A Learning Theory of Ranking Aggregation

Transcription

1 A Learning Theory of Ranking Aggregation France/Japan Machine Learning Workshop Anna Korba, Stephan Clémençon, Eric Sibony November 14, 2017 Télécom ParisTech

2 Outline 1. The Ranking Aggregation Problem 2. Statistical Framework 3. Results 1

3 The Ranking Aggregation Problem

4 The Ranking Aggregation Problem Framework n items: {1,..., n}. N rankings on the n items (from most preferred to last): i 1 i 2 i n. i 1 i n permutation σ on {1,..., n} s.t. σ(i j ) = j. 2

5 The Ranking Aggregation Problem Framework n items: {1,..., n}. N rankings on the n items (from most preferred to last): i 1 i 2 i n. i 1 i n permutation σ on {1,..., n} s.t. σ(i j ) = j. 2

6 The Ranking Aggregation Problem Framework n items: {1,..., n}. N rankings on the n items (from most preferred to last): i 1 i 2 i n. i 1 i n permutation σ on {1,..., n} s.t. σ(i j ) = j. Suppose we have a dataset of rankings/permutations (σ 1,..., σ N ) S N n. 2

7 The Ranking Aggregation Problem Framework n items: {1,..., n}. N rankings on the n items (from most preferred to last): i 1 i 2 i n. i 1 i n permutation σ on {1,..., n} s.t. σ(i j ) = j. Suppose we have a dataset of rankings/permutations (σ 1,..., σ N ) S N n. Consensus Ranking We want to find a global order ( consensus ) σ on the n items that best represents the dataset. 2

8 Methods for Ranking Aggregation Copeland Rule. Sort the items according to their Copeland score, defined for each item i by: s C (i) = 1 N n I[σ t (i) < σ t (j)] N t=1 j=1 j i 3

9 Methods for Ranking Aggregation Copeland Rule. Sort the items according to their Copeland score, defined for each item i by: s C (i) = 1 N n I[σ t (i) < σ t (j)] N Kemeny s rule (1959). Find the solution of : t=1 min σ Sn j=1 j i N i=1 where d is the Kendall s tau distance: d τ (σ, σ ) = i<j d(σ, σ t ) (1) I{(σ(i) σ(j))(σ (i) σ (j)) < 0}, Kemeny s consensus has a lot of interesting properties, but it is NP-hard to compute, even for n = 4 (see Dwork et al., 2001). 3

10 Statistical Framework

11 Statistical Reformulation Suppose the dataset is composed of N i.i.d. copies Σ 1,..., Σ N of a r.v. Σ P. A (Kemeny) median of P w.r.t. d is solution of: min E Σ P [d(σ, σ)], σ S n where L(σ) = E Σ P [d(σ, σ)] is the risk of σ. 4

12 Statistical Reformulation Suppose the dataset is composed of N i.i.d. copies Σ 1,..., Σ N of a r.v. Σ P. A (Kemeny) median of P w.r.t. d is solution of: min E Σ P [d(σ, σ)], σ S n where L(σ) = E Σ P [d(σ, σ)] is the risk of σ. Let L N (σ) = 1 N N t=1 d(σ t, σ). 4

13 Statistical Reformulation Suppose the dataset is composed of N i.i.d. copies Σ 1,..., Σ N of a r.v. Σ P. A (Kemeny) median of P w.r.t. d is solution of: min E Σ P [d(σ, σ)], σ S n where L(σ) = E Σ P [d(σ, σ)] is the risk of σ. Let L N (σ) = 1 N N t=1 d(σ t, σ). Goal of our analysis: Study the performance of Kemeny empirical medians, i.e. solutions σ N of: min σ S n LN (σ), through the excess of risk L( σ N ) L. We establish links with Copeland method. 4

14 Risk of Ranking Aggregation The risk of a median σ is L(σ) = E Σ P [d(σ, σ)], where d is the Kendall s tau distance: d(σ, σ ) = {(σ(i) σ(j))(σ (i) σ (j)) < 0} {i,j} n Let p i,j = P[Σ(i) < Σ(j)] the probability that item i is preferred to item j. 5

15 Risk of Ranking Aggregation The risk of a median σ is L(σ) = E Σ P [d(σ, σ)], where d is the Kendall s tau distance: d(σ, σ ) = {(σ(i) σ(j))(σ (i) σ (j)) < 0} {i,j} n Let p i,j = P[Σ(i) < Σ(j)] the probability that item i is preferred to item j. The risk can be rewritten: L(σ) = i<j p i,j I{σ(i) > σ(j)} + i<j (1 p i,j )I{σ(i) < σ(j)}. 5

16 Risk of Ranking Aggregation The risk of a median σ is L(σ) = E Σ P [d(σ, σ)], where d is the Kendall s tau distance: d(σ, σ ) = {(σ(i) σ(j))(σ (i) σ (j)) < 0} {i,j} n Let p i,j = P[Σ(i) < Σ(j)] the probability that item i is preferred to item j. The risk can be rewritten: L(σ) = i<j p i,j I{σ(i) > σ(j)} + i<j (1 p i,j )I{σ(i) < σ(j)}. So if there exists a permutation σ verifying: i < j s.t. p i,j 1/2, it would be necessary of median for P. (σ(j) σ(i)) (p i,j 1/2) > 0, (2) 5

17 Preference cycles p 1,2 > 1/2 1 p 3,1 > 1/2 2 3 p 2,3 > 1/2 6

18 Preference cycles p 1,2 > 1/2 1 p 3,1 > 1/2 2 3 p 2,3 > 1/2 No permutation can satisfy this condition for each pair of items! 6

19 Preference cycles p 1,2 > 1/2 1 p 3,1 > 1/2 2 3 p 2,3 > 1/2 No permutation can satisfy this condition for each pair of items! Definition P on S n is stochastically transitive if : (i, j, k) n 3, p i,j 1/2 and p j,k 1/2 p i,k 1/2. Moreover, if p i,j 1/2 for all i < j, P is strictly stochastically transitive. 6

20 Results

21 Optimality Theorem If P is stochastically transitive, there exists σ S n verifying: i < j s.t. p i,j 1/2, is verified. (σ(j) σ(i)) (p i,j 1/2) > 0, The Copeland score of an item i, that is: s (i) = 1 + k i I{p i,k < 1 2 } defines a permutation s S n and is the unique median of P iff P is strictly stochastically transitive. 7

22 Optimality Theorem If P is stochastically transitive, there exists σ S n verifying: i < j s.t. p i,j 1/2, is verified. (σ(j) σ(i)) (p i,j 1/2) > 0, The Copeland score of an item i, that is: s (i) = 1 + k i I{p i,k < 1 2 } defines a permutation s S n and is the unique median of P iff P is strictly stochastically transitive. 7

23 Universal Rates The excess risk of σ N is upper bounded: (i) In expectation by E [L( σ N ) L n(n 1) ] 2 N (ii) With probability higher than 1 δ for any δ (0, 1) by L( σ N ) L n(n 1) 2 log(n(n 1)/δ). 2 N 8

24 Conditions for Fast Rates Suppose that P verifies: the Stochastic Transitivity condition: p i,j 1/2 and p j,k 1/2 p i,k 1/2. the Low-Noise condition NA(h) for some h > 0: min p i,j 1/2 h. i<j 9

25 Conditions for Fast Rates Suppose that P verifies: the Stochastic Transitivity condition: p i,j 1/2 and p j,k 1/2 p i,k 1/2. the Low-Noise condition NA(h) for some h > 0: min p i,j 1/2 h. i<j Introduced for binary classification (see Koltchinskii and Beznosova, 2005). Used for estimation of matrix of pairwise probabilities (see Shah et al., 2016). 9

26 Fast rates Let α h = 1 2 log ( 1/(1 4h 2 ) ). Assume that P satisfies the previous conditions. (i) For any empirical Kemeny median σ N, we have: E [L( σ N ) L ] n2 (n 1) 2 e αhn. 8 (ii) With probability at least 1 (n(n 1)/4)e αhn, the empirical Copeland score ŝ N (i) = 1 + k i I{ p i,k < 1 2 } for 1 i n belongs to S n and is the unique solution of Kemeny empirical minimization. In practice: under the needed conditions, Copeland method (O(N ( n 2) )) outputs the Kemeny consensus (NP-hard) with high prob. 10

27 Conclusion and future directions We introduced a general statistical framework for ranking aggregation and established rates of convergence. 11

28 Conclusion and future directions We introduced a general statistical framework for ranking aggregation and established rates of convergence. We can write the empirical risk when one observe pairwise comparisons only instead of full rankings 11

29 Conclusion and future directions We introduced a general statistical framework for ranking aggregation and established rates of convergence. We can write the empirical risk when one observe pairwise comparisons only instead of full rankings Conditions for fast rates have good chance to be satisfied in a homogeneous population so Copeland method could be useful for local methods 11

30 Conclusion and future directions We introduced a general statistical framework for ranking aggregation and established rates of convergence. We can write the empirical risk when one observe pairwise comparisons only instead of full rankings Conditions for fast rates have good chance to be satisfied in a homogeneous population so Copeland method could be useful for local methods 11