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The IID bootstrap can fail at the kink of the absolute mean

Statement refuted

For every finite-variance IID law and every Borel plug-in map $g$ whose sampling root $\sqrt{n} {g ({\overset{ˉ}{X}}_{n}) - g (E X_{1})}$ has a continuous weak limit $L$ , the conditional empirical-bootstrap root $\sqrt{n} {g ({\overset{ˉ}{X}}_{n}^{*}) - g ({\overset{ˉ}{X}}_{n})}$ has CDF converging in probability to the CDF of $L$ at every continuity point, even when $g$ is not differentiable at $E X_{1}$ .

Facts & Assumptions

Given: Let $X_{i}$ be IID Rademacher random variables with $P (X_{i} = 1) = P (X_{i} = - 1) = 1 / 2$ , and use their empirical bootstrap with $g (x) = ∣ x ∣$ . Thus $μ = 0$ and $Var (X_{i}) = 1$ .

[A1]

AC is assumed by the IID CLTs and standard normal law used in this counterexample. The Axiom of Choice

[F1]

For the empirical bootstrap mean with positive projected variance, its conditional CDF converges uniformly in probability to the corresponding normal CDF. Conditional IID bootstrap CLT for an empirical law

[F2]

The centered sample mean of a real IID law with mean zero and positive finite variance obeys the Lindeberg–Levy CLT. Lindeberg-Levy iid central limit theorem

[F3]

Continuous maps preserve convergence in distribution. Continuous mapping theorem

[F4]

If $U_{n} \Rightarrow U$ and $V_{n} \to c$ in probability, then $U_{n} + V_{n} \Rightarrow U + c$ . Slutsky's theorem for real random variables

[F5]

The standard normal CDF $Φ$ is continuous and strictly increasing. Standard normal and normal laws The standard normal CDF is continuous and strictly increasing

Counterexample

1.1A1F2F3given

By [F2], $Z_{n} = \sqrt{n} {\overset{ˉ}{X}}_{n} \Rightarrow Z$ with $Z \sim N (0, 1)$ . Since absolute value is continuous, [F3] gives $\sqrt{n} ∣ {\overset{ˉ}{X}}_{n} ∣ \Rightarrow ∣ Z ∣$ . The limiting CDF of $∣ Z ∣$ is continuous at $0$ and equals $P (∣ Z ∣ \leq 0) = 0$ .

1.2A1F1givenalgebra

Let $V_{n}^{*} = \sqrt{n} ({\overset{ˉ}{X}}_{n}^{*} - {\overset{ˉ}{X}}_{n})$ . Its conditional CDF $F_{n}^{*}$ satisfies $D_{n} : = \sup_{t} ∣ F_{n}^{*} (t) - Φ (t) ∣ \to 0$ in probability by [F1]. The bootstrap root for $g (x) = ∣ x ∣$ is $R_{n}^{*} = \sqrt{n} (∣ {\overset{ˉ}{X}}_{n}^{*} ∣ - ∣ {\overset{ˉ}{X}}_{n} ∣) = ∣ Z_{n} + V_{n}^{*} ∣ - ∣ Z_{n} ∣ .$ For fixed data value $Z_{n} = z \geq 0$ , the event $R_{n}^{*} \leq 0$ is $- 2 z \leq V_{n}^{*} \leq 0$ ; for $z < 0$ , it is $0 \leq V_{n}^{*} \leq - 2 z$ .

2.1F3F4F5step 1.1step 1.2given

For any interval $[u, v]$ , the difference between its conditional bootstrap probability and its standard normal probability is at most $2 D_{n}$ : use $F_{n}^{*} (v) - F_{n}^{*} (u -)$ and take left limits in the uniform CDF bound, using continuity of $Φ$ . Therefore $Q_{n} : = P^{*} (R_{n}^{*} \leq 0 ∣ X_{1}, \dots, X_{n}) = Φ (2 ∣ Z_{n} ∣) - \frac{1}{2} + o_{P} (1) .$ Indeed, the normal mass of either interval in step 1.2 is $Φ (2 ∣ z ∣) - 1 / 2$ . By [F3] and [F4], $Q_{n} \Rightarrow Φ (2 ∣ Z ∣) - 1 / 2$ . This limit is positive with positive probability because $P (∣ Z ∣ \geq 1) > 0$ and [F5] gives $Φ (2) - 1 / 2 > 0$ . Thus $Q_{n}$ does not converge in probability to zero.

3.1F1F5step 1.1step 2.1∎

If the conditional bootstrap law consistently estimated the sampling law of $\sqrt{n} ∣ {\overset{ˉ}{X}}_{n} ∣$ , its conditional CDF at the continuity point $0$ would converge in probability to the sampling limit CDF there, namely zero by step 1.1 and [F5]. Step 2.1 contradicts this. The failure occurs because $g (x) = ∣ x ∣$ is not differentiable at the true mean $0$ .

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Sources

P. J. Bickel, F. Götze, and W. R. van Zwet, Resampling Fewer Than n Observations: Gains, Losses, and Remedies for Losses, Statistica Sinica 7 (1997), §2.3, Example 6 (standard reference, not scraped)
University of Washington STAT 581, Chapter 8: Bootstrap and Jackknife Estimation of Sampling Distributions, §§4–5 (standard reference, not scraped)