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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.

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