\section{Explicit corrections}
\label{sec:examples}

These formulas illustrate the correction identity; the algorithm evaluates
the complex score directly and does not require symbolic integration.

\paragraph{Linear prediction with squared loss.}
Let $\theta,Z,W,U,V\in\mathbb R^p$, with $W=Z+U$ and independent
$U,V\sim N(0,\Omega)$. For a fixed response $y$ and the squared
linear-prediction loss
$\ell(\theta;z)=\tfrac12(\theta^\top z-y)^2$, the clean score is
\[
g(\theta;z)=(\theta^\top z-y)z.
\]
The complex-corrected score and its conditional mean over the simulated
perturbation are
\[
G(\theta;W,V)
=\operatorname{Re}\bigl[(\theta^\top(W+iV)-y)(W+iV)\bigr]
=(\theta^\top W-y)W-(\theta^\top V)V,
\]
\[
\widetilde G(\theta;W)
=\mathbb E_V[G(\theta;W,V)\mid W]
=(\theta^\top W-y)W-\Omega\theta.
\]
In particular, $\mathbb E\widetilde G(\theta;W)=\nabla F(\theta)$.
The identity is exact for every $\theta$. Indeed,
\[
D_zg^{\mathbb C}(\theta;q)
=q\theta^\top+(\theta^\top q-y)I_p,
\qquad
\|D_zg^{\mathbb C}(\theta;q)\|_{\rm op}
\le2\|\theta\|\|q\|+|y|.
\]
Thus Assumption~\ref{ass:envelope} holds with $a=1$ and
$M(\theta)=2\|\theta\|+|y|$. If
$H=\mathbb E[ZZ^\top]\succ0$, let $\mu=\lambda_{\min}(H)$ and
$\theta^*=H^{-1}y\mathbb EZ$. Then
$\nabla F(\theta)=H(\theta-\theta^*)$ and
\[
M(\theta)^2\le
2(2\|\theta^*\|+|y|)^2+8\mu^{-2}\|\nabla F(\theta)\|^2.
\]
The clean score is affine in $\theta$ with square-integrable random
coefficients, so its centered variance also has this relative form.
Consequently the main results apply without bounding the parameter set.

\paragraph{Nonlinear analytic loss.}
Let $\theta,z\in\mathbb R^p$ and
$\ell(\theta;z)=\tfrac12\|\theta-\sin z\|^2$, where $\sin$ acts
coordinatewise. Then $g(\theta;z)=\theta-\sin z$, and the complex update is
\[
G(\theta;W,V)=\theta-\sin W\odot\cosh V.
\]
Since $V_j\sim N(0,\Omega_{jj})$, integrating out $V$ gives
\[
\widetilde G(\theta;W)
=\theta-\bigl(e^{\Omega_{jj}/2}\sin W_j\bigr)_{j=1}^p,
\qquad
\mathbb E\widetilde G(\theta;W)=\nabla F(\theta).
\]
Here $D_zg^{\mathbb C}(\theta;q)=-\operatorname{diag}(\cos q_j)$
has an exponential bound independent of $\theta$, while
$\nabla^2F=I_p$. The assumptions therefore hold on all of $\mathbb R^p$.