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I use the function auc in the R package pROC to calculate the auc value in a simulation study. test.y is the observed response, y_pred_mle is the predicted response.

> test.y
   test.y
1      -1
2      -1
3      -1
4      -1
5      -1
6      -1
7       1
8       1
9       1
10      1
11      1
12      1

> y_pred_mle
        [,1]
76 -166.7094
53 -203.4014
52 -220.0880
51 -189.4703
95 -222.5294
72 -207.0304
24  722.8809
44  727.5353
12  770.5053
42  783.7437
27  733.3144
3   773.2688


> out_auc_mle<- auc( test.y, y_pred_mle  )
> out_auc_mle
[1] Inf

I wonder how can this generate Inf value? In my understanding on auc, the range of auc should be 0 to 1. I am not asking an coding question. I just wonder how the auc value can be Inf. Can anybody explain me about is it possible that the auc is Inf? Then how to interpret it?

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1 Answer 1

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The AUC should be 1 for this data because all 1 examples are ranked higher than all -1 examples. This is because AUC corresponds exactly to the probability that a randomly-selected positive is ranked more highly than a randomly-selected negative. Matthew Drury has a nice write-up here. There must be either an error in usage or a bug, probably the former.

A useful CV thread on AUC, what it is, and how to interpret it is here.

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  • $\begingroup$ Thank you for your answer. Your answer inspires me to solve this problem. But I still wonder about what you said The AUC should be 1 for this data because all 1 examples are ranked higher than all -1 examples (note the signs) , why if all 1 examples are ranked higher than all -1 examples, we can directly said "AUC" =1? Could you give me some explanation? Many thanks. $\endgroup$
    – user93892
    Aug 17, 2017 at 7:50
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    $\begingroup$ I've added more detail. $\endgroup$
    – Sycorax
    Aug 18, 2017 at 4:46
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    $\begingroup$ I worte a less technical proof of the AUC equivelence here: madrury.github.io/jekyll/update/statistics/2017/06/21/… $\endgroup$ Aug 18, 2017 at 5:03

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