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IID terminal influence limit from a separately established expansion

Statement

Assume the Axiom of Choice. Let $d \geq 1$ , let $(O, A, P)$ be a probability space, and let $O_{1}, O_{2}, \dots$ be IID with law $P$ . Let ${\hat{θ}}_{n}$ be a measurable $R^{d}$ -valued estimator based on $O_{1}, \dots, O_{n}$ for a fixed $θ \in R^{d}$ . Suppose $ϕ : O \to R^{d}$ is measurable, $E_{P} ϕ = 0$ , and $E_{P} ∥ ϕ ∥_{2}^{2} < \infty$ , and set $Σ = E_{P} [ϕ ϕ^{T}]$ . Assume the separately established expansion, with remainder $r_{n}$ on the sample probability space,

${\hat{θ}}_{n} - θ = \frac{1}{n} \sum_{i = 1}^{n} ϕ (O_{i}) + r_{n}, r_{n} = o_{P} (n^{- 1 / 2}),$

Then

$\sqrt{n} ({\hat{θ}}_{n} - θ) \Rightarrow N_{d} (0, Σ) .$

The remainder notation means $\sqrt{n} r_{n} \to_{P} 0$ , as in Asymptotically linear estimators and influence functions. Thus $ϕ$ is the estimator influence function in the asymptotically linear sense: the theorem assumes the expansion and does not establish it. The covariance may be singular, including zero.

Facts & Assumptions

Given: Assume AC. The subjects $O_{i}$ are IID with law $P$ ; the measurable influence function $ϕ$ is centered and square-integrable; $Σ = E_{P} [ϕ ϕ^{T}]$ ; and the estimator has the displayed expansion with $\sqrt{n} r_{n} \to_{P} 0$ . Prove the displayed distributional limit.

[A1]

The Axiom of Choice: Assume the Axiom of Choice (AC).

[F1]

Identical distribution and IID families: IID means a common law and mutual independence of the whole family.

[F2]

Independent families of event classes: Independence means that for every finite selection of distinct indices and one event from each selected class, the probability of their intersection is the product of their probabilities.

[F3]

Asymptotically linear estimators and influence functions: For an IID sample, an asymptotically linear estimator has expansion ${\hat{ψ}}_{n} - ψ (P) = (P_{n} - P) ϕ + o_{P} (n^{- 1 / 2})$ , with a mean-zero square-integrable influence function; this definition alone does not assert asymptotic normality.

[F4]

Covariance matrix of a finite-dimensional random vector: For a finite-dimensional square-integrable vector $X$ with mean $μ$ , $Cov (X) = E [(X - μ) (X - μ)^{T}]$ .

[F5]

Multivariate iid central limit theorem: Under AC, IID centered $R^{d}$ vectors with finite second moment and covariance $Σ$ have normalized sums converging to $N_{d} (0, Σ)$ ; $Σ$ may be singular.

[F6]

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

[F7]

Vector Slutsky theorem: If $X_{n} \Rightarrow X$ and $Y_{n} \to_{P} c$ , then every continuous function of $(X_{n}, Y_{n})$ converges in distribution to the same function of $(X, c)$ .

Proof

1.1F1F2given

For each Borel set $B \subseteq R^{d}$ , $ϕ^{- 1} (B) \in A$ . The common law of the $O_{i}$ gives a common image law, and [F2] applied to finitely many preimage events gives independence of the images. Thus $ϕ (O_{1}), ϕ (O_{2}), \dots$ are IID.

2.1A1F4F5step 1.1

The vectors in step 1.1 have mean zero and covariance $Σ$ by the assumptions and [F4]. Thus [F5] gives $S_{n} : = n^{- 1 / 2} \sum_{i = 1}^{n} ϕ (O_{i}) \Rightarrow N_{d} (0, Σ)$ .

3.1F3F6F7step 2.1∎

The assumed expansion gives $\sqrt{n} ({\hat{θ}}_{n} - θ) = S_{n} + \sqrt{n} r_{n}$ , and the asymptotically linear convention [F3] makes its scaled remainder tend to zero in probability; [F6] identifies this as $o_{P} (1)$ . Apply [F7] with $h (x, y) = x + y$ and $c = 0$ to step 2.1. This proves the asserted limit, without deriving the influence expansion.

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Sources

Sourav Chatterjee, Stats 310A (Math 230A) Lecture Notes, §§8.5.3, 8.8–8.9, PDF pp. 104–107 / printed pp. 100–103 (standard reference, not scraped)
Rick Durrett, Probability: Theory and Examples, 5th ed., §3.10, Theorems 3.10.5–3.10.7, printed pp. 200–201 (standard reference, not scraped)
Peter Bartlett, Theoretical Statistics, Stat 210B Lecture 16, PDF pp. 9–12 (standard reference, not scraped)