Small fixes

content/method/method.tex

@@ -231,7 +231,7 @@ This yields two versions of our objective. Substituting the ICDF threshold into
 expressions, our estimated TPR at a fixed FPR of $\alpha$ is $1 - \hat{F}_1(\hat{F}_0^{-1}(\alpha))$.
 Therefore, to maximise TPR we minimise:
 \begin{equation}
-  \mathcal{L}_{\text{ROLL-TPR@FPR}}(f_{\theta}(\mathbf{X}) ; \alpha) = \hat{F}_1(\hat{F}_0^{-1}(\alpha))
+  \mathcal{L}_{\text{ROLL-TPR@FPR}}(f_{\theta}(\mathcal{B}) ; \alpha) = \hat{F}_1(\hat{F}_0^{-1}(\alpha))
   \label{eq:roll-tpr-at-fpr}
 \end{equation}
 
@@ -250,7 +250,7 @@ As both loss forms are symmetric in structure, it is sufficient to treat the TPR
 If $y_i = 1$, then $(\mathbf{x}_i, y_i) \in \mathcal{B}_1$ and $f_\theta(\mathbf{x}_i)$ has no
 effect on $\hat{F}_0^{-1}(\alpha)$. Therefore:
 \begin{equation}
-  \left.\frac{\partial \mathcal{L}_{\text{ROLL-TPR@FPR}}(f_\theta(\mathbf{X}) ; \alpha)}
+  \left.\frac{\partial \mathcal{L}_{\text{ROLL-TPR@FPR}}(f_\theta(\mathcal{B}) ; \alpha)}
              {\partial f_\theta(\mathbf{x}_i)}\right|_{y_i = 1}
   = \frac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial f_\theta(\mathbf{x}_i)}
 \end{equation}
@@ -366,14 +366,11 @@ where TODO EXPLAIN ERF AND IERF. Plugging into the general ROLL framework, the t
 $\tau$ achieving FPR $= \alpha$ and the resulting loss are:
 \begin{align}
   \tau &= \mu_0 + \sigma_0\sqrt{2}\,\text{ierf}(2\alpha - 1) \nonumber \\
-  \mathcal{L}_{\text{ROLL-TPR@FPR}}^{\text{GAUSSIAN}}(f_{\theta}(\mathbf{X}) ; \alpha)
+  \mathcal{L}_{\text{ROLL-TPR@FPR}}^{\text{GAUSSIAN}}(f_{\theta}(\mathcal{B}) ; \alpha)
     &= \frac{1}{2}\left[1 + \text{erf}\!\left(\frac{\tau - \mu_1}{\sigma_1\sqrt{2}}\right)\right]
   \label{eq:roll-tpr-at-fpr-gaussian}
 \end{align}
 
-
-
-
 \subsubsection{Gradient Derivation}
 \label{sec:roll-gaussian-backward}