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Before the big switch.

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Aner Zakobar
2026-07-02 22:54:10 +03:00
parent 7ff2ef6835
commit 10f1bd53f1
1 changed files with 5 additions and 5 deletions
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content/method/method.tex
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@@ -526,7 +526,7 @@ Firstly, for the true population, we calculate the derivatives.
\[ \[
\dfrac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial f_{\theta}(\mathbf{x}_i)} = \dfrac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial f_{\theta}(\mathbf{x}_i)} =
\frac{\partial \sum_{\mathbf{x}_j \in \mathcal{B}_1} (\sigma_1(f_{\theta}(\mathbf{x}_j) - \tau))}{\partial f_{\theta}(\mathbf{x}_i)} \frac{\partial \frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1} (\sigma_1(f_{\theta}(\mathbf{x}_j) - \tau))}{\partial f_{\theta}(\mathbf{x}_i)}
\] \]
\[ \[
@@ -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. 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: 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: 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: Applying the chain rule: