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

===== 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□