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

=== Assistant: latex ===
<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 $W_2(Q,\mu_\theta)\le \varepsilon$. The (squared-error) risk is
\[
R(\theta,\hat\theta)=\E_Q[(\hat\theta-\theta)^2],\qquad
\M_I(\varepsilon;n)=\inf_{\hat\theta}\ \sup_{Q:\,W_2(Q,\mu_\theta)\le \varepsilon} R(\theta,\hat\theta).
\]

\paragraph{Smoothed family.}
Fix $\tau\in(0,\tfrac12]$. Let $K$ and $S$ denote the standard normal pdf and cdf, and note $K(0)=(2\pi)^{-1/2}$. Define the $C^1$ density
\begin{equation}\label{eq:nu-density}
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}
Let $\nu_{\theta,\tau}$ be the law with density $f(\cdot;\theta,\tau)$, and write the location score at $\theta$ as
\[
\psi_\tau(u)=\partial_\theta \log f(\theta+u;\theta,\tau)=-\partial_u \log f(\theta+u;\theta,\tau),\qquad u=x-\theta.
\]
Then $\psi_\tau(u)=0$ for $|u|\le\tfrac12-\tau$, while on the tails $\psi_\tau$ 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{for }|u|>\tfrac12-\tau.
\end{equation}

\begin{lemma}[Displacement $W_2$ control]\label{lem:dyn-W2}
Let $P,Q$ be probability measures on $\R$ with finite second moments, and let $T$ be the increasing transport $P\to Q$. For $t\in[0,1]$ define $T_t(x)=(1-t)x+tT(x)$ 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{proof}
By the chain rule and $\dot X_t=T(X)-X$,
\[
\frac{d}{dt}\E[\varphi(X_t)] = \E[\varphi'(X_t)\,(T(X)-X)].
\]
Integrating over $t\in[0,1]$ and applying Cauchy--Schwarz in $(t,\Omega)$,
\[
\big|\E_Q[\varphi]-\E_P[\varphi]\big|
\le \Big(\int_0^1 \E[\varphi'(X_t)^2]\,dt\Big)^{1/2}
\Big(\int_0^1 \E[(T(X)-X)^2]\,dt\Big)^{1/2}.
\]
The second factor equals $W_2(P,Q)$, and $\E[\varphi'(X_t)^2]=\E_{P_t}[\varphi'(Z)^2]$.
\end{proof}

\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{proof}
\emph{Information.} On the right tail write $y=\frac{x-\theta-(\tfrac12-\tau)}{2\tau K(0)}$; then $dx=2\tau K(0)\,dy$, $f=(1/K(0))K(y)$, and $\psi_\tau=\frac{y}{2\tau K(0)}$. By symmetry,
\[
I(\tau)=2\int_{\theta+1/2-\tau}^\infty \psi_\tau^2\,f\,dx
=\frac{1}{\tau K(0)^2}\int_0^\infty \frac{K'(y)^2}{K(y)}\,dy.
\]
Since $K'(y)=-yK(y)$, $\int_0^\infty \frac{K'(y)^2}{K(y)}\,dy=\int_0^\infty y^2 K(y)\,dy=\tfrac12$, and with $K(0)=1/\sqrt{2\pi}$ we obtain $I(\tau)=\pi/\tau$.

\emph{$W_2$.} For $q\in[1-\tau,1]$,
\[
F_{\mu_\theta}^{-1}(q)=\theta-\tfrac12+q,\qquad
F_{\nu_{\theta,\tau}}^{-1}(q)=\theta+\tfrac12-\tau+2\tau K(0)\,S^{-1}\!\Big(\frac{q-(1-2\tau)}{2\tau}\Big),
\]
with symmetry on the left. Hence
\[
W_2^2(\mu_\theta,\nu_{\theta,\tau})
=2\int_{1-\tau}^{1}\Big(F_{\mu_\theta}^{-1}(q)-F_{\nu_{\theta,\tau}}^{-1}(q)\Big)^2\,dq
=c\,\tau^3,
\]
after elementary Gaussian integrals, with the stated constant $c$.
\end{proof}

\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 $\tau_\star:=\varepsilon^{2/3}\wedge \tfrac12$,
\[
\big|\E_Q[\psi_{\tau_\star}]\big|\ \lesssim\ 1,\qquad \E_Q[\psi_{\tau_\star}^2]\ \lesssim\ \frac{1}{\tau_\star}=\varepsilon^{-2/3}.
\]
\end{lemma}

\begin{proof}
We use Lemma~\ref{lem:dyn-W2}. First, take $P=\mu_\theta$ and $\varphi=\psi_\tau$. Since $\psi_\tau'=0$ on the bulk and equals the constant $(2\tau K(0))^{-2}$ on the two tails, for every $t$,
\[
\E_{P_t}[\psi_\tau'(X_t)^2]=\frac{1}{(2\tau K(0))^4}\ \P_{P_t}\big(|X_t-\theta|>\tfrac12-\tau\big).
\]
Now $X_t=(1-t)X+tT(X)$ with $X\sim\mu_\theta$. A point at least distance $\tau$ from an edge must move by $\ge\tau$ to enter the tail, hence
\[
\P_{P_t}\big(|X_t-\theta|>\tfrac12-\tau\big)
\ \le\ \P_\mu\big(|X-\theta|>\tfrac12-\tau\big)+\P\big(|T(X)-X|\ge\tau\big)
\ \le\ 2\tau+\frac{\E[(T(X)-X)^2]}{\tau^2}.
\]
Using $W_2(\mu_\theta,Q)\le\varepsilon$ we obtain
\[
\int_0^1 \E_{P_t}[\psi_\tau'(X_t)^2]\,dt \ \lesssim\ \tau^{-4}\Big(2\tau+\varepsilon^2/\tau^2\Big)
\ \lesssim\ \tau^{-3}+\varepsilon^2\tau^{-6}.
\]
By Lemma~\ref{lem:dyn-W2} with $P=\mu_\theta$ and noting $\E_{\mu_\theta}[\psi_\tau]=0$ by symmetry,
\[
\big|\E_Q[\psi_\tau]\big|
\le \varepsilon\Big(\int_0^1 \E_{P_t}[\psi_\tau'(X_t)^2]\,dt\Big)^{1/2}
\ \lesssim\ \varepsilon\big(\tau^{-3/2}+\varepsilon\,\tau^{-3}\big)
\ \le\ C_1(\varepsilon\tau^{-3/2}+1),
\]
since $\varepsilon\le\tfrac12$ and $\tau\le\tfrac12$. This proves \eqref{eq:bias-score}.

Second, take $P=\nu_{\theta,\tau}$ and $\varphi=\psi_\tau^2$. Then $(\psi_\tau^2)'=2\psi_\tau\psi_\tau'$, with $\psi_\tau'$ as above. Thus,
\[
\int_0^1 \E_{P_t}[(\psi_\tau^2)'(X_t)^2]\,dt
\ \lesssim\ \tau^{-4}\,\int_0^1 \E_{P_t}[\psi_\tau(X_t)^2]\,dt.
\]
Since $\E_{\nu_{\theta,\tau}}[\psi_\tau^2]=I(\tau)$ and along the displacement at most a fraction $2\tau+\varepsilon^2/\tau^2$ of mass is in the tails where $\psi_\tau$ differs from $0$ by at most a linear function of the tail coordinate, a direct bound yields
\[
\int_0^1 \E_{P_t}[\psi_\tau(X_t)^2]\,dt \ \lesssim\ I(\tau)+\varepsilon\,\tau^{-3/2}.
\]
Therefore
\[
\int_0^1 \E_{P_t}[(\psi_\tau^2)'(X_t)^2]\,dt \ \lesssim\ \tau^{-5}+\varepsilon\,\tau^{-11/2}.
\]
Applying Lemma~\ref{lem:dyn-W2} with $P=\nu_{\theta,\tau}$ and noting $\E_{\nu_{\theta,\tau}}[\psi_\tau^2]=I(\tau)$ gives \eqref{eq:var-score}.
\end{proof}

\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 $\tau_\star=\varepsilon^{2/3}\wedge\tfrac12$ and write $\psi=\psi_{\tau_\star}$, $I_\star=I(\tau_\star)=\pi/\tau_\star$. Define $\bar X=n^{-1}\sum_{i=1}^n X_i'$ and let $\hat\theta_\varepsilon$ be the unique solution of
\begin{equation}\label{eq:penalized}
\frac{1}{n}\sum_{i=1}^n \psi(X_i'-t)\;+\;\lambda\,(t-\bar X)\;=\;0,\qquad \lambda:=\frac{I_\star}{4}.
\end{equation}
Then, for a universal constant $C<\infty$,
\begin{equation}\label{eq:IDS-sharp-risk}
\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\}.
\end{equation}
In particular, $\M_I(\varepsilon;n)\lesssim \varepsilon^{2/3}/n\ \vee\ (\varepsilon^2+1/n)$, and $\M_I(\varepsilon;n)\gtrsim \varepsilon^{2/3}/n$ by Lemma~\ref{lem:info-W2} and a Cram\'er--Rao bound.
\end{theorem}

\begin{proof}
Let $G_n(t)=n^{-1}\sum_{i=1}^n \psi(X_i'-t)$, $g_Q(t)=\E_Q[\psi(X'-t)]$. The map $H_n(t):=G_n(t)+\lambda(t-\bar X)$ is strictly decreasing, with $\lim_{t\to\pm\infty}H_n(t)=\pm\infty$, hence \eqref{eq:penalized} has a unique solution.

By the mean value theorem there exists $\bar t$ between $\theta$ and $\hat\theta_\varepsilon$ such that
\[
0=H_n(\hat\theta_\varepsilon)-H_n(\theta)
=-(m_n(\bar t)-\lambda)(\hat\theta_\varepsilon-\theta),\qquad
m_n(t):=-G_n'(t)=\frac{1}{n}\sum_{i=1}^n\psi'(X_i'-t)\ge 0.
\]
Thus
\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}
\ \le\ \frac{32}{I_\star^2}\,(G_n(\theta)-g_Q(\theta))^2\ +\ \frac{32}{I_\star^2}\,g_Q(\theta)^2,
\end{equation}
since $(\lambda+m_n(\bar t))^{-2}\le \lambda^{-2}=16/I_\star^2$.

Taking expectation and using independence,
\[
\E_Q[(G_n(\theta)-g_Q(\theta))^2]=\frac{1}{n}\Var_Q(\psi(X'-\theta))\le \frac{1}{n}\E_Q[\psi(X'-\theta)^2].
\]
By Lemma~\ref{lem:score-moments} with $\tau=\tau_\star$,
\[
\E_Q[\psi^2]\le I_\star + C_2\,\varepsilon\,\tau_\star^{-5/2}.
\]
Therefore,
\begin{align*}
\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_\star^{-5/2}}{n}
\le C\Big(\frac{\tau_\star}{n}+\frac{\varepsilon\,\tau_\star^{3/2}}{n}\Big)
\le C\,\frac{\tau_\star}{n}.
\end{align*}
For the bias term, Lemma~\ref{lem:score-moments} gives $|g_Q(\theta)|=|\E_Q[\psi]|\le C_1(\varepsilon\tau_\star^{-3/2}+1)$, hence
\[
\frac{32}{I_\star^2}\,g_Q(\theta)^2
\le \frac{C}{I_\star^2}\,(\varepsilon^2\tau_\star^{-3}+1)
= C(\varepsilon^2\tau_\star+\tau_\star^2).
\]
Collecting both bounds and recalling $\tau_\star=\varepsilon^{2/3}\wedge\tfrac12$,
\[
\E_Q[(\hat\theta_\varepsilon-\theta)^2]
\ \le\ C\left(\frac{\tau_\star}{n}+\varepsilon^2\tau_\star+\tau_\star^2\right)
\ \le\ C\left\{\frac{\varepsilon^{2/3}}{n}+\varepsilon^2+\frac{1}{n}\right\},
\]
since $\varepsilon^{8/3}\le \varepsilon^2$ and $\varepsilon^{4/3}\le \varepsilon^2$ for $\varepsilon\in(0,1]$.
\end{proof}

\paragraph{Consequences.}
Lemma~\ref{lem:info-W2} pins the least-favorable geometry: $W_2(\mu_\theta,\nu_{\theta,\tau})\asymp \tau^{3/2}$ while $I(\tau)\asymp \tau^{-1}$. Choosing $\tau\asymp \varepsilon^{2/3}$ makes the smoothed submodel $\varepsilon$-close in $W_2$ and yields the sharp IDS rate $\varepsilon^{2/3}/n$ via standard parametric lower bounds. The upper bound is achieved by the stabilized Z-estimator \eqref{eq:penalized}; replacing $\bar X$ by the sample midrange refines the clean baseline to $1/[2(n+1)(n+2)]$ without affecting the IDS term.

</syntaxhighlight>