From 10f1bd53f1ed36184c698c5debf53e0c17d74ffe Mon Sep 17 00:00:00 2001 From: Aner Zakobar Date: Thu, 2 Jul 2026 22:54:10 +0300 Subject: [PATCH] Before the big switch. --- content/method/method.tex | 10 +++++----- 1 file changed, 5 insertions(+), 5 deletions(-) diff --git a/content/method/method.tex b/content/method/method.tex index c6ef4ce..3261aee 100644 --- a/content/method/method.tex +++ b/content/method/method.tex @@ -526,7 +526,7 @@ Firstly, for the true population, we calculate the derivatives. \[ \dfrac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial f_{\theta}(\mathbf{x}_i)} = - \frac{\partial \sum_{\mathbf{x}_j \in \mathcal{B}_1} (\sigma_1(f_{\theta}(\mathbf{x}_j) - \tau))}{\partial f_{\theta}(\mathbf{x}_i)} + \frac{\partial \frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1} (\sigma_1(f_{\theta}(\mathbf{x}_j) - \tau))}{\partial f_{\theta}(\mathbf{x}_i)} \] \[ @@ -541,11 +541,11 @@ Now, for the false population. As stated in the roll framework equation above (T \[ - \frac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial \hat{F}_0^{-1}(\alpha)} = \frac{\partial \sum_{\mathbf{x}_j \in \mathcal{B}_1} (\sigma_1(f_{\theta}(\mathbf{x}_j) - \tau))}{\partial \tau} + \frac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial \hat{F}_0^{-1}(\alpha)} = \frac{\partial \frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1} (\sigma_1(f_{\theta}(\mathbf{x}_j) - \tau))}{\partial \tau} \] \[ - = -\frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_i \in \mathcal{B}_1}\sigma_1'(f_{\theta}(\mathbf{x}_i) - \tau) + = -\frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1}\sigma_1'(f_{\theta}(\mathbf{x}_j) - \tau) \] At this point, we must calculate the derivative of the threshold w.r.t each score from the false population. @@ -569,13 +569,13 @@ Therefore, we must invert the function in order to calculate $\mathbf{x}_i$ give This inverse equation is given by: \[ - \sigma^{-1}\left( |\mathcal{B}_0|\cdot\alpha - \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma(\tau - f_{\theta}(\mathbf{x}_j))\right) + \tau + \tau - \sigma^{-1}\left( |\mathcal{B}_0|\cdot\alpha - \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma(\tau - f_{\theta}(\mathbf{x}_j))\right) \] Applying the inverse derivative rule to this inverse: \[ - \frac{\partial \hat{F}_0^{-1}(\alpha ; \mathcal{B}_0)}{\partial f_\theta(\mathbf{x}_i)} = \frac{1}{\frac{\partial \sigma^{-1}\left( |\mathcal{B}_0|\cdot \alpha - \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma(\tau - f_{\theta}(\mathbf{x}_j))\right) + \tau}{\partial \tau}} + \frac{\partial \hat{F}_0^{-1}(\alpha ; \mathcal{B}_0)}{\partial f_\theta(\mathbf{x}_i)} = \frac{1}{\frac{\partial \left(\tau - \sigma^{-1}\left( |\mathcal{B}_0|\cdot \alpha - \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma(\tau - f_{\theta}(\mathbf{x}_j))\right)\right)}{\partial \tau}} \] Applying the chain rule: