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The title is my whole question. FPR is the false positive rate. TPR is the true positive rate.

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If you classify a fraction $k$ of your cases as positive then, because of the randomness, the same fraction $k$ of cases which should be positive will be classified positive (true positives), and the same fraction $k$ of cases which should be negative will be classified positive (false positives).

So the true positive rate and the false positive rate are the same.

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  • $\begingroup$ can you possibly write that with probabilistic statements? $\endgroup$ – ℕʘʘḆḽḘ May 29 '18 at 18:59
  • $\begingroup$ @ℕʘʘḆḽḘ Call $T$ the event that a case is actually positive and $R$ the event that it is predicted to be positive with this random predictor. Then $\mathbb P(R \mid T) = k = \mathbb P(R \mid T^{c})$ $\endgroup$ – Henry May 29 '18 at 20:30
  • $\begingroup$ thanks sure but I mean could you explain how you derive this result? $\endgroup$ – ℕʘʘḆḽḘ May 29 '18 at 20:42
  • $\begingroup$ @ℕʘʘḆḽḘ "a random classifier" is stated in the title, so the probability of being predicted positive does not depend on any property of the event $\endgroup$ – Henry May 29 '18 at 20:50
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Identity

Let $T$ be the event that a case is positive, and $R$ the event a case is predicted to be positive by a classifier.

Since $T$ and $T^c$ are mutually exclusive and collectively exhaustive, we can decompose $\mathbb{P}(R)$ as follows:

\begin{split} \mathbb{P}(R) & = \mathbb{P}(R|T)\mathbb{P}(T)+\mathbb{P}(R|T^c)\mathbb{P}(T^c)\\ & = \mathbb{P}(R|T)(1-\mathbb{P}(T^c))+\mathbb{P}(R|T^c)\mathbb{P}(T^c)\\ & = [\mathbb{P}(R|T^c)-\mathbb{P}(R|T)]\mathbb{P}(T^c)+\mathbb{P}(R|T). \end{split}

The identity $$\mathbb{P}(R)-\mathbb{P}(R|T) = [\mathbb{P}(R|T^c)-\mathbb{P}(R|T)]\mathbb{P}(T^c),$$ means that $\mathbb{P}(R)-\mathbb{P}(R|T) = 0$ if and only if $\mathbb{P}(R|T^c)-\mathbb{P}(R|T)=0$, for $\mathbb{P}(T^c)>0$.

Random guessing

To begin, inspect the left-hand side condition: $\mathbb{P}(R)=\mathbb{P}(R|T)$. This condition implies the independence between events $R$ and $T$. A classifier that is based on random guessing has to satisfy this condition.

Suppose the classifier is random guessing but it does not satisfy the left-hand side condition, i.e. $\mathbb{P}(R)\ne\mathbb{P}(R|T)$. Then, the guesses are biased for cases that are in $T$, i.e. the classifier guesses differently when encountering a positive case. This contradicts the notion of a random guess.

In other words, if the classifier is random guessing unconditionally, or conditionally on positive cases, it should perform equally well in both cases.

Line y = x

Next, inspect the right-hand side condition: $\mathbb{P}(R|T)=\mathbb{P}(R|T^c)$. This condition means that the true-positive rate equals to the false-positive rate. Geometrically, the line $y=x$ on the ROC graph represents this right-hand side condition. This is because, the ROC graph y-axis and x-axis represent the true-positive rate and false-positive rate respectively.

Equivalance

To conclude, the left-hand side condition represents a random-guessing classifier, and the right-hand side condition represents the line $y=x$. They are in fact equivalent.

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