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

==== 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}

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===== 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.