From fa20fe0eb70212dfcbda1d6976c280dfe87da95c Mon Sep 17 00:00:00 2001 From: Aner Zakobar Date: Thu, 2 Jul 2026 23:24:19 +0300 Subject: [PATCH] Added sections on symmetry, better equation mapping. --- content/method/method.tex | 120 ++++++++++++++++++++++++-------------- 1 file changed, 76 insertions(+), 44 deletions(-) diff --git a/content/method/method.tex b/content/method/method.tex index 3261aee..783f0a3 100644 --- a/content/method/method.tex +++ b/content/method/method.tex @@ -242,10 +242,17 @@ Therefore, to minimise FPR: \label{eq:roll-fpr-at-tpr} \end{equation} -TODO explain the trick where we can only use one version - base off code. +Observe that \eqref{eq:roll-tpr-at-fpr} and \eqref{eq:roll-fpr-at-tpr} share the same +functional structure: each evaluates one empirical CDF at the quantile of the other. This +symmetry means \eqref{eq:roll-fpr-at-tpr} can be reduced to \eqref{eq:roll-tpr-at-fpr} by +negating the model scores and exchanging the class labels. Concretely, replacing +$f_\theta(\mathbf{x})$ with $-f_\theta(\mathbf{x})$ and $y$ with $1 - y$ swaps the roles +of the two populations, turning a TPR@FPR objective into a FPR@TPR one. Because the +two objectives are therefore interchangeable at the implementation level, all derivations +and implementation details below are given for \eqref{eq:roll-tpr-at-fpr} only; +\eqref{eq:roll-fpr-at-tpr} follows by the same transformation applied to the inputs. We now derive the gradient of \eqref{eq:roll-tpr-at-fpr} with respect to $f_\theta(\mathbf{x}_i)$. -As both loss forms are symmetric in structure, it is sufficient to treat the TPR@FPR case. 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: @@ -470,7 +477,6 @@ As before, to calculate the loss, we calculate: \[ \mathcal{L}_{\text{ROLL-TPR@FPR}}^{\text{KDE}}(f_{\theta}(\mathcal{B}) ; \alpha) = \hat{F}_1(\hat{F}_0^{-1}(\alpha)) - \label{eq:roll-tpr-at-fpr} \] To calculate this, we first define the kernel function as $\sigma '$, and the CDF of the kernel function as $\sigma$. The approach used here is to define $\sigma$ as a sigmoid function,. @@ -522,93 +528,119 @@ Importantly, we must save the calculation of $\tau$ for the backward derivation, % Gradients flow through the kernel evaluations back to the model scores. % This is the section that connects to \Cref{sec:roll-backward} motivation. -Firstly, for the true population, we calculate the derivatives. +We derive the concrete form of the two terms from \Cref{eq:roll-gradient} for the KDE instantiation, +where $\tau = \hat{F}_0^{-1}(\alpha ; \mathcal{B}_0)$. + +\paragraph{Case $y_i = 1$ (positive class).} + +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 $\tau$ (which depends only on $\mathcal{B}_0$). Therefore: \[ \dfrac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial f_{\theta}(\mathbf{x}_i)} = - \frac{\partial \frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1} (\sigma_1(f_{\theta}(\mathbf{x}_j) - \tau))}{\partial f_{\theta}(\mathbf{x}_i)} + \frac{\partial \frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1} \sigma_1(f_{\theta}(\mathbf{x}_j) - \tau)}{\partial f_{\theta}(\mathbf{x}_i)} \] \[ - = \frac{\partial \left(\sum_{\mathbf{x}_j \in \mathcal{B}_1 ; j \neq i} \frac{1}{|\mathcal{B}_1|}(\sigma_1(f_{\theta}(\mathbf{x}_j) - \tau)) + \frac{1}{|\mathcal{B}_1|}\sigma_1(f_{\theta}(\mathbf{x}_i) - \tau)\right)}{\partial f_{\theta}(\mathbf{x}_i)} + = \frac{\partial \left(\sum_{\mathbf{x}_j \in \mathcal{B}_1 ; j \neq i} \frac{1}{|\mathcal{B}_1|}\sigma_1(f_{\theta}(\mathbf{x}_j) - \tau) + \frac{1}{|\mathcal{B}_1|}\sigma_1(f_{\theta}(\mathbf{x}_i) - \tau)\right)}{\partial f_{\theta}(\mathbf{x}_i)} \] +\begin{equation} + \frac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial f_{\theta}(\mathbf{x}_i)} = \frac{1}{|\mathcal{B}_1|}\sigma_1'(f_{\theta}(\mathbf{x}_i) - \tau) + \label{eq:kde-grad-y1} +\end{equation} + +\paragraph{Case $y_i = 0$ (negative class).} + +When $\mathbf{x}_i \in \mathcal{B}_0$, the score $f_\theta(\mathbf{x}_i)$ does not appear directly in +$\hat{F}_1$, but influences it through the threshold $\tau$. Per \Cref{eq:roll-gradient}, we compute +each factor of the chain rule separately. + +\medskip\noindent\textit{Derivative of $\hat{F}_1$ with respect to $\tau$.} + \[ - = \frac{1}{|\mathcal{B}_1|}\sigma_1'(f_{\theta}(\mathbf{x}_i) - \tau) + \frac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial \hat{F}_0^{-1}(\alpha)} = \frac{\partial \frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1} \sigma_1(f_{\theta}(\mathbf{x}_j) - \tau)}{\partial \tau} \] -Now, for the false population. As stated in the roll framework equation above (TODO LINK) we must first compute the derivative w.r.t the threshold $\tau = \hat{F}_{0}^{-1}(\alpha)$: - - -\[ - \frac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial \hat{F}_0^{-1}(\alpha)} = \frac{\partial \frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1} (\sigma_1(f_{\theta}(\mathbf{x}_j) - \tau))}{\partial \tau} -\] - -\[ +\begin{equation} = -\frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1}\sigma_1'(f_{\theta}(\mathbf{x}_j) - \tau) -\] + \label{eq:kde-dF1-dtau} +\end{equation} -At this point, we must calculate the derivative of the threshold w.r.t each score from the false population. +\medskip\noindent\textit{Derivative of $\tau$ with respect to $f_\theta(\mathbf{x}_i)$.} -Unfortunately for us, this involves calculating the derivative of a calculation we computed numerically. +This requires differentiating a quantity computed numerically --- $\tau = \hat{F}_0^{-1}(\alpha ; \mathcal{B}_0)$ +is found via Newton's method rather than in closed form. -In order to derive this, we can utilize our earlier computation, along with the inverse derivative rule: +We apply the inverse derivative rule: \[ f'(x) = \frac{1}{f^{-1}'(f(x))} \] -This equation shows the relationship between a function's derivative and its inverse. It shows that we may use the derivative of the -inverse function. However, in order to do this, we must first define what is the inverse we are looking for. - -As a reminder we are trying to calculate $\frac{\partial \hat{F}_0^{-1}(\alpha ; \mathcal{B}_0)}{\partial f_\theta(\mathbf{x}_i)}$. That is, we have a function that takes $\alpha, \mathcal{B}_0$ as parameters and calculates -our threshold $\tau$. We are trying to calculate the derivative of this function w.r.t $\mathbf{x}_i$. - -Therefore, we must invert the function in order to calculate $\mathbf{x}_i$ given $\alpha, \tau$ and all $\mathcal{B}_0 \backslash \{\mathbf{x}_i\}$. - -This inverse equation is given by: +This allows us to use the derivative of the inverse function instead. We need the inverse of +$\hat{F}_0^{-1}$ viewed as a function of $f_\theta(\mathbf{x}_i)$ --- that is, an expression for +$f_\theta(\mathbf{x}_i)$ in terms of $\tau$, $\alpha$, and $\mathcal{B}_0 \setminus \{\mathbf{x}_i\}$. +Solving $|\mathcal{B}_0|\cdot\alpha = \sum_{\mathbf{x}_j \in \mathcal{B}_0}\sigma_0(\tau - f_\theta(\mathbf{x}_j))$ +for $f_\theta(\mathbf{x}_i)$ gives: \[ - \tau - \sigma^{-1}\left( |\mathcal{B}_0|\cdot\alpha - \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma(\tau - f_{\theta}(\mathbf{x}_j))\right) + f_\theta(\mathbf{x}_i) = \tau - \sigma_0^{-1}\left( |\mathcal{B}_0|\cdot\alpha - \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma_0(\tau - f_{\theta}(\mathbf{x}_j))\right) \] Applying the inverse derivative rule to this inverse: \[ - \frac{\partial \hat{F}_0^{-1}(\alpha ; \mathcal{B}_0)}{\partial f_\theta(\mathbf{x}_i)} = \frac{1}{\frac{\partial \left(\tau - \sigma^{-1}\left( |\mathcal{B}_0|\cdot \alpha - \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma(\tau - f_{\theta}(\mathbf{x}_j))\right)\right)}{\partial \tau}} + \frac{\partial \hat{F}_0^{-1}(\alpha ; \mathcal{B}_0)}{\partial f_\theta(\mathbf{x}_i)} = \frac{1}{\frac{\partial \left(\tau - \sigma_0^{-1}\left( |\mathcal{B}_0|\cdot \alpha - \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma_0(\tau - f_{\theta}(\mathbf{x}_j))\right)\right)}{\partial \tau}} \] Applying the chain rule: - \[ - = \frac{1}{\sigma^{-1}'\left( |\mathcal{B}_0|\cdot \alpha - \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma(\tau - f_{\theta}(\mathbf{x}_j))\right)\cdot\left( \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma'(\tau - f_{\theta}(\mathbf{x}_j))\right) + 1} + = \frac{1}{\sigma_0^{-1}{}'\left( |\mathcal{B}_0|\cdot \alpha - \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma_0(\tau - f_{\theta}(\mathbf{x}_j))\right)\cdot\left( \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma_0'(\tau - f_{\theta}(\mathbf{x}_j))\right) + 1} \] -Remembering the calculation of $\alpha$ we get: +Since $|\mathcal{B}_0|\cdot\alpha = \sum_{\mathbf{x}_j \in \mathcal{B}_0}\sigma_0(\tau - f_\theta(\mathbf{x}_j))$, +the argument of $\sigma_0^{-1}{}'$ simplifies: \[ - \alpha = \sum_{\mathbf{x}_j \in \mathcal{B}_0}\sigma(\tau - f_{\theta}(\mathbf{x}_j)) \Rightarrow + |\mathcal{B}_0|\cdot \alpha - \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma_0(\tau - f_{\theta}(\mathbf{x}_j)) = \sigma_0(\tau - f_{\theta}(\mathbf{x}_i)) \] +Therefore: + \[ - |\mathcal{B}_0|\cdot \alpha - \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma(\tau - f_{\theta}(\mathbf{x}_j)) = \sigma(\tau - f_{\theta}(\mathbf{x}_i)) + \frac{\partial \hat{F}_0^{-1}(\alpha ; \mathcal{B}_0)}{\partial f_\theta(\mathbf{x}_i)} = \frac{1}{\sigma_0^{-1}{}'\!\left( \sigma_0(\tau - f_{\theta}(\mathbf{x}_i)) \right)\cdot\left( \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma_0'(\tau - f_{\theta}(\mathbf{x}_j))\right) + 1 } \] -Therefore, our derivative to calculate from before is equal to: +Substituting $\sigma_0^{-1}{}'(\sigma_0(u)) = \frac{1}{\sigma_0'(u)}$ and multiplying numerator and denominator +by $\sigma_0'(\tau - f_{\theta}(\mathbf{x}_i))$: +\begin{equation} + \frac{\partial \hat{F}_0^{-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} -\[ - \frac{\partial \hat{F}_0^{-1}(\alpha ; \mathcal{B}_0)}{\partial f_\theta(\mathbf{x}_i)} = \frac{1}{\sigma^{-1}'\left( \sigma(\tau - f_{\theta}(\mathbf{x}_i)) \right)\cdot\left( \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma'(\tau - f_{\theta}(\mathbf{x}_j))\right) + 1 } -\] +\paragraph{Combined gradient.} -Substituting $\sigma^{-1}'(\sigma(u)) = \frac{1}{\sigma'(u)}$ and multiplying numerator and denominator by $\sigma'(\tau - f_{\theta}(\mathbf{x}_i))$: +Substituting \Cref{eq:kde-grad-y1}, \Cref{eq:kde-dF1-dtau}, and \Cref{eq:kde-grad-tau} +into \Cref{eq:roll-gradient}: -\[ - \frac{\partial \hat{F}_0^{-1}(\alpha ; \mathcal{B}_0)}{\partial f_\theta(\mathbf{x}_i)} = \frac{\sigma'(\tau - f_{\theta}(\mathbf{x}_i))}{\displaystyle\sum_{\mathbf{x}_j \in \mathcal{B}_0}\sigma'(\tau - f_{\theta}(\mathbf{x}_j))} -\] +\begin{equation} + \frac{\partial \mathcal{L}_{\text{ROLL-TPR@FPR}}}{\partial f_\theta(\mathbf{x}_i)} = + \begin{cases} + \dfrac{1}{|\mathcal{B}_1|}\,\sigma_1'(f_{\theta}(\mathbf{x}_i) - \tau) + & \text{if } y_i = 1 \\[14pt] + -\dfrac{1}{|\mathcal{B}_1|}\!\displaystyle\sum_{\mathbf{x}_j \in \mathcal{B}_1}\!\sigma_1'(f_{\theta}(\mathbf{x}_j) - \tau) + \;\cdot\; + \dfrac{\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))} + & \text{if } y_i = 0 + \end{cases} + \label{eq:kde-grad-combined} +\end{equation} -Which we can plug back into the ROLL derivation framework. +where $\tau = \hat{F}_0^{-1}(\alpha ; \mathcal{B}_0)$. %------------------------------------------------