Fm). Choosing m with right side below any prescribed η>0 proves Xn=OP(1) by [F1].

1.2F1givenalgebra

If Un=oP(1) and Vn=oP(1) in one Euclidean space, the triangle inequality gives {∥Un+Vn∥2>ε}⊆{∥Un∥2>ε/2}∪{∥Vn∥2>ε/2}. The union bound and [F1] show that the left event has probability tending to zero.

1.3F1givenchoosealgebra

If Un=OP(1) and Vn=OP(1), choose MU,MV>0 using [F1]. The event {∥Un+Vn∥2>MU+MV} is contained in {∥Un∥2>MU}∪{∥Vn∥2>MV}, so its limiting upper probability is at most the sum of the two chosen tail bounds. Since the bounds can be arbitrarily small, Un+Vn=OP(1).

1.4F1givenalgebra

If C=0, then B(Un,Vn)=0 for all n. Suppose C>0, Un=oP(1), and Vn=OP(1). For any η>0, choose M>0 with lim sup⁡nPn(∥Vn∥2>M)≤η. The norm bound for B gives {∥B(Un,Vn)∥2>ε}⊆{∥Vn∥2>M}∪{∥Un∥2>ε/(CM)}. 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(Un,Vn)=oP(1). The same argument applies when the small factor is in the second coordinate.

2.1F1givenchoosealgebra∎

If both inputs to B are OP(1), choose MU,MV>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(Un,Vn)∥2≤CMUMV. The union bound makes the limiting upper probability above this fixed bound at most 2η. Letting η decrease to zero proves B(Un,Vn)=OP(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⊂(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∣1A(v) on R2, where 1A is zero outside A. This map satisfies ∣B(u,v)∣≤∣u∣∣v∣ for every pair. Taking Un≡1 and Vn=Z gives bounded-in-probability inputs, but {B(Un,Vn)>0}=A is not a measurable event, so the output is not a random variable and has no OP(1) classification under the definition. Taking instead Un≡1/n gives an oP(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.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Vector Slutsky theorem

Statement

Let d,k,m≥1 be fixed. For each n, let Xn∈Rd and Yn∈Rk be random vectors on one probability space; the probability space may vary with n. Let X∈Rd be a random vector on any probability space, and let c∈Rk be constant. If Xn⇒X and Yn−c=oP(1), then (Xn,Yn)⇒(X,c). Consequently, for every continuous h:Rd+k→Rm, h(Xn,Yn)⇒h(X,c). No independence between Xn and Yn is required.

Facts & Assumptions

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

[F1]

Finite-dimensional stochastic-order calculus and tightness: If Xn⇒X in a finite-dimensional Euclidean space, then Xn=OP(1).

[F2]

Stochastic order and convergence notation for finite-dimensional random vectors: Yn−c=oP(1) means Pn(∥Yn−c∥2>ε)→0 for each ε>0.

[F3]

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

[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 Zn⇒Z and g is measurable with a Z-null discontinuity set, then g(Zn)⇒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 Rn: The Euclidean closed ball is the set of vectors whose Euclidean distance from its centre is at most its radius.

[F9]

Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it: On RN, the Euclidean metric is d2(x,y)=∑i<N(xi−yi)2.

Proof

Given: The objects and hypotheses in the statement. Write Pn for the probability measure carrying (Xn,Yn).

1.1givenF1

By [F1], Xn=OP(1). Fix a bounded continuous f:Rd+k→R, and let η>0. The definition of OP(1) gives R>0 with lim sup⁡nPn(∥Xn∥2>R)≤η.

1.2F2F4F5F8F9F10F11givenchoosealgebra

Let K=B‾2(0,R)×B‾2(c,1)⊆Rd+k. The factor balls are closed by [F11]. The coordinate projections are continuous because [F9] gives d2(x,x′)≤d2((x,y),(x′,y′)) and the analogous inequality for y, so K is closed. For (x,y)∈K, [F9] and [F10] give d2((x,y),(0,0))2=∥x∥22+∥y∥22≤R2+(∥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≤1 such that points of K at distance less than δ have f-values differing by less than γ. By [F8], on {∥Xn∥2≤R, ∥Yn−c∥2≤δ/2} both pair values lie in K, and their Euclidean distance is at most δ/2. Therefore ∣Enf(Xn,Yn)−Enf(Xn,c)∣≤γ+2∥f∥∞(Pn(∥Xn∥2>R)+Pn(∥Yn−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], Xn⇒X means weak convergence of its laws. The function x↦f(x,c) is bounded and continuous, so [F3] gives Enf(Xn,c)→Ef(X,c). Together with step 1.2, bounded-continuous-test convergence gives (Xn,Yn)⇒(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(Xn,Yn)⇒h(X,c).

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Estimator sequences in IID experiments

Definition

Let P:Θ→P(X,A) be a parametric model on a statistical experiment, and let ψ:Θ→T be an estimand whose target set T is equipped with a metric d (Statistical experiments, Parametric statistical models, Estimands on a statistical model, Metric space: d(x,y)=0 iff x=y, symmetry, and the triangle inequality; pseudometric and ultrametric). For n≥1, the IID sample-size-n experiment has sample space Xn, product sigma-algebra A⊗n, and sampling law Pθ⊗n at parameter θ (The product sigma-algebra and its finite iterates). The product law is defined recursively from Pθ; each factor is a probability measure and hence sigma-finite, so the finite product theorem supplies its existence, rectangle formula, and uniqueness (For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique).

A nonrandomized estimator sequence of ψ is a sequence of measurable maps, in the sense of A measurable function between measurable spaces, Tn:(Xn,A⊗n)→(T,B(T)),n≥1, where B(T) is the sigma-algebra generated by the open sets of the metric d, that is, the Borel sigma-algebra of its metric topology (The Borel sigma-algebra of a topological space). Its sampling distribution at θ and sample size n is the pushforward of Pθ⊗n by Tn.

This definition fixes IID repeated sampling and data-measurable estimators. It imposes no measurability on the parameter map θ↦Pθ or on ψ.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Pointwise consistency in probability

Definition

Let (Tn) be an estimator sequence of an estimand ψ with metric target (T,d), and fix θ∈Θ. The sequence is consistent in probability at θ if for every ε>0, Pθ⊗n({x∈Xn:d(Tn(x),ψ(θ))>ε})⟶0.

The error event is measurable: by the triangle inequality, t↦d(t,ψ(θ)) is continuous, so its strict superlevel set at ε is open in the target metric topology and hence Borel; its preimage under Tn is measurable.

The consistency is pointwise because θ is fixed before the limit. It does not assert uniform convergence over Θ (Estimator sequences in IID experiments).

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Consistency of the IID sample mean under a finite first moment

Statement

Let X=R with its Borel sigma-algebra, and let P be a parametric statistical experiment on X. Define the mean estimand by μ(θ)=∫x Pθ(dx) and suppose it is finite for every θ. For a fixed θ, suppose an IID sequence (Xi)i≥1 with one-observation law Pθ is given and Eθ∣X1∣<∞. The sample-mean estimator Tn(x1,…,xn)=n−1∑i=1nxi is consistent in probability for μ at θ in the IID sample-size experiments.

The assertion is pointwise in θ. The existence of a common probability space carrying the IID sequence is part of the hypothesis.

Facts & Assumptions

Given: The experiment, finite mean, and IID sequence in the statement.

[F1]

Identical distribution and IID families: A family is IID when it is identically distributed and the whole family is independent; no moment assumption is implicit.

[F2]

Partial sums, row sums and sample means: For real random variables X1,X2,…, Sn=∑i=1nXi and X‾n=Sn/n; these finite sums and means are random variables.

[F3]

Independent random elements have product joint law: A finite independent family of random elements has joint law equal to the finite product of its marginal laws.

[F4]

Khinchin weak law for integrable IID variables: If (Xi) are IID real random variables and E∣X1∣<∞, then Sn/n→EX1 in probability.

[F5]

Estimator sequences in IID experiments: The size-n sample space is Xn with law Pθ⊗n, and the estimator is a measurable map on that space.

[F6]

Pointwise consistency in probability: The estimator is consistent at θ when Pθ⊗n{∣Tn−μ(θ)∣>ε}→0 for every ε>0.

[F7]

Estimands on a statistical model: An estimand is a specified function from the parameter set to its target set.

[F8]

Expectation of a nonnegative or integrable random variable: For an integrable real random variable, expectation is its finite Lebesgue integral against the probability measure.

[F9]

Arithmetic and lattice operations preserve measurability whenever they are defined: Finite sums and scalar multiples of measurable real-valued functions are measurable whenever the operations are pointwise defined.

[F10]

Change of variables for expectation: If a real random variable X has law Pθ, g:R→R is Borel measurable, and g(X) is integrable, then Eθ[g(X)]=∫g(x) Pθ(dx).

[F11]

Statistical experiments: An experiment supplies the observation measurable space and its parameter-indexed family of sampling laws.

[F12]

Parametric statistical models: The parametric model is the map θ↦Pθ on that experiment.

[F13]

The Borel sigma-algebra of a topological space: The Borel sigma-algebra on R is generated by its open subsets.

Proof

Given: The objects and hypotheses in the statement.

1.1F1F2F3F5F7F8F9F10F11F12F13given

The statistical experiment and parametric model in [F11]–[F12] specify the sampling law Pθ. Each coordinate projection on Rn is measurable because the preimage of a Borel set B⊆R is the cylinder rectangle Ri−1×B×Rn−i in the product sigma-algebra [F5]. Thus [F9] makes Tn(x1,…,xn)=n−1∑i=1nxi measurable into the Borel target [F13], so it is the estimator in [F5]; applied to the given sequence, it is the finite sample mean in [F2]. By [F1] and [F3], the law of (X1,…,Xn) is Pθ⊗n. Since X1 has law Pθ and is integrable, [F10] gives EθX1=∫x Pθ(dx)=μ(θ); the mean is the estimand [F7].

1.2F1F4given

The hypotheses of [F4] hold for this IID sequence, so X‾n→EθX1=μ(θ) in probability on its given common probability space.

2.1F5F6F14step 1.1step 1.2∎

For each n, step 1.1 identifies the joint data law with Pθ⊗n; applying the same measurable sample-mean map therefore gives exactly the error probabilities in [F6] for the absolute-value metric [F14]. Step 1.2 makes those probabilities tend to zero for every ε>0, which is consistency at θ. No infinite-product extension is constructed or needed.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Nonidentifiability precludes simultaneous consistency

Statement

Let P:Θ→P(X,A) be a parametric statistical model, let ψ:Θ→(T,d) be a metric-valued estimand, and let (Tn) be a nonrandomized estimator sequence for IID samples from this model. If ψ is not identifiable, then there is a pair of observationally equivalent parameters with distinct target values, and no estimator sequence is consistent in probability at both parameters.

Facts & Assumptions

Given: The model, estimand, estimator sequence, and hypotheses in the statement.

[F1]

Parametric statistical models: A parametric statistical model is the parameter map P:Θ→P(X,A) with P(θ)=Pθ.

[F2]

Observational equivalence: θ∼η exactly when Pθ=Pη as measures on the observation space.

[F3]

Estimands on a statistical model: An estimand is a specified function ψ:Θ→T.

[F4]

Identifiability of an estimand: An estimand is identifiable exactly when θ∼η implies ψ(θ)=ψ(η).

[F5]

Estimator sequences in IID experiments: The size-n experiment has law Pθ⊗n on (Xn,A⊗n), and an estimator is a measurable map Tn on that space.

[F6]

Pointwise consistency in probability: Consistency at θ means Pθ⊗n{d(Tn,ψ(θ))>ε}→0 for every ε>0.

[F7]

The product sigma-algebra and its finite iterates: The finite product sigma-algebra is generated recursively by measurable rectangles.

[F8]

For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique: The product measure exists and gives each measurable rectangle the product of the factor measures.

[F9]

Measures agreeing on a generating pi-system are equal under an increasing finite-measure exhaustion from that pi-system: Measures agreeing on a generating π-system and on an increasing finite-measure exhaustion from it agree on the generated sigma-algebra.

[F10]

Metric space: d(x,y)=0 iff x=y, symmetry, and the triangle inequality; pseudometric and ultrametric: A metric is nonnegative, symmetric, separates distinct points, and satisfies the triangle inequality.

[F11]

A measurable function between measurable spaces: A map is measurable when the preimage of every target-measurable set is measurable.

[F12]

The Borel sigma-algebra of a topological space: The Borel sigma-algebra of a metric space contains every closed subset.

Proof

Given: The objects and hypotheses in the statement.

1.1F4givenchoose

By [F4], nonidentifiability supplies θ0,θ1∈Θ with θ0∼θ1 and ψ(θ0)≠ψ(θ1). Fix this pair.

2.1F1F2F7F8F9givenstep 1.1

Fix n≥1. By [F1], the observation laws in the model are Pθ. Measurable rectangles in Xn form a π-system generating A⊗n by [F7]. For every rectangle A1×⋯×An, [F8] and [F2] give Pθ0⊗n(A1×⋯×An)=∏j=1nPθ0(Aj)=∏j=1nPθ1(Aj)=Pθ1⊗n(A1×⋯×An). The full space is itself a rectangle and is an increasing exhaustion of finite mass one. Applying [F9] shows Pθ0⊗n=Pθ1⊗n on all of A⊗n.

2.2F3F10givenstep 1.1algebra

Put Δ=d(ψ(θ0),ψ(θ1))>0 and r=Δ/3. The two closed target balls Ci={t∈T:d(t,ψ(θi))≤r} are disjoint: a point in both would give Δ≤2r<Δ by [F10].

3.1F5F6F11F12F13givenstep 1.1step 2.2assume-contra

Suppose, for contradiction, that the estimator sequence is consistent at both parameters. For i∈{0,1} let Ei,n=Tn−1(Ci). These events are measurable because each Ci is closed by [F13] and Borel by [F12], while Tn is measurable by [F5] and preimages of Borel sets are measurable by [F11]. By [F6], Pθi⊗n(Ei,n)→1 for each i.

4.1step 2.1step 2.2step 3.1algebradischarge-contradiction: step 3.1∎

For each n, step 2.1 gives one common sample law for both probabilities, and step 2.2 gives disjoint events, so their probabilities sum to at most one. Step 3.1 says both probabilities tend to one, so their sum tends to two, a contradiction. Therefore the estimator sequence cannot be consistent at both parameters.

5 · Examples, counterexamples and false statements

None yet.

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