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

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* 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.