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content/appendixs/appendix_kde_grad.tex

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+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}.