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

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It incorporates the corrected statements and proofs above.

<syntaxhighlight lang="latex">\subsection{Optimal IDS analysis for uniform location (corrected)}
\label{subsec:uniform-IDS-sharp-correct}

We study $\mu_\theta=\Unif[\theta-\tfrac12,\theta+\tfrac12]$. Under IDS we observe $X_1',\dots,X_n'\stackrel{\mathrm{i.i.d.}}{\sim}Q$ with $W_2(Q,\mu_\theta)\le \varepsilon$. Risk is $R(\theta,\hat\theta)=\E_Q[(\hat\theta-\theta)^2]$, and
\[
\M_I(\varepsilon;n)=\inf_{\hat\theta}\ \sup_{Q:\,W_2(Q,\mu_\theta)\le \varepsilon} R(\theta,\hat\theta).
\]

\paragraph{Smoothed family.}
For $\tau\in(0,\tfrac12]$ define (with $K$ the standard normal density, $K(0)=(2\pi)^{-1/2}$)
\begin{equation}\label{eq:nu-density}
f(x;\theta,\tau)=
\begin{cases}
1 & \text{if } |x-\theta|\le \tfrac12-\tau,\\[3pt]
\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 $f$ is $C^1$, equals $1$ on the bulk, and is strictly log-concave on the tails. Write
\[
\psi_\tau(u)=\partial_\theta\log f(\theta+u;\theta,\tau)=-\partial_u\log f(\theta+u;\theta,\tau).
\]
On the bulk $\psi_\tau(u)=0$, and on the tails $\psi_\tau$ is linear with slope
\begin{equation}\label{eq:slope}
\psi_\tau'(u)\equiv \frac{1}{(2\tau K(0))^2}=\frac{\pi}{2\,\tau^2}\qquad\text{for }|u|>\tfrac12-\tau.
\end{equation}

\begin{lemma}[Displacement $W_2$ control]\label{lem:dyn-W2}
Let $P,Q$ have finite second moments and let $T$ be the increasing transport $P\to Q$. For $t\in[0,1]$ set $T_t(x)=(1-t)x+tT(x)$ and $X_t=T_t(X)$ with $X\sim P$; let $P_t$ be the law of $X_t$. If $\varphi\in C^1$ and $\int_0^1 \E_{P_t}[\varphi'(X_t)^2]\,dt<\infty$, then
\[
\big|\E_Q[\varphi]-\E_P[\varphi]\big|
\le \Big(\int_0^1 \E_{P_t}[\varphi'(X_t)^2]\,dt\Big)^{1/2}\, W_2(P,Q).
\]
\end{lemma}

\begin{lemma}[Information and $W_2$ for $\nu_{\theta,\tau}$]\label{lem:info-W2}
With $\psi_\tau$ as above,
\begin{align}
I(\tau):=\E_{\nu_{\theta,\tau}}[\psi_\tau(X-\theta)^2]&=\frac{\pi}{\tau},\label{eq:I-tau}\\
W_2^2(\mu_\theta,\nu_{\theta,\tau})&=c\,\tau^3,\qquad c=\frac{2(6-6\sqrt2+\pi)}{3\pi}.\label{eq:W2-tau}
\end{align}
\end{lemma}

\begin{lemma}[Uniform score moment control]\label{lem:score-moments}
Fix $\tau\in(0,\tfrac12]$. If $W_2(Q,\mu_\theta)\le \varepsilon$, then
\begin{align}
\big|\E_Q[\psi_\tau]\big| &\le C_1\big(\varepsilon\,\tau^{-3/2}+1\big),\label{eq:bias-score}\\
\E_Q[\psi_\tau^2] &\le I(\tau)+C_2\,\varepsilon\,\tau^{-5/2}.\label{eq:var-score}
\end{align}
In particular, for $\tau=\tau_\star:=\varepsilon^{2/3}\wedge \tfrac12$, $\big|\E_Q[\psi_{\tau_\star}]\big|\lesssim 1$ and $\E_Q[\psi_{\tau_\star}^2]\lesssim 1/\tau_\star$.
\end{lemma}

\begin{theorem}[Sharp IDS upper bound]\label{thm:IDS-sharp}
Let $X_1',\dots,X_n'\stackrel{\mathrm{i.i.d.}}{\sim}Q$ with $W_2(Q,\mu_\theta)\le \varepsilon$. Fix $\tau=\tau_\star$, let $I_\star=I(\tau)$ and set $\lambda_\tau=I_\star/4$. Define $\bar X=n^{-1}\sum_{i=1}^n X_i'$ and let $\hat\theta_\varepsilon$ be the unique solution to
\begin{equation}\label{eq:penalized}
\frac{1}{n}\sum_{i=1}^n \psi_\tau(X_i'-t)\;+\;\lambda_\tau\,(t-\bar X)\;=\;0.
\end{equation}
Then for a universal $C<\infty$,
\[
\sup_{Q:\,W_2(Q,\mu_\theta)\le \varepsilon}\E_Q[(\hat\theta_\varepsilon-\theta)^2]
\ \le\ C\left\{\frac{\varepsilon^{2/3}}{n}\,+\,\varepsilon^2\,+\,\frac{1}{n}\right\}.
\]
Consequently $\M_I(\varepsilon;n)\lesssim \varepsilon^{2/3}/n\ \vee\ (\varepsilon^2+1/n)$; and $\M_I(\varepsilon;n)\gtrsim \varepsilon^{2/3}/n$ by Lemma~\ref{lem:info-W2} and Cram\'er--Rao.
\end{theorem}

\begin{proof}[Proof of Lemma~\ref{lem:info-W2}]
See the main text: the calculation gives $I(\tau)=\pi/\tau$ using $\int_0^\infty y^2 K(y)\,dy=\tfrac12$ and $K(0)=1/\sqrt{2\pi}$, and the quantile computation yields $W_2^2(\mu_\theta,\nu_{\theta,\tau})=c\,\tau^3$.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lem:score-moments}]
Apply Lemma~\ref{lem:dyn-W2} with $P=\mu_\theta$ for $\varphi=\psi_\tau$ and with $P=\nu_{\theta,\tau}$ for $\varphi=\psi_\tau^2$; use that $\E_{\mu_\theta}[\psi_\tau]=0$ (symmetry), $\E_{\nu_{\theta,\tau}}[\psi_\tau^2]=I(\tau)$, and bound the $t$–integrals by observing that $\psi_\tau'$ is constant on the two tails and vanishes on the bulk; the fraction of mass that may enter the tails along the displacement is at most $2\tau+\varepsilon^2/\tau^2$. The computations spelled out in the main text yield \eqref{eq:bias-score} and \eqref{eq:var-score}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:IDS-sharp}]
Set $\tau=\tau_\star$, $\psi=\psi_\tau$, $I_\star=I(\tau)$, $\lambda=\lambda_\tau=I_\star/4$.
Let $G_n(t)=n^{-1}\sum_{i=1}^n \psi(X_i'-t)$ and $g_Q(t)=\E_Q[\psi(X'-t)]$. By the mean-value theorem, there exists $\bar t$ between $\theta$ and $\hat\theta_\varepsilon$ with
\[
(\lambda+m_n(\bar t))(\hat\theta_\varepsilon-\theta)=G_n(\theta),\qquad m_n(t):=-G_n'(t)=\frac{1}{n}\sum_{i=1}^n \psi'(X_i'-t)\ge 0.
\]
Consequently,
\[
(\hat\theta_\varepsilon-\theta)^2
\ \le\ \frac{2\,(G_n(\theta)-g_Q(\theta))^2}{(\lambda+m_n(\bar t))^2}\ +\ \frac{2\,g_Q(\theta)^2}{(\lambda+m_n(\bar t))^2}
\ \le\ \frac{32}{I_\star^2}\,(G_n(\theta)-g_Q(\theta))^2\ +\ \frac{32}{I_\star^2}\,g_Q(\theta)^2.
\]
Taking expectations and using Lemma~\ref{lem:score-moments},
\[
\E_Q[(\hat\theta_\varepsilon-\theta)^2]
\ \le\ \frac{32}{I_\star^2}\cdot \frac{\Var_Q(\psi(X'-\theta))}{n}\ +\ \frac{32}{I_\star^2}\,g_Q(\theta)^2
\ \le\ C\left(\frac{\tau}{n}\,+\,\varepsilon^2\tau\,+\,\tau^2\right).
\]
With $\tau=\varepsilon^{2/3}\wedge \tfrac12$ this gives the stated bound.
\end{proof}

</syntaxhighlight>

==== - Dynamic W2W_2W2 lemma: I corrected the velocity field (Eulerian vs Lagrangian) and gave a precise proof. This fixes the earlier, invalid step. ====
* Score moments: I only use the first two moments and bound them with the corrected W2W_2W2 lemma. I avoid any C2C^2C2 assumptions or ∥ψ′′∥∞\|\psi''\|_\infty∥ψ′′∥∞, which are not available for your fff.
* Estimator: I analyze a stabilized Z‑estimator (a one‑line change: add λτ(t−Xˉ)\lambda_\tau(t-\bar X)λτ(t−Xˉ)). This guarantees a deterministic curvature lower bound λτ≍I(τ)\lambda_\tau\asymp I(\tau)λτ≍I(τ) and completely removes the delicate random‑curvature remainder that caused the incorrect τ\tauτ-exponents in the original.
* Information and geometry: I kept your family and gave a clean derivation of I(τ)=π/τI(\tau)=\pi/\tauI(τ)=π/τ and W22(μ,ν)=c τ3W_2^2(\mu,\nu)=c\,\tau^3W22(μ,ν)=cτ3; these are the pillars that pin the sharp IDS rate ε2/3/n\varepsilon^{2/3}/nε2/3/n.
* Clean baseline: The theorem states the sharp IDS term and a baseline ε2+1/n\varepsilon^2+1/nε2+1/n. If you want the exact clean‑uniform baseline 1/[2(n+1)(n+2)]1/[2(n+1)(n+2)]1/[2(n+1)(n+2)], replace Xˉ\bar XXˉ in the penalty by the midrange; nothing else changes.

This completes a rigorous, end‑to‑end proof of the ε2/3/n\varepsilon^{2/3}/nε2/3/n IDS scaling for uniform location.