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Independent-subject terminal influence limits — Examples

1 · Prerequisites

2 · Summary

The first example verifies an exact Bernoulli sample-mean expansion, and the second calculates the covariance for two fixed-weight independent groups. The final example uses a shrinking risk set whose denominator is empty with probability one half, so its root- $n$ error is not tight. These examples distinguish applying a terminal limit theorem from proving the estimator-specific expansion it requires.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Bernoulli sample mean as an exact terminal influence expansion

Example

Assume AC. Let $B_{1}, B_{2}, \dots$ be IID Bernoulli $(p)$ variables, where $0 \leq p \leq 1$ , and estimate $θ = p$ by ${\hat{θ}}_{n} = {\overset{ˉ}{B}}_{n} = n^{- 1} \sum_{i = 1}^{n} B_{i}$ . Take the subject influence function $ϕ (b) = b - p$ .

Facts & Assumptions

Given: The Bernoulli sample and estimator in the Example. Verify its influence expansion and terminal limit.

[A1]

The observations are IID Bernoulli $(p)$ , with $P (B_{i} = 1) = p$ and $P (B_{i} = 0) = 1 - p$ .

[A2]

Assume AC. The Axiom of Choice

[F1]

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

[F2]

Moments, variance, and covariance on a probability space: For a square-integrable real random variable $X$ , $Var (X) = E [(X - E X)^{2}]$ .

[F3]

IID terminal influence limit from a separately established expansion: A separately supplied centered, finite-variance IID influence expansion has its covariance normal limit under AC.

Verification

1.1A1F1F2algebra

For every Borel set $C \subseteq R$ , ${B_{i} - p \in C} = {B_{i} \in C + p}$ , where $C + p$ is Borel. The common law of the $B_{i}$ therefore gives a common law for $B_{i} - p$ , and applying the finite independence identities in [F1] to these preimage events proves that the shifted observations are IID. The variable $ϕ (B_{i}) = B_{i} - p$ has mean $p - p = 0$ and variance $p (1 - p)$ by [A1] and [F2]. The estimator identity ${\overset{ˉ}{B}}_{n} - p = n^{- 1} \sum_{i} (B_{i} - p)$ gives the required expansion exactly, with zero remainder.

2.1A2F3step 1.1∎

Apply [F3] with $d = 1$ and $Σ = p (1 - p)$ to obtain $\sqrt{n} ({\overset{ˉ}{B}}_{n} - p) \Rightarrow N (0, p (1 - p))$ . At $p = 0$ or $p = 1$ , ${\overset{ˉ}{B}}_{n} = p$ almost surely, so the limit is the point mass $N (0, 0)$ covered by the theorem.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Weighted means from two independent subject groups

Example

Assume AC. Let $X_{1 i}$ be IID Bernoulli $(p_{1})$ and $X_{2 i}$ IID Bernoulli $(p_{2})$ , with the two subject groups independent. Let deterministic sample sizes satisfy $n_{1} + n_{2} = n$ , $n_{1} / n \to ρ \in (0, 1)$ , and $n_{2} / n \to 1 - ρ$ . For $g \in {1, 2}$ define ${\overset{ˉ}{X}}_{g} = n_{g}^{- 1} \sum_{i = 1}^{n_{g}} X_{g i}$ . For the fixed target $θ = ρ p_{1} + (1 - ρ) p_{2}$ , use the fixed-allocation-weight estimator ${\hat{θ}}_{n} = ρ {\overset{ˉ}{X}}_{1} + (1 - ρ) {\overset{ˉ}{X}}_{2}$ .

Facts & Assumptions

Given: The two independent Bernoulli groups and estimator in the Example. Find its exact group influence expansion and limit.

[A1]

In group $g$ , $P (X_{g i} = 1) = p_{g}$ and $P (X_{g i} = 0) = 1 - p_{g}$ ; the observations are IID within groups and the two groups are independent.

[A2]

Assume AC. The Axiom of Choice

[F1]

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

[F2]

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

[F3]

Moments, variance, and covariance on a probability space: For a square-integrable real random variable $X$ , $Var (X) = E [(X - E X)^{2}]$ .

[F4]

Independent-group terminal influence limit: For a fixed finite family of independent IID groups, the covariance of the terminal normal limit is $\sum_{g} ρ_{g}^{- 1} Σ_{g}$ , provided the stated group expansion and hypotheses hold.

Verification

1.1A1F1F2F3F4algebra

Put $ϕ_{1} (x) = ρ (x - p_{1})$ and $ϕ_{2} (x) = (1 - ρ) (x - p_{2})$ . The measurable maps preserve within-group identical laws and the finite event-product identities for independence across groups, so their sampled values meet [F1] and [F2], as required by [F4]. By [A1] and [F3], they are centered, square-integrable group contributions with variances $Σ_{1} = ρ^{2} p_{1} (1 - p_{1})$ and $Σ_{2} = (1 - ρ)^{2} p_{2} (1 - p_{2})$ . The identity $\sqrt{n} ({\hat{θ}}_{n} - θ) = \sqrt{n / n_{1}} n_{1}^{- 1 / 2} \sum_{i = 1}^{n_{1}} ϕ_{1} (X_{1 i}) + \sqrt{n / n_{2}} n_{2}^{- 1 / 2} \sum_{i = 1}^{n_{2}} ϕ_{2} (X_{2 i})$ is exact, so its remainder is zero.

2.1A2F4step 1.1algebra∎

The covariance in [F4] is $ρ^{- 1} Σ_{1} + (1 - ρ)^{- 1} Σ_{2} = ρ p_{1} (1 - p_{1}) + (1 - ρ) p_{2} (1 - p_{2})$ . Thus $\sqrt{n} ({\hat{θ}}_{n} - θ)$ converges to the centered normal with this variance. The formula includes $p_{g} \in {0, 1}$ , where the corresponding group variance is zero.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

An empty risk set prevents root-n tightness

Example

Assume AC. Let $T_{1}, T_{2}, \dots$ be IID uniform on $[0, 1]$ event times and set $X_{i} = 1$ for every subject. For each $n \geq 2$ , use the moving terminal time $t_{n} = 2^{- 1 / n}$ and risk indicators $Y_{n i} = 1 {T_{i} \geq t_{n}}$ . Write $K_{n} = \sum_{i = 1}^{n} Y_{n i}$ . The target mean among subjects at risk is the fixed value $θ = 1$ . Define the risk-set ratio

${\hat{θ}}_{n} = {\begin{cases} \frac{\sum_{i = 1}^{n} X_{i} Y_{n i}}{K_{n}}, & K_{n} > 0, \\ 0, & K_{n} = 0. \end{cases}$

The convention makes the estimator defined even when the risk set is empty.

Facts & Assumptions

Given: The IID uniform event times and ratio estimator in the Example. Show that its root- $n$ error is not tight, so the terminal influence theorem cannot be applied without a separate expansion.

[A1]

The event times are IID, their common law satisfies $P (T_{i} < t) = t$ for $0 \leq t \leq 1$ , and subjects are mutually independent.

[A2]

Assume AC. The Axiom of Choice

[F1]

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

[F2]

Independent families of event classes: For finite selections of distinct independent indices, the probability of an intersection is the product of the event probabilities.

[F3]

Stochastic order and convergence notation for finite-dimensional random vectors: $Z_{n} = O_{P} (1)$ requires that its tail probabilities can be made uniformly small in the limit by choosing a sufficiently large bound.

[F4]

Finite-dimensional stochastic-order calculus and tightness: Any finite-dimensional sequence converging in distribution is bounded in probability.

[F5]

IID terminal influence limit from a separately established expansion: If this estimator had a centered finite-variance IID influence expansion, its root- $n$ error would converge in distribution under AC.

Verification

1.1A1F1F2algebra

For each $n \geq 2$ , [A1] gives $P (Y_{n i} = 0) = t_{n} = 2^{- 1 / n}$ . The events ${Y_{n i} = 0}$ are independent by the finite event-product identity in [F2], hence $P (K_{n} = 0) = (2^{- 1 / n})^{n} = 1 / 2$ . Since $X_{i} = 1$ , the ratio is $1$ when $K_{n} > 0$ and is $0$ otherwise; therefore $\sqrt{n} ({\hat{θ}}_{n} - 1)$ is $0$ when $K_{n} > 0$ and $- \sqrt{n}$ when $K_{n} = 0$ .

2.1A2F3F4F5step 1.1∎

For every fixed $M > 0$ and every integer $n > M^{2}$ , $P (∣ \sqrt{n} ({\hat{θ}}_{n} - 1) ∣ > M) = P (K_{n} = 0) = 1 / 2$ . Taking $η = 1 / 4$ in [F3] shows that the root- $n$ errors are not $O_{P} (1)$ . If they converged in distribution, [F4] would make them $O_{P} (1)$ , a contradiction. Under [A2], the expansion required by [F5] would imply a weak limit, so it cannot hold for this estimator. The example shows why a consumer must control its own random denominator before claiming a terminal influence limit.