CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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A finite-mean functional without a square-integrable influence curve

Statement refuted

Every finite-mean law has a finite-variance influence curve for its mean functional in the dominated local model generated by bounded mean-zero score tilts.

Facts & Assumptions

Given: The statement above and the definitions of DQM paths and parameter influence curves.

[F1]

A dominated differentiable-in-quadratic-mean submodel: a one-sided common-dominated path is DQM when its square-root density has the stated first-order $L^{2}$ expansion.

[F2]

Pathwise differentiability and parameter influence curves: a finite-variance parameter influence curve is a vector in $L_{0}^{2} (P)$ representing path derivatives by the $L^{2} (P)$ pairing.

[F3]

Cauchy-Schwarz inequality for $L^{2}$ : $∣ P [f g] ∣ \leq ∥ f ∥_{2} ∥ g ∥_{2}$ for square-integrable $f, g$ .

[F4]

A uniform derivative bound gives a uniform Taylor remainder bound: if the second derivative is bounded on an interval, the first-order Taylor polynomial has a remainder bounded by a constant times the square of the displacement.

Counterexample

Given: Let $X = N_{\geq 1}$ , $P {j} = 2^{- j}$ , $f (j) = 2^{j / 2}$ , and $ψ (Q) = Q f$ . Allow all local tilts $d P_{t, s} = (1 + t s) d P$ with bounded $s$ satisfying $P s = 0$ .

1.1givenalgebra

The masses sum to one. The geometric series gives $P f = \sum_{j \geq 1} 2^{- j / 2} < \infty$ , while $P [f^{2}] = \sum_{j \geq 1} 1 = \infty$ . Thus $μ = P f$ is finite and $u = f - μ$ is integrable, centered, but not square-integrable.

1.2F1F4givenalgebra

For any bounded centered $s$ , choose $t \geq 0$ so small that $∣ t s ∣ \leq 1 / 2$ everywhere. Then $1 + t s \geq 0$ and $P (1 + t s) = 1$ . On $[- 1 / 2, 1 / 2]$ , $g (x) = \sqrt{1 + x}$ has $g^{''} (x) = - 1 / (4 (1 + x)^{3 / 2})$ , so $∣ g^{''} (x) ∣ \leq 1$ . [F4] with $n = 1$ gives $\sqrt{1 + t s} = 1 + \frac{t}{2} s + O (t^{2} ∥ s ∥_{\infty}^{2})$ uniformly; thus the tilt is DQM with score $s$ . Since $f \in L^{1} (P)$ and $s$ is bounded, $ψ (P_{t, s}) = P f + t P [f s]$ and ${\dot{ψ}}_{P} (s) = P [u s]$ .

1.3givenalgebra

For $M > 0$ , put $u_{M} = u 1_{{∣ u ∣ \leq M}}$ , $c_{M} = P u_{M}$ , and $s_{M} = u_{M} - c_{M}$ . Then $s_{M}$ is bounded and centered, $D_{M} : = P [u^{2} 1_{{∣ u ∣ \leq M}}] = P [u s_{M}]$ , and $∥ s_{M} ∥_{2}^{2} = D_{M} - c_{M}^{2} \leq D_{M}$ . For all sufficiently large $j$ , $u (j) \geq 2^{j / 2} / 2$ , so each such included atom contributes at least $1 / 4$ to $D_{M}$ ; as $M \to \infty$ , arbitrarily many atoms are included and $D_{M} \to \infty$ . For large $M$ , $s_{M} \neq 0$ because the truncation vanishes on a positive-mass tail and is nonzero on included atoms. Hence $∣ {\dot{ψ}}_{P} (s_{M}) ∣ / ∥ s_{M} ∥_{2} \geq \sqrt{D_{M}} \to \infty$ .

2.1F2F3step 1.3given∎

If a finite-variance influence curve $ϕ \in L_{0}^{2} (P)$ represented these derivatives, [F3] would give $∣ {\dot{ψ}}_{P} (s_{M}) ∣ = ∣ P [ϕ s_{M}] ∣ \leq ∥ ϕ ∥_{2} ∥ s_{M} ∥_{2}$ for every $M$ , contradicting step 1.3. Thus the mean functional has no square-integrable influence curve in this model, refuting the statement.

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