Before the big switch.
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@@ -526,7 +526,7 @@ Firstly, for the true population, we calculate the derivatives.
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\[
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\dfrac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial f_{\theta}(\mathbf{x}_i)} =
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\frac{\partial \sum_{\mathbf{x}_j \in \mathcal{B}_1} (\sigma_1(f_{\theta}(\mathbf{x}_j) - \tau))}{\partial f_{\theta}(\mathbf{x}_i)}
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\frac{\partial \frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1} (\sigma_1(f_{\theta}(\mathbf{x}_j) - \tau))}{\partial f_{\theta}(\mathbf{x}_i)}
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\]
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\[
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@@ -541,11 +541,11 @@ Now, for the false population. As stated in the roll framework equation above (T
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\[
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\frac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial \hat{F}_0^{-1}(\alpha)} = \frac{\partial \sum_{\mathbf{x}_j \in \mathcal{B}_1} (\sigma_1(f_{\theta}(\mathbf{x}_j) - \tau))}{\partial \tau}
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\frac{\partial \hat{F}_1(\hat{F}_0^{-1}(\alpha))}{\partial \hat{F}_0^{-1}(\alpha)} = \frac{\partial \frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1} (\sigma_1(f_{\theta}(\mathbf{x}_j) - \tau))}{\partial \tau}
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\]
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\[
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= -\frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_i \in \mathcal{B}_1}\sigma_1'(f_{\theta}(\mathbf{x}_i) - \tau)
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= -\frac{1}{|\mathcal{B}_1|}\sum_{\mathbf{x}_j \in \mathcal{B}_1}\sigma_1'(f_{\theta}(\mathbf{x}_j) - \tau)
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\]
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At this point, we must calculate the derivative of the threshold w.r.t each score from the false population.
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@@ -569,13 +569,13 @@ Therefore, we must invert the function in order to calculate $\mathbf{x}_i$ give
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This inverse equation is given by:
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\[
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\sigma^{-1}\left( |\mathcal{B}_0|\cdot\alpha - \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma(\tau - f_{\theta}(\mathbf{x}_j))\right) + \tau
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\tau - \sigma^{-1}\left( |\mathcal{B}_0|\cdot\alpha - \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma(\tau - f_{\theta}(\mathbf{x}_j))\right)
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\]
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Applying the inverse derivative rule to this inverse:
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\[
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\frac{\partial \hat{F}_0^{-1}(\alpha ; \mathcal{B}_0)}{\partial f_\theta(\mathbf{x}_i)} = \frac{1}{\frac{\partial \sigma^{-1}\left( |\mathcal{B}_0|\cdot \alpha - \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma(\tau - f_{\theta}(\mathbf{x}_j))\right) + \tau}{\partial \tau}}
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\frac{\partial \hat{F}_0^{-1}(\alpha ; \mathcal{B}_0)}{\partial f_\theta(\mathbf{x}_i)} = \frac{1}{\frac{\partial \left(\tau - \sigma^{-1}\left( |\mathcal{B}_0|\cdot \alpha - \sum_{\mathbf{x}_j \in \mathcal{B}_0 ; j \neq i}\sigma(\tau - f_{\theta}(\mathbf{x}_j))\right)\right)}{\partial \tau}}
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\]
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Applying the chain rule:
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