Active Bipartite Ranking
Authors: James Cheshire
Vincent Laurent
Stephan Clémençon
Venue: NeurIPS 2023 poster (2023)
Keywords: bipartite ranking
multi armed bandits
active learning
TL;DR: In this paper, we develop an active learning framework for the bipartite ranking problem.

Abstract: In this paper, we develop an active learning framework for the bipartite ranking problem.
Motivated by numerous applications, ranging from supervised anomaly detection to credit-scoring through the design of medical diagnosis support systems, and usually formulated as the problem of optimizing (a scalar summary of) the ROC curve, bipartite ranking has been the subject of much attention in the passive context. Various dedicated algorithms have been recently proposed and studied by the machine-learning community. In contrast, active bipartite ranking rule is poorly documented in the literature. Due to its global nature, a strategy for labeling sequentially data points that are difficult to rank w.r.t. to the others is required. This learning task is much more complex than binary classification, for which many active algorithms have been designed. It is the goal of this article to provide a rigorous formulation of such a selective sampling approach. We propose a dedicated algorithm, referred to as active-rank, which aims to minimise the distance between the ROC curve of the ranking function built and the optimal one, w.r.t. the sup norm. We show that, for a fixed confidence level $\epsilon$ and probability $\delta$,  active-rank is PAC$(\epsilon,\delta)$. In addition, we provide a problem dependent upper bound on the expected sampling time of  active-rank and also demonstrate a problem dependent lower bound on the expected sampling time of any PAC$(\epsilon,\delta)$ algorithm. Beyond the theoretical analysis carried out, numerical results are presented, providing strong empirical evidence of the performance of the algorithm proposed, which compares favorably with more naive approaches.

Active Bipartite Ranking
James Cheshire
Stephan Clemencon
Telecom ParisTech
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Vincent Laurent
ENS Paris Saclay
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Abstract
In this paper, we develop active learning framework for the bipartite ranking problem. Motivated by numerous applications, ranging from supervised anomaly detection credit-scoring through the design of medical diagnosis support systems, and usually formulated as the problem of optimizing (a scalar summary of) the ROC curve, bipartite ranking has been the subject of much attention in the passive context. Various dedicated algorithms have been recently proposed and studied by the machine-learning community. In contrast, active bipartite ranking rule is poorly documented in the literature. Due to its global nature, a strategy for labeling sequentially data points that are difficult rank w.r.t. to the others is required. This learning task is much more complex than binary classification, for which many active algorithms have been designed. It is the goal of this article provide a rigorous formulation of such selective sampling approach. We propose dedicated algorithm, referred distance between the ROC curve of the ranking function built and the optimal one, w.r.t. the sup norm. We show that, for a fixed confidence level e and probability 8, active-ranke is PAC(E, 8). In addition, we provide problem dependent upper bound on expected sampling time of |active-ranhe and also demonstrate a problem dependent lower bound on the expected sampling time of any PAC(e, 8) algorithm. Beyond the theoretical analysis carried out, numerical results are presented, providing strong empirical evidence of the performance of the algorithm proposed, which compares favorably with more naive approaches.
lactive-rande which aims to minimise the
Introduction
In bipartite ranking, the statistical framework is exactly the same as that in standard binary classifi- cation, the flagship problem in statistical learning theory. One observes
independent copies D,, = {(X1, Y1). (X,, Yx), } of generic random pair (X, Y) with (unknown) distribution P, where is binary random label, valued in {-1, +1} say, and X is a high dimensional random vector, taking its values in X Rd with d > 1, that models some information hopefully useful to predict Y. In contrast to binary classification, the goal pursued is of global (and not local) nature. not assign a label, positive negative, any of (temporarily unlabeled) observations X ...., X,, by means of (measurable) scoring function X - R, that those with positive label appear on top of the list (i.e. those with the highest scores) with high probability. More formally, the accuracy of any scoring rule can be evaluated through the ROC curve criterion its popular scalar summary, the AUC (standing for the Area Un- der the ROC Curve), and, as expected, optimal scoring functions w.r.t. these performance measures can be shown to be increasing transforms of the posterior probability nTx)) = IP +1 X x).
input observation X but to rank any set
37th Conference on Neural Information Processing Systems (NcurIPS 2023).

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I E X. Though easy to formulate, this problem encompasses many applications, ranging from credit risk screening to the design of decision support tools for medical diagnosis through (supervised) anomaly detection. Hence, motivated by a wide variety of applications, bipartite ranking has received much attention these last few years. Many approaches to this global learning problem (i.e. the problem of learning a preorder on the input space X' based on a binary feedback) have been proposed and investigated. In Clémençon and Vayatis (2009), optimization of the (empirical) ROC curve in sup norm is considered via nonlinear approximation techniques, while this functional optimization problem is viewed as a superposition of cost-sensitive binary classification problems in Clémençon and Vayatis (2008) so as propose alternative method. In Cléémençon et al. (2008) (see also Agarwal et al. (2005)), ) empirical maximization of the empirical AUC criterion is considered, bipartite ranking being viewed as a pairwise classification problem, plug-ti in approach to bipartite ranking is developed in Clémençon and Robbiano (201 1) and scalar performance criteria other than the AUC have been recently reviewed in Menon and Williamsson (016). Whereas the vast majority of dedicated articles consider the batch situation solely, where the learning procedure fully relies on a set Dr of training examples given in advance, the goal of this paper is to develop an active learning framework for bipartite ranking, in other words to investigate this problem in an iterative context, where the learning procedure can formulate queries in a sequential manner, so as to observe the labels at new data points in order to refine progressively the scoring/ranking model. Precisely, the challenge consists in determining an incremental experimental design to label the data points in X' that would permit to improve the ROC curve progressively, with statistical guarantees.
Our contributions We describe an algorithm, active-rank which sequentially queries points of the feature space X. Given a confidence level E > O and probability 6 >> O, the goal of lactive-rank is, in few queries possible, to output a ranking of X, such that the induced ROC curve of said ranking is within of the optimal ROC curve, in terms of the sup norm, with probability greater than 6- We restrict our selves to dimension 1, i.c. X' [0, 1] and make a single key assumption that the posterior 1) is piecewise constant on a grid of size K. Theorem 3.2 then shows that active-rani satisfies the above statistical guarantee and furthermore, provides an upper bound it's expected total number of queries. In Theorem 3.3 we provide a lower bound on the expected number of queries for any possible policy, which satisfies confidence level with probability greater than 1 - ō. The aforementioned bounds are problem dependent, in the sense that they depend on features of the posterior v). Finally we conduct practical analysis of active-ranki on synthetic data, comparing it to several naive approaches.
The article is structured as follows. In Section 2 we formally define our setting as well recalling some key notions related to bipartite ranking and ROC analysis. In Section 2 we also cover some of the existing literature in active learning, of relevance to bipartite ranking. We then describe the lactive-ratt in Section 3.2 Lastly, the experiments are displayed in Section 4 We then conclude and discuss some perspectives for future research in Section Technical details and proofs are deferred to the Supplementary Material.
algorithm in Section 3.1. Following this our theoretical results are presented
2 Background and preliminaries
2.1 Notation
Here we introduce several dedicated notions that will be extensively used in the subsequent analysis. For any integer n > 1, we set [n]] := {1, .... n). denote by 6, the symmetric group of permutations on {1, ..., 12,, by I,, the identity map of G,, By is meant the Lebesgue measure on [0, 1]. Given two probability distributions P and Q on measurable space (2, F), we write P & Q when P is absolutely continuous w.r.t. Q. For any a, b in J0, 1). Ber(a) refers to the Bernoulli distribution with mean a and kl(a, b) to the Kullback Leibler divergence between the Bernoulli distributions Ber(a) and Ber(b). For any a, b E 0,, 1). the Chernoff Information between the distributions Ber(a) and Ber(b) is defined as, kl" (a. b) = kl(x", a) kl(x", b). where x² is the unique x E [0, 1] such that kl(x, a) = (a b). The indicator function of any event E is denoted by I(8), the Dirac mass at any point by dg and the incuro-inverse of any cdf k(u) on R by K (t) (v E R: <(v) 2 t).

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2.2 Setting
The bipartite ranking problem A rigorous formulation of bipartite ranking involves functional performance measures. Let S be the set of all scoring functions, any S defines preorder is on X:: for all (x,z²) E X2, x 1s x' @ s(x) I s(x'). From a quantitative perspective. the
accuracy of any scoring rule can be evaluated through the ROC curve criterion, namely the PP- plot E R + (1 - H.)(), 1 - (!),,, where H. = (((((X) < Y + =-1 and G.(t) P(x(X) St t Y = +1}, for all t E R. The curve can also be viewed the graph of the cåd-låg function a E (0. + (OC(s = G2 o (1 - xx. The notion of ROC curve defines a partial order on the set of all scoring functions (respectively, the set of all preorders more accurate than 52 when ROC(s2, a) & ROC(s),aa a) for all a E (0, 1). As can be proved by straightforward Jeyman-Pearson argument, the set S* of optimal scoring functions is composed of increasing transforms of the posterior probability n)))) = P[Y -y=a +1 X = æ). E X. We have S {s : V(z. x') E X2, n12) < m(xx) s (zx (z')) and
X): 81 is
(ss S 8*, Ya E 1), ROC(s, a: I (OC ROC(S", o.)
The ranking performance of a candidate S can be thus measured by the distance in sup- sup-norm between its ROC curve and ROC", namely (s, == ((.)) (ROC* (a) - ROC(ss })} An alternative convention to represent the ROC of a scoring function 8, which we will use for the remainder of this paper. is to consider the broken line ROC(s..). .). which arises from connecting the PP-plot by line segments at each possible jump of the edf H From here on out when referring to the ROC of a scoring function 8, we refer to the broken line ROC(s,.).
The active learning setting Whereas in the batch mode, the construction of a nearly optimal scoring function (i.e. a function S such that deo s*) is 'small' with high probability) is based on a collection of independent training examples given in advance, the objective of an active learner is to formulate queries in order to recover sequentially the optimal preorder defined by the supposedly unknown function 1). That is, the active learner plays a game with multiple time steps, where, at time each step n. they must choose point a, E X' to query, so to observe the random label Y, ~ (n)(a), ) and refine the scoring model incrementally. After a sufficient number of rounds has elapsed, chosen at the learner's discretion, final scoring function 8, is output.
the feature space X
Piecewise constant scoring functions. Here we consider the simplest scoring functions, measurable functions that are constant pieces of the input space X' forming a partition. As shown in ClCemmnnno and Vayatis| (2009) (see subsection 2.3 therein), when smooth enough, ROC can be accurately approximated by the (stepwise) ROC curve of piecewise constant scoring function. Because the goal of this paper is to highlight the nature of active bipartite ranking rather than treating the problem in full generality, various simplifying assumptions
made in the subsequent analysis. For simplicity we suppose that X = [0. 1) and introduce the grid points [G1..., Gx } = {i/K i= =, .... K 1}. where K > 1. A preorder on X can be then naturally defined by means of permutation o E GR. Consider indeed the scoring function
K
S (z)
i 11xx E [Go(i)) Go(s41))}-
1=1
We denote by Sx the set of all functions of type 1 To avoid dealing with model bias here, we assume that the optimal preorder, that induced by ()x)) namely, can be defined by a scoring function in Sk.
Assumption 2.1. There exist a permutation o E Gx and distinct constants μis .... Px in (0, such that
K
()xx) = Σ ps He E (Go()) Go(s41))) for all x E (0,
We write k Eie[k] Hi. We point out that, as 7) may remain constant over multiple sections of the grid, the permutation satisfying assumption 2.1 is not necessarily unique. In the subsequent analysis, the parameter K is supposed to be known, in contrast with the μu; 's, which have to be learned by means of an active strategy. As we assume no structure between the grid points, one can easily consider Assumption 2.1 in higher dimensions, simply consider the d dimensional grid of size K As such, all results presented in this paper extend trivially to higher dimensions.
3

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Policies and fixed confidence regime. We denote the outputted scoring function of the learne E Sk- The way the learner interacts with the environment i.c. their choice of points query, how many samples to draw in total and their choice of & E Sk, we term the policy of the learner. We write C. for the set of all possible policies of the learner. For policy æ E C and problem v E B we denote random variable r as the stopping time of policy æ. We write 85 for the scoring function outputted by policy x on problem v. Where obvious we may drop the dependency on n, v in the notation, referring to the scoring function outputted by the learner as simply 3. We write F2,a as the distribution on all samples gathered by a policy т on problem v. We similarly define Ey,x.
For the duration of this paper we will work in the fixed confidence regime. For a confidence level E, define, Sk : {s Sk dx (s. n) S =}. A policy ι said be PAC (probably approximately correct), on the class of problems B, if, Av E B, Pu E Skl 2 1 - 8. The goal of the learner to then obtain PAC(Ó, policy n, such that the expected stopping time in the worst case, SuPveB E [r5], is minimised.
Defining problem complexity The expected minimum number of samples a policy must draw on certain problem, E B, to be PAC(5, E) is a quantity which depends upon the features of v, specifically, the shape of the posterior 17. When defining our measure of problem complexity we must capture this dependence as succinctly as possible. To build intuition for our definition we first explore a naive strategy and introduce some informative Lemmas. active bipartite ranking problem, is to treat each pair of points on the grid, i, E [K] as a separate classification problem. To correctly distinguish the situations, H² = μ4; M4j H3 14 M.j. with probability greater than 1 5, it is well known, see e.g. Kaufmann et al. 2014), that for small
naive approach to the
log[178] 6, the minimum number of samples required is of the order R[ where we remind the reader
kl the the Chernoff Information, closely related to the kl divergence, see Section 2.1 Thus, the learner wished to output scoring function in 5 the sample complexity would be of the order,
DiE[K] (ninge(k](kK (s)) Of course, distinguishing between Hid and Hi" is impractical when μ4; and My are very close, or even equal. However, in our regime, the learner is not required to correctly rank every pair of points j,j E [K]. only to output scoring function existing in Sj. Intuition indicates that the learner may be irreverent to the ranking within certain groups of points on the gird, as long as there posterior values are sufficiently close. For instance, consider a partition of [0, 1]. P {C1, C2, Cs), and increasing sequence (61, $2, 83) E [0, 1]s with
18g(1/8)
n(x) = = 1(x C.),, 5(x) 1(x 1 E Ci) + +11(x E (C2 (32))
(2)
Where the scoring function function essentially. treats all points of C2, C3 of the same rank. See Figure [ for the ROC curves of 1 and 8. Via simple calculation, we have the following, do (1, 8) A(Ca) 8/)2 which suggests that, whether or not there exists scoring function P 1=(25+P))/2*
in Sk. which treats the groups C2, C3 of the same rank, depends upon two things, the size of the groups and also their position on the ROC curve. Specifically, if B3- B2 2
2=x(1((Bt++bb)2)) then A(Ca)
& Sk. The following lemma formalises this intuition, the proof follows via a direct generalisation of the above example.
Lemma 2.2. Let A > 0, E [K] and define ((,A)) as the set of scoring functions such that, for all K one has that Vj |H, Hsl 2 A, sign(s(i) - s(])) = sign( 45)) There exist a B, such that on problem v, doo (1, 8) A10:1m-- p(1 p.)
8 E s(i.A)
5 E S s(A
Lemma 2.2 suggests that the for i E [K] the learner must be at least able to distinguish H:d vs H,
for all j: |ps / < Å, where, A == max x > 0: Eiz =T(( - 1وا S x) < Kep(1 - um))
The following Lemma shows that Å is not only upper bound on the necessary order of the confidence level around μı but also a lower bound, the proof of which can be found in the proof of Theorem[32
Lemma 2.3. For a problem v E B. let s E Sx be a scoring function such that the following holds, Vi E [ Vj |pg Hal 2 A./4, sign(s(i) s(i)) sign(p: 145)

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1.0
1.0
25
True positive rate
0.8
0.8
20
0.6
Complexity
0.4
0.2
ROC
0.2
ROC
0.0
0.00
0.25
0.50
0.75
1.00
0
2
3
False positive rate
Cell index
Figure 1: The ROC curves of and 8, defined in Equations (2), with (81, 882 3a) (0.1, 0.2, 0.6) and C1 C2 C3 equally sized.
Figure The complexity of points E [K]. for defined in Scenario of the experi- ments, see Section
then E Sj-
Under the conditions where Kep > and for grid point i E [K], 14 is close to one, a certain phenomenon occurs. Specifically, when Å; 2 1 jk, for the learner to know the value of μa up to an eroor of Å, is no longer sufficient, as in such a case, to the best of the learners knowledge, the value of μu; may be arbitrarily close to 1, and thus have arbitrarily large effect the regret. With this in mind, we define A; as follows,
145 xx) < Kep(1 - mi) A (1 ps)
A = max I
>
0:
Σ [
aT
Note that in the case where Ksp < we have A, = Å. In the case where bi 2 A,, we thus define the complexity at point E [K] as H.
klpp.pe A.)kll If μ4 S A,, H!" [(....:33.)) For a problem v E B, the total problem complexity is then given as Σie[k] H!!) As an illustrative example, Figure for the complexity of E [K] with 17 defined as in Scenario 1 of the experiments, see section
There is a natural comparison to multi armed bandits, where the problem complexity is typically given the summation across the individual complexity of each arm. However, in most multi armed settings the complexity of single arm is dependent upon its distance to single other arm, e.g. the optimal arm, whereas in our setting the complexity of a single grid point [K] has a more complex dependency on the shape of the posterior around G..
2.3 Performance of the passive approach
At this point we can consider how a uniform sampling strategy would perform, that is, the learner simply draws sample from each section of the grid in turn. This would be essentially a passive approach, in line with the classical batch setting. For a uniform/passive sampling strategy to be PAC(e, 6) one would have draw samples until the width of the confidence interval all points is less than A. Therefore, a uniform sampling strategy, with an appropriate stopping rule, would have the following tight upper bound on its expected sampling time, up log terms, cK | H(p.)" for some absolute constants c, > 0. The improvement we hope to gain in the active setting is to replace the max with weighted summation across the grid. Thus, in settings where the A; are relatively constant across large sections of the grid, then the theoretical performance of passive approach can be close to optimal. In contrast, in cases where very small section of the interval is hard to rank and the rest is easy i.e. K max k((us** is much greater than Σie[K] AA.J passive approach will fail. Such settings involve large K where the gaps A; on the majority of cells are large, with a relatively small number of cells with small gaps A. Incidentally, point out that this corresponds to many situations of interest in practice (in
5

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information retrieval, for specific request, the vast majority of the documents are equally irrelevant, while the ranking of a very small fraction of relevant documents is challenging; the same phenomenon is also observed in credit-risk screening). In these cases the benefit to the practitioner, will be that they quickly focus on the interesting sections of the feature space.
2.4 Related literature in active learning
While, to the best of our knowledge, Bipartite Ranking has not yet been considered under active learning, there are several related settings. Firstly, it is important to note, that we assume the posterior 7) is piecewise constant on the grid of size K, we can view our problems as a K armed bandit. In the case where K the bipartite ranking problem becomes akin best arm identification (BAI) for the two armed bandit, also known as A/B-Testing. In BAI for the two armed bandit, the learner sequentially draws samples from two distributions VA, VB with respective means J B- Their objective is then carry out the hypothesis test, Ho == (μΛ S (B). Hi == (Po >> p.)) in as few samples as possible. A/B Testing considered in an active, fix confidence regime, KKau'mann and Kalyanakrishnan (2013), wherein they prove a lower bound on the expected sampling time of any PAC(e, 6) as of the order log(1/8)
[[( Note that, in the case where K = 2 active-ravt matches said lower bound up logarithmic terms. In the fixed confidence regime, the BAI problem has also been generalised for larger K > 2 see Garivier and Kaufmann 2016)]Jamicss along with the TopM problem. where the learner must output the M best arms see Kaufmann et al.) (2014). Kalyanakrishnan et al. (2012), however, for K > 2 both BAI and TopM problems are no longer comparable our setting.
and Talwalkar (2016)
Aside from BAI and the TopM problem, there are several other settings in active learning that, while not directly comparable to own, are worth mentioning. The first is active clustering. Several works have considered clustering in an online framework, see CeeoCrommaas and Monteleoni (2012). Liberty et al. 2016). Cohen-Addad et al. (202T) and Khaleghi et al. (2012). In the above works new observations from certain arms become available to the learner at each time step, however the learner does not actively choose which arms to pull and therefore the flavour of the above literature is very different to our setting. Much closer is the work of Yang et al. (2022) in which the authors consider active clustering problem, represented as K armed d dimensional bandit, where the arms are split into M clusters. They work in a PAC(8) setup where their goal is to recover the entire clustering of the arms, with probability greater than 5 in as few samples as possible. Comparing to our setting, if one is to view a section of the grid on which is constant as single cluster, by retrieving the clustering of the arms one can then easily do ranking. Their results differ to our own in several key ways though. Firstly their algorithm takes the number of clusters M as a parameter, this highlights the main difference between their setting and our own. In the Bipartite ranking problem, assuming 8 is not very small, one does not have to recover exactly all the clusters to ensure regret under the do norm is less than E. Therefore we do not need to know the number of clusters and our algorithm must be able to exploit larger to achieve smaller stopping times. The second key difference is that the results of Yang et al. (2022) are hold only in the asymptotics, that is as - O. Their algorithm employs forced exploration phase, which ensures each arm is pulled at at least sub linear rate. Essentially, this means that in such an asymptotic setting, the means of the arms are known to the learner, which naturally drastically changes the nature of their results. Extension bounds for fixed 5 > 0 would be none trivial, noted as potential future work in Yang et al. (0222) and essential if one were compare to our confidence setting.
Also of note is active multi class classification. In Krishnamurthy et al. (017)) they consider a cost sensitive classification problem, where the learner receives input examples x E X and cost vectors c E K where c(y) is the cost of predicting label y on x. For each input example received the learner is able to query subset of labels. The objective is to then train classifier with minimal expected loss in as few queries as possible. The results of Krishnamurthy et al. (2017) cannot be directly compared to own, as in our setting there is no such thing the cost of classifier at given point x E X, as the cost miss ranking a section of [0, 1] is dependent on our ranking of the entire feature space.

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3 Our results
3.1 The active-rank Algorithm
Our algorithm active-rank maintains maintann active set of grid points across several rounds. At the beginning of each round jactive-ranke draws a sample, uniformly, from all points of the grid in the active and at the end of each round, eliminates points from the active criterion. We track the empirical mean of all samples drawn from the ith point of the grid, G.+1) up round as jt. At the beginning of each round for each grid point i [ we will maintain an upper and lower confidence bound, on μ4i which we term the UCB and LCB index respectively. At time I, for each grid point E [K] and exploration parameter ((1, : N [0, 1] R,, remaining in the active set, we then define the LCB index,
based on a specific
LCB((, == minn E [0,4]] kl(2),)) - 3(1,8)
(3)
t
and the UCB index,
UCB(t, i) == max q [At, kl 9) - B(t,8) t
(4)
Let S. denote the active set at the beginning round 4, via careful choice of exploration parameter the following Lemma holds, the proof of which can be found in Section A of the supplementary material.
Lemma 3.1. We have that, the event
n {uu E (LCB((,)) UCB(L,)))
BEN
occurs with probability greater than 1-
- Fol- For E [K], time and 2 E IR define, U2,((2) == {j +
lowing Lemma 2.2 our intuition would be to then remove a point i from the active set if, (UCB(i))
eKep(1 ( for some well chosen constant e 0. How- ever, due to the technical difficulty the proof, we make the following concession. For > 0, let S, be the list S' at time I. At time let Au) = maxies, (UCB(L, - LCB(L, i)). Furthermore, at time t define the set of grid points,
LCB(i, S (UCB(G,)--CCB....
1
KEP:
A1 A1)a- (1)(111))) mi)
Q
= i E A0) S 24 (T,)(GA
(5)
If a point exists in Qt, we remove it from the active set. Note that active ranee-rankeee does not take the average the posterior p as a parameter. Instead we show it is possible use an estimate P. which updates round by round.
Elimination algorithms such as active-rank have seen wide usage in the literature for BAI, see
In Paulson (964)/Mannor and T sitsikkis (2004). Even-Dar et al. (2002) and -aa et al. (2006).
However, closer work is the Racing algorithm Kaufmann and Kalyanakrishnan (2013). designed for the TopM problem, where, as in our approach, the confidence bounds used are based on the kl divergence as opposed to Hoeffdings. Their elimination criterion, however, differs considerably to our own. For simplicity let us consider the Top1 problem, that is BAI the following arguments can be extended in the case of TopM. The racing algorithm Kaufmann and Kalyanakrishnan (2013) eliminates an arm i E [K] from the active set, at time t, when, the positive gap the lower confidence bound around the highest empirical mean and the upper confidence bound point is greater than e. However, due to the global nature of the ranking problem, in our setting, the decision to remove point from the active set is not made based on the distance to another single point. We rather consider a condition on the local smoothness of the posterior around the point i. An additional difficulty that arises here is that the local smoothness around point can potentially depend upon points no longer in the active set and once point is no longer in the active set, we essentially have no control of the width of confidence interval.
3.2 Theoretical bounds
7

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Proving active-rank is PAC(E, 6) and up- per bounding the expected sampling time Theorem 3.2 demonstrates that our algorithm active-rank PAC(e, and provides problem dependent upper bound on it's ex- pected sampling time. The proof can be found in Section A of the supplementary ma- terial. Theorem 3.2 makes no assumption the posterior n, aside from it being piece- wise constant on the grid of size K, i.c. As-
Algorithm active rank
Initialise: S [K] repeat
for do
Sample point drawn uniformly from [G; Gi+1)
end for
Let pr be a point drawn uniformly from [0, 1]
and update P:- = ((t 1) + Pc)/t
sumption 2.1 For E [K], set 11,"
for E S do
if Q. (n) S 0t/4 then
maxje(K] ((155 V kllpesps
A./8)
S S {+} end if end for = 1++1 until S =
Theorem 3.2. For e, 0. 7,7> 480/ log( K)) with ((t, 6) = c. log(to (2t((=///) where c, constant depending only on 7 all problems E B, on execution of active-rank with output 8, we have that,
Let 6; E GK be the permutation sorting the list
(K])εTK] into ascending order.
d (3, E
Output: 8 Eiek i. il1 E [Ga(s), Ca(+1))))
with probability greater than 1-ô. Furthermore, the expected stopping time of aattaee is upper bounded by the following,
H log (H K1)).
KE[K]
where dnp dy are constants depending only on 7
Lower bound Theorem 3.3 provides problem dependent lower bound on the expected sam- pling time of any PAC policy. The proof Theorem 3.3 can be found in Section of the supplementary material.
Theorem 3.3. Let 1/4), < < 1 exp(-1/8) and v E B such that Kep < 1/8. For any PAC(e, 5) policy æ, there exists a problem D E B such that, for all E [K] A 2 A./2 where Å, the gap of the ith grid point on problem D, where the expected stopping time of policy on problem D bounded as follows,
Ep +
Σ
(1)
KE
where e' > is an absolute constant and H.") () is the complexity of point on problem D. Furthermore we have that,
Σ
A Z Σ H
*E[R]
xE[K]
where e is an absolute constant.
The proof of Theorem 3.3 follows from novel application of Fano type inequality on a well chosen
set of problems.
Gap between upper and lower bound There are essentially two components in the gap between the bounds of Theorems, 3.3 and 3.2 The first is the additional logarithmic dependency upon K present in our upper bound. Despite being logarithmic this dependence is potentially significant, as in practical situations, the size of grid needed, for the assumption that the posterior v) is piecewise constant, may be very large. The second component, in the gap between upper and lower bounds is the
difference in the H("1 and H (2) terms. The reason H. (2) appears in Theorem 3.2 is that, the decision to remove a point
[K] from our active set made based on the minimum width of confidence interval across the entire grid, A(r) opposed to the local width at UCB(i, - Li, (), see Equation (5). As we are dealing with Bernoulli distributions and divergence based confidence bounds, for fixed number of samples, points close to one will have tighter confidence bounds

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increasing family (14)) Our study scenarios then as and thus may be sampled more than is necessary. If one were to assume that the posterior exists solely in the interval (n, 1 - > for some 7 > then for all E [K], 11,1) and H, H!)") will be with constant factor of each other, with that constant depending on 7.
In authors opinion, both the logarithmic dependency K and usage of (n)) may be removed. However, this would require several non trivial modifications to the proof of Theorem [.2 see the discussion in Section of the supplementary material for details. As, to the best of our knowledge, this is the first work to consider the bipartite ranking problem in an active learning setting, we present Theorem 3.2 as it stands and leave the aforementioned improvements for future works.
Experiments
In this section we discuss practical cases based on synthetic data. For all experiments 8 is fixed at 0.01 and the constants used are smaller than their theoretical counterparts, which are typically overestimated, furthermore is calculated with all previous samples. As represented in Figure each cell is assigned a level value μi, so that follows the Assumption [2 Without loss of generality, we assume that be described
follows:
Scenario I: (14)ST(K) 0.,. 0.38) and K 16;
Scenario 2: ([4)sE[K] (0.8((i - 1)/K) + 0.1) and K 64;
Scenario 3: (μt))E]K] is sub-sampled (with replacement) of ((i ( (E()0))) and
K = 64
Scenario 4: μ4 = 0.8((1 1)/K) + O.1VV E [K]M7788 with μ47 = 0.8(6/K) + 0.3, Ms 0.8(7/K) 0.1 and K 16
The objective these scenarios is evaluate the capacity of the algorithm on different cases. As shown in Figure scenarios 1 and 3 will have variable jumps and cell sizes.
Competing algorithms To our knowledge there is no algorithm dealing with the active learning for bipartite ranking problem, thus compare to the following naive approaches. Passive rank: each new point a, is drawn uniformly on [0, 1]. Naive rank: each new point a, is sampled in P, s.t. = arg maX UCB(t, It, i), this algorithm reduces the bias in undifferentiated way without considering the problem global (requiring peer-to-peer comparison). Active classification: this algorithm we consider binary classification with threshold 0.5. The set of active cells S, as defined in Algorithm is then S, {i [K] 0.5 [ LCB(t, i); At UCB(t, h)]} U
i; a:agminn(0.... (11) As the competing algorithms do not output stopping time, for a single <E[K]
77 algorithm active-ranki is run across several values of E, following a geometric sequence of common ratio 0.99 and initial value 1. The values of then plotted against the respective stopping
times of aactiverank
0.6
0.2
Scenario
Scenario
Scenario
Scenario
Figure 3: Different scenarios chosen for the experiments
Interpretation of results On scenario 1, the simplest case, and to an extent scenario 2, active-ranki and passive suffer near identical regret for larger sample sizes. On all other see- narios active-rank outperforms all competitors. The fact that active-rank does not reach extremely small values of regret suggests that analysis for much higher sample size may be interest- ing, however this creates issues in computation time, the same goes for larger K. Also, the uniform

---

naive-active-rank
passive-rank
active-classfficttion
active-rank
1.0
0.9
0.8
0.5
0.4
1000
250
500
Figure 4: Regret estimated by Monte Carlo for 100 realizations of each algorithm, corresponding to the scenarios of the figure respectively.
approach still performs relatively well, and appears difficult to fool. Some more work may be needed to find setting in which uniform sampling suffers considerably.
5
Conclusion
To the best of our knowledge we have developed the first rigorous framework for the active bipartite ranking problem and our algorithm, lactive-rank] is the first to tackle said problem. Our upper bound on performance of factive-ranks matches lower bound up to logarithmic terms, in the case where the posterior is not very close to 0 or I at any point. As well as theoretical guarantees have demonstrated good practical performance of lative-rank settings. We conclude with some perspectives for future research.
synthetic data, in various
An obvious path for future research, is to replace the Assumption 2. with smoothness assumption on the posterior 77, e.g. Hölder condition. The setting would then be equivalent continuous armed bandit as opposed to finite armed bandit. Assuming the learner has knowledge of the Hölder coefficient, a standard approach in continuous armed bandits to first discretise and then apply classic techniques from finite armed bandits, carefully choosing the discretisation level to balance the discretisation error and classical regret. It is of our opinion that such an approach would not be sufficient in must vary the level discretisation across the feature space, based on the flatness of the posterior function 1) and placement the ROC curve. Also, under such Hölder condition, the extension to higher dimensions would no longer be immediate.
We conjecture that to achieve optimal or optimal performance the learner
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A Proof of theorem 3.2
Before the proof of Theorem 3.2 we prove some initial lemmas, the first of which is Lemma 3.1 which lower bounds the probability of our favourable event & by 1 - 8. We prove slightly extended version of Lemma 3.I At time t, define LCB index of as,
LCB(t, 0) = min {
((t,8)
g E [0, Pel kl 9) < 1
(6)
and the UCB index,
UCB(1, = max E (Pts 1] kl(fr. S 3(t,8)
(7)
Note that as Pr E [min(()) max(p))), we have that,
UCB(0, t) LCB(O, 1) (1)
The extended version of Lemma 3.1 is then as follows,
Lemma A.1. We have that, the event
& = n n {ax UCB(,,
LeNic(S,)u(()
occurs with probability greater than 8.
Proof. Via Chernoff's inequality, for i [K] U (0), at time t we have that,
(I,.CB(t, k) Z m4) S exp(-8(t,
and,
P(UCB(t, S mb) - exp(-3(t,
then remains to note that via our choice of exploration parameter and a union bound,
2 Σ
exp(-3(t,6)) 8
AEN kE(K]U(()
and the result follows.
The following Lemma shows that, on rounds which point is removed from the active set, our estimate P remains within constant factor of the true p.
Proposition A.2. On event E we have that, for all rounds t such that a point is removed from the active set, 2p/3 S P. S 4p/3.
Proof. Let be round in which a point is removed from the active set. On event E, for all e [K],
1ls A - A()
we thus have Ip 1p-2 øl < A()) which, in combination with the fact ( S $/4, implies A(o) S (5+ + /3)3 and thus A(o) S p/3.
For the purposes of the proof split Theorem 3.2 into parts. The first part is covered by the following lemma, which states that active rank is PAC(s, 5).
Lemma A.3. For E, 8 > 0, 1/K 2 ep on all problems v B, on execution of active-rant with output we have that,
doo Se,
with probability greater than 1 -. 8.
12

---

Proof. For the remainder of this proof we will work under the following event,
= n n {or E (L) UCB(L, (CC(,
LEN KE[K]
Proposition A.4. Let i E [K] be removed from St time t, such that {j : 10 pyl A.) c St. For all j |p Hyl S A we have that, vk E St 1 Hyl 2 Aj,
sign( - [1x) = sign (13g - pk)
Proof. Firstly note that, for all s 2 t, (s) S (m) Thus for all j E St, 12 Myl $ ( and therefore for any point j (a I Aj. we have that, for all h E Se : l - pul S Aj.
sign( ( = sign(H - pk)
Thus to prove our result we must show that, for all j I1; Usl S Aj. Dj S 45(0). Now, if is removed from S, at time I, we have that,
( S 24
1
KeP:
(1)
(1-M)) m,)
(b)()
thus,
(4) S 1 (1- m
1
(8)
24
and therefore via Proposition A.2
(6, - Kepi(1 - 7
Kep(1 - (m))
Kep(1 - ps)
S
24
12A (1)
6 (1)
and we have that (1))) S Dj. Now let j: M4j Mil S 4A(10, (0), we then have, [i; pul - 2A(0)
(0) and
thus (o)) c (6)) and furthermore A()) 122 (1 T2)) therefore, via proposition A.2
(4A)) < KEPi
(3)
Kep(1 - 7
(ep(1-11)
240(0) (1)
120 (1)
(1
and result A S (A())- Now take j |kg Msl (1) If D I 4D(r) done. Thus
assume A; 2 (n) In this case, we have that |p4 M₁l 2 Ay. Thus we have shown that, for
point j E St, if [44 /ق] < A,, then ((1) I A and the result follows.
Proposition A.5. At time t, let i E So, we have that, Vj E [K]] \ S : l Mal Ai.
sign(A - M.) = sign(p; - p:)
and furthemoree,, Vj (k: muy 1 & Sc. Wk E [K] 1 Mbl I Dj,
sig - M = sign( - pk)
Proof. The proof will follow by induction. Assume time t that Vi S we have that, Vj [K] \ S 1p4- M1 2 A,
sign(A - M. = sign( - p4)
and that, Vj E [K] {k: Iky m E Si-1, Vk 1ike Mn S 4
sign(A? - ( = sign (ng - pk)
Now let i St, and j E [K] |m; Uأ} 2 Ag. we must show that,
sign( m ) sign(M} mg)
and that, Vj [K] (k: |4g m & S.. Vi la Myl Aj.
sign(Az - m sign( ph)
13

---

For the first statement, assume j E St-1 and that { Iag -mel C St-1, as otherwise we are done, via the inductive assumption. The point is then removed from the active set the end of round ++ 1 and the statement thus follows via Proposition A.4
For the second statement, for j E St-1 assume (k: : mn m&l} C St-1, otherwise are done via the inductive assumption. Therefore, there must exist such an k. removed from the active set at the end of round t 1, such that lus- Myl I Aj. The result then follows via Proposition A.4
For set C C [0, 1]. define,
x(C) X(C) / n(x) dr.
1
Let E [0, 1), define the subset Za C [0, such that, PP(X E ZaIY -1) = a that is,
X(Za) (1 x(Za)) =a
1-p p
and such that, for some ia E [K],
:j: Pi. Pj c Za Vj: < Mi, pjЛ =
(9)
We then have ROC (a) P(X E Z, +1) MZ,)(((2)) necessarily unique, and 7) me be constant across multiple sections of the grid & is also not necessarily unique, in this case we take arbitrary Zo. ix. Now define the subset 2. [0, 1] such that,
The choice of Zr is not
P
X x(Za) (1 x(Za)) =a 1-p
and,
Ve E Za, y & 2n, sp(a) 2 8p(1),
so ROC(s, = (n(23
Again 2, is not necessarily unique, in which case we choose have that, Vj [K] = ( 14, Ai..
arbitrarily. Via Propositions
and A.5
and
Lgg (T.)) sign(u} - pt..)
(10)
Let
= Za (n(x) Mial 5 Aiah,
E = to E 20 : :n((x) Mial S Aia)
Via Equation 10 we have that,
ROC(a, 77) - ROC(O, [p)((,sp)= = \(Z.)((Zz))
((2))((2))
p
Before finalising the proof we must lower bound A x(2x) and x² x(2)). We first lower bound R( (2))
k x(2.) 2 Min Din
x(Za), S Pia + A.a
(11)
We will now lower bound x(2))
x(2%)
1 1- k
< Hi. + Din
(12)
X(Z() x2(2))
Min Ain
Via combinations of Equations
II and 12). we have,
((Za)s(Z'a)
(222)(22))
< X(Z))(H + +...) - X(%) - -A..))
(13)
p
p
1 (XIZ)AAA. + +A..) X(Z2)
(Mis - Di, )1 Mi. + A.. ))
<
p
1 Pia Ai
(14)
2X(2%)
2X(Z.) in
(15)
p(1 Mis A.
p(1 pi.))
14

---

It remains to remark that, A S
x(2)) by definition, and thus,
Ep
R 7) ROC(a, sp) S E.
As we chose w.l.o.g the proof then follows.
A.1 Proof of stopping time for lactiveraan
We will now prove the second part of Theorem 3.2 the upper bound on the expected sampling time of lackive-rank
Lemma A.6. For 0, > 480/ log(K) K), 1/K 2 pe, with 3(t, 6) = c. log (og((222/00 where e, a constant depending only on 7, on all problems B such that Vi [K], A; < Mix we have that, for >> 1, the expected stopping time of active-rank] is upper bounded by the following,
E SE[K]
HH. log K216)
where Eyp Cit are constants depending only on
Proof. We upper bound E|r] as follows. We denote the number of times a section of the grid [Gt; G(41) has been sampled by the learner, up and including time as, NN(t). Let 7, (T()).
B B
P(T 2 t)
(16)
P(i & 2.)
(17)
Define the event,
EA, == (fj E (UCB(tt. & My + A,/96] U {Vj E S,, 1g - 4,/9))
Upp S p Ac)
Proposition A.7. For i E [K] such that A S 2p, t > 0, we have that, on Ei, E Q-
Proof.
First note that, under event Ei,s (() f A and thus for j U., we have,
|p= Mal < 9 : + 6 A..
thus,
(,,,(6A (1)) < {jj |p 1 my A.
(18)
and furthermore, (c) & p/4 which implies p < 2p. Now, via definition of A.,
Kpe
A1)a 14)
A <
(19)
[í: /14 My < A:
In the case, [[:lu By
Kpe
2 1,
SA.
(1) +
A (1) S
< (1 R.)/48
96
and we have that
Kpe
Kpe
3K
S A()))
[i: 1/4 - 141 2 A:
UT.((A (1;))
(6A())
thus
Kp=
2
(U,
15

---

finally leading to,
A (1) I Kepr(1 (1))
(65
Then let us assume
Kp
S 1. In this case.
Rail E
KEp(1 14:)
KEP((1- ( (1. + (1)) (8)
K - (
(n) 2 4./96 - U (6)(
(65
where the third inequality comes from the fact that (0) S p/4 which implies p - 2p.
In what follows, assume is such that A S 2p. Via combination Proposition A.7 and Equation we have that,
E M
P((..)
We will now upper bound Ei= P(C,,)) For > let,
TV min 6(t,6)
<kl(uss M4 + A,/96) kl ../96)) --/000)) log(K)r
))
min sE[K]
^ log(K)7
We have that, for all > T j E S..
P(UCB(t,)) 2 + 9.696) S P kl(p)) R5 + A./96) S kl(Hg Hy + -,/96)
(20)
log( (og(K))
Let,
= fx (1g, MJ + A,/96) kl( M3 + /96)) = kl My + (log6)
Consider the function ((x) = kl(jy + v. by A, on the interval [0, /96) Via the properties
of the KI. divergence, Φ is convex and Φi /96) 0. As result, o(x) (1 - x) kl(p) A,/96)/(2A,, which implies,
4
(()) 2 My
A
2 Hy
96
log(K)r
192
for log( (K)? 384. We now have, for > Tjj
kl(2)). p4 + A:/96) ki(p5, + As/96)
PPkk((j)) My + A./96) s kl(r()), + +a/96))
log(K)7
PPA r(7))
S exp(
+ kl( r(+)))
log(K)) ((()))((77))
exp
(tt,
kl(p), H1 + 0, 966)
kl(u, M1 + 192
S exp
3(1, 5) kl(145, Hy + A,/96)
exp(
(k),)
where C, is constant depending only on y. Thus, via Equation (0)). for 2 T
P(UCB(+,)) 2 My A.22) < (tt via similar reasoning we have also that,
P(LCB(t.)) SP, 2./2) S exp( (og))
and furthermore, P(e, & [p A + (o)) < exp(- 8)), and now via union bound,
16

---

P(C);)) S cexp(-cy).
(21)
t= +1
where is absolute constant. Thus we have shown that for point i: A, S 2p,
[| S Tu ck exp(-cy)
is then straight forward to note that for any point A; 2 2p. B[m < To + ck exp(-ca),
where 1' any point i' such that Ay s 2p. It remains to upper bound Tj Via definition of the exploration parameter, we have that,
Tá $ H () log (2HH
where ? > O is an absolute constant depending only on Before continuing, we demonstrate the following Lemma-
Lemma A.8. On all problems E B, we have that,
(i A; < 2p}l 2 K/4.
Proof. Firstly assume that,
21 K/4 and define i max A; pJ. We then have that,
Vj A 2 Mí. 11- 1p-- Myl S A.,,
and thus,
Kεp
4 [ 1 - Myl S A
S sp.
Thus have Hi A p}}
K/4. Now assume,
H( A, 2 n3l K/4.
As we must have {i: : μ4 2p} & K/2, we thus have, Hi As < 2pp12 K/4
With Lemma A.8
can then upper bound [[[]] follows,
E(r] 4
B(m] S 4 Σ H.") log(e, H²² K/8) + d K (exp(-c),
KE[K]
and the result follows.
B Proof of Theorem 3.3
Proof. The proof will follow by application of Fano type inequality on well chosen set of problems.
Step 1: Constructing our well chosen set of problems We define a set of grid points Uo, U1,.. recursively as follows, Un arg min( Ai), for 2 0 then define
Un+1 arg min ( A Vj S m, Iuw, Hil 34, + 3D
Let M be the largest integer for which UM exists. Note that the sequence (Aum)mms is monotoni- cally increasing and furthermore, for all E [K].
Im E [M] [3pu.. - 3/4/ 2 AU.. + 2
(22)
We then define the corresponding set of groups, Do. D1,.., DM follows. For i E [K] let m.
[M] be such that, Mu... S μ4 HV. then set i+ m V n and mAn. If i1 pi1 S A, + Act then i E D.+, otherwise i D;-.
17

---

Proposition B.1. For all m E [M] we have that, Vi E Dm, Aun S As.
Proof. W.l.o.g assume μ4 2 Pu.. Let j be such that lpu, - 14l I 33; + 3Au, Via Equation 22 have that pe,,, I μr S no, and by definition of Dm, Dj. m > J. Thus 1j m l Mo, 3 + 3A; and therefore, if A; < Au... then
arg min ( ( Vj S m 1, luu, - 14] 2 33m + +32,} + Un which is a contradiction via the definition of Um.
For m [M]] set Wm = fi: 114 - Huml I I J. }. Proposition B.2. For all m [M].
Hi E Dm 10s Mum! 2 33uum 3 36|Wml
Proof. Firstly, by definition we have that,
A KEp(1- - pUm)
(23)
[Wml
Now, let 2" = arg max (.n. (Ar). As we have that Vi E Dm, 1a4 - Mu 1 34, + 3A the following holds,
Ei Dm
Wm.
mi 1p: -1-- 3A,- 3Au
Now consider the set of adjacent grid points I. C Dm \ Wm such that X X(UiEL Gi+1)) s Ar. for j E L.
Ksp(1 - -K5)
(24)
[T.]
Ksp((1 M4 ) + +34, - -3Avm)
(25)
[
Kep(1 - HlVmm 6
+
(26)
[Z]
[L
Where the final inequality comes from the fact we assume Kep < 1. Thus if we assume [[]
2
Wm. we have that, A; < Kep(1-Hem) contradiction via proposition B.1 and Equation 23
2/Ca]
The proof of the following proposition follows via the same argument as the proof of Proposition
B.2
Proposition B.3, For all m [M].
Wml s 3 (i Wm [p, mum! s Aum
We are now ready to construct our set of problems. Consider a family of problems 2& indexed by Q (0, 1)* where the target function ?2Q corresponding to 22 is defined as follows. Set the coefficients,
(3 3 mm)mE[M)
and for m E [M] set 8Thh = Mu + 4/3Aumm.
32 MU - 4/3AU,,, also define the sets
m,
U
GG Gi+1)
m,
U
[G Gi+1).
iEWm:Qu=l
iEWm:Q.:=11
and C.. = UsEW.. (Gi, Git1) we then have
(0(x)= = Σ
1(x E Cº 18 + 1(x E 202
ME[M]
18

---

Let Q' be the transformation of Finally let C8 = [0, 1 m. 1. :m m E [M]} and note that for all E Co n(z) = 0. The 2
following Lemma shows that, for a problem E B, the gaps and complexity across our family problems, uº indexed by Q, does not differ too much from v.
Lemma B.4. Let Q {0, for all m [M], [Gi, G1+1) E Cm. we have that AU Anm/4 <
A
Q
3Au.. where A (Q) is the gap of point on problem y
Proof. We have immediately that 3A S <3Aum.. definition of Au.. Then, by proposition B.3
A < Kep(1 - (U
KEp(1 - mmm)
Kep(1 - Mum.. "Auu///
4KEPDu
3|(; [4: P0
رركك
[Wml
mm
3 3
which implies, in combination with our assumption Kep 1/8 that,
Kep(1 M
"4Au /3)
Au /6
[m
and thus that for all i W m. A AU.//.
Lemma B.5. Given, B, we have that for all 1/, i [K], 4 2 A./2, where A A is the gap
of point on problem v Furthermore, Σ EH1 2 2 H where HiQ is the complexity of point i on problem v for some absolute constant 0.
Proof. We prove the first statement. For all i:
[G;, Gi+1) E Co we have that A 9 = and The second statement then follows by combination
otherwise the statement follows from Lemma B.5 of Lemma B.5 B.5 and Propositions B.2 and B.1
Step 2: showing that one suffers e regret on well chosen event Q that flips the ith coordinate,
(Qa If k Q: = Qo lf
We remind the reader that we denote the scoring function outputted by the learner S. Now define,
Am min (2) (Umml Cn U (1 x (ummm Cn U (R..)
(1-p)
define the event,
fim = { fx [Gi, G.11) :(x) > 2m3 1 2K
and then the events,
1{(mm) 5 3KA(Om)
KA(Cm))
ET
E
Σ
((
4
(1)
For set C C [0, 1]. define.
<(C) X(C) ( n(x) dr
1
Let 2m [0.1] be the largest such that, Vx 2m, y $ 2m, ((x)(())))) s(y), and,
m-1
Cn U. ) (1-N
m-1
Cn U Cm..1 ))
x(2m)(1 k(2m)) S
U
U n=1
n=1
19

---

Note that 2m is not necessarily unique, in this case we choose an arbitrary such 2m. Furthermore define,
2m
E 2m : Cn
2m
{ar 2m x E Cm..).
n=l
and
U cnl. n=m+1
M
27 =
x E 2m : C
And let G...
-Un=1 Cn 20 Note that under event E1", we have
A(Z))) S K/4
(27)
Now, via definition of 2m, we have the following.
X( (AN,
(28)
Thus, for problem 2 EijG,, Qi KXCC)//2, event ET", via combination of Equations
27 and
(1
x (m,,2)) & 3)(Cm.)) (0mmmm1100mmmm/--mmm0000 M( -00(-K(Gm) n((mm)
(29)
1) (((mmm)))))))))
(30)
where the final inequality comes from the fact x <(Cm.1) 2 x(Gm). To complete Step: 2 we now lower bound (s, m) on event Enn. Firstly note that,
XUN On UCm.) (Um=T Cn U.
ROC (1- (1-2m))(2m)
111 ())
1 1 -
XUNNi Cn (n) Ca) X(Cm)))((Cm.)
P therefore, for problem v2; Exrieo Qi KM(m)/22 on event ETM
)(Gm) ((Cm.1)((Cm..).
x(Cm,1)A(2m)
2(Cmm.2)M(2
doo (8,
P
P
P
p
AGm))
3ACmm.)N((Cma))
x(Cm.))A(2a)
D
4p
AGm))R(Gm)
10.4 X(Gm)) 1 - K< <(Cm,,)
P
4p
p
1 K x(Om,))
3A(Cm..)
1 k(Cmm.))
(<(Cm,1) x(Cm,1)
4p
x(Cmm2)
3A(Cm,,)
(n(Cm,1)
-k((mm.)
4p
k((Cm,2)
1)(Cm,1)
8/3Dum
3X(Cm,1 U Cm..2)
8/3AU
4p
x((Cm,))
Sp
x (m,
where the first inequality follows from Equation (27 and the second from (29p.
Thus, as we assume policy π is PAC(E, on all problems v we must have that, on all problems R EeGGe Qi KA(Cm)//2
Poo (ETh) 5
(31)
Via similar reasoning we can show that on all problems v²; Eijch Q S KX(Cm)/2
we must have that,
P,o = (E 5
(32)
20

---

Step 4: bounding the probability of the sum of 5m"h Now, for m E [M via the Azuma hoeffding inequality applied the martingale,
Σ
(Sr.)) PQ(Ci,m)).
we have that,
[1(5.m) Po(Et.m)] KA(Cm) log (118) )
PQ
Σ
1
- 11-8. (33)
G,1 Q.=0
Thus via combination of Equations (32 and K2(Cm)/2,
33
we must have that, VQ:
Σx
Q: -
Σ DECC
Po(Sim) S M(m))
K + K log
11-3)
)
3
(34)
1 1-5
8
where the second inequality comes from our assumption 8 < - 1e1-exp(-1/11 also have that, VQ: EiNG.ann Qa 2 KXCCm)/2
Via similar reasoning
Σ 1mm
Po(Sim) 2 \(mm)
K
K log
5KX(Cm)
(35)
2
1
8
Step 5: applying a Fano type inequality apply Fano.
We first define the class of problems upon which we will
KA(C
Q
Q: Vm [M].
Σ
Q.
K)(Cm)
Do
Q:Vi Vm [M]
Q:
Σ
We remind the reader that for i E [K], we write T, = -t= 1(a E [G Gi+ ()) and for m E [M]. T(m) E. T,. We see that, for Q E 2o. Qi 0, there exists unique Q Q such that Q'' Q, therefore,
Σ QED mE[M] &G 1.=E0..
Σ
(Qr()im))
Σ Σ
Σ
Po(Eim)
QEDo mE[M]
&:G,, 1)=00.Q
and thus, using the data processing inequality and the convexity of the relative entropy we have,
kl
2
Σ
Po (E
Σ
Po(Em.)
K)(Cm)
K)(Cm)
QED mc[M]
i..EC.....-
QEn mE(M)
i...C.....II
26/8
210/8
Σ
2
EQ[Tm]
kl(wu,. 44/3Amm
Un. HUm +4/3Aum..
KX(Cm)
QED mE[M]
E.P.Cmm.Q==.
Eo| (()) kl(Hum. 4/30 Un. HUm. + 4/33V...) K)(Cm)
QED mE(M]
(o(T()) kl -4/3Au Uni + 44/3Dum)
Σ
max QED ME(M]
K)(Cm))
21

---

Then using the Pinsker inequality <l(a,)) 2 2(x y)? we obtain
E E
po(sim) S
3
max QEO
Σ
E kl(w 44/3Aum Un MUm + 4/3AUmm
1m
K)(Cm)
K)(Cm)
QE mE[M]
mc M
and therefore,
Ee Eo(Tm)) kl(pu, - mm MUm + Am)
Σ me [M]]
QED
K)(Cm))
thus
K)(Cm) Bo( 2 d' Σ kl(wu.. 4/3Au Um. pum 4/3Aum))
Σ
QEN
ME[M]
mE[M]
where è' > 0 is an absolute constant. The proof now follows, as Vm E [M], i: [Gi, C++1) C C..
2 Klpu 1 M )1/
0
22


Reviews:
[{"review_id": "dn1BjIRtkU", "reviewer": "Reviewer_iq8o", "summary": "- This paper studies sample complexity bounds for bipartite ranking under an active learning paradigm. The paper gives an algorithm called “active-rank” to solve the variant of the problem under the assumption that the posterior is a piecewise constant function over a grid of size K. The sample complexity bounds are problem-dependent, for which the authors define an instance-dependent notion of problem complexity where each arm/grid point’s global effect needs to be treated individually. \n\n- The paper also gives a lower bound (not matching with the upper bound), for the class of problem instances where the posterior differs only by a constant over the input range. The authors hope that the dependency of these bounds on $\\Delta$ and the logarithmic gap in the upper bound can be removed but leave it as a future work.", "questions": "- Please answer my question about the theoretical comparison of active vs. passive ranking for the 1D grid.\n\n- The explanation about Lemma 2.2 does not seem correct, even after ignoring the typos in Lines 158 and 159. Delta_i is an upper bound on how big a confidence interval you can have around $\\mu_i$. The inequality should be $|\\mu_i - \\mu_j| \\ge \\Delta_i$. Only for such $j \\neq i$, the scoring function has to get the sign correct. Is that what you meant?\n\n- Line 319 says, “active-rank all competitors”. What is missing? Active-rank outperforms all the competitors? That doesn’t seem true from Figure 3.\n\n- Is the x-axis in Figure 4 stopping time?", "limitations": "Limitations adequately addressed", "rating": 6, "confidence": 4, "soundness": 3, "presentation": 2, "contribution": 2, "strengths": "- Although not new, active learning is an important learning paradigm, and its application to bipartite ranking is a significant contribution.\n\n- The ideas in the paper are well-presented. The discussion about the global nature of the problem compared to other works where only pairwise comparisons, a local property, needs to be estimated with high error tolerance and probability helped access the novelty of this work.", "weaknesses": "- The results in the paper hold only for a carefully designed set of problem instances of bipartite ranking. However, the results contribute toward improving the understanding of active learning in bipartite ranking. So I wouldn’t consider this a weakness.\n\n- However, a theoretical comparison with passive ranking complexity is missing. Are there no existing works studying this? It is necessary to understand how the sample complexity compares between active and passive learning. What does uniform sampling give on the grid? Does active learning give any improvement over passive learning? The authors say before Assumption 2.1 that the assumption helps in understanding the power of active sampling over passive, but I did not see anywhere in the paper a theoretical comparison.\n\n- The presentation of the paper can be improved. The figures are hard to read; the axes are not labelled (Fig 4) and sometimes the legends are missing (Fig 2).", "quality": null, "clarity": null, "significance": null, "originality": null, "strengths_and_weaknesses": "", "was_revised": false, "final_justification": "", "rebuttal": ""}, {"review_id": "VyrHSwxA0b", "reviewer": "Reviewer_WRWn", "summary": "An active bipartite ranking algorithm is proposed in this article for one-dimensional data with a posterior probability $\\eta(x) = {\\rm P}(Y =+1\\vert X =x )$ piecewise constant on a grid of size $K$. An upper bound is provided on the number of queried to achieve a controlled error of type PAC$(\\epsilon,\\delta)$ on the ${\\rm sup}$ norm between the learned ROC curve and the optimal one. Compared to the lower bound of any sampling policy also established in this paper, the upper bound suffers notably from a logarithmic dependence on the size $K$ of the grid. Experiments on synthetic data conforming to the aforementioned setting show a consistent superiority of the proposed algorithm over several baseline. However a really close match is observed between the uniform passive sampling and the proposed active method.", "questions": "* Does there exist an upper bound of the uniform passive sampling under the same setting or a similar one? Seeing the empirical results, I suspect that the passive sampling achieves already nearly optimal performance, possibly due to the fact that the posterior is piecewise constant on a uniform partition of the space and data points inside each interval are statistically equivalent.", "limitations": "The authors have discussed the extensions to smooth posteriors and to higher dimensions.", "rating": 6, "confidence": 3, "soundness": 3, "presentation": 3, "contribution": 2, "strengths": "* The studied problem, active bipartite ranking, is practically interesting and original.\n* The theoretical framework is clearly presented with a brief explanation of its relation to the best arm identification (BAI) problem.\n* Statistical guarantees are provided to support the proposed algorithm and to lay the ground for future investigation.\n* The proposed algorithm is tested on synthetic data.", "weaknesses": "* The data setting is quite restrictive in a one-dimensional space with the posterior probability piecewise constant on a grid of a known size $K$.\n* The passive sampling performs comparably to the proposed active algorithm under various scenarios conforming to the data setting underlying the proposed algorithm.\n* There is no testing on real data.", "quality": null, "clarity": null, "significance": null, "originality": null, "strengths_and_weaknesses": "", "was_revised": false, "final_justification": "", "rebuttal": ""}, {"review_id": "ZM2dXT8g3e", "reviewer": "Reviewer_RSb9", "summary": "This paper develops an active learning framework for the bipartite ranking problem. A selective sampling procedure or strategy is necessary for lebeling data points sequentially. The proposed method is called active-rank, which aims to minimize the distance between the ROC curve of the ranking function built and the optimal one, w.r.t., the sup norm. Theoretical results as well as some associated numerical results based on synthetic data demonstrate strong empirical evidence of performance.\n", "questions": "- Can you describe the pros and cons for practitioners, when working on real datasets where the scenario situation described in Section 4 is not clear?", "limitations": "no concern", "rating": 7, "confidence": 2, "soundness": 3, "presentation": 2, "contribution": 3, "strengths": "- The paper is well-written with clear notation.\n- Concrete theoretical results are provided and discussed for the proposed algorithm.\n- The performance of the proposed active-rank method is demonstrated by experiments that are in line with the theoretical considerations in the theory and active sampling procedure.", "weaknesses": "- Given the elimination nature of the proposed method, the experiment lacks behavioral observations of the proposed method in real dataset, that may be helpful for the practitioner's reference.", "quality": null, "clarity": null, "significance": null, "originality": null, "strengths_and_weaknesses": "", "was_revised": false, "final_justification": "", "rebuttal": ""}, {"review_id": "5sTKzpuzRz", "reviewer": "Reviewer_rRJf", "summary": "The paper proposes an active learning for the bipartite ranking problem in which the goal is to provide a ranking for a set of points rather than given positive or negative labels. The proposed algorithm called active-rank tries to optimize the ROC curve, evaluated by the distance to the optimal one. The authors provided detailed theoretical analysis that guarantees the distance can be arbitrary small with high probability. The evaluation is performed on synthetic data.", "questions": "- The authors assumes the piecewise constant model. Is it a reasonable assumption for the bipartite ranking setting?\n\n- K is assumed to be known. How is it selected in practice?\n\n- In section 4, scenario 1 is 'K = 16'. Is it 'K = 4'?", "limitations": "Some limitations are discussed in the conclusion section.", "rating": 4, "confidence": 1, "soundness": 3, "presentation": 2, "contribution": 2, "strengths": "- The paper appears to be technically sound, and the theoretical analysis is in depth.\n\n- According to the authors, the paper is the first rigorous framework for active bipartite ranking.", "weaknesses": "- The paper is difficult to follow for those who are not familiar with this topic.\n\n- The dimension of the input is restricted to 1, for which significance is not clear to me. The authors should have clarified it more. \n\n- The experiments are only for synthetic data. ", "quality": null, "clarity": null, "significance": null, "originality": null, "strengths_and_weaknesses": "", "was_revised": false, "final_justification": "", "rebuttal": ""}, {"review_id": "NcmNXm3Czo", "reviewer": "Reviewer_jBg3", "summary": "The paper proposes an elimination active learning criterion for the bipartite ranking problem. The criterion account for the local smoothness of a data sample’s posterior distribution. The author proves the sample complexity and the PAC learnability of the proposed algorithm under some assumptions. \n", "questions": "My questions are listed in the weakness section.", "limitations": "None", "rating": 7, "confidence": 2, "soundness": 3, "presentation": 3, "contribution": 3, "strengths": "1The proposed methodology is theoretically sound. The removal of a point in the active set corresponds to the learning objective, which is to minimize the gap between the current ROC and the optimal ROC. Beyond that, the PAC learning bound has been derived.", "weaknesses": "1The proposed method seems not to be practically sound. Most learning problems nowadays have high input dimensions. It is not clear whether the proposed algorithm can be easily extended to solve those real-world problems or if it only enjoys the nice properties for one dimension data.\n2It is not clear how the criterion will change after the removal of a data point. If the impact on the posterior around data point i is dependent on other points, is it still a fair criterion?\n", "quality": null, "clarity": null, "significance": null, "originality": null, "strengths_and_weaknesses": "", "was_revised": false, "final_justification": "", "rebuttal": ""}]

Author rebuttal:
We thank all reviewers for the time taken to read our paper and for their constructive feedback. Some topics come up for several reviewers. 

Performance of passive approach in comparison to our active algorithm: 

For a uniform/passive sampling strategy to be PAC$(\epsilon,\delta)$ one would have to draw samples until the width of the confidence interval at all points $i$ is less than $\Delta_i$. Therefore, a uniform sampling strategy, with an appropriate stopping rule, would have the following tight upper bound on its expected sampling time, up to log terms, $cK\max_{i\in [K]}\frac{1}{kl(\mu_i,\mu_i + c'\Delta_i)},$ for some absolute constants $c,c'>0$. Essentially, our improvement in the active setting is to replace the $\max$ with a weighted summation across the grid. Thus, in settings where the $\Delta_i$ are relatively constant across large sections of the grid then the theoretical performance of a passive approach can be close to optimal. In contrast, in cases where a very small section of the interval is hard to rank and the rest is easy - i.e. $K\max_{i\in [K]}\frac{1}{kl(\mu_i,\mu_i + \Delta_i)}$ is much greater than $\sum_{i\in [K]}\frac{1}{kl(\mu_i,\mu_i + \Delta_i)},$ a passive approach will fail. Such settings involve large $K$  where the gaps $\Delta_i$ on the majority of cells are large, with a relatively small number of cells with small gaps $\Delta_i$. Incidentally, we point out that this corresponds to many situations of interest in practice (in information retrieval, for a specific request, the vast majority of the documents are equally irrelevant, while the ranking of a very small fraction of relevant documents is challenging; the same phenomenon is also observed in credit-risk screening). In these cases the benefit to the practitioner will then be that they quickly focus on the interesting sections of the feature space. For such a setting one needs very small tolerance $\epsilon$ as well as large $K$. This results in computational difficulties for experiments and as noted in the paper, some work is required to reveal a setting where the active rank show a large improvement over a uniform approach. We agree with the reviewers that comparison to a passive approach is lacking, a final version of the paper would include additional discussion in line with the above. 

Extension to higher dimensions: 

Our analysis extends immediately to higher dimensions. Indeed one may consider a function piece wise constant on a $d$ dimensional grid of size $K$ and as we can always project a $d$ dimensional grid onto the $[0,1]$ interval, our results remain unchanged. Indeed, for better illustration, our experiments are carried out in dimension 2. A more interesting question is to include a sparsity assumption with increasing dimension $d$, as this arises in many practical situations. This would fundamentally change the nature of the problem and is beyond the scope of this paper. A discussion will be added in the final version for the sake of clarity.

The piece wise constant assumption and assumed knowledge of $K$:

For active learning in general such an assumption is very reasonable, first and foremost as it is essential for the majority of the literature in finite multi armed bandits. Indeed our Bipartite ranking problem can be viewed as a finite multi armed bandit where the objective of the learner is to rank the $K$ arms. Other pure exploration problems, such as best arm identification, have seen extensive interest, in the case of fixed known $K$. We believe ranking the arms is as natural a question as finding the best arm and the potential for practical application is no less. 

In practice one can replace knowledge of $K$ with an upper bound, although this will then appear in the bounds. Selecting $K$ is a tricky question, ideally one would have expert knowledge, however, in the absence of this it may be better to remove the piece wise constant assumption and view the problem under a continuity assumption, e.g. $\beta$-Hölder smoothness constraint. This is akin to the extension of the finite multi armed bandit to the infinite armed bandit. Now the question becomes, how does one proceed without knowledge of $\beta$? Ideas from the bandit literature may be applied to our setting, e.g. Carpentier, Valko, 2015, where $\beta$ is estimated. This question goes beyond the scope of our paper and will also be especially challenging in higher dimensions. 

Reviewer 5:

Indeed, due to the interconnected nature of the problem the decision to remove a point may depend on points already removed from the active set, a problem not found in multi armed bandits, see the discussion at the end of section 3.1. For this reason a point is removed from the active set only when the number of points appearing within a certain distance of it is sufficiently small. This distance is $\Delta_{(t)}$, multiplied by some large constant. Thus any point $j$ not yet clearly distinguished from the removed point also has the guarantee that $\Delta_j \leq c\Delta_{(t)}$, for some constant $c$, thus ensuring that our criterion is consistent. See the proof of Lemma A.3 for details.

Reviewer 1:

Line 159 is a typo, it should read "for all $j: |\mu_i - \mu_k| \geq \Delta_i$". The x-axis in figure 4 is indeed stopping time, this will be clarified. Aside from scenario one, for larger stopping times, i.e. greater than 500 active-rank outperforms the competitors, although by an admittedly small margin in some cases. The typo will be corrected to say something to this effect. 

Reviewer 4:

The definition of scenario 1 was wrong, the correct version is,

Scenario 1: $\mu_i = 0$ when $i\leq 11$, $\mu_{12}=\mu_{13} = 0.23$, $\mu_{14}=\mu_{15} = 0.33$, $\mu_{16}=0.35$    and $K = 16$.

This scenario aims at testing the algorithm when a significant section of the feature space is constant.

Meta-review: The paper proposes Active rank, a new ranking algorithm under the Active learning framework, for bi-partite ranking. It provides rigorous bounds on sample complexity for this result. Overall this should be of interest to ranking community and hence to Neurips audience.
Decision: Accept (poster)