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Let $\mathcal{P}_i$ be the set of probability density functions to which $f_i$ belongs, $(i=0,1)$. Furthermore assume that $$L(y)=\frac{f_1(y)}{f_0(y)}$$ is an increasing function for any chosen $f_1$ and $f_2$. Let the support of the densities be a compact set in reals defined by $\mathbb{K}$.

For a given threshold $\tau\in\mathbb{K}$ one can calculate the probability of false alarm and probability of miss detection as follows:


$$P_M(\tau)=\int_{-\infty}^\tau f_1(y)dy$$

ROC:=$(P_F(\tau),P_M(\tau))$ forms a curve in $[0,1]$ which is convex.

(ROC, for those who don't know, stands for Receiver Operator Characteristic).

Here is an example:



with $\sigma_0=\sigma_1=1$ and $\mu_0=0$ and $\mu_1=1$. Then we have the following figure for $(P_F(\tau),P_M(\tau))$ when $\tau$ is changed from $-\infty$ to $\infty$, ($\mathbb{K}=\mathbb{R}$).

enter image description here

As known and can be seen from the figure, the blue curve is convex.

For any chosen pair of densities $(f_0,f_1)\in \mathcal{P}_0\times \mathcal{P}_1$. The ROC curve (the blue one) $(P_F(\tau),P_M(\tau))$ when $\tau\in (-\infty,\infty)$ will lie in the butterfly given in the figure with red lines assuming that the point $\theta=P_F=P_M$ is common for all densities in $\mathcal{P}_0\times\mathcal{P}_1$ (in the figure $\theta \approx 0.3$)


Assume that all densities $(f_0,f_1)\in\mathcal{P}_0\times\mathcal{P}_1$ are known to have a particular $\theta$ in their ROC. In other words, let $\mathcal{P}_0\times\mathcal{P}_1$ define only the pair of densities that have $\theta$ in their ROC and furthermore let one choose any pair of density from $\mathcal{P}_0\times\mathcal{P}_1$ with equal probability.

What is the probabilty that a single point of the ROC that we obtain by this selection will lie in the green sector?

Once again the green sector is the intersection of the butterfly with the area under the line which passes through $\theta$ and $f_1/f_0$ is increasing as defined before. One can assume any $\mathbb{K}$ for example $\mathbb{K}=[0,1]$ or ($\mathbb{K}=\mathbb{R}$).

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@Andres Caicedo you sure that this question doesnt have anything to do with set theory? – Seyhmus Güngören Nov 5 '12 at 16:10
Yes. Completely sure. In fact, I would suggest to remove "infinities" from the title. – Andrés E. Caicedo Nov 5 '12 at 16:11
Is the black line tangent to your ROC? Also as an unrelated aside, this would be a fairly poor ROC. You'd be much better off with $1-$ your classifier! Such a curve would be concave upward, so I'm not sure if the question would be relevant or different in such a case. – Emily Nov 5 '12 at 16:14
@Andres Caicedo Then you will give me a hint how to deal with my sets with infinite elements and some have even more and I am interested in how much more? – Seyhmus Güngören Nov 5 '12 at 16:15
I'd actually add the tags signal-processing and possible machine-learning to this to attract attention of folks who work with this more often. – Emily Nov 5 '12 at 16:15
up vote 1 down vote accepted

If the black line is tangent, and the blue curve is convex, then there is only a single point of the blue line contained in the green area. This is because the green area is defined by the tangent line, and convexity guaranteed that the blue curve will not intersect the black line, and hence the green region, at any other point.

If you're looking for the probability that a single realization will land in this region, simply compute the area of the region. The ROC curve defines the "dividing line" of classification; however, the unit square is still your global probability space.

If you want to know the probability that a different ROC curve intersects this green region, then you can employ a few different conditions. First, assume that any other ROC curve is convex and continuous. Then, the curve defined by $$ R = \left\{ \left(P_F(\tau),P_M(\tau)\right) \right\}$$ is continuous and monotonic and maps from $[0,1]$ to $[0,1]$.

Therefore, this curve has a fixed point in $[0,1]$, namely the point where

$$P_F(\tau) = P_M(\tau).$$

These fixed points lie on the line $y=x$. Obviously, any monotonic curve whose fixed point is $x' > \theta$ will not intersect the green region.

Conversely, any curve with a fixed point $x' \le \theta$ will pass through the green triangle.

Therefore, your probability is $P(x' \le \theta)$ and is uniformly distributed, so your probability in question is therefore exactly $\theta$, which agrees with my previous assessment.

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you are not right. you can draw infinitely many ROC intersecting the black line. Of course I am not talking the point $\theta=P_M=P_F$ as it is always and trivially in the green region. – Seyhmus Güngören Nov 5 '12 at 16:31
I have no idea in what context you're using ROC. Are you talking about drawing more blue curves that possible intersect the black line? Are you talking about using a receiver with this characteristic and wanting to find the probability of landing in this region? – Emily Nov 5 '12 at 16:34
I calculated the area of green area and divided it to the whole area of the butterfly. This will give you exactly $\theta$. However I dont think that this is the solution. Thats the reason why I posted the question here. – Seyhmus Güngören Nov 5 '12 at 16:34
Yes. Think that you are given all probabilty densities which have $P_F=P_M=0.3$, then clear that we have infinitely many of them. We put them in a box and pick one pair with equal probabily. Then lets draw the ROC curve for the chosen pair. We will get a convex curve in the butterfly (can or cannot intersect the black line). Question is what is the probability that it will intersect? Again excluding the point $\theta$ – Seyhmus Güngören Nov 5 '12 at 16:37
Ok, I think I understand what you're asking. I am editing my answer. – Emily Nov 5 '12 at 16:43

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