Editing Openai/6922876a-7988-8007-9c62-5f71772af6aa (section)

===== We will need only the first two score moments under QQQ, and in upper‑bound form. =====
(The stabilized estimator \eqref{eq:penalized-M-est} avoids any random‑curvature remainder.)

\begin{lemma}[Uniform moment controls]\label{lem:score-moments-correct}
Fix τ∈(0,12]\tau\in(0,\tfrac12]τ∈(0,21]. For every QQQ with W2(Q,μθ)≤εW_2(Q,\mu_\theta)\le \varepsilonW2(Q,μθ)≤ε,
\begin{align}
\big|\E_Q[\psi_\tau]\big| &;\le; C_1\Big(\varepsilon,\tau^{-3/2}+1\Big),\label{eq:bias-score-correct}\
\E_Q[\psi_\tau^2] &;\le; I(\tau);+; C_2,\varepsilon,\tau^{-5/2}.\label{eq:var-score-correct}
\end{align}
The constants C1,C2C_1,C_2C1,C2 are absolute. In particular, for τ=τ⋆=ε2/3\tau=\tau_\star=\varepsilon^{2/3}τ=τ⋆=ε2/3,

∣\EQ[ψτ⋆]∣ ≲ 1,\EQ[ψτ⋆2] ≲ 1τ⋆ = ε−2/3.\big|\E_Q[\psi_{\tau_\star}]\big|\ \lesssim\ 1,\qquad
\E_Q[\psi_{\tau_\star}^2]\ \lesssim\ \frac{1}{\tau_\star}\ =\ \varepsilon^{-2/3}.\EQ[ψτ⋆] ≲ 1,\EQ[ψτ⋆2] ≲ τ⋆1 = ε−2/3.
\end{lemma}

\begin{proof}
We apply Lemma \ref{lem:dyn-W2-correct} with P=μθP=\mu_\thetaP=μθ (so W2(P,Q)≤εW_2(P,Q)\le \varepsilonW2(P,Q)≤ε).
For φ=ψτ\varphi=\psi_\tauφ=ψτ, φ′=ψτ′\varphi'=\psi_\tau'φ′=ψτ′ equals the tail‑slope (2τK(0))−2(2\tau K(0))^{-2}(2τK(0))−2 on {∣u∣>12−τ}\{|u|>\tfrac12-\tau\}{∣u∣>21−τ} and 000 on the bulk. For every t∈[0,1]t\in[0,1]t∈[0,1],

\EPt[ψτ′(Xt)2] ≤ 1(2τK(0))4 ¶Pt(∣Xt−θ∣>12−τ).\E_{P_t}[\psi_\tau'(X_t)^2]\ \le\ \frac{1}{(2\tau K(0))^4}\,\P_{P_t}\big(|X_t-\theta|>\tfrac12-\tau\big).\EPt[ψτ′(Xt)2] ≤ (2τK(0))41¶Pt(∣Xt−θ∣>21−τ).
Since Xt=(1−t)X+tT(X)X_t=(1-t)X+tT(X)Xt=(1−t)X+tT(X) with X∼μθX\sim\mu_\thetaX∼μθ, a point starting at least distance τ\tauτ from the edge must move by at least τ\tauτ to enter the tail. By Markov,
¶(∣T(X)−X∣≥τ)≤\E[(T(X)−X)2]/τ2≤ε2/τ2,\P(|T(X)-X|\ge \tau)\le \E[(T(X)-X)^2]/\tau^2\le \varepsilon^2/\tau^2,¶(∣T(X)−X∣≥τ)≤\E[(T(X)−X)2]/τ2≤ε2/τ2,
and ¶μ(∣X−θ∣>12−τ)=2τ\P_\mu(|X-\theta|>\tfrac12-\tau)=2\tau¶μ(∣X−θ∣>21−τ)=2τ. Hence

sup⁡t∈[0,1]¶Pt(∣Xt−θ∣>12−τ) ≤ 2τ+ε2/τ2.\sup_{t\in[0,1]}\P_{P_t}\big(|X_t-\theta|>\tfrac12-\tau\big)\ \le\ 2\tau+\varepsilon^2/\tau^2.t∈[0,1]sup¶Pt(∣Xt−θ∣>21−τ) ≤ 2τ+ε2/τ2.
Therefore

∫01\EPt[ψτ′(Xt)2] dt ≲ τ−4 (2τ+ε2/τ2) ≲ τ−3+ε2τ−6.\int_0^1 \E_{P_t}[\psi_\tau'(X_t)^2]\,dt\ \lesssim\ \tau^{-4}\,(2\tau+\varepsilon^2/\tau^2)\ \lesssim\ \tau^{-3}+\varepsilon^2\tau^{-6}.∫01\EPt[ψτ′(Xt)2]dt ≲ τ−4(2τ+ε2/τ2) ≲ τ−3+ε2τ−6.
Applying Lemma \ref{lem:dyn-W2-correct} with P=μθP=\mu_\thetaP=μθ gives

∣\EQ[ψτ]−\Eμθ[ψτ]∣ ≲ ε(τ−3/2+ε τ−3).\big|\E_Q[\psi_\tau]-\E_{\mu_\theta}[\psi_\tau]\big|
\ \lesssim\ \varepsilon\left(\tau^{-3/2}+\varepsilon\,\tau^{-3}\right).\EQ[ψτ]−\Eμθ[ψτ] ≲ ε(τ−3/2+ετ−3).
A direct symmetry check shows \Eμθ[ψτ]=0\E_{\mu_\theta}[\psi_\tau]=0\Eμθ[ψτ]=0 (the two interior edge contributions cancel), yielding \eqref{eq:bias-score-correct}.
For φ=ψτ2\varphi=\psi_\tau^2φ=ψτ2, we have φ′=2ψτψτ′\varphi'=2\psi_\tau\psi_\tau'φ′=2ψτψτ′. Using ψτ′=(2τK(0))−2\psi_\tau'= (2\tau K(0))^{-2}ψτ′=(2τK(0))−2 on tails and 000 on bulk together with \Eνθ,τ[ψτ2]=I(τ)\E_{\nu_{\theta,\tau}}[\psi_\tau^2]=I(\tau)\Eνθ,τ[ψτ2]=I(τ) and the same tail‑mass bound as above,

∫01\EPt[(ψτ2)′(Xt)2] dt ≲ τ−4 ∫01\EPt[ψτ(Xt)2] dt ≲ τ−4 (I(τ)+ε τ−3/2) ≲ τ−5+ε τ−11/2,\int_0^1 \E_{P_t}[(\psi_\tau^2)'(X_t)^2]\,dt
\ \lesssim\ \tau^{-4}\,\int_0^1\E_{P_t}[\psi_\tau(X_t)^2]\,dt
\ \lesssim\ \tau^{-4}\,\Big(I(\tau)+\varepsilon\,\tau^{-3/2}\Big)
\ \lesssim\ \tau^{-5}+\varepsilon\,\tau^{-11/2},∫01\EPt[(ψτ2)′(Xt)2]dt ≲ τ−4∫01\EPt[ψτ(Xt)2]dt ≲ τ−4(I(τ)+ετ−3/2) ≲ τ−5+ετ−11/2,
and Lemma \ref{lem:dyn-W2-correct} with P=νθ,τP=\nu_{\theta,\tau}P=νθ,τ (so Eν[ψτ2]=I(τ)E_{\nu}[\psi_\tau^2]=I(\tau)Eν[ψτ2]=I(τ)) yields \eqref{eq:var-score-correct}.
\end{proof}