This appendix gives the full proof of \Cref{prop:gradient-balance}.

\begin{proof}
We treat each class separately via a uniform-shift argument.

\medskip\noindent\textit{Sum over $\mathcal{B}_0$.}
From \Cref{eq:roll-gradient}, for $y_i = 0$:
\[
  \frac{\partial \mathcal{L}}{\partial f_\theta(\mathbf{x}_i)}
  = \frac{\partial \hat{F}_1(\tau)}{\partial \tau}
    \cdot \frac{\partial \tau}{\partial f_\theta(\mathbf{x}_i)}
\]
We claim $\sum_{\mathbf{x}_i \in \mathcal{B}_0}\frac{\partial \tau}{\partial f_\theta(\mathbf{x}_i)} = 1$.
Shift all negative scores uniformly: $f_\theta(\mathbf{x}_i) \to f_\theta(\mathbf{x}_i) + \delta$ for
$\mathbf{x}_i \in \mathcal{B}_0$. By translation invariance, $\hat{F}_0$ simply translates, so
$\tau = \hat{F}_0^{-1}(1-\alpha)$ shifts by exactly $\delta$. Differentiating with respect to $\delta$ at
$\delta = 0$:
\[
  1 = \frac{\mathrm{d}\tau}{\mathrm{d}\delta}\bigg|_{\delta=0}
    = \sum_{\mathbf{x}_i \in \mathcal{B}_0} \frac{\partial \tau}{\partial f_\theta(\mathbf{x}_i)}
\]
Summing the $y_i = 0$ gradients therefore gives:
\begin{equation}
  \sum_{\mathbf{x}_i \in \mathcal{B}_0}
    \frac{\partial \mathcal{L}}{\partial f_\theta(\mathbf{x}_i)}
  = \frac{\partial \hat{F}_1(\tau)}{\partial \tau} \cdot 1
  = \frac{\partial \hat{F}_1(\tau)}{\partial \tau}
  \label{eq:grad-balance-b0}
\end{equation}

\medskip\noindent\textit{Sum over $\mathcal{B}_1$.}
For $y_i = 1$, $\frac{\partial \mathcal{L}}{\partial f_\theta(\mathbf{x}_i)} = \frac{\partial \hat{F}_1(\tau)}{\partial f_\theta(\mathbf{x}_i)}$.
Shifting all positive scores by $\delta$ and applying translation invariance of $\hat{F}_1$:
$\hat{F}_1(\tau\,;\,\mathbf{s}+\delta\mathbf{1}) = \hat{F}_1(\tau-\delta\,;\,\mathbf{s})$.
Differentiating at $\delta = 0$:
\[
  \sum_{\mathbf{x}_i \in \mathcal{B}_1}
    \frac{\partial \hat{F}_1(\tau)}{\partial f_\theta(\mathbf{x}_i)}
  = -\frac{\partial \hat{F}_1(\tau)}{\partial \tau}
\]
Combining with \eqref{eq:grad-balance-b0} gives \eqref{eq:grad-balance}.
\end{proof}