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

==== We study μθ=\Unif[θ−12,θ+12]\mu_\theta=\Unif[\theta-\tfrac12,\theta+\tfrac12]μθ=\Unif[θ−21,θ+21]. ====
Under IDS we observe 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.
Risk is squared error R(θ,θ^)=\EQ[(θ^−θ)2]R(\theta,\hat\theta)=\E_Q[(\hat\theta-\theta)^2]R(θ,θ^)=\EQ[(θ^−θ)2], and

\MI(ε;n)=inf⁡θ^ sup⁡Q: W2(Q,μθ)≤εR(θ,θ^).\M_I(\varepsilon;n)=\inf_{\hat\theta}\ \sup_{Q:\,W_2(Q,\mu_\theta)\le \varepsilon} R(\theta,\hat\theta).\MI(ε;n)=θ^inf Q:W2(Q,μθ)≤εsupR(θ,θ^).
===== Let K(x)=(2π)−1/2e−x2/2K(x)=(2\pi)^{-1/2}e^{-x^2/2}K(x)=(2π)−1/2e−x2/2 denote the standard normal density, SSS its CDF, and fix a smoothing parameter τ∈(0,12]\tau\in(0,\tfrac12]τ∈(0,21]. =====
We define a location family {νθ,τ}θ∈R\{\nu_{\theta,\tau}\}_{\theta\in\R}{νθ,τ}θ∈R by replacing the two edges of the uniform with Gaussian tails; its density is
\begin{equation}\label{eq:nu-density-correct}
f(x;\theta,\tau)=
\begin{cases}
1 & \text{if } |x-\theta|\le \tfrac12-\tau,\[4pt]
\dfrac{1}{K(0)},K!\left(\dfrac{|x-\theta|-(\tfrac12-\tau)}{2\tau K(0)}\right) & \text{if } |x-\theta|>\tfrac12-\tau.
\end{cases}
\end{equation}
This density is C1C^1C1, equals 1 on the bulk, and is strictly log‑concave on the two tails.
Write

ψτ(u)  =  ∂θlog⁡f(θ+u;θ,τ)  =  −∂ulog⁡f(θ+u;θ,τ)\psi_\tau(u) \;=\; \partial_\theta \log f(\theta+u;\theta,\tau)\;=\;-\partial_u\log f(\theta+u;\theta,\tau)ψτ(u)=∂θlogf(θ+u;θ,τ)=−∂ulogf(θ+u;θ,τ)
for the location score at θ\thetaθ, and ψτ′(u)=−∂u2log⁡f(θ+u;θ,τ)\psi_\tau'(u)=-\partial_u^2\log f(\theta+u;\theta,\tau)ψτ′(u)=−∂u2logf(θ+u;θ,τ).
On the bulk {∣u∣≤12−τ}\{|u|\le \tfrac12-\tau\}{∣u∣≤21−τ}, ψτ(u)=0\psi_\tau(u)=0ψτ(u)=0; on each tail ψτ\psi_\tauψτ is linear with constant slope
\begin{equation}\label{eq:slope}
\psi_\tau'(u)\equiv \frac{1}{(2\tau K(0))^2}=\frac{\pi}{2,\tau^2}\qquad\text{whenever }|u|>\tfrac12-\tau,
\end{equation}
and ψτ\psi_\tauψτ is continuous across the junctions.

===== 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.

===== \begin{lemma}[Displacement W2W_2W2 control]\label{lem:dyn-W2-correct} =====
Let P,QP,QP,Q be probability laws on R\RR with finite second moments, and TTT the increasing transport map pushing PPP to QQQ. For t∈[0,1]t\in[0,1]t∈[0,1], 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) with X∼PX\sim PX∼P; write PtP_tPt for the law of XtX_tXt.
For any C1C^1C1 function φ\varphiφ such that ∫01\EPt[φ′(Xt)2] dt<∞\int_0^1 \E_{P_t}[\varphi'(X_t)^2]\,dt<\infty∫01\EPt[φ′(Xt)2]dt<∞,

∣\EQ[φ]−\EP[φ]∣  ≤  (∫01\EPt[φ′(Xt)2] dt)1/2 W2(P,Q).\big|\E_Q[\varphi]-\E_P[\varphi]\big|
\;\le\;
\Big(\int_0^1 \E_{P_t}[\varphi'(X_t)^2]\,dt\Big)^{1/2}\, W_2(P,Q).\EQ[φ]−\EP[φ]≤(∫01\EPt[φ′(Xt)2]dt)1/2W2(P,Q).
\end{lemma}

\begin{proof}
By the chain rule and the fact that X˙t=T(X)−X\dot X_t=T(X)-XX˙t=T(X)−X,

ddt\E[φ(Xt)]=\E[φ′(Xt) (T(X)−X)].\frac{d}{dt}\E[\varphi(X_t)]=\E[\varphi'(X_t)\,(T(X)-X)].dtd\E[φ(Xt)]=\E[φ′(Xt)(T(X)−X)].
Integrating over t∈[0,1]t\in[0,1]t∈[0,1] and using Cauchy–Schwarz in (t,Ω)(t,\Omega)(t,Ω) gives

∣\EQ[φ]−\EP[φ]∣≤(∫01\E[φ′(Xt)2] dt)1/2(∫01\E[(T(X)−X)2] dt)1/2.\big|\E_Q[\varphi]-\E_P[\varphi]\big|
\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}.\EQ[φ]−\EP[φ]≤(∫01\E[φ′(Xt)2]dt)1/2(∫01\E[(T(X)−X)2]dt)1/2.
The second factor equals W2(P,Q)W_2(P,Q)W2(P,Q), and \E[φ′(Xt)2]=\EPt[φ′(Z)2]\E[\varphi'(X_t)^2]=\E_{P_t}[\varphi'(Z)^2]\E[φ′(Xt)2]=\EPt[φ′(Z)2].
\end{proof}

===== \begin{lemma}[Information and W2W_2W2 scalings]\label{lem:info-W2} =====
Let ψτ\psi_\tauψτ denote the location score of νθ,τ\nu_{\theta,\tau}νθ,τ. Then, for all θ∈R\theta\in\Rθ∈R and τ∈(0,12]\tau\in(0,\tfrac12]τ∈(0,21],
\begin{align}
I(\tau)\equiv \E_{\nu_{\theta,\tau}}!\big[\psi_\tau(X-\theta)^2\big]&=\frac{\pi}{\tau},\label{eq:I-tau-correct}\
W_2^2!\left(\mu_\theta,\nu_{\theta,\tau}\right)&=c,\tau^3,\qquad c=\frac{2\left(6-6\sqrt2+\pi\right)}{3\pi}.\label{eq:W2-tau-correct}
\end{align}
\end{lemma}

\begin{proof}
On the right tail x≥θ+12−τx\ge \theta+\tfrac12-\taux≥θ+21−τ, write

y=x−θ−(12−τ)2τK(0),ψτ(x−θ)=−∂∂xlog⁡f(x;θ,τ)=y2τK(0).y=\frac{x-\theta-(\tfrac12-\tau)}{2\tau K(0)},\qquad
\psi_\tau(x-\theta)=-\frac{\partial}{\partial x}\log f(x;\theta,\tau)
=\frac{y}{2\tau K(0)}.y=2τK(0)x−θ−(21−τ),ψτ(x−θ)=−∂x∂logf(x;θ,τ)=2τK(0)y.
By symmetry (two tails),

I(τ)=2∫θ+1/2−τ∞ψτ2 f dx.I(\tau)=2\int_{\theta+1/2-\tau}^\infty \psi_\tau^2\, f\,dx.I(τ)=2∫θ+1/2−τ∞ψτ2fdx.
With the change of variables dx=2τK(0) dydx=2\tau K(0)\,dydx=2τK(0)dy and f=(1/K(0))K(y)f=(1/K(0))K(y)f=(1/K(0))K(y),

I(τ)=2(2τK(0))2K(0)∫θ+1/2−τ∞K′(y)2/K(y) dx=1τK(0)2∫0∞K′(y)2K(y) dy.I(\tau)=\frac{2}{(2\tau K(0))^2 K(0)}\int_{\theta+1/2-\tau}^\infty K'(y)^2/K(y)\,dx
=\frac{1}{\tau K(0)^2}\int_0^\infty \frac{K'(y)^2}{K(y)}\,dy.I(τ)=(2τK(0))2K(0)2∫θ+1/2−τ∞K′(y)2/K(y)dx=τK(0)21∫0∞K(y)K′(y)2dy.
Since K′(y)=−yK(y)K'(y)=-yK(y)K′(y)=−yK(y), K′(y)2K(y)=y2K(y)\frac{K'(y)^2}{K(y)}=y^2K(y)K(y)K′(y)2=y2K(y), so ∫0∞y2K(y) dy=12\int_0^\infty y^2K(y)\,dy=\tfrac12∫0∞y2K(y)dy=21. With K(0)=1/2πK(0)=1/\sqrt{2\pi}K(0)=1/2π we obtain I(τ)=π/τI(\tau)=\pi/\tauI(τ)=π/τ.
For W2W_2W2, it suffices to compare quantiles: F1−1(q)=θ−1/2+qF_1^{-1}(q)=\theta-1/2+qF1−1(q)=θ−1/2+q (uniform) and, by inverting the right‑tail CDF of νθ,τ\nu_{\theta,\tau}νθ,τ,

F2−1(q)=θ+12−τ+2τK(0) S−1 ⁣(q−(1−2τ)2τ)for q∈[1−τ,1],F_2^{-1}(q)=\theta+\tfrac12-\tau+2\tau K(0)\,S^{-1}\!\left(\frac{q-(1-2\tau)}{2\tau}\right)
\quad\text{for } q\in[1-\tau,1],F2−1(q)=θ+21−τ+2τK(0)S−1(2τq−(1−2τ))for q∈[1−τ,1],
with symmetry on the left. A direct computation gives

W22(μθ,νθ,τ)=2∫1−τ1 ⁣(F1−1−F2−1)2 dq=c τ3,W_2^2(\mu_\theta,\nu_{\theta,\tau})
=2\int_{1-\tau}^1\!\big(F_1^{-1}-F_2^{-1}\big)^2\,dq
=c\,\tau^3,W22(μθ,νθ,τ)=2∫1−τ1(F1−1−F2−1)2dq=cτ3,
with ccc as claimed (elementary Gaussian integrals).
\end{proof}

===== 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}

===== Set τ=τ⋆\tau=\tau_\starτ=τ⋆ and abbreviate ψ=ψτ\psi=\psi_\tauψ=ψτ, I⋆=I(τ)=π/τI_\star=I(\tau)=\pi/\tauI⋆=I(τ)=π/τ, λ=λτ=I⋆/4\lambda=\lambda_\tau=I_\star/4λ=λτ=I⋆/4. =====
Define

Gn(t):=1n∑i=1nψ(Xi′−t),gQ(t):=\EQ[ψ(X′−t)].G_n(t):=\frac{1}{n}\sum_{i=1}^n \psi(X_i'-t),\qquad
g_Q(t):=\E_Q[\psi(X'-t)].Gn(t):=n1i=1∑nψ(Xi′−t),gQ(t):=\EQ[ψ(X′−t)].
The stabilized score map
Hn(t):=Gn(t)+λ(t−Xˉ)H_n(t):=G_n(t)+\lambda(t-\bar X)Hn(t):=Gn(t)+λ(t−Xˉ)
is strictly decreasing (because GnG_nGn is decreasing and λ>0\lambda>0λ>0 adds a linear drift), with lim⁡t→±∞Hn(t)=±∞\lim_{t\to\pm\infty}H_n(t)=\pm\inftylimt→±∞Hn(t)=±∞, hence \eqref{eq:penalized-M-est} has a unique solution.

By the mean‑value theorem, there exists a random tˉ\bar ttˉ on the segment between θ\thetaθ and θ^ε\hat\theta_\varepsilonθ^ε such that

0=Hn(θ^ε)−Hn(θ)=(Gn(θ^ε)−Gn(θ))+λ(θ^ε−θ)=−(mn(tˉ)−λ)(θ^ε−θ),0=H_n(\hat\theta_\varepsilon)-H_n(\theta)
=\big(G_n(\hat\theta_\varepsilon)-G_n(\theta)\big)+\lambda(\hat\theta_\varepsilon-\theta)
=-(m_n(\bar t)-\lambda)(\hat\theta_\varepsilon-\theta),0=Hn(θ^ε)−Hn(θ)=(Gn(θ^ε)−Gn(θ))+λ(θ^ε−θ)=−(mn(tˉ)−λ)(θ^ε−θ),
where mn(t):=−Gn′(t)=n−1∑iψ′(Xi′−t)≥0m_n(t):=-G_n'(t)=n^{-1}\sum_i \psi'(X_i'-t)\ge 0mn(t):=−Gn′(t)=n−1∑iψ′(Xi′−t)≥0. Hence
\begin{equation}\label{eq:MV-identity}
(\lambda+m_n(\bar t))(\hat\theta_\varepsilon-\theta)=G_n(\theta).
\end{equation}
Add and subtract gQ(θ)g_Q(\theta)gQ(θ) on the right:

Gn(θ)=(Gn(θ)−gQ(θ))+gQ(θ).G_n(\theta)=(G_n(\theta)-g_Q(\theta))+g_Q(\theta).Gn(θ)=(Gn(θ)−gQ(θ))+gQ(θ).
Therefore
\begin{equation}\label{eq:basic-sq}
(\hat\theta_\varepsilon-\theta)^2
\ \le\ \frac{2,(G_n(\theta)-g_Q(\theta))^2}{(\lambda+m_n(\bar t))^2}
\ +\ \frac{2,g_Q(\theta)^2}{(\lambda+m_n(\bar t))^2}.
\end{equation}
Since mn(tˉ)≥0m_n(\bar t)\ge 0mn(tˉ)≥0, we have (λ+mn(tˉ))−2≤λ−2=16/I⋆2(\lambda+m_n(\bar t))^{-2}\le \lambda^{-2}=16/I_\star^2(λ+mn(tˉ))−2≤λ−2=16/I⋆2. Taking expectation,
\begin{align}
\E_Q[(\hat\theta_\varepsilon-\theta)^2]
&\le \frac{32}{I_\star^2},\E_Q!\left[(G_n(\theta)-g_Q(\theta))^2\right]
* \frac{32}{I_\star^2},g_Q(\theta)^2.\label{eq:two-terms} \end{align} We now bound the two terms separately.

\medskip\noindent
\emph{(i) Empirical fluctuation term.}
By independence,
\EQ[(Gn(θ)−gQ(θ))2]=\VarQ(ψ(X′−θ))/n.\E_Q[(G_n(\theta)-g_Q(\theta))^2]=\Var_Q(\psi(X'-\theta))/n.\EQ[(Gn(θ)−gQ(θ))2]=\VarQ(ψ(X′−θ))/n.
Lemma \ref{lem:score-moments-correct} gives \EQ[ψ2]≤I⋆+C ε τ−5/2\E_Q[\psi^2]\le I_\star + C\,\varepsilon\,\tau^{-5/2}\EQ[ψ2]≤I⋆+Cετ−5/2; thus
\begin{equation}\label{eq:fluctuation}
\frac{32}{I_\star^2},\E_Q[(G_n(\theta)-g_Q(\theta))^2]
\ \le\ \frac{C}{I_\star^2}\cdot \frac{I_\star + C \varepsilon \tau^{-5/2}}{n}
\ \le\ C\Big(\frac{\tau}{n},+,\frac{\varepsilon,\tau^{3/2}}{n}\Big)
\ \le\ C,\frac{\tau}{n},
\end{equation}
since ε≤12\varepsilon\le \tfrac12ε≤21 and τ≤12\tau\le \tfrac12τ≤21.

\medskip\noindent
\emph{(ii) Population bias term.}
By Lemma \ref{lem:score-moments-correct}, ∣gQ(θ)∣=∣\EQ[ψ]∣≤C1(ετ−3/2+1)|g_Q(\theta)|=|\E_Q[\psi]|\le C_1(\varepsilon\tau^{-3/2}+1)∣gQ(θ)∣=∣\EQ[ψ]∣≤C1(ετ−3/2+1). Hence
\begin{equation}\label{eq:bias}
\frac{32}{I_\star^2},g_Q(\theta)^2
\ \le\ \frac{C}{I_\star^2},\Big(\varepsilon^2 \tau^{-3}+1\Big)
\ =\ C\Big(\varepsilon^2 \tau\ +\ \tau^2\Big).
\end{equation}
Combining \eqref{eq:two-terms}, \eqref{eq:fluctuation}, \eqref{eq:bias} we conclude that, uniformly over QQQ with W2(Q,μθ)≤εW_2(Q,\mu_\theta)\le \varepsilonW2(Q,μθ)≤ε,

\EQ[(θ^ε−θ)2] ≤ C(τn + ε2τ + τ2).\E_Q[(\hat\theta_\varepsilon-\theta)^2]\ \le\ C\left(\frac{\tau}{n}\ +\ \varepsilon^2 \tau\ +\ \tau^2\right).\EQ[(θ^ε−θ)2] ≤ C(nτ + ε2τ + τ2).
Finally take τ=τ⋆=ε2/3∧12\tau=\tau_\star=\varepsilon^{2/3}\wedge \tfrac12τ=τ⋆=ε2/3∧21. If ε≤12\varepsilon\le \tfrac12ε≤21, then τ⋆=ε2/3\tau_\star=\varepsilon^{2/3}τ⋆=ε2/3 and the three terms scale as

τ⋆n=ε2/3n,ε2τ⋆=ε8/3,τ⋆2=ε4/3.\frac{\tau_\star}{n}=\frac{\varepsilon^{2/3}}{n},\qquad
\varepsilon^2\tau_\star=\varepsilon^{8/3},\qquad
\tau_\star^2=\varepsilon^{4/3}.nτ⋆=nε2/3,ε2τ⋆=ε8/3,τ⋆2=ε4/3.
Since ε8/3≤ε2\varepsilon^{8/3}\le \varepsilon^2ε8/3≤ε2 and ε4/3≤ε2\varepsilon^{4/3}\le \varepsilon^2ε4/3≤ε2 for ε∈(0,1]\varepsilon\in(0,1]ε∈(0,1], we obtain \eqref{eq:IDS-sharp-risk-correct}.
(When ε>12\varepsilon>\tfrac12ε>21, the trivial ε2\varepsilon^2ε2 baseline dominates.)
□\quad\Box□

===== By Lemma \ref{lem:info-W2}, the least‑favorable τ\tauτ satisfies W2(μθ,νθ,τ)≍τ3/2W_2(\mu_\theta,\nu_{\theta,\tau})\asymp \tau^{3/2}W2(μθ,νθ,τ)≍τ3/2. Choosing τ≍ε2/3\tau\asymp \varepsilon^{2/3}τ≍ε2/3 makes the submodel {νθ,τ}\{\nu_{\theta,\tau}\}{νθ,τ} ε\varepsilonε–close (in W2W_2W2) to μθ\mu_\thetaμθ, while its Fisher information is I(τ)≍τ−1≍ε−2/3I(\tau)\asymp \tau^{-1}\asymp \varepsilon^{-2/3}I(τ)≍τ−1≍ε−2/3. A standard parametric (Cramér–Rao/Le Cam) lower bound thus gives =====

\MI(ε;n) ≳ 1n I(τ) ≍ ε2/3n,\M_I(\varepsilon;n)\ \gtrsim\ \frac{1}{n\,I(\tau)}\ \asymp\ \frac{\varepsilon^{2/3}}{n},\MI(ε;n) ≳ nI(τ)1 ≍ nε2/3,
matching the leading term of Theorem \ref{thm:IDS-sharp-correct}.