diff --git a/content/method/method.tex b/content/method/method.tex
index 783f0a3..a6aee0b 100644
--- a/content/method/method.tex
+++ b/content/method/method.tex
@@ -348,15 +348,15 @@ Firstly, we must estimate the parameters to fit the gaussian distributions
 to the true and false populations of the score output.
 
 \begin{equation}
-  \mu_0 = \frac{\sum_{\mathbf{x}_0 \in \mathcal{B}_0}{f_{\theta}(\mathbf{x}_0)}}{|\mathcal{B}_0|}
+  \mu_0 = \frac{1}{|\mathcal{B}_0|}\sum_{\mathbf{x}_j \in \mathcal{B}_0}{f_{\theta}(\mathbf{x}_j)}
   \qquad
-  \mu_1 = \frac{\sum_{\mathbf{x}_0 \in \mathcal{B}_1}{f_{\theta}(\mathbf{x}_1)}}{|\mathcal{B}_1|}
+  \mu_1 = \frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1}{f_{\theta}(\mathbf{x}_j)}
 \end{equation}
 
 \begin{equation}
-  \sigma_0^2 = \frac{1}{|\mathcal{B}_0|}\sum_{\mathbf{x}_0 \in \mathcal{B}_0}(\mathbf{x}_0 - \mu_0)^2
+  \sigma_0^2 = \frac{1}{|\mathcal{B}_0|}\sum_{\mathbf{x}_j \in \mathcal{B}_0}(f_\theta(\mathbf{x}_j) - \mu_0)^2
   \qquad
-  \sigma_1^2 = \frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_1 \in \mathcal{B}_1}(\mathbf{x}_1 - \mu_1)^2
+  \sigma_1^2 = \frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1}(f_\theta(\mathbf{x}_j) - \mu_1)^2
 \end{equation}
 
 Once these parameters are estimated, we can then define our ROLL objective as required.
@@ -369,7 +369,11 @@ and its inverse (quantile function) is:
 \[
   F^{-1}(p) = \mu + \sigma\sqrt{2}\,\text{ierf}(2p - 1)
 \]
-where TODO EXPLAIN ERF AND IERF. Plugging into the general ROLL framework, the threshold
+where $\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}\,\mathrm{d}t$ is the Gauss
+error function and $\operatorname{ierf}$ denotes its functional inverse, satisfying
+$\operatorname{ierf}(\operatorname{erf}(x)) = x$. Neither has a closed form, but both are
+available in standard numerical libraries (e.g.\ \texttt{torch.erf} and
+\texttt{torch.erfinv}). Plugging into the general ROLL framework, the threshold
 $\tau$ achieving FPR $= \alpha$ and the resulting loss are:
 \begin{align}
   \tau &= \mu_0 + \sigma_0\sqrt{2}\,\text{ierf}(2\alpha - 1) \nonumber \\
@@ -395,7 +399,7 @@ Firstly, for $y_i = 1$.
 \]
 
 \[
-  = \frac{1}{\sqrt{2}\sigma_1}\exp\left(\frac{\hat{F}_0^{-1}(\alpha) - \mu_1}{\sqrt{2}\sigma_1}\right)
+  = -\frac{1}{\sigma_1\sqrt{2\pi}}\exp\!\left(-\frac{(\hat{F}_0^{-1}(\alpha) - \mu_1)^2}{2\sigma_1^2}\right)
 \]
 
 \[
@@ -408,28 +412,75 @@ Firstly, for $y_i = 1$.
 \]
 
 \[
-  = \frac{\hat{F}_0^{-1}(\alpha)-\mu_1}{\sqrt{2\pi}\sigma_1^2} \exp\left( - {\left( \frac{\hat{F}_0^{-1}(\alpha) - \mu_1}{\sqrt{2}\sigma_1} \right)}^2 \right)
+  = \frac{\mu_1 - \hat{F}_0^{-1}(\alpha)}{\sigma_1^2\sqrt{2\pi}} \exp\!\left( -\frac{(\hat{F}_0^{-1}(\alpha) - \mu_1)^2}{2\sigma_1^2} \right)
 \]
 \[
-  \frac{\partial \sigma_1}{\partial f_{\theta}(\mathbf{x}_i)} = \frac{\mathbf{x}_i - \mu_1}{|\mathcal{B}_1| \sigma_1}
+  \frac{\partial \sigma_1}{\partial f_{\theta}(\mathbf{x}_i)} = \frac{f_\theta(\mathbf{x}_i) - \mu_1}{|\mathcal{B}_1| \sigma_1}
 \]
 
-The final derivation, given $y_i = 1$, is given as:
+Combining via the chain rule, the gradient for $y_i = 1$ is:
+\begin{equation}
+  \dfrac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial f_{\theta}(\mathbf{x}_i)}
+  = -\frac{1}{|\mathcal{B}_1|\,\sigma_1\sqrt{2\pi}}
+    \exp\!\left(-\frac{(\tau - \mu_1)^2}{2\sigma_1^2}\right)
+    \left(1 + \frac{(\tau - \mu_1)(f_\theta(\mathbf{x}_i) - \mu_1)}{\sigma_1^2}\right)
+  \label{eq:gauss-grad-y1}
+\end{equation}
+where $\tau = \hat{F}_0^{-1}(\alpha) = \mu_0 + \sigma_0\sqrt{2}\,\operatorname{ierf}(2\alpha - 1)$.
 
+Given $y_i = 0$, the score $f_\theta(\mathbf{x}_i)$ affects the loss only through the threshold
+$\tau = \hat{F}_0^{-1}(\alpha)$, which depends on $\mu_0$ and $\sigma_0$.
+Applying the chain rule:
 \[
-  \dfrac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial f_{\theta}(\mathbf{x}_i)} = \dfrac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial \mu_1} \cdot \frac{1}{|\mathcal{B}_1|} +
-  \dfrac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial \sigma_1} \cdot \frac{\partial \sigma_1}{\partial f_{\theta}(\mathbf{x}_i)} = \frac{\mathbf{x}_i - \mu_1}{|\mathcal{B}_1| \sigma_1}
+  \frac{\partial \hat{F}_1(\tau)}{\partial f_\theta(\mathbf{x}_i)}
+  = \underbrace{\frac{\partial \hat{F}_1(\tau)}{\partial \tau}}_{\text{PDF of class 1 at }\tau}
+    \cdot \frac{\partial \tau}{\partial f_\theta(\mathbf{x}_i)}
 \]
 
-TODO - above equation is based off of mathexchange derivation. Link and reference properly!
-
-Given $y_i = 0$ we first must derive:
-
+The first factor is the Gaussian PDF evaluated at $\tau$:
 \[
-  \dfrac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial \hat{F}_0^{-1}(\alpha)} = \frac{1}{\sqrt{2 \pi \sigma_1^2}}\exp\left( -{\left( \frac{\hat{F}_0^{-1}(\alpha) - \mu_1}{2\sigma_1^2}\right)}^2\right)
+  \frac{\partial \hat{F}_1(\tau)}{\partial \tau}
+  = \frac{1}{\sigma_1\sqrt{2\pi}}\exp\!\left(-\frac{(\tau - \mu_1)^2}{2\sigma_1^2}\right)
 \]
 
-TODO finish somehow - later!
+For the second factor, $\tau = \mu_0 + \sigma_0\sqrt{2}\,\operatorname{ierf}(2\alpha - 1)$, so
+$\frac{\partial \tau}{\partial \mu_0} = 1$ and
+$\frac{\partial \tau}{\partial \sigma_0} = \sqrt{2}\,\operatorname{ierf}(2\alpha - 1) = \frac{\tau - \mu_0}{\sigma_0}$.
+Combined with $\frac{\partial \mu_0}{\partial f_\theta(\mathbf{x}_i)} = \frac{1}{|\mathcal{B}_0|}$ and
+$\frac{\partial \sigma_0}{\partial f_\theta(\mathbf{x}_i)} = \frac{f_\theta(\mathbf{x}_i) - \mu_0}{|\mathcal{B}_0|\sigma_0}$:
+\[
+  \frac{\partial \tau}{\partial f_\theta(\mathbf{x}_i)}
+  = \frac{1}{|\mathcal{B}_0|}\left(1 + \frac{(\tau - \mu_0)(f_\theta(\mathbf{x}_i) - \mu_0)}{\sigma_0^2}\right)
+\]
+
+Therefore, the gradient for $y_i = 0$ is:
+\begin{equation}
+  \dfrac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial f_\theta(\mathbf{x}_i)}
+  = \frac{1}{|\mathcal{B}_0|\,\sigma_1\sqrt{2\pi}}
+    \exp\!\left(-\frac{(\tau - \mu_1)^2}{2\sigma_1^2}\right)
+    \left(1 + \frac{(\tau - \mu_0)(f_\theta(\mathbf{x}_i) - \mu_0)}{\sigma_0^2}\right)
+  \label{eq:gauss-grad-y0}
+\end{equation}
+
+\paragraph{Combined gradient.}
+
+Substituting \Cref{eq:gauss-grad-y1} and \Cref{eq:gauss-grad-y0} into \Cref{eq:roll-gradient},
+and letting $\tau = \hat{F}_0^{-1}(\alpha) = \mu_0 + \sigma_0\sqrt{2}\,\operatorname{ierf}(2\alpha-1)$:
+
+\begin{equation}
+  \frac{\partial \mathcal{L}_{\text{ROLL-TPR@FPR}}^{\text{GAUSSIAN}}}{\partial f_\theta(\mathbf{x}_i)} =
+  \begin{cases}
+    \dfrac{-1}{|\mathcal{B}_1|\,\sigma_1\sqrt{2\pi}}
+    \exp\!\left(-\dfrac{(\tau-\mu_1)^2}{2\sigma_1^2}\right)
+    \!\left(1 + \dfrac{(\tau-\mu_1)(f_\theta(\mathbf{x}_i)-\mu_1)}{\sigma_1^2}\right)
+      & \text{if } y_i = 1 \\[18pt]
+    \dfrac{1}{|\mathcal{B}_0|\,\sigma_1\sqrt{2\pi}}
+    \exp\!\left(-\dfrac{(\tau-\mu_1)^2}{2\sigma_1^2}\right)
+    \!\left(1 + \dfrac{(\tau-\mu_0)(f_\theta(\mathbf{x}_i)-\mu_0)}{\sigma_0^2}\right)
+      & \text{if } y_i = 0
+  \end{cases}
+  \label{eq:gauss-grad-combined}
+\end{equation}
 
 \subsection{Beta ROLL}
 \label{sec:roll-beta}