I'm studying the problem of combining two sensors for anomaly detection. I want to analyze its performance by the ROC curve. Now I have obtained a parametric equation about the ROC curve:
$$(x,y) = (FPR,TPR) = (\Phi(c(a-t)), \Phi(c(b-t)))$$
where $a,b,c$ are non-negative real numbers, $t$ is the threshold of the anomaly detection method, and $\Phi$ is the cumulative distribution function (CDF) of a standard normal distribution.
Is it possible to analytically express the ROC curve in a function form like $y=f(x)$?
I have drawn the ROC curve (see figure). It seems like an exponential function (maybe $y=x^\alpha$ ?), but I still have no idea whether it is possible to get an analytic form of $f$.
We have $$x=\Phi(c(a-t))=\frac{1}{2} \left(1+\text{erf}\left(\frac{c (a-t)}{\sqrt{2}}\right)\right)$$ $$y=\Phi(c(b-t))=\frac{1}{2} \left(1+\text{erf}\left(\frac{c (b-t)}{\sqrt{2}}\right)\right)$$ So, $$t=a-\frac{\sqrt{2}}{c}\, \text{erf}^{-1}(2 x-1)$$ and then $$y=\frac 12 \Bigg[1+\text{erf}\left(\frac{c (b-a)}{\sqrt{2}}+\text{erf}^{-1}(2 x-1)\right) \Bigg]$$ From a formal point of view, it is done.
From a practical point of view, you need now to find a "simple" approximation of the inverse of the error function.
If you look at my answer to this question, you will find approximations of the error function which can easily be inversed. Using $P_1(x)$, it is at the price of a quadratic equation; it is just immediate with $P_0$ (which could be sufficient).
Otherwise, you can use the series represntations given here.
Simple would be $$ \text{erf}^{-1}(z)=u+\frac{1}{3}u^3+\frac{7 }{30}u^5+\frac{127 }{630}u^7+\frac{4369 }{22680} u^9+O\left(u^{11}\right)$$ where $u=\frac{\sqrt{\pi } }{2}z$
Edit
Starting from the series, a quite good approximation could be obtained with the $[5,4]$ Padé approximant $$ \text{erf}^{-1}(z)=u\,\frac{1-\alpha_, u^2+\beta\, u^4 } { 1-\gamma\, u^2+\delta\, u^4}\qquad \text{with}\qquad u=\frac{\sqrt{\pi } }{2}z$$ whose error is $\frac{u^{11}}{250}$.
The required coefficients are $$\alpha=\frac{4397}{4338}\qquad \beta=\frac{111547}{910980} \qquad \gamma=\frac{5843}{4338} \qquad \delta=\frac{20533}{60732}$$ The absolute error is less than $0.01$ as long as $z \leq 0.90$.
Worked example
Using for $(a,b,c)$ the values used to generate the plot in the post, that is to say $(0,1,\sqrt{2})$ and all the elements given in this answer, the absolute error on $y$ is smaller than $0.001$ as soon as $x>0.0854215$.
To give an idea : using the norm $$\Psi(a)=\frac 14\int_a^1\Bigg[\text{erf}\left(1-\text{erf}^{-1}(1-2 x)\right)-\text{erf}\left(\text{approximation} \right)\Bigg]^2\,dx$$
$$\left( \begin{array}{cc} a & \Psi(a) \\ 0.00 & 1.24386\times 10^{-4} \\ 0.01 & 2.50636\times 10^{-5} \\ 0.02 & 7.26261\times 10^{-6} \\ 0.03 & 2.37431\times 10^{-6} \\ 0.04 & 8.33009\times 10^{-7} \\ 0.05 & 3.06603\times 10^{-7} \\ 0.06 & 1.16904\times 10^{-7} \\ 0.07 & 4.58574\times 10^{-8} \\ 0.08 & 1.84804\times 10^{-8} \\ 0.09 & 7.70450\times 10^{-9} \\ 0.10 & 3.39594\times 10^{-9} \\ \end{array} \right)$$