nips nips2003 nips2003-3 knowledge-graph by maker-knowledge-mining

3 nips-2003-AUC Optimization vs. Error Rate Minimization


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Author: Corinna Cortes, Mehryar Mohri

Abstract: The area under an ROC curve (AUC) is a criterion used in many applications to measure the quality of a classification algorithm. However, the objective function optimized in most of these algorithms is the error rate and not the AUC value. We give a detailed statistical analysis of the relationship between the AUC and the error rate, including the first exact expression of the expected value and the variance of the AUC for a fixed error rate. Our results show that the average AUC is monotonically increasing as a function of the classification accuracy, but that the standard deviation for uneven distributions and higher error rates is noticeable. Thus, algorithms designed to minimize the error rate may not lead to the best possible AUC values. We show that, under certain conditions, the global function optimized by the RankBoost algorithm is exactly the AUC. We report the results of our experiments with RankBoost in several datasets demonstrating the benefits of an algorithm specifically designed to globally optimize the AUC over other existing algorithms optimizing an approximation of the AUC or only locally optimizing the AUC. 1 Motivation In many applications, the overall classification error rate is not the most pertinent performance measure, criteria such as ordering or ranking seem more appropriate. Consider for example the list of relevant documents returned by a search engine for a specific query. That list may contain several thousand documents, but, in practice, only the top fifty or so are examined by the user. Thus, a search engine’s ranking of the documents is more critical than the accuracy of its classification of all documents as relevant or not. More generally, for a binary classifier assigning a real-valued score to each object, a better correlation between output scores and the probability of correct classification is highly desirable. A natural criterion or summary statistic often used to measure the ranking quality of a classifier is the area under an ROC curve (AUC) [8].1 However, the objective function optimized by most classification algorithms is the error rate and not the AUC. Recently, several algorithms have been proposed for maximizing the AUC value locally [4] or maximizing some approximations of the global AUC value [9, 15], but, in general, these algorithms do not obtain AUC values significantly better than those obtained by an algorithm designed to minimize the error rates. Thus, it is important to determine the relationship between the AUC values and the error rate. ∗ This author’s new address is: Google Labs, 1440 Broadway, New York, NY 10018, corinna@google.com. 1 The AUC value is equivalent to the Wilcoxon-Mann-Whitney statistic [8] and closely related to the Gini index [1]. It has been re-invented under the name of L-measure by [11], as already pointed out by [2], and slightly modified under the name of Linear Ranking by [13, 14]. True positive rate ROC Curve. AUC=0.718 (1,1) True positive rate = (0,0) False positive rate = False positive rate correctly classified positive total positive incorrectly classified negative total negative Figure 1: An example of ROC curve. The line connecting (0, 0) and (1, 1), corresponding to random classification, is drawn for reference. The true positive (negative) rate is sometimes referred to as the sensitivity (resp. specificity) in this context. In the following sections, we give a detailed statistical analysis of the relationship between the AUC and the error rate, including the first exact expression of the expected value and the variance of the AUC for a fixed error rate.2 We show that, under certain conditions, the global function optimized by the RankBoost algorithm is exactly the AUC. We report the results of our experiments with RankBoost in several datasets and demonstrate the benefits of an algorithm specifically designed to globally optimize the AUC over other existing algorithms optimizing an approximation of the AUC or only locally optimizing the AUC. 2 Definition and properties of the AUC The Receiver Operating Characteristics (ROC) curves were originally developed in signal detection theory [3] in connection with radio signals, and have been used since then in many other applications, in particular for medical decision-making. Over the last few years, they have found increased interest in the machine learning and data mining communities for model evaluation and selection [12, 10, 4, 9, 15, 2]. The ROC curve for a binary classification problem plots the true positive rate as a function of the false positive rate. The points of the curve are obtained by sweeping the classification threshold from the most positive classification value to the most negative. For a fully random classification, the ROC curve is a straight line connecting the origin to (1, 1). Any improvement over random classification results in an ROC curve at least partially above this straight line. Fig. (1) shows an example of ROC curve. The AUC is defined as the area under the ROC curve and is closely related to the ranking quality of the classification as shown more formally by Lemma 1 below. Consider a binary classification task with m positive examples and n negative examples. We will assume that a classifier outputs a strictly ordered list for these examples and will denote by 1X the indicator function of a set X. Lemma 1 ([8]) Let c be a fixed classifier. Let x1 , . . . , xm be the output of c on the positive examples and y1 , . . . , yn its output on the negative examples. Then, the AUC, A, associated to c is given by: m n i=1 j=1 1xi >yj (1) A= mn that is the value of the Wilcoxon-Mann-Whitney statistic [8]. Proof. The proof is based on the observation that the AUC value is exactly the probability P (X > Y ) where X is the random variable corresponding to the distribution of the outputs for the positive examples and Y the one corresponding to the negative examples [7]. The Wilcoxon-Mann-Whitney statistic is clearly the expression of that probability in the discrete case, which proves the lemma [8]. Thus, the AUC can be viewed as a measure based on pairwise comparisons between classifications of the two classes. With a perfect ranking, all positive examples are ranked higher than the negative ones and A = 1. Any deviation from this ranking decreases the AUC. 2 An attempt in that direction was made by [15], but, unfortunately, the authors’ analysis and the result are both wrong. Threshold θ k − x Positive examples x Negative examples n − x Negative examples m − (k − x) Positive examples Figure 2: For a fixed number of errors k, there may be x, 0 ≤ x ≤ k, false negative examples. 3 The Expected Value of the AUC In this section, we compute exactly the expected value of the AUC over all classifications with a fixed number of errors and compare that to the error rate. Different classifiers may have the same error rate but different AUC values. Indeed, for a given classification threshold θ, an arbitrary reordering of the examples with outputs more than θ clearly does not affect the error rate but leads to different AUC values. Similarly, one may reorder the examples with output less than θ without changing the error rate. Assume that the number of errors k is fixed. We wish to compute the average value of the AUC over all classifications with k errors. Our model is based on the simple assumption that all classifications or rankings with k errors are equiprobable. One could perhaps argue that errors are not necessarily evenly distributed, e.g., examples with very high or very low ranks are less likely to be errors, but we cannot justify such biases in general. For a given classification, there may be x, 0 ≤ x ≤ k, false positive examples. Since the number of errors is fixed, there are k − x false negative examples. Figure 3 shows the corresponding configuration. The two regions of examples with classification outputs above and below the threshold are separated by a vertical line. For a given x, the computation of the AUC, A, as given by Eq. (1) can be divided into the following three parts: A1 + A2 + A3 A= , with (2) mn A1 = the sum over all pairs (xi , yj ) with xi and yj in distinct regions; A2 = the sum over all pairs (xi , yj ) with xi and yj in the region above the threshold; A3 = the sum over all pairs (xi , yj ) with xi and yj in the region below the threshold. The first term, A1 , is easy to compute. Since there are (m − (k − x)) positive examples above the threshold and n − x negative examples below the threshold, A1 is given by: A1 = (m − (k − x))(n − x) (3) To compute A2 , we can assign to each negative example above the threshold a position based on its classification rank. Let position one be the first position above the threshold and let α1 < . . . < αx denote the positions in increasing order of the x negative examples in the region above the threshold. The total number of examples classified as positive is N = m − (k − x) + x. Thus, by definition of A2 , x A2 = (N − αi ) − (x − i) (4) i=1 where the first term N − αi represents the number of examples ranked higher than the ith example and the second term x − i discounts the number of negative examples incorrectly ranked higher than the ith example. Similarly, let α1 < . . . < αk−x denote the positions of the k − x positive examples below the threshold, counting positions in reverse by starting from the threshold. Then, A3 is given by: x A3 = (N − αj ) − (x − j) (5) j=1 with N = n − x + (k − x) and x = k − x. Combining the expressions of A1 , A2 , and A3 leads to: A= A1 + A2 + A3 (k − 2x)2 + k ( =1+ − mn 2mn x i=1 αi + mn x j=1 αj ) (6) Lemma 2 For a fixed x, the average value of the AUC A is given by: < A >x = 1 − x n + k−x m 2 (7) x Proof. The proof is based on the computation of the average values of i=1 αi and x j=1 αj for a given x. We start by computing the average value < αi >x for a given i, 1 ≤ i ≤ x. Consider all the possible positions for α1 . . . αi−1 and αi+1 . . . αx , when the value of αi is fixed at say αi = l. We have i ≤ l ≤ N − (x − i) since there need to be at least i − 1 positions before αi and N − (x − i) above. There are l − 1 possible positions for α1 . . . αi−1 and N − l possible positions for αi+1 . . . αx . Since the total number of ways of choosing the x positions for α1 . . . αx out of N is N , the average value < αi >x is: x N −(x−i) l=i < αi >x = l l−1 i−1 N −l x−i (8) N x Thus, x < αi >x = x i=1 i=1 Using the classical identity: x < αi >x = N −(x−i) l−1 l i−1 l=i N x u p1 +p2 =p p1 N l=1 l N −1 x−1 N x i=1 N −l x−i v p2 = = N l=1 = u+v p N (N + 1) 2 x l−1 i=1 i−1 N x l N −l x−i (9) , we can write: N −1 x−1 N x = x(N + 1) 2 (10) Similarly, we have: x < αj >x = j=1 x Replacing < i=1 αi >x and < Eq. (10) and Eq. (11) leads to: x j=1 x (N + 1) 2 (11) αj >x in Eq. (6) by the expressions given by (k − 2x)2 + k − x(N + 1) − x (N + 1) =1− 2mn which ends the proof of the lemma. < A >x = 1 + x n + k−x m 2 (12) Note that Eq. (7) shows that the average AUC value for a given x is simply one minus the average of the accuracy rates for the positive and negative classes. Proposition 1 Assume that a binary classification task with m positive examples and n negative examples is given. Then, the expected value of the AUC A over all classifications with k errors is given by: < A >= 1 − k (n − m)2 (m + n + 1) − m+n 4mn k−1 m+n x=0 x k m+n+1 x=0 x k − m+n (13) Proof. Lemma 2 gives the average value of the AUC for a fixed value of x. To compute the average over all possible values of x, we need to weight the expression of Eq. (7) with the total number of possible classifications for a given x. There are N possible ways of x choosing the positions of the x misclassified negative examples, and similarly N possible x ways of choosing the positions of the x = k − x misclassified positive examples. Thus, in view of Lemma 2, the average AUC is given by: < A >= k N x=0 x N x (1 − k N x=0 x N x k−x x n+ m 2 ) (14) r=0.05 r=0.01 r=0.1 r=0.25 0.0 0.1 0.2 r=0.5 0.3 Error rate 0.4 0.5 .00 .05 .10 .15 .20 .25 0.5 0.6 0.7 0.8 0.9 1.0 Mean value of the AUC Relative standard deviation r=0.01 r=0.05 r=0.1 0.0 0.1 r=0.25 0.2 0.3 Error rate r=0.5 0.4 0.5 Figure 3: Mean (left) and relative standard deviation (right) of the AUC as a function of the error rate. Each curve corresponds to a fixed ratio of r = n/(n + m). The average AUC value monotonically increases with the accuracy. For n = m, as for the top curve in the left plot, the average AUC coincides with the accuracy. The standard deviation decreases with the accuracy, and the lowest curve corresponds to n = m. This expression can be simplified into Eq. (13)3 using the following novel identities: k X N x x=0 k X N x x x=0 ! N x ! ! N x ! = = ! k X n+m+1 x x=0 (15) ! k X (k − x)(m − n) + k n + m + 1 2 x x=0 (16) that we obtained by using Zeilberger’s algorithm4 and numerous combinatorial ’tricks’. From the expression of Eq. (13), it is clear that the average AUC value is identical to the accuracy of the classifier only for even distributions (n = m). For n = m, the expected value of the AUC is a monotonic function of the accuracy, see Fig. (3)(left). For a fixed ratio of n/(n + m), the curves are obtained by increasing the accuracy from n/(n + m) to 1. The average AUC varies monotonically in the range of accuracy between 0.5 and 1.0. In other words, on average, there seems nothing to be gained in designing specific learning algorithms for maximizing the AUC: a classification algorithm minimizing the error rate also optimizes the AUC. However, this only holds for the average AUC. Indeed, we will show in the next section that the variance of the AUC value is not null for any ratio n/(n + m) when k = 0. 4 The Variance of the AUC 2 Let D = mn + (k−2x) +k , a = i=1 αi , a = j=1 αj , and α = a + a . Then, by 2 Eq. (6), mnA = D − α. Thus, the variance of the AUC, σ 2 (A), is given by: (mn)2 σ 2 (A) x x = < (D − α)2 − (< D > − < α >)2 > = < D2 > − < D >2 + < α2 > − < α >2 −2(< αD > − < α >< D >) (17) As before, to compute the average of a term X over all classifications, we can first determine its average < X >x for a fixed x, and then use the function F defined by: F (Y ) = k N N x=0 x x k N N x=0 x x Y (18) and < X >= F (< X >x ). A crucial step in computing the exact value of the variance of x the AUC is to determine the value of the terms of the type < a2 >x =< ( i=1 αi )2 >x . 3 An essential difference between Eq. (14) and the expression given by [15] is the weighting by the number of configurations. The authors’ analysis leads them to the conclusion that the average AUC is identical to the accuracy for all ratios n/(n + m), which is false. 4 We thank Neil Sloane for having pointed us to Zeilberger’s algorithm and Maple package. x Lemma 3 For a fixed x, the average of ( i=1 αi )2 is given by: x(N + 1) < a2 > x = (3N x + 2x + N ) 12 (19) Proof. By definition of a, < a2 >x = b + 2c with: x x α2 >x i b =< c =< αi αj >x (20) 1≤i

Reference: text


Summary: the most important sentenses genereted by tfidf model

sentIndex sentText sentNum sentScore

1 com Abstract The area under an ROC curve (AUC) is a criterion used in many applications to measure the quality of a classification algorithm. [sent-4, score-0.094]

2 However, the objective function optimized in most of these algorithms is the error rate and not the AUC value. [sent-5, score-0.129]

3 We give a detailed statistical analysis of the relationship between the AUC and the error rate, including the first exact expression of the expected value and the variance of the AUC for a fixed error rate. [sent-6, score-0.189]

4 Our results show that the average AUC is monotonically increasing as a function of the classification accuracy, but that the standard deviation for uneven distributions and higher error rates is noticeable. [sent-7, score-0.178]

5 Thus, algorithms designed to minimize the error rate may not lead to the best possible AUC values. [sent-8, score-0.112]

6 We show that, under certain conditions, the global function optimized by the RankBoost algorithm is exactly the AUC. [sent-9, score-0.051]

7 We report the results of our experiments with RankBoost in several datasets demonstrating the benefits of an algorithm specifically designed to globally optimize the AUC over other existing algorithms optimizing an approximation of the AUC or only locally optimizing the AUC. [sent-10, score-0.138]

8 1 Motivation In many applications, the overall classification error rate is not the most pertinent performance measure, criteria such as ordering or ranking seem more appropriate. [sent-11, score-0.189]

9 Consider for example the list of relevant documents returned by a search engine for a specific query. [sent-12, score-0.092]

10 That list may contain several thousand documents, but, in practice, only the top fifty or so are examined by the user. [sent-13, score-0.038]

11 Thus, a search engine’s ranking of the documents is more critical than the accuracy of its classification of all documents as relevant or not. [sent-14, score-0.202]

12 More generally, for a binary classifier assigning a real-valued score to each object, a better correlation between output scores and the probability of correct classification is highly desirable. [sent-15, score-0.016]

13 A natural criterion or summary statistic often used to measure the ranking quality of a classifier is the area under an ROC curve (AUC) [8]. [sent-16, score-0.225]

14 1 However, the objective function optimized by most classification algorithms is the error rate and not the AUC. [sent-17, score-0.129]

15 Thus, it is important to determine the relationship between the AUC values and the error rate. [sent-19, score-0.066]

16 1 The AUC value is equivalent to the Wilcoxon-Mann-Whitney statistic [8] and closely related to the Gini index [1]. [sent-22, score-0.075]

17 It has been re-invented under the name of L-measure by [11], as already pointed out by [2], and slightly modified under the name of Linear Ranking by [13, 14]. [sent-23, score-0.058]

18 718 (1,1) True positive rate = (0,0) False positive rate = False positive rate correctly classified positive total positive incorrectly classified negative total negative Figure 1: An example of ROC curve. [sent-26, score-0.621]

19 The line connecting (0, 0) and (1, 1), corresponding to random classification, is drawn for reference. [sent-27, score-0.018]

20 The true positive (negative) rate is sometimes referred to as the sensitivity (resp. [sent-28, score-0.106]

21 In the following sections, we give a detailed statistical analysis of the relationship between the AUC and the error rate, including the first exact expression of the expected value and the variance of the AUC for a fixed error rate. [sent-30, score-0.189]

22 2 We show that, under certain conditions, the global function optimized by the RankBoost algorithm is exactly the AUC. [sent-31, score-0.051]

23 We report the results of our experiments with RankBoost in several datasets and demonstrate the benefits of an algorithm specifically designed to globally optimize the AUC over other existing algorithms optimizing an approximation of the AUC or only locally optimizing the AUC. [sent-32, score-0.138]

24 2 Definition and properties of the AUC The Receiver Operating Characteristics (ROC) curves were originally developed in signal detection theory [3] in connection with radio signals, and have been used since then in many other applications, in particular for medical decision-making. [sent-33, score-0.015]

25 Over the last few years, they have found increased interest in the machine learning and data mining communities for model evaluation and selection [12, 10, 4, 9, 15, 2]. [sent-34, score-0.014]

26 The ROC curve for a binary classification problem plots the true positive rate as a function of the false positive rate. [sent-35, score-0.264]

27 The points of the curve are obtained by sweeping the classification threshold from the most positive classification value to the most negative. [sent-36, score-0.179]

28 For a fully random classification, the ROC curve is a straight line connecting the origin to (1, 1). [sent-37, score-0.083]

29 Any improvement over random classification results in an ROC curve at least partially above this straight line. [sent-38, score-0.065]

30 The AUC is defined as the area under the ROC curve and is closely related to the ranking quality of the classification as shown more formally by Lemma 1 below. [sent-41, score-0.162]

31 Consider a binary classification task with m positive examples and n negative examples. [sent-42, score-0.21]

32 We will assume that a classifier outputs a strictly ordered list for these examples and will denote by 1X the indicator function of a set X. [sent-43, score-0.122]

33 , xm be the output of c on the positive examples and y1 , . [sent-48, score-0.121]

34 Then, the AUC, A, associated to c is given by: m n i=1 j=1 1xi >yj (1) A= mn that is the value of the Wilcoxon-Mann-Whitney statistic [8]. [sent-52, score-0.147]

35 The proof is based on the observation that the AUC value is exactly the probability P (X > Y ) where X is the random variable corresponding to the distribution of the outputs for the positive examples and Y the one corresponding to the negative examples [7]. [sent-54, score-0.345]

36 The Wilcoxon-Mann-Whitney statistic is clearly the expression of that probability in the discrete case, which proves the lemma [8]. [sent-55, score-0.142]

37 With a perfect ranking, all positive examples are ranked higher than the negative ones and A = 1. [sent-57, score-0.241]

38 Threshold θ k − x Positive examples x Negative examples n − x Negative examples m − (k − x) Positive examples Figure 2: For a fixed number of errors k, there may be x, 0 ≤ x ≤ k, false negative examples. [sent-60, score-0.433]

39 3 The Expected Value of the AUC In this section, we compute exactly the expected value of the AUC over all classifications with a fixed number of errors and compare that to the error rate. [sent-61, score-0.125]

40 Different classifiers may have the same error rate but different AUC values. [sent-62, score-0.093]

41 Indeed, for a given classification threshold θ, an arbitrary reordering of the examples with outputs more than θ clearly does not affect the error rate but leads to different AUC values. [sent-63, score-0.263]

42 Similarly, one may reorder the examples with output less than θ without changing the error rate. [sent-64, score-0.122]

43 We wish to compute the average value of the AUC over all classifications with k errors. [sent-66, score-0.062]

44 Our model is based on the simple assumption that all classifications or rankings with k errors are equiprobable. [sent-67, score-0.049]

45 One could perhaps argue that errors are not necessarily evenly distributed, e. [sent-68, score-0.032]

46 , examples with very high or very low ranks are less likely to be errors, but we cannot justify such biases in general. [sent-70, score-0.097]

47 For a given classification, there may be x, 0 ≤ x ≤ k, false positive examples. [sent-71, score-0.094]

48 Since the number of errors is fixed, there are k − x false negative examples. [sent-72, score-0.149]

49 The two regions of examples with classification outputs above and below the threshold are separated by a vertical line. [sent-74, score-0.14]

50 Since there are (m − (k − x)) positive examples above the threshold and n − x negative examples below the threshold, A1 is given by: A1 = (m − (k − x))(n − x) (3) To compute A2 , we can assign to each negative example above the threshold a position based on its classification rank. [sent-78, score-0.438]

51 Let position one be the first position above the threshold and let α1 < . [sent-79, score-0.071]

52 < αx denote the positions in increasing order of the x negative examples in the region above the threshold. [sent-82, score-0.236]

53 The total number of examples classified as positive is N = m − (k − x) + x. [sent-83, score-0.137]

54 Thus, by definition of A2 , x A2 = (N − αi ) − (x − i) (4) i=1 where the first term N − αi represents the number of examples ranked higher than the ith example and the second term x − i discounts the number of negative examples incorrectly ranked higher than the ith example. [sent-84, score-0.334]

55 < αk−x denote the positions of the k − x positive examples below the threshold, counting positions in reverse by starting from the threshold. [sent-88, score-0.261]

56 Combining the expressions of A1 , A2 , and A3 leads to: A= A1 + A2 + A3 (k − 2x)2 + k ( =1+ − mn 2mn x i=1 αi + mn x j=1 αj ) (6) Lemma 2 For a fixed x, the average value of the AUC A is given by: < A >x = 1 − x n + k−x m 2 (7) x Proof. [sent-90, score-0.239]

57 The proof is based on the computation of the average values of i=1 αi and x j=1 αj for a given x. [sent-91, score-0.051]

58 We start by computing the average value < αi >x for a given i, 1 ≤ i ≤ x. [sent-92, score-0.062]

59 We have i ≤ l ≤ N − (x − i) since there need to be at least i − 1 positions before αi and N − (x − i) above. [sent-100, score-0.064]

60 Since the total number of ways of choosing the x positions for α1 . [sent-108, score-0.107]

61 (11) leads to: x j=1 x (N + 1) 2 (11) αj >x in Eq. [sent-113, score-0.013]

62 (6) by the expressions given by (k − 2x)2 + k − x(N + 1) − x (N + 1) =1− 2mn which ends the proof of the lemma. [sent-114, score-0.034]

63 (7) shows that the average AUC value for a given x is simply one minus the average of the accuracy rates for the positive and negative classes. [sent-116, score-0.259]

64 Proposition 1 Assume that a binary classification task with m positive examples and n negative examples is given. [sent-117, score-0.281]

65 Then, the expected value of the AUC A over all classifications with k errors is given by: < A >= 1 − k (n − m)2 (m + n + 1) − m+n 4mn k−1 m+n x=0 x k m+n+1 x=0 x k − m+n (13) Proof. [sent-118, score-0.073]

66 Lemma 2 gives the average value of the AUC for a fixed value of x. [sent-119, score-0.087]

67 To compute the average over all possible values of x, we need to weight the expression of Eq. [sent-120, score-0.072]

68 (7) with the total number of possible classifications for a given x. [sent-121, score-0.016]

69 There are N possible ways of x choosing the positions of the x misclassified negative examples, and similarly N possible x ways of choosing the positions of the x = k − x misclassified positive examples. [sent-122, score-0.32]

70 Thus, in view of Lemma 2, the average AUC is given by: < A >= k N x=0 x N x (1 − k N x=0 x N x k−x x n+ m 2 ) (14) r=0. [sent-123, score-0.037]

71 0 Mean value of the AUC Relative standard deviation r=0. [sent-146, score-0.055]

72 5 Figure 3: Mean (left) and relative standard deviation (right) of the AUC as a function of the error rate. [sent-157, score-0.067]

73 Each curve corresponds to a fixed ratio of r = n/(n + m). [sent-158, score-0.063]

74 The average AUC value monotonically increases with the accuracy. [sent-159, score-0.092]

75 For n = m, as for the top curve in the left plot, the average AUC coincides with the accuracy. [sent-160, score-0.097]

76 The standard deviation decreases with the accuracy, and the lowest curve corresponds to n = m. [sent-161, score-0.093]

77 (13), it is clear that the average AUC value is identical to the accuracy of the classifier only for even distributions (n = m). [sent-171, score-0.099]

78 For n = m, the expected value of the AUC is a monotonic function of the accuracy, see Fig. [sent-172, score-0.041]

79 For a fixed ratio of n/(n + m), the curves are obtained by increasing the accuracy from n/(n + m) to 1. [sent-174, score-0.065]

80 The average AUC varies monotonically in the range of accuracy between 0. [sent-175, score-0.104]

81 In other words, on average, there seems nothing to be gained in designing specific learning algorithms for maximizing the AUC: a classification algorithm minimizing the error rate also optimizes the AUC. [sent-178, score-0.109]

82 Indeed, we will show in the next section that the variance of the AUC value is not null for any ratio n/(n + m) when k = 0. [sent-180, score-0.063]

83 4 The Variance of the AUC 2 Let D = mn + (k−2x) +k , a = i=1 αi , a = j=1 αj , and α = a + a . [sent-181, score-0.072]

84 A crucial step in computing the exact value of the variance of x the AUC is to determine the value of the terms of the type < a2 >x =< ( i=1 αi )2 >x . [sent-185, score-0.086]

85 (14) and the expression given by [15] is the weighting by the number of configurations. [sent-187, score-0.035]

86 The authors’ analysis leads them to the conclusion that the average AUC is identical to the accuracy for all ratios n/(n + m), which is false. [sent-188, score-0.087]

87 4 We thank Neil Sloane for having pointed us to Zeilberger’s algorithm and Maple package. [sent-189, score-0.02]

88 x Lemma 3 For a fixed x, the average of ( i=1 αi )2 is given by: x(N + 1) < a2 > x = (3N x + 2x + N ) 12 (19) Proof. [sent-190, score-0.037]


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same-paper 1 0.99999934 3 nips-2003-AUC Optimization vs. Error Rate Minimization

Author: Corinna Cortes, Mehryar Mohri

Abstract: The area under an ROC curve (AUC) is a criterion used in many applications to measure the quality of a classification algorithm. However, the objective function optimized in most of these algorithms is the error rate and not the AUC value. We give a detailed statistical analysis of the relationship between the AUC and the error rate, including the first exact expression of the expected value and the variance of the AUC for a fixed error rate. Our results show that the average AUC is monotonically increasing as a function of the classification accuracy, but that the standard deviation for uneven distributions and higher error rates is noticeable. Thus, algorithms designed to minimize the error rate may not lead to the best possible AUC values. We show that, under certain conditions, the global function optimized by the RankBoost algorithm is exactly the AUC. We report the results of our experiments with RankBoost in several datasets demonstrating the benefits of an algorithm specifically designed to globally optimize the AUC over other existing algorithms optimizing an approximation of the AUC or only locally optimizing the AUC. 1 Motivation In many applications, the overall classification error rate is not the most pertinent performance measure, criteria such as ordering or ranking seem more appropriate. Consider for example the list of relevant documents returned by a search engine for a specific query. That list may contain several thousand documents, but, in practice, only the top fifty or so are examined by the user. Thus, a search engine’s ranking of the documents is more critical than the accuracy of its classification of all documents as relevant or not. More generally, for a binary classifier assigning a real-valued score to each object, a better correlation between output scores and the probability of correct classification is highly desirable. A natural criterion or summary statistic often used to measure the ranking quality of a classifier is the area under an ROC curve (AUC) [8].1 However, the objective function optimized by most classification algorithms is the error rate and not the AUC. Recently, several algorithms have been proposed for maximizing the AUC value locally [4] or maximizing some approximations of the global AUC value [9, 15], but, in general, these algorithms do not obtain AUC values significantly better than those obtained by an algorithm designed to minimize the error rates. Thus, it is important to determine the relationship between the AUC values and the error rate. ∗ This author’s new address is: Google Labs, 1440 Broadway, New York, NY 10018, corinna@google.com. 1 The AUC value is equivalent to the Wilcoxon-Mann-Whitney statistic [8] and closely related to the Gini index [1]. It has been re-invented under the name of L-measure by [11], as already pointed out by [2], and slightly modified under the name of Linear Ranking by [13, 14]. True positive rate ROC Curve. AUC=0.718 (1,1) True positive rate = (0,0) False positive rate = False positive rate correctly classified positive total positive incorrectly classified negative total negative Figure 1: An example of ROC curve. The line connecting (0, 0) and (1, 1), corresponding to random classification, is drawn for reference. The true positive (negative) rate is sometimes referred to as the sensitivity (resp. specificity) in this context. In the following sections, we give a detailed statistical analysis of the relationship between the AUC and the error rate, including the first exact expression of the expected value and the variance of the AUC for a fixed error rate.2 We show that, under certain conditions, the global function optimized by the RankBoost algorithm is exactly the AUC. We report the results of our experiments with RankBoost in several datasets and demonstrate the benefits of an algorithm specifically designed to globally optimize the AUC over other existing algorithms optimizing an approximation of the AUC or only locally optimizing the AUC. 2 Definition and properties of the AUC The Receiver Operating Characteristics (ROC) curves were originally developed in signal detection theory [3] in connection with radio signals, and have been used since then in many other applications, in particular for medical decision-making. Over the last few years, they have found increased interest in the machine learning and data mining communities for model evaluation and selection [12, 10, 4, 9, 15, 2]. The ROC curve for a binary classification problem plots the true positive rate as a function of the false positive rate. The points of the curve are obtained by sweeping the classification threshold from the most positive classification value to the most negative. For a fully random classification, the ROC curve is a straight line connecting the origin to (1, 1). Any improvement over random classification results in an ROC curve at least partially above this straight line. Fig. (1) shows an example of ROC curve. The AUC is defined as the area under the ROC curve and is closely related to the ranking quality of the classification as shown more formally by Lemma 1 below. Consider a binary classification task with m positive examples and n negative examples. We will assume that a classifier outputs a strictly ordered list for these examples and will denote by 1X the indicator function of a set X. Lemma 1 ([8]) Let c be a fixed classifier. Let x1 , . . . , xm be the output of c on the positive examples and y1 , . . . , yn its output on the negative examples. Then, the AUC, A, associated to c is given by: m n i=1 j=1 1xi >yj (1) A= mn that is the value of the Wilcoxon-Mann-Whitney statistic [8]. Proof. The proof is based on the observation that the AUC value is exactly the probability P (X > Y ) where X is the random variable corresponding to the distribution of the outputs for the positive examples and Y the one corresponding to the negative examples [7]. The Wilcoxon-Mann-Whitney statistic is clearly the expression of that probability in the discrete case, which proves the lemma [8]. Thus, the AUC can be viewed as a measure based on pairwise comparisons between classifications of the two classes. With a perfect ranking, all positive examples are ranked higher than the negative ones and A = 1. Any deviation from this ranking decreases the AUC. 2 An attempt in that direction was made by [15], but, unfortunately, the authors’ analysis and the result are both wrong. Threshold θ k − x Positive examples x Negative examples n − x Negative examples m − (k − x) Positive examples Figure 2: For a fixed number of errors k, there may be x, 0 ≤ x ≤ k, false negative examples. 3 The Expected Value of the AUC In this section, we compute exactly the expected value of the AUC over all classifications with a fixed number of errors and compare that to the error rate. Different classifiers may have the same error rate but different AUC values. Indeed, for a given classification threshold θ, an arbitrary reordering of the examples with outputs more than θ clearly does not affect the error rate but leads to different AUC values. Similarly, one may reorder the examples with output less than θ without changing the error rate. Assume that the number of errors k is fixed. We wish to compute the average value of the AUC over all classifications with k errors. Our model is based on the simple assumption that all classifications or rankings with k errors are equiprobable. One could perhaps argue that errors are not necessarily evenly distributed, e.g., examples with very high or very low ranks are less likely to be errors, but we cannot justify such biases in general. For a given classification, there may be x, 0 ≤ x ≤ k, false positive examples. Since the number of errors is fixed, there are k − x false negative examples. Figure 3 shows the corresponding configuration. The two regions of examples with classification outputs above and below the threshold are separated by a vertical line. For a given x, the computation of the AUC, A, as given by Eq. (1) can be divided into the following three parts: A1 + A2 + A3 A= , with (2) mn A1 = the sum over all pairs (xi , yj ) with xi and yj in distinct regions; A2 = the sum over all pairs (xi , yj ) with xi and yj in the region above the threshold; A3 = the sum over all pairs (xi , yj ) with xi and yj in the region below the threshold. The first term, A1 , is easy to compute. Since there are (m − (k − x)) positive examples above the threshold and n − x negative examples below the threshold, A1 is given by: A1 = (m − (k − x))(n − x) (3) To compute A2 , we can assign to each negative example above the threshold a position based on its classification rank. Let position one be the first position above the threshold and let α1 < . . . < αx denote the positions in increasing order of the x negative examples in the region above the threshold. The total number of examples classified as positive is N = m − (k − x) + x. Thus, by definition of A2 , x A2 = (N − αi ) − (x − i) (4) i=1 where the first term N − αi represents the number of examples ranked higher than the ith example and the second term x − i discounts the number of negative examples incorrectly ranked higher than the ith example. Similarly, let α1 < . . . < αk−x denote the positions of the k − x positive examples below the threshold, counting positions in reverse by starting from the threshold. Then, A3 is given by: x A3 = (N − αj ) − (x − j) (5) j=1 with N = n − x + (k − x) and x = k − x. Combining the expressions of A1 , A2 , and A3 leads to: A= A1 + A2 + A3 (k − 2x)2 + k ( =1+ − mn 2mn x i=1 αi + mn x j=1 αj ) (6) Lemma 2 For a fixed x, the average value of the AUC A is given by: < A >x = 1 − x n + k−x m 2 (7) x Proof. The proof is based on the computation of the average values of i=1 αi and x j=1 αj for a given x. We start by computing the average value < αi >x for a given i, 1 ≤ i ≤ x. Consider all the possible positions for α1 . . . αi−1 and αi+1 . . . αx , when the value of αi is fixed at say αi = l. We have i ≤ l ≤ N − (x − i) since there need to be at least i − 1 positions before αi and N − (x − i) above. There are l − 1 possible positions for α1 . . . αi−1 and N − l possible positions for αi+1 . . . αx . Since the total number of ways of choosing the x positions for α1 . . . αx out of N is N , the average value < αi >x is: x N −(x−i) l=i < αi >x = l l−1 i−1 N −l x−i (8) N x Thus, x < αi >x = x i=1 i=1 Using the classical identity: x < αi >x = N −(x−i) l−1 l i−1 l=i N x u p1 +p2 =p p1 N l=1 l N −1 x−1 N x i=1 N −l x−i v p2 = = N l=1 = u+v p N (N + 1) 2 x l−1 i=1 i−1 N x l N −l x−i (9) , we can write: N −1 x−1 N x = x(N + 1) 2 (10) Similarly, we have: x < αj >x = j=1 x Replacing < i=1 αi >x and < Eq. (10) and Eq. (11) leads to: x j=1 x (N + 1) 2 (11) αj >x in Eq. (6) by the expressions given by (k − 2x)2 + k − x(N + 1) − x (N + 1) =1− 2mn which ends the proof of the lemma. < A >x = 1 + x n + k−x m 2 (12) Note that Eq. (7) shows that the average AUC value for a given x is simply one minus the average of the accuracy rates for the positive and negative classes. Proposition 1 Assume that a binary classification task with m positive examples and n negative examples is given. Then, the expected value of the AUC A over all classifications with k errors is given by: < A >= 1 − k (n − m)2 (m + n + 1) − m+n 4mn k−1 m+n x=0 x k m+n+1 x=0 x k − m+n (13) Proof. Lemma 2 gives the average value of the AUC for a fixed value of x. To compute the average over all possible values of x, we need to weight the expression of Eq. (7) with the total number of possible classifications for a given x. There are N possible ways of x choosing the positions of the x misclassified negative examples, and similarly N possible x ways of choosing the positions of the x = k − x misclassified positive examples. Thus, in view of Lemma 2, the average AUC is given by: < A >= k N x=0 x N x (1 − k N x=0 x N x k−x x n+ m 2 ) (14) r=0.05 r=0.01 r=0.1 r=0.25 0.0 0.1 0.2 r=0.5 0.3 Error rate 0.4 0.5 .00 .05 .10 .15 .20 .25 0.5 0.6 0.7 0.8 0.9 1.0 Mean value of the AUC Relative standard deviation r=0.01 r=0.05 r=0.1 0.0 0.1 r=0.25 0.2 0.3 Error rate r=0.5 0.4 0.5 Figure 3: Mean (left) and relative standard deviation (right) of the AUC as a function of the error rate. Each curve corresponds to a fixed ratio of r = n/(n + m). The average AUC value monotonically increases with the accuracy. For n = m, as for the top curve in the left plot, the average AUC coincides with the accuracy. The standard deviation decreases with the accuracy, and the lowest curve corresponds to n = m. This expression can be simplified into Eq. (13)3 using the following novel identities: k X N x x=0 k X N x x x=0 ! N x ! ! N x ! = = ! k X n+m+1 x x=0 (15) ! k X (k − x)(m − n) + k n + m + 1 2 x x=0 (16) that we obtained by using Zeilberger’s algorithm4 and numerous combinatorial ’tricks’. From the expression of Eq. (13), it is clear that the average AUC value is identical to the accuracy of the classifier only for even distributions (n = m). For n = m, the expected value of the AUC is a monotonic function of the accuracy, see Fig. (3)(left). For a fixed ratio of n/(n + m), the curves are obtained by increasing the accuracy from n/(n + m) to 1. The average AUC varies monotonically in the range of accuracy between 0.5 and 1.0. In other words, on average, there seems nothing to be gained in designing specific learning algorithms for maximizing the AUC: a classification algorithm minimizing the error rate also optimizes the AUC. However, this only holds for the average AUC. Indeed, we will show in the next section that the variance of the AUC value is not null for any ratio n/(n + m) when k = 0. 4 The Variance of the AUC 2 Let D = mn + (k−2x) +k , a = i=1 αi , a = j=1 αj , and α = a + a . Then, by 2 Eq. (6), mnA = D − α. Thus, the variance of the AUC, σ 2 (A), is given by: (mn)2 σ 2 (A) x x = < (D − α)2 − (< D > − < α >)2 > = < D2 > − < D >2 + < α2 > − < α >2 −2(< αD > − < α >< D >) (17) As before, to compute the average of a term X over all classifications, we can first determine its average < X >x for a fixed x, and then use the function F defined by: F (Y ) = k N N x=0 x x k N N x=0 x x Y (18) and < X >= F (< X >x ). A crucial step in computing the exact value of the variance of x the AUC is to determine the value of the terms of the type < a2 >x =< ( i=1 αi )2 >x . 3 An essential difference between Eq. (14) and the expression given by [15] is the weighting by the number of configurations. The authors’ analysis leads them to the conclusion that the average AUC is identical to the accuracy for all ratios n/(n + m), which is false. 4 We thank Neil Sloane for having pointed us to Zeilberger’s algorithm and Maple package. x Lemma 3 For a fixed x, the average of ( i=1 αi )2 is given by: x(N + 1) < a2 > x = (3N x + 2x + N ) 12 (19) Proof. By definition of a, < a2 >x = b + 2c with: x x α2 >x i b =< c =< αi αj >x (20) 1≤i

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same-paper 1 0.86649561 3 nips-2003-AUC Optimization vs. Error Rate Minimization

Author: Corinna Cortes, Mehryar Mohri

Abstract: The area under an ROC curve (AUC) is a criterion used in many applications to measure the quality of a classification algorithm. However, the objective function optimized in most of these algorithms is the error rate and not the AUC value. We give a detailed statistical analysis of the relationship between the AUC and the error rate, including the first exact expression of the expected value and the variance of the AUC for a fixed error rate. Our results show that the average AUC is monotonically increasing as a function of the classification accuracy, but that the standard deviation for uneven distributions and higher error rates is noticeable. Thus, algorithms designed to minimize the error rate may not lead to the best possible AUC values. We show that, under certain conditions, the global function optimized by the RankBoost algorithm is exactly the AUC. We report the results of our experiments with RankBoost in several datasets demonstrating the benefits of an algorithm specifically designed to globally optimize the AUC over other existing algorithms optimizing an approximation of the AUC or only locally optimizing the AUC. 1 Motivation In many applications, the overall classification error rate is not the most pertinent performance measure, criteria such as ordering or ranking seem more appropriate. Consider for example the list of relevant documents returned by a search engine for a specific query. That list may contain several thousand documents, but, in practice, only the top fifty or so are examined by the user. Thus, a search engine’s ranking of the documents is more critical than the accuracy of its classification of all documents as relevant or not. More generally, for a binary classifier assigning a real-valued score to each object, a better correlation between output scores and the probability of correct classification is highly desirable. A natural criterion or summary statistic often used to measure the ranking quality of a classifier is the area under an ROC curve (AUC) [8].1 However, the objective function optimized by most classification algorithms is the error rate and not the AUC. Recently, several algorithms have been proposed for maximizing the AUC value locally [4] or maximizing some approximations of the global AUC value [9, 15], but, in general, these algorithms do not obtain AUC values significantly better than those obtained by an algorithm designed to minimize the error rates. Thus, it is important to determine the relationship between the AUC values and the error rate. ∗ This author’s new address is: Google Labs, 1440 Broadway, New York, NY 10018, corinna@google.com. 1 The AUC value is equivalent to the Wilcoxon-Mann-Whitney statistic [8] and closely related to the Gini index [1]. It has been re-invented under the name of L-measure by [11], as already pointed out by [2], and slightly modified under the name of Linear Ranking by [13, 14]. True positive rate ROC Curve. AUC=0.718 (1,1) True positive rate = (0,0) False positive rate = False positive rate correctly classified positive total positive incorrectly classified negative total negative Figure 1: An example of ROC curve. The line connecting (0, 0) and (1, 1), corresponding to random classification, is drawn for reference. The true positive (negative) rate is sometimes referred to as the sensitivity (resp. specificity) in this context. In the following sections, we give a detailed statistical analysis of the relationship between the AUC and the error rate, including the first exact expression of the expected value and the variance of the AUC for a fixed error rate.2 We show that, under certain conditions, the global function optimized by the RankBoost algorithm is exactly the AUC. We report the results of our experiments with RankBoost in several datasets and demonstrate the benefits of an algorithm specifically designed to globally optimize the AUC over other existing algorithms optimizing an approximation of the AUC or only locally optimizing the AUC. 2 Definition and properties of the AUC The Receiver Operating Characteristics (ROC) curves were originally developed in signal detection theory [3] in connection with radio signals, and have been used since then in many other applications, in particular for medical decision-making. Over the last few years, they have found increased interest in the machine learning and data mining communities for model evaluation and selection [12, 10, 4, 9, 15, 2]. The ROC curve for a binary classification problem plots the true positive rate as a function of the false positive rate. The points of the curve are obtained by sweeping the classification threshold from the most positive classification value to the most negative. For a fully random classification, the ROC curve is a straight line connecting the origin to (1, 1). Any improvement over random classification results in an ROC curve at least partially above this straight line. Fig. (1) shows an example of ROC curve. The AUC is defined as the area under the ROC curve and is closely related to the ranking quality of the classification as shown more formally by Lemma 1 below. Consider a binary classification task with m positive examples and n negative examples. We will assume that a classifier outputs a strictly ordered list for these examples and will denote by 1X the indicator function of a set X. Lemma 1 ([8]) Let c be a fixed classifier. Let x1 , . . . , xm be the output of c on the positive examples and y1 , . . . , yn its output on the negative examples. Then, the AUC, A, associated to c is given by: m n i=1 j=1 1xi >yj (1) A= mn that is the value of the Wilcoxon-Mann-Whitney statistic [8]. Proof. The proof is based on the observation that the AUC value is exactly the probability P (X > Y ) where X is the random variable corresponding to the distribution of the outputs for the positive examples and Y the one corresponding to the negative examples [7]. The Wilcoxon-Mann-Whitney statistic is clearly the expression of that probability in the discrete case, which proves the lemma [8]. Thus, the AUC can be viewed as a measure based on pairwise comparisons between classifications of the two classes. With a perfect ranking, all positive examples are ranked higher than the negative ones and A = 1. Any deviation from this ranking decreases the AUC. 2 An attempt in that direction was made by [15], but, unfortunately, the authors’ analysis and the result are both wrong. Threshold θ k − x Positive examples x Negative examples n − x Negative examples m − (k − x) Positive examples Figure 2: For a fixed number of errors k, there may be x, 0 ≤ x ≤ k, false negative examples. 3 The Expected Value of the AUC In this section, we compute exactly the expected value of the AUC over all classifications with a fixed number of errors and compare that to the error rate. Different classifiers may have the same error rate but different AUC values. Indeed, for a given classification threshold θ, an arbitrary reordering of the examples with outputs more than θ clearly does not affect the error rate but leads to different AUC values. Similarly, one may reorder the examples with output less than θ without changing the error rate. Assume that the number of errors k is fixed. We wish to compute the average value of the AUC over all classifications with k errors. Our model is based on the simple assumption that all classifications or rankings with k errors are equiprobable. One could perhaps argue that errors are not necessarily evenly distributed, e.g., examples with very high or very low ranks are less likely to be errors, but we cannot justify such biases in general. For a given classification, there may be x, 0 ≤ x ≤ k, false positive examples. Since the number of errors is fixed, there are k − x false negative examples. Figure 3 shows the corresponding configuration. The two regions of examples with classification outputs above and below the threshold are separated by a vertical line. For a given x, the computation of the AUC, A, as given by Eq. (1) can be divided into the following three parts: A1 + A2 + A3 A= , with (2) mn A1 = the sum over all pairs (xi , yj ) with xi and yj in distinct regions; A2 = the sum over all pairs (xi , yj ) with xi and yj in the region above the threshold; A3 = the sum over all pairs (xi , yj ) with xi and yj in the region below the threshold. The first term, A1 , is easy to compute. Since there are (m − (k − x)) positive examples above the threshold and n − x negative examples below the threshold, A1 is given by: A1 = (m − (k − x))(n − x) (3) To compute A2 , we can assign to each negative example above the threshold a position based on its classification rank. Let position one be the first position above the threshold and let α1 < . . . < αx denote the positions in increasing order of the x negative examples in the region above the threshold. The total number of examples classified as positive is N = m − (k − x) + x. Thus, by definition of A2 , x A2 = (N − αi ) − (x − i) (4) i=1 where the first term N − αi represents the number of examples ranked higher than the ith example and the second term x − i discounts the number of negative examples incorrectly ranked higher than the ith example. Similarly, let α1 < . . . < αk−x denote the positions of the k − x positive examples below the threshold, counting positions in reverse by starting from the threshold. Then, A3 is given by: x A3 = (N − αj ) − (x − j) (5) j=1 with N = n − x + (k − x) and x = k − x. Combining the expressions of A1 , A2 , and A3 leads to: A= A1 + A2 + A3 (k − 2x)2 + k ( =1+ − mn 2mn x i=1 αi + mn x j=1 αj ) (6) Lemma 2 For a fixed x, the average value of the AUC A is given by: < A >x = 1 − x n + k−x m 2 (7) x Proof. The proof is based on the computation of the average values of i=1 αi and x j=1 αj for a given x. We start by computing the average value < αi >x for a given i, 1 ≤ i ≤ x. Consider all the possible positions for α1 . . . αi−1 and αi+1 . . . αx , when the value of αi is fixed at say αi = l. We have i ≤ l ≤ N − (x − i) since there need to be at least i − 1 positions before αi and N − (x − i) above. There are l − 1 possible positions for α1 . . . αi−1 and N − l possible positions for αi+1 . . . αx . Since the total number of ways of choosing the x positions for α1 . . . αx out of N is N , the average value < αi >x is: x N −(x−i) l=i < αi >x = l l−1 i−1 N −l x−i (8) N x Thus, x < αi >x = x i=1 i=1 Using the classical identity: x < αi >x = N −(x−i) l−1 l i−1 l=i N x u p1 +p2 =p p1 N l=1 l N −1 x−1 N x i=1 N −l x−i v p2 = = N l=1 = u+v p N (N + 1) 2 x l−1 i=1 i−1 N x l N −l x−i (9) , we can write: N −1 x−1 N x = x(N + 1) 2 (10) Similarly, we have: x < αj >x = j=1 x Replacing < i=1 αi >x and < Eq. (10) and Eq. (11) leads to: x j=1 x (N + 1) 2 (11) αj >x in Eq. (6) by the expressions given by (k − 2x)2 + k − x(N + 1) − x (N + 1) =1− 2mn which ends the proof of the lemma. < A >x = 1 + x n + k−x m 2 (12) Note that Eq. (7) shows that the average AUC value for a given x is simply one minus the average of the accuracy rates for the positive and negative classes. Proposition 1 Assume that a binary classification task with m positive examples and n negative examples is given. Then, the expected value of the AUC A over all classifications with k errors is given by: < A >= 1 − k (n − m)2 (m + n + 1) − m+n 4mn k−1 m+n x=0 x k m+n+1 x=0 x k − m+n (13) Proof. Lemma 2 gives the average value of the AUC for a fixed value of x. To compute the average over all possible values of x, we need to weight the expression of Eq. (7) with the total number of possible classifications for a given x. There are N possible ways of x choosing the positions of the x misclassified negative examples, and similarly N possible x ways of choosing the positions of the x = k − x misclassified positive examples. Thus, in view of Lemma 2, the average AUC is given by: < A >= k N x=0 x N x (1 − k N x=0 x N x k−x x n+ m 2 ) (14) r=0.05 r=0.01 r=0.1 r=0.25 0.0 0.1 0.2 r=0.5 0.3 Error rate 0.4 0.5 .00 .05 .10 .15 .20 .25 0.5 0.6 0.7 0.8 0.9 1.0 Mean value of the AUC Relative standard deviation r=0.01 r=0.05 r=0.1 0.0 0.1 r=0.25 0.2 0.3 Error rate r=0.5 0.4 0.5 Figure 3: Mean (left) and relative standard deviation (right) of the AUC as a function of the error rate. Each curve corresponds to a fixed ratio of r = n/(n + m). The average AUC value monotonically increases with the accuracy. For n = m, as for the top curve in the left plot, the average AUC coincides with the accuracy. The standard deviation decreases with the accuracy, and the lowest curve corresponds to n = m. 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Indeed, we will show in the next section that the variance of the AUC value is not null for any ratio n/(n + m) when k = 0. 4 The Variance of the AUC 2 Let D = mn + (k−2x) +k , a = i=1 αi , a = j=1 αj , and α = a + a . Then, by 2 Eq. (6), mnA = D − α. Thus, the variance of the AUC, σ 2 (A), is given by: (mn)2 σ 2 (A) x x = < (D − α)2 − (< D > − < α >)2 > = < D2 > − < D >2 + < α2 > − < α >2 −2(< αD > − < α >< D >) (17) As before, to compute the average of a term X over all classifications, we can first determine its average < X >x for a fixed x, and then use the function F defined by: F (Y ) = k N N x=0 x x k N N x=0 x x Y (18) and < X >= F (< X >x ). A crucial step in computing the exact value of the variance of x the AUC is to determine the value of the terms of the type < a2 >x =< ( i=1 αi )2 >x . 3 An essential difference between Eq. (14) and the expression given by [15] is the weighting by the number of configurations. The authors’ analysis leads them to the conclusion that the average AUC is identical to the accuracy for all ratios n/(n + m), which is false. 4 We thank Neil Sloane for having pointed us to Zeilberger’s algorithm and Maple package. x Lemma 3 For a fixed x, the average of ( i=1 αi )2 is given by: x(N + 1) < a2 > x = (3N x + 2x + N ) 12 (19) Proof. By definition of a, < a2 >x = b + 2c with: x x α2 >x i b =< c =< αi αj >x (20) 1≤i

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(1) can be divided into the following three parts: A1 + A2 + A3 A= , with (2) mn A1 = the sum over all pairs (xi , yj ) with xi and yj in distinct regions; A2 = the sum over all pairs (xi , yj ) with xi and yj in the region above the threshold; A3 = the sum over all pairs (xi , yj ) with xi and yj in the region below the threshold. The first term, A1 , is easy to compute. Since there are (m − (k − x)) positive examples above the threshold and n − x negative examples below the threshold, A1 is given by: A1 = (m − (k − x))(n − x) (3) To compute A2 , we can assign to each negative example above the threshold a position based on its classification rank. Let position one be the first position above the threshold and let α1 < . . . < αx denote the positions in increasing order of the x negative examples in the region above the threshold. The total number of examples classified as positive is N = m − (k − x) + x. Thus, by definition of A2 , x A2 = (N − αi ) − (x − i) (4) i=1 where the first term N − αi represents the number of examples ranked higher than the ith example and the second term x − i discounts the number of negative examples incorrectly ranked higher than the ith example. Similarly, let α1 < . . . < αk−x denote the positions of the k − x positive examples below the threshold, counting positions in reverse by starting from the threshold. Then, A3 is given by: x A3 = (N − αj ) − (x − j) (5) j=1 with N = n − x + (k − x) and x = k − x. Combining the expressions of A1 , A2 , and A3 leads to: A= A1 + A2 + A3 (k − 2x)2 + k ( =1+ − mn 2mn x i=1 αi + mn x j=1 αj ) (6) Lemma 2 For a fixed x, the average value of the AUC A is given by: < A >x = 1 − x n + k−x m 2 (7) x Proof. The proof is based on the computation of the average values of i=1 αi and x j=1 αj for a given x. We start by computing the average value < αi >x for a given i, 1 ≤ i ≤ x. Consider all the possible positions for α1 . . . αi−1 and αi+1 . . . αx , when the value of αi is fixed at say αi = l. We have i ≤ l ≤ N − (x − i) since there need to be at least i − 1 positions before αi and N − (x − i) above. There are l − 1 possible positions for α1 . . . αi−1 and N − l possible positions for αi+1 . . . αx . Since the total number of ways of choosing the x positions for α1 . . . αx out of N is N , the average value < αi >x is: x N −(x−i) l=i < αi >x = l l−1 i−1 N −l x−i (8) N x Thus, x < αi >x = x i=1 i=1 Using the classical identity: x < αi >x = N −(x−i) l−1 l i−1 l=i N x u p1 +p2 =p p1 N l=1 l N −1 x−1 N x i=1 N −l x−i v p2 = = N l=1 = u+v p N (N + 1) 2 x l−1 i=1 i−1 N x l N −l x−i (9) , we can write: N −1 x−1 N x = x(N + 1) 2 (10) Similarly, we have: x < αj >x = j=1 x Replacing < i=1 αi >x and < Eq. (10) and Eq. (11) leads to: x j=1 x (N + 1) 2 (11) αj >x in Eq. (6) by the expressions given by (k − 2x)2 + k − x(N + 1) − x (N + 1) =1− 2mn which ends the proof of the lemma. < A >x = 1 + x n + k−x m 2 (12) Note that Eq. (7) shows that the average AUC value for a given x is simply one minus the average of the accuracy rates for the positive and negative classes. Proposition 1 Assume that a binary classification task with m positive examples and n negative examples is given. Then, the expected value of the AUC A over all classifications with k errors is given by: < A >= 1 − k (n − m)2 (m + n + 1) − m+n 4mn k−1 m+n x=0 x k m+n+1 x=0 x k − m+n (13) Proof. Lemma 2 gives the average value of the AUC for a fixed value of x. To compute the average over all possible values of x, we need to weight the expression of Eq. (7) with the total number of possible classifications for a given x. There are N possible ways of x choosing the positions of the x misclassified negative examples, and similarly N possible x ways of choosing the positions of the x = k − x misclassified positive examples. Thus, in view of Lemma 2, the average AUC is given by: < A >= k N x=0 x N x (1 − k N x=0 x N x k−x x n+ m 2 ) (14) r=0.05 r=0.01 r=0.1 r=0.25 0.0 0.1 0.2 r=0.5 0.3 Error rate 0.4 0.5 .00 .05 .10 .15 .20 .25 0.5 0.6 0.7 0.8 0.9 1.0 Mean value of the AUC Relative standard deviation r=0.01 r=0.05 r=0.1 0.0 0.1 r=0.25 0.2 0.3 Error rate r=0.5 0.4 0.5 Figure 3: Mean (left) and relative standard deviation (right) of the AUC as a function of the error rate. Each curve corresponds to a fixed ratio of r = n/(n + m). The average AUC value monotonically increases with the accuracy. For n = m, as for the top curve in the left plot, the average AUC coincides with the accuracy. The standard deviation decreases with the accuracy, and the lowest curve corresponds to n = m. This expression can be simplified into Eq. (13)3 using the following novel identities: k X N x x=0 k X N x x x=0 ! N x ! ! N x ! = = ! k X n+m+1 x x=0 (15) ! k X (k − x)(m − n) + k n + m + 1 2 x x=0 (16) that we obtained by using Zeilberger’s algorithm4 and numerous combinatorial ’tricks’. From the expression of Eq. (13), it is clear that the average AUC value is identical to the accuracy of the classifier only for even distributions (n = m). For n = m, the expected value of the AUC is a monotonic function of the accuracy, see Fig. (3)(left). For a fixed ratio of n/(n + m), the curves are obtained by increasing the accuracy from n/(n + m) to 1. The average AUC varies monotonically in the range of accuracy between 0.5 and 1.0. In other words, on average, there seems nothing to be gained in designing specific learning algorithms for maximizing the AUC: a classification algorithm minimizing the error rate also optimizes the AUC. However, this only holds for the average AUC. Indeed, we will show in the next section that the variance of the AUC value is not null for any ratio n/(n + m) when k = 0. 4 The Variance of the AUC 2 Let D = mn + (k−2x) +k , a = i=1 αi , a = j=1 αj , and α = a + a . Then, by 2 Eq. (6), mnA = D − α. Thus, the variance of the AUC, σ 2 (A), is given by: (mn)2 σ 2 (A) x x = < (D − α)2 − (< D > − < α >)2 > = < D2 > − < D >2 + < α2 > − < α >2 −2(< αD > − < α >< D >) (17) As before, to compute the average of a term X over all classifications, we can first determine its average < X >x for a fixed x, and then use the function F defined by: F (Y ) = k N N x=0 x x k N N x=0 x x Y (18) and < X >= F (< X >x ). A crucial step in computing the exact value of the variance of x the AUC is to determine the value of the terms of the type < a2 >x =< ( i=1 αi )2 >x . 3 An essential difference between Eq. (14) and the expression given by [15] is the weighting by the number of configurations. The authors’ analysis leads them to the conclusion that the average AUC is identical to the accuracy for all ratios n/(n + m), which is false. 4 We thank Neil Sloane for having pointed us to Zeilberger’s algorithm and Maple package. x Lemma 3 For a fixed x, the average of ( i=1 αi )2 is given by: x(N + 1) < a2 > x = (3N x + 2x + N ) 12 (19) Proof. By definition of a, < a2 >x = b + 2c with: x x α2 >x i b =< c =< αi αj >x (20) 1≤i

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