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

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\subsection{Optimal IDS analysis for uniform location}\label{subsec:uniform-IDS-sharp}

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

We introduce a smoothed location family $\{\nu_{\theta,\tau}:\theta\in\R,\; 0<\tau\le \tfrac12\}$ by replacing the uniform's two edges by Gaussian kernel tails. Let $K(x)=(2\pi)^{-1/2}e^{-x^2/2}$, $S$ its CDF. 
Define the density
\begin{align}\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{align}
By construction $f(\cdot;\theta,\tau)$ is $C^1$, equals 
unity on the bulk, and is strictly log-concave on the two edge strips of width $\tau$; the choice $K'(0)=0$ guarantees $C^1$ matching at the junctions.

We provide an upper bound achieved by a Z-estimator in the smoothed family.

\begin{theorem}[Sharp IDS risk]\label{thm:IDS-sharp}
Let $X_1',\dots,X_n'$ be i.i.d.\ from $Q$ with $W_2(Q,\mu_\theta)\le \ep$. Define
\[
\tau_\star=\ep^{2/3}\wedge \frac12,\qquad \psi_{\tau_\star}(x-\theta)=\partial_\theta \log f(x;\theta,\tau_\star),
\]
with $f$ as in \eqref{eq:nu-density}. Let $\hat\theta_\ep$ be a solution to the estimating equation
\begin{align}\label{eq:M-est}
\frac{1}{n}\sum_{i=1}^n \psi_{\tau_\star}(X_i'-t)=0.
\end{align}
It follows from our analysis that this solution is unique. 
There exists a constant $C<\infty$ such that, for all $n\ge 1$ and $\ep\in(0,\tfrac12]$,s
\begin{align}\label{eq:IDS-sharp-risk}
\sup_{Q:W_2(Q,\mu_\theta)\le \ep} \E_Q\big[(\hat\theta_\ep-\theta)^2\big]
\;\le\; C\left(\,\frac{\ep^{2/3}}{n}\;\;\vee\;\;\left[\ep^2+\frac{1}{2(n+1)(n+2)}\right]\right).
\end{align}
Consequently,
\begin{align}\label{eq:IDS-sandwich}
\M_I(\ep;n)\;\asymp\; \frac{\ep^{2/3}}{n}\ \vee\ \M_C(\ep;n).
\end{align}
\end{theorem}

\paragraph{Discussion.} The estimator \eqref{eq:M-est} is a one-dimensional $M$-estimator with a strictly decreasing $C^1$ score; 
as we will show, 
existence and uniqueness follow from strict concavity of the smoothed log-likelihood. 
% Tuning uses only $\tau_\star\asymp \ep^{2/3}$. If $\ep$ is unknown one may plug an upper bound or use a Lepski grid over $\tau$; rates are unaffected. \ed{?}

The proof uses three ingredients. First, a standard $M$-estimation expansion reduces the risk to a variance term of order $1/\{n I(\tau_\star)\}$ plus a squared bias term governed by the population score under $Q$. Second, a one-dimensional dynamic formulation of $W_2$ controls the deviation of score moments between $Q$ and the reference model $\mu_\theta$ through the $L^2$ size of derivatives of the score; in our construction these derivatives concentrate on the edge strips and have explicit $\tau$-scaling. Third, Lemma~\ref{lem:info-W2} identifies $I(\tau_\star)=\pi/\tau_\star$ and the least-favorable scaling $\tau_\star\asymp \ep^{2/3}$.

We next formalize the transport control we will use.

\begin{lemma}[Dynamic $W_2$ control of expectation differences]\label{lem:dyn-W2}
Let $P,Q$ be Borel probability measures on $\R$ with finite second moments, and let $T$ be the monotone transport pushing $P$ to $Q$. For $t\in[0,1]$ set $X_t=(1-t)X+tT(X)$ with $X\sim P$, and let $P_t$ denote the law of $X_t$. Then, for any $C^1$ function $\varphi$ such that $\int_0^1 \E_{P_t}[\varphi'(X_t)^2]\,dt<\infty$,
\[
\left|\E_Q[\varphi]-\E_P[\varphi]\right|
\le \left(\int_0^1 \E_{P_t}[\varphi'(X_t)^2]\,dt\right)^{1/2}\, W_2(P,Q).
\]
\end{lemma}

\begin{proof}
By the fundamental theorem of calculus,
\[
\E_Q[\varphi]-\E_P[\varphi]=\int_0^1 \frac{d}{dt}\E_{P_t}[\varphi]\,dt
=\int_0^1 \E_{P_t}[\varphi'(X_t)\,v_t(X_t)]\,dt,
\]
where $v_t(x)=T(x)-x$ is the velocity along the monotone geodesic. Cauchy–Schwarz yields
\[
\left|\E_Q[\varphi]-\E_P[\varphi]\right|
\le \left(\int_0^1 \E_{P_t}[\varphi'(X_t)^2]\,dt\right)^{1/2}\left(\int_0^1 \E_{P_t}[v_t(X_t)^2]\,dt\right)^{1/2}.
\]
Finally $\int_0^1 \E_{P_t}[v_t(X_t)^2]\,dt=\E[(T(X)-X)^2]=W_2(P,Q)^2$ in one dimension with monotone $T$.
\end{proof}

We now bound, uniformly over the IDS ball, the score moments that appear in the $M$-estimation risk.
\begin{lemma}[Uniform control of score moments on the $W_2$ ball, benchmarked at $\nu_{\theta,\tau}$]\label{lem:score-moments}
Fix $\tau\in(0,\tfrac12]$ and let $\psi_\tau$ be the score of $\nu_{\theta,\tau}$. There exist constants $C_1,C_2,C_3<\infty$ such that for all $Q$ with $W_2(Q,\mu_\theta)\le \ep$,
\begin{align}
\left|\E_Q[\psi_\tau]-\E_{\nu_{\theta,\tau}}[\psi_\tau]\right| &\le C_1\,\frac{\ep}{\sqrt{\tau}},\label{eq:bias-score}\\
\left|\E_Q[\psi_\tau']-\E_{\nu_{\theta,\tau}}[\psi_\tau']\right| &\le C_2\,\frac{\ep}{\sqrt{\tau}},\label{eq:curv-score}\\
\left|\E_Q[\psi_\tau^2]-\E_{\nu_{\theta,\tau}}[\psi_\tau^2]\right| &\le C_3\,\frac{\ep}{\tau^{3/2}}.\label{eq:var-score}
\end{align}
In particular, since for the smoothed family the information identity gives $\E_{\nu_{\theta,\tau}}[\psi_\tau']=\E_{\nu_{\theta,\tau}}[\psi_\tau^2]=I(\tau)$ with $I(\tau)=\pi/\tau$ from \eqref{eq:I-tau}, we obtain
\[
\E_Q[\psi_\tau']=I(\tau)+O\!\left(\frac{\ep}{\sqrt{\tau}}\right),\qquad
\E_Q[\psi_\tau^2]=I(\tau)+O\!\left(\frac{\ep}{\tau^{3/2}}\right).
\]
\end{lemma}

\begin{proof}
Apply Lemma~\ref{lem:dyn-W2} with $P=\nu_{\theta,\tau}$, $Q$ as given, and successively $\varphi=\psi_\tau$, $\varphi=\psi_\tau'$, and $\varphi=\psi_\tau^2$. This yields
\[
\left|\E_Q[\varphi]-\E_{\nu_{\theta,\tau}}[\varphi]\right|
\le \left(\int_0^1 \E_{P_t}[\varphi'(X_t)^2]\,dt\right)^{1/2}\, W_2(Q,\nu_{\theta,\tau}),
\]
where $X_t=(1-t)X+tT(X)$ with $X\sim \nu_{\theta,\tau}$ and $T$ the monotone transport from $\nu_{\theta,\tau}$ to $Q$. By the triangle inequality and Lemma~\ref{lem:info-W2},
\[
W_2(Q,\nu_{\theta,\tau}) \;\le\; W_2(Q,\mu_\theta)+W_2(\mu_\theta,\nu_{\theta,\tau}) \;\le\; \ep + c^{1/2}\tau^{3/2}.
\]
We next bound the $t$–integrals. By construction, $\psi_\tau$ vanishes on the bulk $\{|u|\le \tfrac12-\tau\}$ and is smooth on the two edge regions. Differentiating the explicit tail expression in \eqref{eq:nu-density} and using a change of variables $y=\frac{u-(1/2-\tau)}{2\tau K(0)}$ on each edge, one checks that there are absolute constants $A_1,A_2,A_3$ such that, for all $t\in[0,1]$,
\[
\E_{P_t}\!\big[\psi_\tau'(X_t-\theta)^2\big]\le \frac{A_1}{\tau},\qquad
\E_{P_t}\!\big[\psi_\tau''(X_t-\theta)^2\big]\le \frac{A_2}{\tau^2},\qquad
\E_{P_t}\!\big[((\psi_\tau^2)'(X_t-\theta))^2\big]\le \frac{A_3}{\tau^3}.
\]
Intuitively, these come from the facts that: (i) all these derivatives vanish on the bulk and are nonzero only on the two edge regions; (ii) the $\nu_{\theta,\tau}$-mass of each edge region equals $\tau$ (the two edges carry total mass $2\tau$); and (iii) the derivatives scale, respectively, like $\tau^{-1/2}$, $\tau^{-1}$, and $\tau^{-3/2}$ in $L^\infty$ over the edge regions generated by the kernel rescaling in \eqref{eq:nu-density}. Integrating over $t\in[0,1]$ preserves these orders.

Putting the pieces together,
\[
\left(\int_0^1 \E_{P_t}[\psi_\tau'(X_t-\theta)^2]\,dt\right)^{1/2}\!\!\lesssim \tau^{-1/2},\,
\left(\int_0^1 \E_{P_t}[\psi_\tau''(X_t-\theta)^2]\,dt\right)^{1/2}\!\!\lesssim \tau^{-1},
\]
and $\left(\int_0^1 \E_{P_t}[((\psi_\tau^2)'(X_t-\theta))^2]\,dt\right)^{1/2}\!\!\lesssim \tau^{-3/2}$.
Since $W_2(Q,\nu_{\theta,\tau})\le \ep + c^{1/2}\tau^{3/2}$ and $\tau\le \tfrac12$, the $\ep$ term controls the product for the regimes of interest; absorbing harmless constants yields the stated bounds \eqref{eq:bias-score}–\eqref{eq:var-score}. Finally, the “in particular” lines follow by inserting $\E_{\nu_{\theta,\tau}}[\psi_\tau']=\E_{\nu_{\theta,\tau}}[\psi_\tau^2]=I(\tau)$ from \eqref{eq:I-tau}.
\end{proof}

We can now prove Theorem~\ref{thm:IDS-sharp}.

\begin{proof}[Proof of Theorem~\ref{thm:IDS-sharp}]
Fix $\tau=\tau_\star$ and write $\psi:=\psi_{\tau_\star}$, $I_\star=I(\tau_\star)=\pi/\tau_\star$. Let $m(Q)=-\partial_t \E_Q[\psi(X'-t)]\big|_{t=\theta}=\E_Q[\psi'(X'-\theta)]$ denote the population curvature under $Q$.
Next, we claim that
\begin{equation}\label{mq}
m(Q)=I_\star + O\!\left(\frac{\ep}{\sqrt{\tau_\star}}\right).    
\end{equation}
Indeed, 
for the location family $\{\nu_{\theta,\tau}:\theta\in\R\}$ with score $s_\theta(x)=\psi_\tau(x-\theta)$, the information identity
\(
-E_{\nu_{\theta,\tau}}\!\big[\partial_\theta s_\theta(X)\big] \;=\; E_{\nu_{\theta,\tau}}\!\big[s_\theta(X)^2\big]
\)
holds whenever differentiation under the integral is justified (true here because $f(\cdot;\theta,\tau)$ is $C^1$ and piecewise real analytic on the two edge strips, with bounded score and score derivative).
Since $s_\theta(x)$ depends on $\theta$ only via $x-\theta$, we have $\partial_\theta s_\theta(x)=-\psi_\tau'(x-\theta)$, hence
\begin{equation}\label{eq:E-nu-psi'-is-I}
\E_{\nu_{\theta,\tau}}\!\big[\psi_\tau'(X-\theta)\big]\;=\; \E_{\nu_{\theta,\tau}}\!\big[\psi_\tau(X-\theta)^2\big]\;=\; I(\tau).
\end{equation}

\paragraph{Step 2: curvature under $Q$.}
By Lemma~\ref{lem:score-moments} with $\varphi=\psi_\tau'$ and $P=\nu_{\theta,\tau}$, we have
\[
m(Q)\equiv \E_Q[\psi_\tau'(X'-\theta)]
=\E_{\nu_{\theta,\tau}}[\psi_\tau']+O\!\left(\frac{\ep}{\sqrt{\tau}}\right)
=I(\tau)+O\!\left(\frac{\ep}{\sqrt{\tau}}\right).
\]
With $\tau=\tau_\star$ this is exactly \eqref{mq}.

Since $\tau_\star=\ep^{2/3}$, the error term in \eqref{mq} equals $O(\ep^{2/3})$, hence $m(Q)\ge \tfrac12 I_\star$ for all $\ep\in(0,\ep_0]$ with $\ep_0$ fixed absolute. 
For larger $\ep$,
the other term \eqref{eq:IDS-sharp-risk} will dominate the risk,
so we may assume $m(Q)\ge \tfrac12 I_\star$ without loss of generality below.

Standard $M$-estimation analysis yields
\begin{align}\label{eq:risk-decomp}
\E_Q[(\hat\theta_\ep-\theta)^2]
\;\le\; \frac{1}{n}\,\frac{\mathrm{Var}_Q(\psi)}{m(Q)^2}\;+\; 
\frac{\E_Q[\psi]^2}{m(Q)^2}
+\frac{C}{n}\,\frac{(\E_Q[\psi^4])^{1/2}\,\|\psi'\|_\infty^2}{m(Q)^4}.
\end{align}
See Section \ref{mearg} for a self-contained argument.
In our construction $\psi$ is bounded by $C\tau_\star^{-1/2}$, so $\E_Q[\psi^4]\lesssim \tau_\star^{-2}$.
Moreover, $\|\psi'\|_\infty^2 \lesssim \tau_\star^{-2}$
and, since $m(Q)\gtrsim I_\star\asymp \tau_\star^{-1}$,
the last term is of order
$\frac{\tau_\star^{3/2}}{n}$.

By Lemma \ref{lem:score-moments}, $\mathrm{Var}Q(\psi)\le \E_Q[\psi^2]\le I\star + C\ep/\tau_\star^{3/2}$, and $m(Q)=I_\star+O(\ep/\sqrt{\tau_\star})$
Therefore the first term in \eqref{eq:risk-decomp} is bounded by $C/(n I_\star)= C\,\tau_\star/(\pi n)$. The second term is a squared bias term; Lemma~\ref{lem:score-moments} gives $|\E_Q[\psi]|\le C\,\ep/\sqrt{\tau_\star}$ and hence
\[
\frac{\E_Q[\psi]^2}{m(Q)^2}\ \le\ C\,\frac{\ep^2\,\tau_\star}{I_\star^2}
= C\,\ep^2\,\tau_\star^3
= C\,\ep^2\left(\ep^2\right)
= C\,\frac{\ep^4}{c}.
\]
In the small-shift regime $\ep\le 1$ this is dominated by the variance term $C/(n I_\star)$ provided $\ep\ll n^{-1/2}$; when $\ep\gtrsim n^{-1/2}$, the CDS baseline $\ep^2$ (and the uniform noncontaminated term $1/[2(n+1)(n+2)]$) already dominates. 
Finally, the remainder term $\frac{\tau_\star^{3/2}}{n}$ is of smaller order than to $1/(n I_\star)$.

Combining the pieces and recalling $\tau_\star=\ep^{2/3}$, $I_\star=\pi/\tau_\star$, we obtain
\[
\E_Q[(\hat\theta_\ep-\theta)^2]
\ \le\ C\left(\frac{\tau_\star}{n}\right)\ \vee\ \left(\ep^2+\frac{1}{2(n+1)(n+2)}\right)
= C\left(\frac{\ep^{2/3}}{n}\right)\ \vee\ \M_C(\ep;n),
\]
uniformly over $Q$ with $W_2(Q,\mu_\theta)\le \ep$, as claimed in \eqref{eq:IDS-sharp-risk}. 
Then, \eqref{eq:IDS-sandwich} follows by combining this upper bound with the lower bound obtained from Lemma~\ref{lem:info-W2} by taking $\tau=\ep^{2/3}$ in the least-favorable submodel $\{\nu_{\theta,\tau}\}$ and invoking the Cram\'er–Rao bound $1/\{n I(\tau)\}$. This completes the proof.
\end{proof}

\paragraph{Consequences.} The IDS difficulty is pinned to the cubic edge geometry: the least-favorable $\tau$ satisfies $W_2(\mu_\theta,\nu_{\theta,\tau})\asymp \ep$ and $I(\tau)\asymp \tau^{-1}\asymp \ep^{-2/3}$, hence the sharp rate $\ep^{2/3}/n$.

\begin{lemma}[Information and $W_2$ distance for $\nu_{\theta,\tau}$]\label{lem:info-W2}
Let $\psi_\tau(x-\theta)=\partial_\theta \log f(x;\theta,\tau)=-\partial_x \log f(x;\theta,\tau)$ denote the score of the location family $\{\nu_{\theta,\tau}\}$. Then, for all $\theta\in\R$ and $\tau\in(0,\tfrac12]$,
\begin{align}
I(\tau)\equiv \E_{\nu_{\theta,\tau}}[\psi_\tau(X-\theta)^2]=\frac{\pi}{\tau},\label{eq:I-tau}\\
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}
\end{align}
\end{lemma}

\begin{proof}
The score vanishes on the bulk and is supported in the edge strips. On $x\ge \theta+\tfrac12-\tau$ one has
\[
\psi_\tau(x-\theta)=-\frac{\partial}{\partial x}\log f(x;\theta,\tau)
=-\frac{K'\!\left(\frac{x-\theta-(\tfrac12-\tau)}{2\tau K(0)}\right)}{2\tau K(0)\,K\!\left(\frac{x-\theta-(\tfrac12-\tau)}{2\tau K(0)}\right)},
\]
with the analogous expression on the left edge by symmetry. Using the substitution $y=(x-\theta-(\tfrac12-\tau))/(2\tau K(0))$ and symmetry of the two edges,
\begin{align*}
I(\tau)
&=2\int_{\theta+1/2-\tau}^\infty \psi_\tau(x-\theta)^2\, f(x;\theta,\tau)\,dx\\
&=\frac{2}{(2\tau K(0))^2 K(0)}\int_{\theta+1/2-\tau}^\infty \frac{K'(y)^2}{K(y)}\,dx
=\frac{1}{\tau K(0)^2}\int_0^\infty \frac{K'(y)^2}{K(y)}\,dy
=\frac{\pi}{\tau},
\end{align*}
where in the last equality we used $K(0)=1/\sqrt{2\pi}$ and the Gaussian integral $\int_0^\infty K'(y)^2/K(y)\,dy=1$. For $W_2$, work with quantile functions: $F_1^{-1}(q)=\theta-1/2+q$ for $\mu_\theta$, and on $q\in[1-\tau,1]$ a direct inversion of the right-edge CDF yields
\[
F_2^{-1}(q)=\theta+\tfrac12-\tau+2\tau K(0)\,S^{-1}\!\left(\frac{q-(1-2\tau)}{2\tau}\right).
\]
By symmetry,
\begin{align*}
W_2^2(\mu_\theta,\nu_{\theta,\tau})
&=2\int_{1-\tau}^1\Big(F_1^{-1}(q)-F_2^{-1}(q)\Big)^2\,dq\\
&= 2\int_{1-\tau}^1 \Big((q+\tau-1)-2\tau K(0) S^{-1}\!\big(\tfrac{q-(1-2\tau)}{2\tau}\big)\Big)^2\,dq\\
&=\tau^3\left(\frac{2}{3}+16K(0)^2\int_{0}^\infty u^2 K(u)\,du -16K(0)\int_0^\infty (2S(u)-1)\,u\,K(u)\,du\right).
\end{align*}
Evaluating the Gaussian integrals gives \eqref{eq:W2-tau} with the stated constant $c$.
\end{proof}

\subsubsection{M-estimation argument}
\label{mearg}
Here, we show \eqref{eq:risk-decomp}.
Let $G_n(t)=n^{-1}\sum_{i=1}^n\psi(X_i'-t)$, $g_Q(t)=\E_Q[\psi(X'-t)]$, 
so that 
$m(Q)=-g_Q'(\theta)=\E_Q[\psi'(X'-\theta)]$. We assume:

\begin{enumerate}
\item[(A1)] $\psi$ is $C^1$, strictly decreasing, and 
has 
bounded derivatives: $\|\psi\|_\infty<\infty$, $\|\psi'\|_\infty<\infty$, $\|\psi''\|_\infty<\infty$.
\item[(A2)] $g_Q$ is differentiable in a neighborhood of $\theta$ and $m(Q)>0$.
\end{enumerate}

In our model these hold: $\psi$ vanishes on the bulk and is smooth, monotone, and bounded on the two edge strips.
By \eqref{mq},
we have $m(Q)\ge \tfrac12 I(\tau_\star)>0$.

\medskip
\noindent\textbf{Mean-value expansion of the estimating equation.}
By definition $\hat\theta_\ep$ solves $G_n(\hat\theta_\ep)=0$. By the mean-value theorem there exists a (random) $\bar t$ on the line segment $[\theta,\hat\theta_\ep]$ such that
\[
0 \;=\; G_n(\hat\theta_\ep) \;=\; G_n(\theta) + (\hat\theta_\ep-\theta)\,\frac{d}{dt}G_n(t)\big|_{t=\bar t}
\;=\; G_n(\theta) - (\hat\theta_\ep-\theta)\,m_n(\bar t),
\]
where we set $m_n(t):=-G_n'(t)=n^{-1}\sum_{i=1}^n \psi'(X_i'-t)$. Hence
\begin{equation}\label{eq:basic-mvt}
\hat\theta_\ep-\theta \;=\; \frac{G_n(\theta)}{m_n(\bar t)}.
\end{equation}
Add and subtract expectations in the numerator:
\[
G_n(\theta) \;=\; \underbrace{\big(G_n(\theta)-g_Q(\theta)\big)}_{\text{empirical fluctuation}}
\;+\; \underbrace{g_Q(\theta)}_{\text{population bias}},
\]
and write
\begin{equation*}
\frac{1}{m_n(\bar t)} \;=\; \frac{1}{m(Q)} \;+\; \Big(\frac{1}{m_n(\bar t)}-\frac{1}{m(Q)}\Big).
\end{equation*}
Multiplying this by $G_n(\theta)$ and substituting into \eqref{eq:basic-mvt} yields the exact decomposition
\begin{align}
\hat\theta_\ep-\theta
&= \frac{1}{m(Q)}\Big(G_n(\theta)-g_Q(\theta)\Big) \;+\; \frac{g_Q(\theta)}{m(Q)} \;+\; \mathrm{Rem}, \\
\mathrm{where}\quad
\mathrm{Rem}
&:= \Big(G_n(\theta)-g_Q(\theta)\Big)\Big(\frac{1}{m_n(\bar t)}-\frac{1}{m(Q)}\Big)
+ g_Q(\theta)\Big(\frac{1}{m_n(\bar t)}-\frac{1}{m(Q)}\Big).\label{eq:Rem-def}
\end{align}

Next, we claim that
\begin{equation}\label{eq:Rem-quad-goal}
\E_Q[\mathrm{Rem}^2]
\ \le\ \frac{C}{m(Q)^4}\,\E_Q\!\left[\Big(G_n-g_Q\Big)^2\big(m_n(\bar t)-m(Q)\big)^2\right]
\ +\ \frac{C}{m(Q)^4}\,\big(\E_Q[\psi^4]\big)^{1/2}\cdot \frac{\|\psi'\|_\infty^2}{n},
\end{equation}
By the triangle inequality, 
\begin{align}
\mathrm{Rem}^2
&\le 2\Big(G_n(\theta)-g_Q(\theta)\Big)^2\Big(\frac{1}{m_n(\bar t)}-\frac{1}{m(Q)}\Big)^2
+ 2\,g_Q(\theta)^2\Big(\frac{1}{m_n(\bar t)}-\frac{1}{m(Q)}\Big)^2.\label{eq:Rem-square-raw}
\end{align}
Next, for any $a,b>0$, $|a^{-1}-b^{-1}|=|a-b|/(ab)$. Apply this with $a=m_n(\bar t)$, $b=m(Q)$. 
Define the <code>good curvature'' event
\(
\mathcal A:=\{\,m_n(\bar t)\ge m(Q)/2\,\}.
\)
On $\mathcal A$,
\begin{equation}\label{eq:recip-good}
\Big|\frac{1}{m_n(\bar t)}-\frac{1}{m(Q)}\Big|
=\frac{|\,m_n(\bar t)-m(Q)\,|}{m_n(\bar t)\,m(Q)}
\le \frac{2}{m(Q)^2}\,|\,m_n(\bar t)-m(Q)\,|,
\end{equation}
and on $\mathcal A^c$ we have deterministically
\begin{equation}\label{eq:recip-bad}
\Big|\frac{1}{m_n(\bar t)}-\frac{1}{m(Q)}\Big|
\le \frac{1}{m_n(\bar t)}+\frac{1}{m(Q)}
\le \frac{2}{m(Q)}.
\end{equation}

Taking expectations in \eqref{eq:Rem-square-raw} and splitting over $\mathcal A$ and $\mathcal A^c$, we obtain
\begin{align}
\E_Q[\mathrm{Rem}^2]
&\le 2\,\E_Q\!\left[\Big(G_n(\theta)-g_Q(\theta)\Big)^2\Big(\frac{1}{m_n(\bar t)}-\frac{1}{m(Q)}\Big)^2\mathbf 1_{\mathcal A}\right]\nonumber\\
&\quad + 2\,\E_Q\!\left[\Big(G_n(\theta)-g_Q(\theta)\Big)^2\Big(\frac{1}{m_n(\bar t)}-\frac{1}{m(Q)}\Big)^2\mathbf 1_{\mathcal A^c}\right]\nonumber\\
&\quad + 2\,g_Q(\theta)^2\,\E_Q\!\left[\Big(\frac{1}{m_n(\bar t)}-\frac{1}{m(Q)}\Big)^2\mathbf 1_{\mathcal A}\right]
+ 2\,g_Q(\theta)^2\,\E_Q\!\left[\Big(\frac{1}{m_n(\bar t)}-\frac{1}{m(Q)}\Big)^2\mathbf 1_{\mathcal A^c}\right].\label{eq:four-terms}
\end{align}

\paragraph{Bounds on the </code>good'' part $\mathcal A$.}
Use \eqref{eq:recip-good} inside the expectations:
\begin{align}
2\,\E_Q\!\left[\Big(G_n-g_Q\Big)^2\Big(\frac{1}{m_n(\bar t)}-\frac{1}{m(Q)}\Big)^2\mathbf 1_{\mathcal A}\right]
&\le \frac{8}{m(Q)^4}\,\E_Q\!\left[\Big(G_n-g_Q\Big)^2\big(m_n(\bar t)-m(Q)\big)^2\right],\label{eq:good1}\\
2\,g_Q(\theta)^2\,\E_Q\!\left[\Big(\frac{1}{m_n(\bar t)}-\frac{1}{m(Q)}\Big)^2\mathbf 1_{\mathcal A}\right]
&\le \frac{8\,g_Q(\theta)^2}{m(Q)^4}\,\E_Q\!\left[\big(m_n(\bar t)-m(Q)\big)^2\right].\label{eq:good2}
\end{align}

\paragraph{Bounds on the `bad'' part $\mathcal A^c$.}
Use \eqref{eq:recip-bad}:
\begin{align}
2\,\E_Q\!\left[\Big(G_n-g_Q\Big)^2\Big(\frac{1}{m_n(\bar t)}-\frac{1}{m(Q)}\Big)^2\mathbf 1_{\mathcal A^c}\right]
&\le \frac{8}{m(Q)^2}\,\E_Q\!\left[\Big(G_n-g_Q\Big)^2\mathbf 1_{\mathcal A^c}\right],\label{eq:bad1}\\
2\,g_Q(\theta)^2\,\E_Q\!\left[\Big(\frac{1}{m_n(\bar t)}-\frac{1}{m(Q)}\Big)^2\mathbf 1_{\mathcal A^c}\right]
&\le \frac{8\,g_Q(\theta)^2}{m(Q)^2}\,\mathrm{P}_Q(\mathcal A^c).\label{eq:bad2}
\end{align}
We now control the two factors $\E_Q[(G_n-g_Q)^2\mathbf 1_{\mathcal A^c}]$ and $\mathrm{P}_Q(\mathcal A^c)$.

\medskip
\noindent\emph{(a) Bounding $\mathrm{P}_Q(\mathcal A^c)$ via Chebyshev.}
By definition $\mathcal A^c=\{\,m_n(\bar t)<m(Q)/2\,\}$, hence
\[
\mathrm{P}_Q(\mathcal A^c)\;\le\; \mathrm{P}_Q\big(|m_n(\bar t)-m(Q)|\ge m(Q)/2\big)\;\le\; \frac{4}{m(Q)^2}\,\E_Q\big[(m_n(\bar t)-m(Q))^2\big].\label{eq:Ac-prob}
\]
Therefore,
\begin{equation}\label{eq:bad2-bnd}
\eqref{eq:bad2}\ \le\ \frac{32\,g_Q(\theta)^2}{m(Q)^4}\,\E_Q\!\left[(m_n(\bar t)-m(Q))^2\right].
\end{equation}

\medskip
\noindent\emph{(b) Bounding $\E_Q[(G_n-g_Q)^2\mathbf 1_{\mathcal A^c}]$ via Cauchy--Schwarz and a 4th moment bound.}
By Cauchy--Schwarz,
\begin{equation}\label{eq:CS-bad1}
\E_Q\!\left[\Big(G_n-g_Q\Big)^2\mathbf 1_{\mathcal A^c}\right]
\le \Big(\E_Q\big[\big(G_n-g_Q\big)^4\big]\Big)^{1/2}\ \mathrm{P}_Q(\mathcal A^c)^{1/2}.
\end{equation}
We bound the fourth moment of the empirical mean $G_n(\theta)=n^{-1}\sum_{i=1}^n Y_i$ with $Y_i=\psi(X_i'-\theta)$, which are i.i.d.\ under $Q$, centered at $g_Q(\theta)=\E_Q[Y]$. A standard inequality (e.g., the Marcinkiewicz--Zygmund inequality or a direct expansion) yields
\begin{equation}\label{eq:Gn-fourth}
\E_Q\!\left[\Big(G_n(\theta)-g_Q(\theta)\Big)^4\right]
\ \le\ \frac{C}{n^2}\,\E_Q\!\left[|Y-g_Q(\theta)|^4\right]
\ \le\ \frac{C}{n^2}\,\E_Q\!\left[|Y|^4\right]
\ \le\ \frac{C}{n^2}\,\E_Q[\psi(X'-\theta)^4],
\end{equation}
where we used $|g_Q(\theta)|\le(\E_Q[Y^2])^{1/2}$ and then $(a+b)^4\le 8(a^4+b^4)$ to absorb $g_Q(\theta)$ into the constant $C$. Combining \eqref{eq:CS-bad1}, \eqref{eq:Gn-fourth}, and \eqref{eq:Ac-prob} gives
\begin{equation}\label{eq:bad1-bnd}
\eqref{eq:bad1}\ \le\ \frac{8}{m(Q)^2}\cdot \frac{C^{1/2}}{n}\,\big(\E_Q[\psi^4]\big)^{1/2}\ \cdot\ \frac{2}{m(Q)}\Big(\E_Q[(m_n(\bar t)-m(Q))^2]\Big)^{1/2}.
\end{equation}

\paragraph{Assembling the pieces.}
Summing \eqref{eq:good1}, \eqref{eq:good2}, \eqref{eq:bad2-bnd}, and \eqref{eq:bad1-bnd} we obtain
\begin{align}
\E_Q[\mathrm{Rem}^2]
&\le \frac{8}{m(Q)^4}\,\E_Q\!\left[\Big(G_n-g_Q\Big)^2\big(m_n(\bar t)-m(Q)\big)^2\right]
+ \frac{C\,g_Q(\theta)^2}{m(Q)^4}\,\E_Q\!\left[(m_n(\bar t)-m(Q))^2\right]\nonumber\\
&\quad + \frac{C}{m(Q)^3 n}\,\big(\E_Q[\psi^4]\big)^{1/2}\cdot \Big(\E_Q[(m_n(\bar t)-m(Q))^2]\Big)^{1/2}.\label{eq:sum-before-CS}
\end{align}
By the inequality 
$ab\le \frac12(a^2+b^2)$ applied to the last term,
\begin{align}
\frac{C}{m(Q)^3 n}\,\big(\E_Q[\psi^4]\big)^{1/2}\cdot \Big(\E_Q[(m_n-m(Q))^2]\Big)^{1/2}
&\le \frac{C}{2m(Q)^4}\,\E_Q[(m_n-m(Q))^2]\ +\ \frac{C}{2m(Q)^2 n^2}\,\E_Q[\psi^4].\label{eq:CS-last}
\end{align}
Insert \eqref{eq:CS-last} into \eqref{eq:sum-before-CS} and absorb constants:
\begin{align}
\E_Q[\mathrm{Rem}^2]
&\le \frac{C}{m(Q)^4}\,\E_Q\!\left[\Big(G_n-g_Q\Big)^2\big(m_n(\bar t)-m(Q)\big)^2\right]\nonumber\\
&\quad + \frac{C}{m(Q)^4}\,\Big(g_Q(\theta)^2+1\Big)\,\E_Q\!\left[(m_n(\bar t)-m(Q))^2\right]
+ \frac{C}{m(Q)^2 n^2}\,\E_Q[\psi^4].\label{eq:pre-final}
\end{align}

\paragraph{Bounding $g_Q(\theta)^2$ and reducing to \eqref{eq:Rem-quad-goal}.}
By Cauchy--Schwarz,
\[
g_Q(\theta)^2=\big(\E_Q[\psi(X'-\theta)]\big)^2\le \E_Q[\psi(X'-\theta)^2]\le \big(\E_Q[\psi(X'-\theta)^4]\big)^{1/2}.
\]
Therefore the second line of \eqref{eq:pre-final} is bounded by
\[
\frac{C}{m(Q)^4}\,\big(\E_Q[\psi^4]\big)^{1/2}\,\E_Q\!\left[(m_n(\bar t)-m(Q))^2\right]\ +\ \frac{C}{m(Q)^2 n^2}\,\E_Q[\psi^4].
\]
Finally, we use the standard variance bound for the empirical derivative at $t=\theta$:
\begin{equation}\label{eq:mn-theta-var}
\E_Q\!\left[(m_n(\theta)-m(Q))^2\right]\ \le\ \frac{1}{n}\mathrm{Var}_Q\!\big(\psi'(X'-\theta)\big)\ \le\ \frac{\|\psi'\|_\infty^2}{n}.
\end{equation}
Replacing $\bar t$ by $\theta$ only weakens the bound (since $\E[(m_n(\bar t)-m(Q))^2]\le 2\E[(m_n(\theta)-m(Q))^2]+ 2\E[(m_n(\bar t)-m_n(\theta))^2]$, and the second term can be handled by a Lipschitz bound in $t$; any such term proportional to $\E_Q[(\hat\theta_\ep-\theta)^2]$ will be absorbed as explained below). Using \eqref{eq:mn-theta-var} and $(\E_Q[\psi^4])^{1/2}\ge 1$, the second line of \eqref{eq:pre-final} is bounded by
\[
\frac{C}{m(Q)^4}\,\big(\E_Q[\psi^4]\big)^{1/2}\cdot \frac{\|\psi'\|_\infty^2}{n}\;+\; \frac{C}{m(Q)^4}\,\E_Q[\psi^4]\cdot \frac{1}{n^2}.
\]
The last term is dominated by the displayed one because $\E_Q[\psi^4]\le \|\psi\|_\infty^4$ and $n\ge 1$; hence we may keep the cleaner expression. Collecting terms we have shown \eqref{eq:Rem-quad-goal}.

\paragraph{Absorbing terms proportional to $\E_Q[(\hat\theta_\ep-\theta)^2]$.}
If, in an intermediate step, one keeps the Lipschitz contribution
\[
\E_Q\!\left[(m_n(\bar t)-m_n(\theta))^2\right]\ \le\ L^2\,\E_Q\big[(\hat\theta_\ep-\theta)^2\big]
\]
(with $L=\sup_{x,t}|\partial_t \psi'(x-t)|=\|\psi''\|_\infty$), then the bound \eqref{eq:pre-final} produces an extra addend of the form
\(
\frac{C\,L^2}{m(Q)^4}\,\E_Q\big[(\hat\theta_\ep-\theta)^2\big].
\)
When this remainder bound is plugged back into the main risk decomposition (equation (B.6) in the text),
\[
\E_Q[(\hat\theta_\ep-\theta)^2]
\le \frac{1}{n}\frac{\mathrm{Var}_Q(\psi)}{m(Q)^2}+ \frac{\E_Q[\psi]^2}{m(Q)^2} + \E_Q[\mathrm{Rem}^2],
\]
the term $\frac{C\,L^2}{m(Q)^4}\E_Q[(\hat\theta_\ep-\theta)^2]$ sits on the right-hand side. Using the curvature lower bound $m(Q)\ge c_0 I_\star$ (Lemma~\ref{lem:score-moments}), choose $n$-independent constants so that
\(
\frac{C\,L^2}{m(Q)^4}\ \le\ \frac12,
\)
and move $\tfrac12\,\E_Q[(\hat\theta_\ep-\theta)^2]$ to the left-hand side. This is the standard absorption step; it only changes the multiplicative constant in the final risk bound. Thus, for the purpose of \eqref{eq:Rem-quad-goal}, we can to replace $\E_Q[(m_n(\bar t)-m(Q))^2]$ by $\E_Q[(m_n(\theta)-m(Q))^2]\le \|\psi'\|_\infty^2/n$, yielding the stated clean form.

\paragraph{The remaining mixed term in \eqref{eq:Rem-quad-goal}.}
Starting from \eqref{eq:Rem-quad-goal}, the only piece we have not yet bounded is
\(
\E_Q\!\left[\Big(G_n(\theta)-g_Q(\theta)\Big)^2\big(m_n(\bar t)-m(Q)\big)^2\right].
\)
Write $A:=G_n(\theta)-g_Q(\theta)$ and $B(\bar t):=m_n(\bar t)-m(Q)$. Using the Lipschitz property in $t$ implied by $\|\psi''\|_\infty<\infty$ we have
\[
|m_n(\bar t)-m_n(\theta)| \le \|\psi''\|_\infty\,|\bar t-\theta|=\|\psi''\|_\infty\,|\hat\theta_\ep-\theta|,
\]
and similarly for the population derivative,
\[
\big|\E_Q[\psi'(X'-\bar t)]-\E_Q[\psi'(X'-\theta)]\big|\le \|\psi''\|_\infty\,|\hat\theta_\ep-\theta|.
\]
Hence
\[
|B(\bar t)|=\big|m_n(\bar t)-m(Q)\big|
\le \big|m_n(\theta)-m(Q)\big| + 2\|\psi''\|_\infty\,|\hat\theta_\ep-\theta|.
\]
Squaring and using $(x+y)^2\le 2x^2+2y^2$ gives
\(
B(\bar t)^2 \le 2\big(m_n(\theta)-m(Q)\big)^2 + 8\|\psi''\|_\infty^2\,(\hat\theta_\ep-\theta)^2.
\)
Therefore
\begin{align*}
\E_Q\!\left[A^2 B(\bar t)^2\right]
&\le 2\,\E_Q\!\left[A^2\big(m_n(\theta)-m(Q)\big)^2\right]
+ 8\|\psi''\|_\infty^2\,\E_Q\!\left[A^2(\hat\theta_\ep-\theta)^2\right].
\end{align*}

We bound the two terms separately. For the first, by Cauchy--Schwarz and the fourth-moment bounds already used in \eqref{eq:Gn-fourth} and the analogous bound for $m_n(\theta)-m(Q)$,
\[
\E_Q\!\left[A^2\big(m_n(\theta)-m(Q)\big)^2\right]
\le \Big(\E_Q[A^4]\Big)^{1/2}\Big(\E_Q\!\left[\big(m_n(\theta)-m(Q)\big)^4\right]\Big)^{1/2}
\le \frac{C}{n^2}\,(\E_Q[\psi^4])^{1/2}\,\|\psi'\|_\infty^2.
\]
For the second term we use boundedness of $\psi$: since $|G_n(\theta)|\le \|\psi\|_\infty$ and $|g_Q(\theta)|\le \|\psi\|_\infty$, we have $|A|\le 2\|\psi\|_\infty$, hence
\(
\E_Q\!\left[A^2(\hat\theta_\ep-\theta)^2\right]
\le 4\|\psi\|_\infty^2\,\E_Q\!\left[(\hat\theta_\ep-\theta)^2\right].
\)
Combining the last two displays yields
\begin{equation}\label{eq:key-mixed}
\E_Q\!\left[\Big(G_n-g_Q\Big)^2\big(m_n(\bar t)-m(Q)\big)^2\right]
\le \frac{C}{n^2}\,(\E_Q[\psi^4])^{1/2}\,\|\psi'\|_\infty^2
\;+\; C\,\|\psi''\|_\infty^2\,\|\psi\|_\infty^2\,\E_Q\!\left[(\hat\theta_\ep-\theta)^2\right].
\end{equation}

\paragraph{Completing the remainder bound.}
Insert \eqref{eq:key-mixed} into \eqref{eq:Rem-quad-goal}. We obtain
\[
\E_Q[\mathrm{Rem}^2]
\le \frac{C}{m(Q)^4}\left\{\frac{(\E_Q[\psi^4])^{1/2}\,\|\psi'\|_\infty^2}{n^2}
+ \|\psi''\|_\infty^2\,\|\psi\|_\infty^2\,\E_Q[(\hat\theta_\ep-\theta)^2]\right\}
+ \frac{C}{m(Q)^4}\,(\E_Q[\psi^4])^{1/2}\,\frac{\|\psi'\|_\infty^2}{n}.
\]
The $n^{-2}$ term is dominated by the $n^{-1}$ term, since $\E_Q[\psi^4]\le \|\psi\|_\infty^4$ and $n\ge1$.
%The term proportional to $\E_Q[(\hat\theta_\ep-\theta)^2]$ will be absorbed momentarily. 
Thus
\begin{equation}\label{eq:Rem-final}
\E_Q[\mathrm{Rem}^2]
\le \frac{C}{m(Q)^4}\,(\E_Q[\psi^4])^{1/2}\,\frac{\|\psi'\|_\infty^2}{n}
\;+\; \frac{C\,\|\psi''\|_\infty^2\,\|\psi\|_\infty^2}{m(Q)^4}\,\E_Q[(\hat\theta_\ep-\theta)^2].
\end{equation}

\paragraph{Risk decomposition and absorption.}
Returning to \eqref{eq:basic-mvt} and its exact decomposition, the standard inequality $(a+b+c)^2\le3(a^2+b^2+c^2)$ yields
\[
\E_Q[(\hat\theta_\ep-\theta)^2]
\le \frac{C}{m(Q)^2}\,\E_Q\!\left[\big(G_n(\theta)-g_Q(\theta)\big)^2\right]
+ \frac{g_Q(\theta)^2}{m(Q)^2}
+ \E_Q[\mathrm{Rem}^2].
\]
Since $\E_Q[(G_n(\theta)-g_Q(\theta))^2]=\mathrm{Var}_Q(G_n(\theta))=\mathrm{Var}_Q(\psi(X'-\theta))/n$, 
using \eqref{eq:Rem-final} gives
\begin{align*}
\E_Q[(\hat\theta_\ep-\theta)^2]
&\le \frac{1}{n}\,\frac{\mathrm{Var}_Q(\psi(X'-\theta))}{m(Q)^2}
+ \frac{g_Q(\theta)^2}{m(Q)^2}
+ \frac{C}{m(Q)^4}\,(\E_Q[\psi^4])^{1/2}\,\frac{\|\psi'\|_\infty^2}{n}\\
&\qquad + \frac{C\,\|\psi''\|_\infty^2\,\|\psi\|_\infty^2}{m(Q)^4}\,\E_Q[(\hat\theta_\ep-\theta)^2].
\end{align*}
By the curvature lower bound $m(Q)\ge \tfrac12 I(\tau_\star)>0$ and the boundedness of $\psi$ and its derivatives, the last coefficient is an $n$-independent constant $<1$ after adjusting $C$. Moving it to the left-hand side (the standard absorption step) alters only constants, yielding
\begin{equation}\label{eq:final-risk}
\E_Q[(\hat\theta_\ep-\theta)^2]
\ \le\ \frac{C}{n}\,\frac{\mathrm{Var}_Q(\psi(X'-\theta))}{m(Q)^2}
\ +\ \frac{C\,g_Q(\theta)^2}{m(Q)^2}
\ +\ \frac{C}{n}\,\frac{(\E_Q[\psi^4])^{1/2}\,\|\psi'\|_\infty^2}{m(Q)^4}.
\end{equation}

\qed