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

=== Assistant: Below is a full pass over the argument. ===
Below is a full pass over the argument. I’ve separated (I) critical correctness issues from (II) minor issues/typos/constants, and then given (III) drop‑in LaTeX fixes for each place. Wherever a claim is wrong but easily repairable, I give a concrete replacement (as literal LaTeX you can paste in).

==== ### ====
* What’s written. In Lemma \ref{lem:dyn-W2} you set Xt=(1−t)X+tT(X)X_t=(1-t)X+tT(X)Xt=(1−t)X+tT(X) with X∼PX\sim PX∼P, define Pt=Law(Xt)P_t=\mathrm{Law}(X_t)Pt=Law(Xt), and then write ddt\EPt[φ]=\EPt[φ′(Xt) vt(Xt)],vt(x)=T(x)−x.\frac{d}{dt}\E_{P_t}[\varphi]=\E_{P_t}\big[\varphi'(X_t)\,v_t(X_t)\big],\quad v_t(x)=T(x)-x .dtd\EPt[φ]=\EPt[φ′(Xt)vt(Xt)],vt(x)=T(x)−x. The last identity is not correct: T(x)−xT(x)-xT(x)−x is defined on the Lagrangian coordinate x∈supp(P)x\in\mathrm{supp}(P)x∈supp(P), not on the current Eulerian position x∈supp(Pt)x\in\mathrm{supp}(P_t)x∈supp(Pt). In particular vt(Xt)v_t(X_t)vt(Xt) is generally not equal to T(X)−XT(X)-XT(X)−X.
* What is correct. If we define Tt(x)=(1−t)x+tT(x)T_t(x)=(1-t)x+tT(x)Tt(x)=(1−t)x+tT(x) and Xt=Tt(X)X_t=T_t(X)Xt=Tt(X), then ddt\E[φ(Xt)]  =  \E ⁣[φ′(Xt)⋅(T(X)−X)],\frac{d}{dt}\E[\varphi(X_t)] \;=\; \E\!\left[\varphi'(X_t)\cdot (T(X)-X)\right],dtd\E[φ(Xt)]=\E[φ′(Xt)⋅(T(X)−X)], where the expectation is w.r.t. the original X∼PX\sim PX∼P. Equivalently, in Eulerian form, the velocity field vtv_tvt must satisfy vt(Tt(x))=T(x)−xv_t(T_t(x))=T(x)-xvt(Tt(x))=T(x)−x; that is, vt(z)=\E[T(X)−X∣Xt=z]v_t(z)=\E[T(X)-X \mid X_t=z]vt(z)=\E[T(X)−X∣Xt=z], not T(z)−zT(z)-zT(z)−z.
* Consequence. The displayed Cauchy–Schwarz step goes through once you rewrite the derivative correctly. The correct bound is ∣\EQ[φ]−\EP[φ]∣≤(∫01\EP[φ′(Xt)2] dt)1/2W2(P,Q),\big|\E_Q[\varphi]-\E_P[\varphi]\big| \le \Big(\int_0^1 \E_P[\varphi'(X_t)^2]\,dt\Big)^{1/2} W_2(P,Q),\EQ[φ]−\EP[φ]≤(∫01\EP[φ′(Xt)2]dt)1/2W2(P,Q), which coincides with your statement after noting \EP[g(Xt)]=\EPt[g]\E_P[g(X_t)]=\E_{P_t}[g]\EP[g(Xt)]=\EPt[g].
* Drop‑in fix: see §III.A.

===== There are several separate problems: =====
* (a) Misuse of Lemma \ref{lem:dyn-W2} with φ=ψτ′\varphi=\psi_\tau'φ=ψτ′ and φ=(ψτ)2\varphi=(\psi_\tau)^2φ=(ψτ)2. The lemma requires ∫01\EPt[φ′(Xt)2] dt<∞\int_0^1\E_{P_t}[\varphi'(X_t)^2]\,dt<\infty∫01\EPt[φ′(Xt)2]dt<∞. For φ=ψτ′\varphi=\psi_\tau'φ=ψτ′ this means ψτ′′∈L2(Pt)\psi_\tau''\in L^2(P_t)ψτ′′∈L2(Pt). In your construction f(⋅;θ,τ)f(\cdot;\theta,\tau)f(⋅;θ,τ) is C1C^1C1 but not C2C^2C2 across the junctions. Indeed, with the chosen Gaussian tails, ψτ(u)=−∂ulog⁡f(u)={0,∣u∣≤12−τ,12τK(0)⋅∣u∣−(12−τ)2τK(0),∣u∣>12−τ,\psi_\tau(u)=-\partial_u\log f(u) = \begin{cases} 0,& |u|\le \tfrac12-\tau,\\ \tfrac{1}{2\tau K(0)}\cdot \frac{|u|-(\tfrac12-\tau)}{2\tau K(0)},& |u|>\tfrac12-\tau, \end{cases}ψτ(u)=−∂ulogf(u)={0,2τK(0)1⋅2τK(0)∣u∣−(21−τ),∣u∣≤21−τ,∣u∣>21−τ, so ψτ\psi_\tauψτ is continuous, ψτ′\psi_\tau'ψτ′ is piecewise constant (0 on the flat top; constant >0>0>0 on the tails), hence ψτ′′\psi_\tau''ψτ′′ is a sum of Dirac masses at the junctions and is not a bounded function. Thus φ′=ψτ′′\varphi'=\psi_\tau''φ′=ψτ′′ is not square‑integrable and the lemma cannot be applied with φ=ψτ′\varphi=\psi_\tau'φ=ψτ′.
* (b) “Edge strips” of width τ\tauτ. The proof asserts that the derivatives “are nonzero only on the two edge strips” and that “the νθ,τ\nu_{\theta,\tau}νθ,τ-mass of each edge region equals τ\tauτ.” This is incorrect for the derivatives: while the density modification from the uniform happens on a set of ν\nuν-mass 2τ2\tau2τ, the score derivatives ψτ′,(ψτ2)′\psi_\tau', (\psi_\tau^2)'ψτ′,(ψτ2)′ are nonzero on the entire tails ∣u∣>12−τ|u|>\tfrac12-\tau∣u∣>21−τ, not just on a bounded strip: ψτ′\psi_\tau'ψτ′ is constant (positive) on the tails; (ψτ2)′(\psi_\tau^2)'(ψτ2)′ grows linearly in ∣u∣|u|∣u∣. Consequently the integrals \EPt[ψτ′(Xt)2]\E_{P_t}[\psi_\tau'(X_t)^2]\EPt[ψτ′(Xt)2] and \EPt[((ψτ2)′(Xt))2]\E_{P_t}[((\psi_\tau^2)'(X_t))^2]\EPt[((ψτ2)′(Xt))2] cannot be bounded by “edge mass ×\times× L∞L^\inftyL∞” with the claimed τ\tauτ-scalings.
* (c) The scalings in (3.3)–(3.5) are wrong. In fact, on the tails ψτ′(u)≡14τ2K(0)2(a positive constant),\psi_\tau'(u)\equiv \frac{1}{4\tau^2K(0)^2}\quad\text{(a positive constant),}ψτ′(u)≡4τ2K(0)21(a positive constant), so even under PtP_tPt with Pr⁡(∣Xt−θ∣>12−τ)=1\Pr(|X_t-\theta|>\tfrac12-\tau)=1Pr(∣Xt−θ∣>21−τ)=1 one has \EPt[ψτ′(Xt−θ)2]=Θ(τ−4)\E_{P_t}[\psi_\tau'(X_t-\theta)^2]=\Theta(\tau^{-4})\EPt[ψτ′(Xt−θ)2]=Θ(τ−4), not O(τ−1)O(\tau^{-1})O(τ−1). Similarly, (ψτ2)′=2ψτψτ′(\psi_\tau^2)'=2\psi_\tau\psi_\tau'(ψτ2)′=2ψτψτ′ grows linearly in ∣u∣|u|∣u∣ on the tails, so its L2L^2L2 under a general PtP_tPt depends on higher moments that are not controlled by a W2W_2W2 constraint.
* (d) Dependence on 4th moments of QQQ. Later you use \EQ[ψ4]≲τ⋆−2\E_Q[\psi^4]\lesssim \tau_\star^{-2}\EQ[ψ4]≲τ⋆−2 and ∥ψ∥∞≲τ⋆−1/2\|\psi\|_\infty\lesssim \tau_\star^{-1/2}∥ψ∥∞≲τ⋆−1/2. Neither is true for the current ψ\psiψ. For the Gaussian tails we have ∣ψ(u)∣≍∣u∣/(τ2)|\psi(u)|\asymp |u|/(\tau^2)∣ψ(u)∣≍∣u∣/(τ2) for large ∣u∣|u|∣u∣; thus ψ\psiψ is unbounded, and \EQ[ψ4]\E_Q[\psi^4]\EQ[ψ4] can be arbitrarily large—even with W2(Q,μθ)≤εW_2(Q,\mu_\theta)\le\varepsilonW2(Q,μθ)≤ε—by sending a tiny mass δ\deltaδ far out (cost ∼δR2\sim \delta R^2∼δR2 in W22W_2^2W22), while \EQ[ψ4]∼δR4/τ8→∞\E_Q[\psi^4]\sim \delta R^4/\tau^8\to\infty\EQ[ψ4]∼δR4/τ8→∞ as R→∞R\to\inftyR→∞. So the uniform bound in \eqref{eq:risk-decomp} that relies on \EQ[ψ4]\E_Q[\psi^4]\EQ[ψ4] is not valid over the whole W2W_2W2-ball.
* Bottom line. Lemma \ref{lem:score-moments} as stated is false. Its proof contains (i) an inapplicable use of Lemma \ref{lem:dyn-W2}, (ii) incorrect support/scaling claims, and (iii) reliance on moment bounds that fail uniformly over the W2W_2W2-ball.
* Two viable fixes (choose one and adjust the rest accordingly): Fix A (localize / clip the score). Replace ψτ\psi_\tauψτ by a clipped score ψ~τ,B\tilde\psi_{\tau,B}ψ~τ,B that coincides with ψτ\psi_\tauψτ on the region where ∣u∣≤12−τ+c τ|u|\le \tfrac12-\tau + c\,\tau∣u∣≤21−τ+cτ and is truncated outside so that ψ~τ,B\tilde\psi_{\tau,B}ψ~τ,B and ψ~τ,B′\tilde\psi_{\tau,B}'ψ~τ,B′ are bounded and Lipschitz uniformly in uuu. Then Lemma \ref{lem:dyn-W2} applies with ∫01\EPt[ψ~τ,B′(Xt)2] dt≲τ−1,\int_0^1\E_{P_t}[\tilde\psi_{\tau,B}'(X_t)^2]\,dt \lesssim \tau^{-1},∫01\EPt[ψ~τ,B′(Xt)2]dt≲τ−1, and one can prove ∣\EQ[ψ~τ,B]−\Eνθ,τ[ψ~τ,B]∣≲ετ,∣\EQ[ψ~τ,B′]−\Eνθ,τ[ψ~τ,B′]∣≲ετ,\big|\E_Q[\tilde\psi_{\tau,B}]-\E_{\nu_{\theta,\tau}}[\tilde\psi_{\tau,B}]\big| \lesssim \frac{\varepsilon}{\sqrt{\tau}},\quad \big|\E_Q[\tilde\psi_{\tau,B}']-\E_{\nu_{\theta,\tau}}[\tilde\psi_{\tau,B}']\big| \lesssim \frac{\varepsilon}{\sqrt{\tau}},\EQ[ψ~τ,B]−\Eνθ,τ[ψ~τ,B]≲τε,\EQ[ψ~τ,B′]−\Eνθ,τ[ψ~τ,B′]≲τε, uniformly over Q:W2(Q,μθ)≤εQ:W_2(Q,\mu_\theta)\le\varepsilonQ:W2(Q,μθ)≤ε. The clipped estimator coincides with the unclipped one whenever no observation falls in the extreme tails; for the W2W_2W2-ball, the clipping error can be made negligible at the τ\tauτ-scales of interest, and all fourth‑moment issues vanish. Fix B (smooth everywhere, not only the edges). Replace the “flat‑top + Gaussian tails” fff by a globally smoothed log‑concave kernel (e.g. convolve μθ\mu_\thetaμθ with a C∞C^\inftyC∞ compactly supported bump of width ≍τ\asymp\tau≍τ); then ψτ∈C∞\psi_\tau\in C^\inftyψτ∈C∞, ψτ′,ψτ′′\psi_\tau',\psi_\tau''ψτ′,ψτ′′ are bounded and supported in a strip of width O(τ)O(\tau)O(τ), and Lemma \ref{lem:dyn-W2} yields exactly your τ\tauτ-scalings. This also resolves uniqueness (see 3) below). I include drop‑in text for Fix A (clipping) in §III.B and for Fix B (global smoothing) in §III.C. Either path repairs the entire chain of arguments and preserves the claimed ε2/3/n\varepsilon^{2/3}/nε2/3/n rate and constants‑up‑to‑≍\asymp≍.

===== - What’s written. “It follows from our analysis that this solution is unique.” =====
* What happens. With your fff, log⁡f(⋅;θ,τ)\log f(\cdot;\theta,\tau)logf(⋅;θ,τ) is flat (constant) on ∣x−θ∣≤12−τ|x-\theta|\le\tfrac12-\tau∣x−θ∣≤21−τ. For samples X1′,…,Xn′X_1',\dots,X_n'X1′,…,Xn′ with max⁡iXi′−min⁡iXi′≤1−2τ\max_i X_i'-\min_i X_i'\le 1-2\taumaxiXi′−miniXi′≤1−2τ, the set I=⋂i=1n[ Xi′−(12−τ), Xi′+(12−τ)]\mathcal{I}=\bigcap_{i=1}^n\big[\,X_i'-(\tfrac12-\tau),\, X_i'+(\tfrac12-\tau)\big]I=i=1⋂n[Xi′−(21−τ),Xi′+(21−τ)] is a nonempty interval, and for all t∈It\in\mathcal{I}t∈I one has ψτ(Xi′−t)=0\psi_\tau(X_i'-t)=0ψτ(Xi′−t)=0, hence Gn(t)=0G_n(t)=0Gn(t)=0. Thus every t∈It\in\mathcal{I}t∈I solves \eqref{eq:M-est}; the solution need not be unique. This event has positive probability (1−2τ)n(1-2\tau)^n(1−2τ)n under Q=μθQ=\mu_\thetaQ=μθ.
* Fix. Either: (i) tie‑break the zero set by a deterministic rule (e.g. choose the midpoint of the interval of roots, or the maximizer of the concave log‑likelihood, or the smallest root), and state uniqueness is not guaranteed; or (ii) adopt Fix B above, which makes log⁡f\log flogf strictly concave in ttt and restores uniqueness. See §III.D for a drop‑in sentence/footnote.

===== - As noted, ψτ′\psi_\tau'ψτ′ has a jump at the two junctions; ψτ′′\psi_\tau''ψτ′′ is not a bounded function, so the step ∣mn(tˉ)−mn(θ)∣≤∥ψ′′∥∞ ∣tˉ−θ∣|m_n(\bar t)-m_n(\theta)|\le \|\psi''\|_\infty\,|\bar t-\theta|∣mn(tˉ)−mn(θ)∣≤∥ψ′′∥∞∣tˉ−θ∣ is not justified. With either Fix A (clipped score is globally Lipschitz) or Fix B (globally smoothed fff so ψ′′\psi''ψ′′ bounded), the bound becomes correct and the subsequent “absorption” step is valid as written. See §III.B/§III.C. =====

===== - You write (just after \eqref{eq:risk-decomp}): \EQ[ψ]2m(Q)2 ≤ C ε2 τ⋆I⋆2  =  C ε2 τ⋆3  =  C ε4/c.\frac{\E_Q[\psi]^2}{m(Q)^2}\ \le\ C\,\frac{\varepsilon^2\,\tau_\star}{I_\star^2} \;=\; C\,\varepsilon^2\,\tau_\star^3 \;=\; C\,\varepsilon^4/c.m(Q)2\EQ[ψ]2 ≤ CI⋆2ε2τ⋆=Cε2τ⋆3=Cε4/c. The equality (ε2 τ⋆)/I⋆2=ε2 τ⋆3(\varepsilon^2\,\tau_\star)/I_\star^2 = \varepsilon^2\,\tau_\star^3(ε2τ⋆)/I⋆2=ε2τ⋆3 is wrong since I⋆=π/τ⋆I_\star=\pi/\tau_\starI⋆=π/τ⋆ entails 1/I⋆2=τ⋆2/π21/I_\star^2=\tau_\star^2/\pi^21/I⋆2=τ⋆2/π2. The correct scaling is \EQ[ψ]2m(Q)2 ≲ ε2τ⋆,\frac{\E_Q[\psi]^2}{m(Q)^2}\ \lesssim\ \varepsilon^2 \tau_\star,m(Q)2\EQ[ψ]2 ≲ ε2τ⋆, i.e. O(ε8/3)O(\varepsilon^{8/3})O(ε8/3) when τ⋆=ε2/3\tau_\star=\varepsilon^{2/3}τ⋆=ε2/3. (This does not hurt the final rate comparison.) =====
* Drop‑in correction: see §III.E.

==== 1. Typo in Theorem statement. “for all n≥1n\ge 1n≥1 and ε∈(0,12]\varepsilon\in(0,\tfrac12]ε∈(0,21],s” → trailing “s” should be removed. (§III.F) ====
# Lemma \ref{lem:info-W2}, Gaussian integral constant. You write “∫0∞K′(y)2/K(y) dy=1\int_0^\infty K'(y)^2/K(y)\,dy=1∫0∞K′(y)2/K(y)dy=1” but use π/τ\pi/\tauπ/τ downstream. The correct value is ∫0∞K′(y)2K(y) dy=∫0∞y2K(y) dy=12.\int_0^\infty \frac{K'(y)^2}{K(y)}\,dy=\int_0^\infty y^2 K(y)\,dy=\frac{1}{2}.∫0∞K(y)K′(y)2dy=∫0∞y2K(y)dy=21. With K(0)=1/2πK(0)=1/\sqrt{2\pi}K(0)=1/2π, this yields I(τ)=π/τI(\tau)=\pi/\tauI(τ)=π/τ as claimed. (§III.G)
# Language about “edge strips of width τ\tauτ”. It is true that νθ,τ\nu_{\theta,\tau}νθ,τ assigns mass τ\tauτ to each side beyond θ±(12−τ)\theta\pm(\tfrac12-\tau)θ±(21−τ), but the tails extend to +∞+\infty+∞. When discussing supports of derivatives, say “outside the bulk” rather than “on strips of width τ\tauτ”.
# “Strictly decreasing C1C^1C1 score”. For the present fff, ψτ\psi_\tauψτ is continuous and piecewise linear in uuu with a derivative jump at the junctions; it is not C1C^1C1. Replace by “continuous, nonincreasing, and piecewise C1C^1C1” (or adopt Fix B so it is C1C^1C1). (§III.H)

==== Below are literal replacements you can paste into your file. ====

===== Replace Lemma \ref{lem:dyn-W2} by: =====

<syntaxhighlight lang="latex">\begin{lemma}[Displacement $W_2$ control of expectation differences]\label{lem:dyn-W2}
Let $P,Q$ be Borel probability measures on $\R$ with finite second moments, and let $T$ be the monotone transport pushing $P$ to $Q$. For $t\in[0,1]$ set $T_t(x)=(1-t)x+tT(x)$ and $X_t=T_t(X)$ with $X\sim P$, and let $P_t$ denote the law of $X_t$. Then, for any $C^1$ function $\varphi$ such that $\int_0^1 \E_{P_t}[\varphi'(X_t)^2]\,dt<\infty$,
\[
\left|\E_Q[\varphi]-\E_P[\varphi]\right|
\le \left(\int_0^1 \E_{P_t}[\varphi'(X_t)^2]\,dt\right)^{1/2}\, W_2(P,Q).
\]
\end{lemma}

\begin{proof}
By the fundamental theorem of calculus and the chain rule,
\[
\E_Q[\varphi]-\E_P[\varphi]
=\int_0^1 \frac{d}{dt}\E[\varphi(X_t)]\,dt
=\int_0^1 \E\big[\varphi'(X_t)\cdot (T(X)-X)\big]\,dt.
\]
Applying Cauchy--Schwarz in $(t,\Omega)$ gives
\[
\left|\E_Q[\varphi]-\E_P[\varphi]\right|
\le \Big(\int_0^1 \E[\varphi'(X_t)^2]\,dt\Big)^{1/2}
\Big(\int_0^1 \E[(T(X)-X)^2]\,dt\Big)^{1/2}.
\]
Since $\int_0^1 \E[(T(X)-X)^2]\,dt=\E[(T(X)-X)^2]=W_2(P,Q)^2$, the claim follows. Finally, $\E[\varphi'(X_t)^2]=\E_{P_t}[\varphi'(Z)^2]$ by definition of $P_t$.
\end{proof}

</syntaxhighlight>

===== Insert after (2.1) defining ψτ\psi_\tauψτ: =====

<syntaxhighlight lang="latex">\medskip\noindent\textbf{Clipped score.}
For $B\ge 1$, define the clipped score
\[
\tilde\psi_{\tau,B}(u)\ :=\ \mathrm{clip}\big(\psi_\tau(u),\,[-B/\sqrt{\tau},\,B/\sqrt{\tau}]\big),
\]
and let $\tilde\psi'_{\tau,B}$ be any selection of its (a.e.) derivative; then $\|\tilde\psi_{\tau,B}\|_\infty\le B/\sqrt{\tau}$ and $\|\tilde\psi'_{\tau,B}\|_\infty\le C/\tau$ uniformly in $B$.
We henceforth use $\tilde\psi_{\tau_\star,B}$ in the estimating equation \eqref{eq:M-est}, and write $\hat\theta_\ep$ for any solution. The clipping level $B$ is fixed (e.g. $B=10$); on the $W_2$-ball all bounds below are uniform in $B$ and coincide with the unclipped estimator whenever no sample point falls in the extreme tails.

</syntaxhighlight>

Replace Lemma \ref{lem:score-moments} by the following (stated and proved for the clipped score):

<syntaxhighlight lang="latex">\begin{lemma}[Uniform control of clipped score moments]\label{lem:score-moments-clipped}
Fix $\tau\in(0,\tfrac12]$ and let $\tilde\psi_\tau:=\tilde\psi_{\tau,B}$ be the clipped score defined above. There exist constants $C_1,C_2,C_3<\infty$ (independent of $B$) such that for all $Q$ with $W_2(Q,\mu_\theta)\le \ep$,
\begin{align}
\left|\E_Q[\tilde\psi_\tau]-\E_{\nu_{\theta,\tau}}[\tilde\psi_\tau]\right| &\le C_1\,\frac{\ep}{\sqrt{\tau}},\label{eq:bias-score-clipped}\\
\left|\E_Q[\tilde\psi'_\tau]-\E_{\nu_{\theta,\tau}}[\tilde\psi'_\tau]\right| &\le C_2\,\frac{\ep}{\sqrt{\tau}},\label{eq:curv-score-clipped}\\
\left|\E_Q[\tilde\psi_\tau^2]-\E_{\nu_{\theta,\tau}}[\tilde\psi_\tau^2]\right| &\le C_3\,\frac{\ep}{\tau^{3/2}}.\label{eq:var-score-clipped}
\end{align}
\end{lemma}

\begin{proof}
Apply Lemma~\ref{lem:dyn-W2} with $P=\nu_{\theta,\tau}$, $\varphi=\tilde\psi_\tau$, $\varphi=\tilde\psi'_\tau$, and $\varphi=\tilde\psi_\tau^2$, respectively. Since $\|\tilde\psi'_\tau\|_\infty\lesssim \tau^{-1}$ and $\|(\tilde\psi_\tau^2)'\|_\infty\lesssim \tau^{-3/2}$, we have
\[
\int_0^1 \E_{P_t}[(\tilde\psi'_\tau(X_t-\theta))^2]\,dt \lesssim \tau^{-1},\quad
\int_0^1 \E_{P_t}[((\tilde\psi_\tau^2)'(X_t-\theta))^2]\,dt \lesssim \tau^{-3},
\]
uniformly over $Q$ with $W_2(Q,\mu_\theta)\le \ep$. The claim follows using $W_2(Q,\nu_{\theta,\tau})\le \ep + W_2(\mu_\theta,\nu_{\theta,\tau})$ and Lemma~\ref{lem:info-W2}.
\end{proof}

</syntaxhighlight>

Adjust the M‑estimation remainder bounds (replace all occurrences of ψ,ψ′,ψ′′\psi,\psi',\psi''ψ,ψ′,ψ′′ by ψ~τ,ψ~τ′\tilde\psi_\tau,\tilde\psi'_\tauψ~τ,ψ~τ′, and remove every use of ∥ψ′′∥∞\|\psi''\|_\infty∥ψ′′∥∞. Where you currently use Lipschitz in ttt via ∥ψ′′∥∞\|\psi''\|_\infty∥ψ′′∥∞, replace it by the uniform bound ∥ψ~τ′∥∞≲τ−1\|\tilde\psi'_\tau\|_\infty\lesssim\tau^{-1}∥ψ~τ′∥∞≲τ−1 and the simple inequality

∣mn(tˉ)−mn(θ)∣≤1n∑i=1n∣ψ~τ′(Xi′−tˉ)−ψ~τ′(Xi′−θ)∣≤∥ψ~τ′∥Lip ∣tˉ−θ∣≲τ−1 ∣tˉ−θ∣.|m_n(\bar t)-m_n(\theta)| \le \frac{1}{n}\sum_{i=1}^n\big|\tilde\psi'_\tau(X_i'-\bar t)-\tilde\psi'_\tau(X_i'-\theta)\big|
\le \|\tilde\psi'_\tau\|_\mathrm{Lip}\,|\bar t-\theta|
\lesssim \tau^{-1}\,|\bar t-\theta|.∣mn(tˉ)−mn(θ)∣≤n1i=1∑nψ~τ′(Xi′−tˉ)−ψ~τ′(Xi′−θ)≤∥ψ~τ′∥Lip∣tˉ−θ∣≲τ−1∣tˉ−θ∣.
Finally, every appearance of \EQ[ψ4]\E_Q[\psi^4]\EQ[ψ4] can be replaced by ∥ψ~τ∥∞4≲τ−2\|\tilde\psi_\tau\|_\infty^4\lesssim \tau^{-2}∥ψ~τ∥∞4≲τ−2, uniformly over the W2W_2W2-ball.

===== If you prefer to avoid clipping, replace \eqref{eq:nu-density} by a globally smoothed bump (e.g. convolution with a compactly supported C∞C^\inftyC∞ kernel κτ\kappa_\tauκτ): =====

<syntaxhighlight lang="latex">\medskip
Define $\kappa\in C^\infty_c(\R)$, $\kappa\ge0$, $\int \kappa=1$, and set $\kappa_\tau(x)=\tau^{-1}\kappa(x/\tau)$. Let
\[
f(x;\theta,\tau) = \big(\mathbf{1}_{[\theta-\frac12,\ \theta+\frac12]} * \kappa_\tau\big)(x).
\]
Then $f(\cdot;\theta,\tau)\in C^\infty$, strictly log‑concave, and $\psi_\tau(u)=-\partial_u\log f(u)$ together with its derivatives satisfy $\|\psi_\tau\|_\infty\lesssim \tau^{-1/2}$, $\|\psi_\tau'\|_\infty\lesssim \tau^{-1}$, $\|\psi_\tau''\|_\infty\lesssim \tau^{-3/2}$. All calculations below carry through with the same $\tau$‑scalings \eqref{eq:bias-score}–\eqref{eq:var-score}.

</syntaxhighlight>

With this modification, Lemma \ref{lem:score-moments} (your original statement) becomes correct; you can keep it as‑is but replace the justification paragraph by the above C∞C^\inftyC∞‐bounds rather than “edge strips” heuristics.

===== Replace the sentence right after \eqref{eq:M-est} by: =====

<syntaxhighlight lang="latex">Let $\hat\theta_\ep$ be any measurable selection from the (nonempty) solution set of \eqref{eq:M-est}.
For the globally smoothed model (Remark~\ref{rem:global-smooth}) this solution is unique by strict concavity of the log‑likelihood; for the flat‑top model the solution need not be unique on the event $\max_i X_i'-\min_i X_i'\le 1-2\tau_\star$, in which case we choose the midpoint of the interval of roots.

</syntaxhighlight>

(If you adopt III.C, add a short Remark saying uniqueness holds a.s. because the log‑likelihood is strictly concave in ttt.)

===== Replace the bias line by: =====

<syntaxhighlight lang="latex">By Lemma~\ref{lem:score-moments-clipped}, $|\E_Q[\psi]|\lesssim \ep/\sqrt{\tau_\star}$ and $m(Q)\asymp I_\star=\pi/\tau_\star$, hence
\[
\frac{\E_Q[\psi]^2}{m(Q)^2}\ \lesssim\ \frac{\ep^2/\tau_\star}{(\pi/\tau_\star)^2}
\ =\ \frac{\ep^2\,\tau_\star}{\pi^2}
\ =\ O\!\big(\ep^{8/3}\big)\quad\text{when } \tau_\star=\ep^{2/3}.
\]

</syntaxhighlight>

(This is strictly smaller than the variance term C/(nI⋆)C/(n I_\star)C/(nI⋆) in the regime where the IDS term dominates.)

===== Replace line in Theorem \ref{thm:IDS-sharp} by: =====

<syntaxhighlight lang="latex">There exists a constant $C<\infty$ such that, for all $n\ge 1$ and $\ep\in(0,\tfrac12]$,

</syntaxhighlight>

===== Replace the end of the computation of I(τ)I(\tau)I(τ) by: =====

<syntaxhighlight lang="latex">=\frac{1}{\tau K(0)^2}\int_0^\infty \frac{K'(y)^2}{K(y)}\,dy
=\frac{1}{\tau K(0)^2}\int_0^\infty y^2 K(y)\,dy
=\frac{1}{\tau K(0)^2}\cdot \frac{1}{2}
=\frac{\pi}{\tau},

</syntaxhighlight>

and modify the explanatory sentence to:

<syntaxhighlight lang="latex">where we used $K(0)=1/\sqrt{2\pi}$ and the Gaussian identity $\int_0^\infty y^2 K(y)\,dy=\tfrac12$.

</syntaxhighlight>

===== Replace (A1) by a version that is true under either III.B or III.C: =====

<syntaxhighlight lang="latex">\item[(A1)] The score $\psi$ is continuous, nonincreasing, and piecewise $C^1$ with bounded derivative; in particular $\|\psi\|_\infty<\infty$ and $\|\psi'\|_\infty<\infty$. 
% (Under the globally smoothed model, $\psi\in C^2$ and $\|\psi''\|_\infty<\infty$ as well.)

</syntaxhighlight>

If you adopt III.C, you may keep the stronger $C^2$ version and the uses of $|\psi''|\infty$; if you adopt III.B (clipping), simply remove every use of $|\psi''|\infty$ and replace with the Lipschitz bound in ψ′\psi'ψ′ as indicated above.

==== With either fix (clipping or global smoothing), you recover: ====
* m(Q)=I(τ)+O(ε/τ)m(Q)=I(\tau)+O(\varepsilon/\sqrt{\tau})m(Q)=I(τ)+O(ε/τ) uniformly on the W2W_2W2-ball,
* \VarQ(ψ)≤I(τ)+O(ε/τ3/2)\Var_Q(\psi)\le I(\tau)+O(\varepsilon/\tau^{3/2})\VarQ(ψ)≤I(τ)+O(ε/τ3/2),
* all remainder terms are O(τ/n)O\big(\tau/n\big)O(τ/n) or smaller,

so optimizing at τ⋆≍ε2/3\tau_\star\asymp \varepsilon^{2/3}τ⋆≍ε2/3 yields the stated

sup⁡Q:W2(Q,μθ)≤ε\EQ[(θ^ε−θ)2] ≲ ε2/3n ∨ (ε2+12(n+1)(n+2)).\sup_{Q:W_2(Q,\mu_\theta)\le\varepsilon}\E_Q[(\hat\theta_\varepsilon-\theta)^2]
\ \lesssim\ \frac{\varepsilon^{2/3}}{n}\ \vee\ \Big(\varepsilon^2+\tfrac{1}{2(n+1)(n+2)}\Big).Q:W2(Q,μθ)≤εsup\EQ[(θ^ε−θ)2] ≲ nε2/3 ∨ (ε2+2(n+1)(n+2)1).
(The only change is the corrected bias calculation, which is strictly smaller and therefore harmless for the rate comparison.)

==== - In Lemma \ref{lem:info-W2} you can slightly streamline the W2W_2W2 computation by the change of variables q=(1−2τ)+2τS(u)q=(1-2\tau)+2\tau S(u)q=(1−2τ)+2τS(u), which gives the “cubic constant” in one line: W22(μθ,νθ,τ)=8τ3∫0∞(S(u)−12−K(0)u)2 K(u) du,W_2^2(\mu_\theta,\nu_{\theta,\tau}) = 8\tau^3 \int_0^\infty \Big(S(u)-\tfrac12 - K(0)u\Big)^2\,K(u)\,du,W22(μθ,νθ,τ)=8τ3∫0∞(S(u)−21−K(0)u)2K(u)du, from which your expression for ccc follows after evaluating ∫u2K(u)du\int u^2K(u)du∫u2K(u)du and ∫u(2S(u)−1)K(u) du\int u(2S(u)-1)K(u)\,du∫u(2S(u)−1)K(u)du. ====
* If you prefer to state the theorem for the unclipped score and the present fff, add a succinct assumption that rules out adversarial fourth‑moment explosions, e.g. “assume QQQ has finite fourth moment uniformly on the W2W_2W2-ball” or “assume QQQ is dominated by a sub‑Gaussian envelope,” but this weakens the minimax statement. The clipping (III.B) avoids this altogether.

===== 1. Fix Lemma \ref{lem:dyn-W2} (velocity field) — §III.A. =====
# Replace Lemma \ref{lem:score-moments} by a valid version. Easiest is clipping (§III.B) or use a globally smoothed fff (§III.C).
# Acknowledge/resolve non‑uniqueness (tie‑break or global smoothing) — §III.D.
# Correct the bias-term algebra (use ε2τ⋆\varepsilon^2\tau_\starε2τ⋆, not ε2τ⋆3\varepsilon^2\tau_\star^3ε2τ⋆3) — §III.E.
# Fix constants/typos — §III.F–G, and relax (A1) as per §III.H.

With these amendments, the proof becomes sound and delivers the claimed sharp IDS rate ε2/3/n\varepsilon^{2/3}/nε2/3/n (up to constants).