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Motivated by this reference, it states under ROC Space

When evaluating a binary classifier, we often use a Confusion Matrix...however here we need only TPR and FPR

I'd feel more comfortable if the ROC were instead called positive-ROC suggesting there is an alternative ROC curve that is used from TNR and FNR. But that doesn't seem to be the case. Everywhere I've looked, ROC refers to using TPR and FPR, specifically.

Is there a version of the ROC curve that instead accounts to average the success of positive and negative example classification, say $\frac{1}{2}(ROC_{+} + ROC_-)$?, where $ROC_+$ is the popular metric using TPR/FPR and $ROC_-$ is the unpopular version of the same curve using TNR/FNR?

My question is, what makes TPR, FPR so much better than chosing TNR, FNR for the ROC curve? Why is it more important ( that is, ROC is a popular metric) to consider classification of positive examples than negative?

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The true positive rate is one minus the false negative rate; the false positive rate is one minus the true negative rate: so it doesn't really make any difference which one you pick from each pair to talk about. You're certainly not in any sense giving more importance to classification of positive cases over negative cases (how could you when the classification is dichotomous?) by making this arbitrary choice.

(As a matter of fact the abscissa of an ROC curve is often labelled as one minus specificity—a.k.a. true negative rate.)

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    $\begingroup$ This answer makes some sense to me but surely you're losing information to only consider positive examples, no? That is, what if the dataset included 5 positive examples and 100 negative examples? The ROC curve, which is only in terms of positive examples, could surely be calculated more accurately by using the negative examples, as in $\frac{1}{2}(ROC_+ + ROC_-)$, yes? $\endgroup$ – user27886 Mar 30 '15 at 16:01
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    $\begingroup$ Does the calculation of true positive rate & false positive rate $not$ use the negative examples? $\endgroup$ – Scortchi - Reinstate Monica♦ Mar 30 '15 at 16:03
  • $\begingroup$ I see your point. $\endgroup$ – user27886 Mar 30 '15 at 16:05

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