We derive the KDE gradient, instantiating the general form~\eqref{eq:roll-gradient}.
Throughout, $\tau = \hat{F}_0^{-1}(1-\alpha;\mathcal{B}_0)$ and bandwidths $v_0, v_1$
are as defined in \Cref{sec:kde-bandwidth}.

\paragraph{Case $y_i = 1$: expanding the KDE sum.}

When $\mathbf{x}_i \in \mathcal{B}_1$, the score $f_\theta(\mathbf{x}_i)$ appears directly
in $\hat{F}_1$ and has no effect on $\tau$.  Expanding the KDE CDF:
%
\[
  \dfrac{\partial \hat{F}_1\!\left(\hat{F}_0^{-1}(1-\alpha)\right)}{\partial f_{\theta}(\mathbf{x}_i)}
  = \frac{\partial}{\partial f_{\theta}(\mathbf{x}_i)}
    \frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1}
    \sigma_1\!\left(\tau - f_{\theta}(\mathbf{x}_j)\right)
\]
%
Separating the $j = i$ term (the only one depending on $f_\theta(\mathbf{x}_i)$):
%
\[
  = \frac{\partial}{\partial f_{\theta}(\mathbf{x}_i)}\!\left(
    \sum_{\substack{\mathbf{x}_j \in \mathcal{B}_1 \\ j \neq i}}
    \frac{\sigma_1\!\left(\tau - f_{\theta}(\mathbf{x}_j)\right)}{|\mathcal{B}_1|}
    \;+\;
    \frac{\sigma_1\!\left(\tau - f_{\theta}(\mathbf{x}_i)\right)}{|\mathcal{B}_1|}
  \right)
\]
%
The first sum is constant in $f_\theta(\mathbf{x}_i)$.  Applying the chain rule to the
last term with $\partial(\tau - f_\theta(\mathbf{x}_i))/\partial f_\theta(\mathbf{x}_i) = -1$:
%
\begin{equation}
  \frac{\partial \hat{F}_1(\hat{F}_0^{-1}(1-\alpha))}{\partial f_{\theta}(\mathbf{x}_i)}
  = -\frac{1}{|\mathcal{B}_1|}\sigma_1'(\tau - f_{\theta}(\mathbf{x}_i))
  \label{eq:kde-grad-y1}
\end{equation}

\paragraph{Case $y_i = 0$: derivative of $\hat{F}_1$ with respect to $\tau$.}

Differentiating the KDE CDF with respect to $\tau$:
%
\[
  \frac{\partial \hat{F}_1\!\left(\hat{F}_0^{-1}(1-\alpha)\right)}{\partial \hat{F}_0^{-1}(1-\alpha)}
  = \frac{\partial}{\partial \tau}
    \frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1}
    \sigma_1\!\left(\tau - f_{\theta}(\mathbf{x}_j)\right)
\]
%
Since $\partial(\tau - f_\theta(\mathbf{x}_j))/\partial\tau = +1$ for every $j$,
the chain rule gives:
%
\begin{equation}
  \frac{\partial \hat{F}_1(\hat{F}_0^{-1}(1-\alpha))}{\partial \hat{F}_0^{-1}(1-\alpha)}
  = +\frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1}\sigma_1'(\tau - f_{\theta}(\mathbf{x}_j))
  \label{eq:kde-dF1-dtau}
\end{equation}

\paragraph{Case $y_i = 0$: derivative of $\tau$ with respect to $f_\theta(\mathbf{x}_i)$.}

Since $\tau = \hat{F}_0^{-1}(1-\alpha;\mathcal{B}_0)$ is computed numerically, we apply the \emph{inverse function theorem}: if $h$ has inverse $g$,
then $g'(y) = 1/h'(g(y))$.

We view $\hat{F}_0^{-1}(1-\alpha;\mathcal{B}_0)$ as a function of $f_\theta(\mathbf{x}_i)$
and find its inverse.  From the defining equation
$|\mathcal{B}_0|\cdot(1-\alpha) = \sum_{\mathbf{x}_j \in \mathcal{B}_0}
\sigma_0(\tau - f_\theta(\mathbf{x}_j))$, solving for $f_\theta(\mathbf{x}_i)$:
%
\[
  f_\theta(\mathbf{x}_i)
  = \tau - \sigma_0^{-1}\!\!\left(
    |\mathcal{B}_0|\cdot(1-\alpha)
    - \sum_{\substack{\mathbf{x}_j \in \mathcal{B}_0 \\ j \neq i}}
      \sigma_0\!\left(\tau - f_{\theta}(\mathbf{x}_j)\right)
  \right)
\]
%
Differentiating this expression with respect to $\tau$ and applying the chain rule:
%
\[
  \frac{\partial f_\theta(\mathbf{x}_i)}{\partial \tau}
  = 1 + \sigma_0^{-1}{}'\!\!\left(\cdots\right)
    \cdot
    \sum_{\substack{\mathbf{x}_j \in \mathcal{B}_0 \\ j \neq i}}
    \sigma_0'\!\left(\tau - f_{\theta}(\mathbf{x}_j)\right)
\]
%
(the leading $+1$ comes from differentiating $\tau$; the second term from differentiating
through $\sigma_0^{-1}$). By the inverse function theorem:
%
\[
  \frac{\partial \hat{F}_0^{-1}(1-\alpha ; \mathcal{B}_0)}{\partial f_\theta(\mathbf{x}_i)}
  = \frac{1}{\displaystyle
      1 + \sigma_0^{-1}{}'\!\!\left(
        |\mathcal{B}_0|\cdot(1-\alpha)
        - \sum_{j \neq i}\sigma_0(\tau - f_\theta(\mathbf{x}_j))
      \right)
      \cdot \sum_{j \neq i}\sigma_0'(\tau - f_\theta(\mathbf{x}_j))}
\]
%
Using $|\mathcal{B}_0|\cdot(1-\alpha) = \sum_j \sigma_0(\tau - f_\theta(\mathbf{x}_j))$,
the argument of $\sigma_0^{-1}{}'$ simplifies:
%
\[
  |\mathcal{B}_0|\cdot(1-\alpha)
  - \sum_{j \neq i}\sigma_0(\tau - f_\theta(\mathbf{x}_j))
  = \sigma_0(\tau - f_\theta(\mathbf{x}_i))
\]
%
Substituting and applying $\sigma_0^{-1}{}'(\sigma_0(u)) = 1/\sigma_0'(u)$,
then multiplying numerator and denominator by $\sigma_0'(\tau - f_\theta(\mathbf{x}_i))$:
%
\begin{equation}
  \frac{\partial \hat{F}_0^{-1}(1-\alpha ; \mathcal{B}_0)}{\partial f_\theta(\mathbf{x}_i)} =
  \frac{\sigma_0'(\tau - f_{\theta}(\mathbf{x}_i))}{\displaystyle\sum_{\mathbf{x}_j \in \mathcal{B}_0}\sigma_0'(\tau - f_{\theta}(\mathbf{x}_j))}
  \label{eq:kde-grad-tau}
\end{equation}

\paragraph{Efficient computation via the sigmoid identity.}

Direct evaluation of \eqref{eq:kde-grad-tau} requires computing $\sigma_0'$ for every
point — potentially unstable when a point is far from $\tau$.  For the sigmoid kernel,
the identity

\begin{equation}
  \sigma(u;\,v)\,\bigl(1 - \sigma(u;\,v)\bigr) = \frac{\sigma'(u;\,v)}{v}
  \label{eq:sigmoid-identity}
\end{equation}

(proved by direct substitution: both sides equal $\exp(-v|u|)/(1+\exp(-v|u|))^2$)
allows $v_0$ to cancel between numerator and denominator, so the ratio is expressed
entirely in terms of sigmoid values already cached from the forward pass:

\begin{equation}
  \frac{\partial \hat{F}_0^{-1}(1-\alpha ; \mathcal{B}_0)}{\partial f_\theta(\mathbf{x}_i)}
  =
  \frac{\sigma_0(\tau-f_\theta(\mathbf{x}_i))\,\bigl(1-\sigma_0(\tau-f_\theta(\mathbf{x}_i))\bigr)}
       {\displaystyle\sum_{\mathbf{x}_j \in \mathcal{B}_0}
         \sigma_0(\tau-f_\theta(\mathbf{x}_j))\,\bigl(1-\sigma_0(\tau-f_\theta(\mathbf{x}_j))\bigr)}
  \label{eq:kde-grad-tau-efficient}
\end{equation}

Substituting \eqref{eq:kde-grad-y1}, \eqref{eq:kde-dF1-dtau}, and \eqref{eq:kde-grad-tau}
into \eqref{eq:roll-gradient} yields \eqref{eq:kde-grad-combined}.