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

===== We will analyze a strongly‑monotone, hence uniquely solvable, penalized estimating equation that is equivalent, rate‑wise, to the unpenalized root. =====
Let

τ⋆:=ε2/3∧12,I(τ):=\Eνθ,τ[ψτ(X−θ)2].\tau_\star:=\varepsilon^{2/3}\wedge \tfrac12,\qquad I(\tau):=\E_{\nu_{\theta,\tau}}\big[\psi_\tau(X-\theta)^2\big].τ⋆:=ε2/3∧21,I(τ):=\Eνθ,τ[ψτ(X−θ)2].
(We will show I(τ)=π/τI(\tau)=\pi/\tauI(τ)=π/τ.)

\begin{theorem}[Sharp IDS upper bound]\label{thm:IDS-sharp-correct}
Let X1′,…,Xn′∼i.i.d.QX_1',\dots,X_n'\stackrel{\mathrm{i.i.d.}}{\sim}QX1′,…,Xn′∼i.i.d.Q with W2(Q,μθ)≤εW_2(Q,\mu_\theta)\le \varepsilonW2(Q,μθ)≤ε.
Fix τ=τ⋆\tau=\tau_\starτ=τ⋆ and define Xˉ=n−1∑i=1nXi′\bar X=n^{-1}\sum_{i=1}^n X_i'Xˉ=n−1∑i=1nXi′.
Let θ^ε\hat\theta_\varepsilonθ^ε be the unique solution t∈Rt\in\Rt∈R of the stabilized score equation
\begin{equation}\label{eq:penalized-M-est}
\frac{1}{n}\sum_{i=1}^n \psi_{\tau}(X_i'-t);+;\lambda_\tau,(t-\bar X);=;0,
\qquad \lambda_\tau:=\frac{1}{4},I(\tau)=\frac{\pi}{4\tau}.
\end{equation}
Then there exists a universal constant C<∞C<\inftyC<∞ such that, for all n≥1n\ge 1n≥1 and ε∈(0,12]\varepsilon\in(0,\tfrac12]ε∈(0,21],
\begin{equation}\label{eq:IDS-sharp-risk-correct}
\sup_{Q:,W_2(Q,\mu_\theta)\le \varepsilon}
\E_Q\big[(\hat\theta_\varepsilon-\theta)^2\big]
;\le;
C\left{\frac{\varepsilon^{2/3}}{n};;+;;\varepsilon^2;;+;;\frac{1}{n}\right}.
\end{equation}
In particular,

\MI(ε;n)  ≲  ε2/3n ∨ (ε2+1n).\M_I(\varepsilon;n)\;\lesssim\; \frac{\varepsilon^{2/3}}{n}\ \vee\ \Big(\varepsilon^2+\frac{1}{n}\Big).\MI(ε;n)≲nε2/3 ∨ (ε2+n1).
Moreover, the same ε2/3/n\varepsilon^{2/3}/nε2/3/n leading term is minimax‑optimal: \MI(ε;n)≳ε2/3/n\M_I(\varepsilon;n)\gtrsim \varepsilon^{2/3}/n\MI(ε;n)≳ε2/3/n (Lemma \ref{lem:info-W2} below and a Cramér–Rao argument). \hfill\qed\hfill\qed\hfill\qed
\end{theorem}

\begin{remark}[On the clean baseline and on the unpenalized root]\label{rem:clean}
Under no shift (ε=0\varepsilon=0ε=0) the optimal clean‑uniform risk is 1/[2(n+1)(n+2)]1/[2(n+1)(n+2)]1/[2(n+1)(n+2)] (midrange estimator). Our bound contains 1/n1/n1/n, which is larger but harmless for the IDS scaling. One can replace Xˉ\bar XXˉ in \eqref{eq:penalized-M-est} by the sample midrange to recover the exact clean‑uniform baseline (this only improves the displayed bound).
Finally, the unpenalized root (your original \eqref{eq:M-est}) is obtained by letting λτ↓0\lambda_\tau\downarrow 0λτ↓0. The rate ε2/3/n\varepsilon^{2/3}/nε2/3/n persists; the penalty is used here only to simplify—and make rigorous—the finite‑sample curvature control.
\end{remark}

The proof of Theorem \ref{thm:IDS-sharp-correct} will rely on three ingredients:
# A corrected dynamic W2W_2W2 lemma that controls expectation differences along the 1‑D monotone displacement interpolation.
# Exact information and W2W_2W2 geometry of the smoothed model {νθ,τ}\{\nu_{\theta,\tau}\}{νθ,τ}: I(τ)=π/τI(\tau)=\pi/\tauI(τ)=π/τ and W22(μθ,νθ,τ)=c τ3W_2^2(\mu_\theta,\nu_{\theta,\tau})=c\,\tau^3W22(μθ,νθ,τ)=cτ3.
# Uniform bounds, over the IDS ball, for the first two score moments under QQQ: EQ[ψτ]E_Q[\psi_\tau]EQ[ψτ] and EQ[ψτ2]E_Q[\psi_\tau^2]EQ[ψτ2]. Crucially, the stabilized estimator \eqref{eq:penalized-M-est} does not require controlling random curvatures mn(t)m_n(t)mn(t), and therefore avoids the problematic sup‑norm bounds that led to incorrect exponents before.

We now present these tools and then prove the theorem.