Soft-margin SVM can address linearly separable problems with outliers

Transcription

1 Non-linear Support Vector Machines Non-linearly separable problems Hard-margin SVM can address linearly separable problems Soft-margin SVM can address linearly separable problems with outliers Non-linearly separable problems need a higher expressive power (i.e. more complex feature combinations) We do not want to loose the advantages of linear separators (i.e. large margin, theoretical guarantees) Solution Map input examples in a higher dimensional feature space Perform linear classification in this higher dimensional space Non-linear Support Vector Machines feature map Φ : X H Φ is a function mapping each example to a higher dimensional space H Examples x are replaced with their feature mapping Φ(x) The feature mapping should increase the expressive power of the representation (e.g. introducing features which are combinations of input features) Examples should be (approximately) linearly separable in the mapped space Feature map Homogeneous (d = 2) Inhomogeneous (d = 2) Polynomial mapping ( ) x1 Φ = x 2 ( ) x1 Φ = x 2 x 2 1 x 1 x 2 x 2 2 x 1 x 2 x 2 1 x 1 x 2 x 2 2 Maps features to all possible conjunctions (i.e. products) of features: 1. of a certain degree d (homogeneous mapping) 2. up to a certain degree (inhomogeneous mapping)

2 Feature map Φ Non-linear Support Vector Machines f(x) Linear separation in feature space SVM algorithm is applied just replacing x with Φ(x): f(x) = w T Φ(x) + w 0 A linear separation (i.e. hyperplane) ( in ) feature space corresponds to a non-linear separation in input space, e.g.: x1 f = sgn(w 1 x w 2 x 1 x 2 + w 3 x w 0 ) x 2 Rationale Retain combination of regularization a data fitting Regularization means smoothness (i.e. smaller weights, lower complexity) of the learned function Use a sparsifying loss to have few support vector 2

3 ɛ-insensitive loss l(f(x), y) = y f(x) ɛ = { 0 if y f(x) ɛ y f(x) ɛ otherwise Tolerate small (ɛ) deviations from the true value (i.e. no penalty) Defines an ɛ-tube of insensitiveness around true values This also allows to trade off function complexity with data fitting (playing onn ɛ value) Optimization problem Note 1 min w X,w 0 IR, ξ,ξ IR m 2 w 2 + C subject to (ξ i + ξi ) w T Φ(x i ) + w 0 y i ɛ + ξ i y i (w T Φ(x i ) + w 0 ) ɛ + ξ i ξ i, ξ i 0 Two constraints for each example for the upper and lower sides of the tube Slack variables ξ i, ξ i penalize predictions out of the ɛ-insensitive tube Lagrangian We include constraints in the minimization function using Lagrange multipliers (α i, α i, β i, βi 0): L = 1 2 w 2 + C (ξ i + ξi ) (β i ξ i + βi ξi ) α i (ɛ + ξ i + y i w T Φ(x i ) w 0 ) αi (ɛ + ξi y i + w T Φ(x i ) + w 0 ) 3

4 Vanishing the derivatives wrt the primal variables we obtain: m w = w (αi α i )Φ(x i ) = 0 w = (αi α i )Φ(x i ) w 0 = (α i αi ) = 0 ξ i = C α i β i = 0 α i [0, C] = C αi βi = 0 αi [0, C] ξ i Substituting in the Lagrangian we get: (αi α i )(αj α j )Φ(x i ) T Φ(x j ) } {{ } w 2 ξ i (C β i α i ) + =0 ξ i (C β i α i ) =0 ɛ (α i + αi ) + y i (αi α i ) + w 0 (αi α i )(αj α j )Φ(x i ) T Φ(x j ) m (α i α i ) } {{ } =0 max α IR m 1 2 subject to (αi α i )(αj α j )Φ(x i ) T Φ(x j ) ɛ (αi + α i ) + y i (αi α i ) (α i αi ) = 0 α i, α i [0, C] i [1, m] 4

5 Regression function f(x) = w T Φ(x) + w 0 = (α i αi )Φ(x i ) T Φ(x) + w 0 Karush-Khun-Tucker conditions (KKT) At the saddle point it holds that for all i: α i (ɛ + ξ i + y i w T Φ(x i ) w 0 ) = 0 αi (ɛ + ξi y i + w T Φ(x i ) + w 0 ) = 0 (C α i )ξ i = 0 (C αi )ξi = 0 Support Vectors All patterns within the ɛ-tube, for which f(x i ) y i < ɛ, have α i, αi estimated function f. = 0 and thus don t contribute to the Patterns for which either 0 < α i < C or 0 < α i < C are on the border of the ɛ-tube, that is f(x i) y i = ɛ. They are the unbound support vectors. The remaining training patterns are margin errors (either ξ i > 0 or ξi > 0), and reside out of the ɛ-insensitive region. They are bound support vectors, with corresponding α i = C or αi = C. Support Vectors 5

6 : example for decreasing ɛ Rationale Characterize a set of examples defining boundaries enclosing them Find smallest hypersphere in feature space enclosing data points Account for outliers paying a cost for leaving examples out of the sphere Usage One-class classification: model a class when no negative examples exist Anomaly/novelty detection: detect test data falling outside of the sphere and return them as novel/anomalous (e.g. intrusion detection systems, Alzheimer s patients monitoring) 6

7 Optimization problem min R IR,o H,ξ IR m R 2 + C ξ i subject to Φ(x i ) o 2 R 2 + ξ i i = 1,..., m ξ i 0, i = 1,..., m Note o is the center of the sphere R is the radius which is minimized slack variables ξ i gather costs for outliers Lagrangian (α i, β i 0) L = R 2 + C ξ i α i (R 2 + ξ i Φ(x i ) o 2 ) β i ξ i Vanishing the derivatives wrt primal variables m R = 2R(1 α i ) = 0 α i = 1 m o = 2 α i (Φ(x i ) o)( 1) = 0 o ξ i = C α i β i = 0 α i [0, C] α i =1 = α i Φ(x i ) R 2 (1 α i ) + ξ i (C α i β i ) =0 =0 + α i (Φ(x i ) α j Φ(x j )) T (Φ(x i ) α h Φ(x h )) j=1 o h=1 o 7

8 α i (Φ(x i ) α j Φ(x j )) T (Φ(x i ) α h Φ(x h )) = j=1 = α i Φ(x i ) T Φ(x i ) α i Φ(x i ) T m α i j=1 α j Φ(x j ) T Φ(x i ) + h=1 m h=1 α i j=1 =1 = α i Φ(x i ) T Φ(x i ) α i Φ(x i ) T α h Φ(x h ) α j Φ(x j ) T m j=1 α j Φ(x j ) m h=1 α h Φ(x h ) = Distance function max α IR m subject to α i Φ(x i ) T Φ(x i ) α i α j Φ(x i ) T Φ(x j ) α i = 1, 0 α i C, i = 1,..., m. The distance of a point from the origin is: R 2 (x) = Φ(x) o 2 = (Φ(x) α i Φ(x i )) T (Φ(x) α j Φ(x j )) = Φ(x) T Φ(x) 2 j=1 α i Φ(x i ) T Φ(x) + α i α j Φ(x i ) T Φ(x j ) Karush-Khun-Tucker conditions (KKT) At the saddle point it holds that for all i: β i ξ i = 0 α i (R 2 + ξ i Φ(x i ) o 2 ) = 0 Support vectors Unbound support vectors (0 < α i < C), whose images lie on the surface of the enclosing sphere. Bound support vectors (α i = C), whose images lie outside of the enclosing sphere, which correspond to outliers. All other points (α = 0) with images inside the enclosing sphere. 8

9 Decision function The radius R of the enclosing sphere can be computed using the distance function on any unbound support vector x: R 2 (x) = Φ(x) T Φ(x) 2 α i Φ(x i ) T Φ(x) + α i α j Φ(x i ) T Φ(x j ) A decision function for novelty detection could be: f(x) = sgn ( R 2 (x) (R ) 2) i.e. positive if the examples lays outside of the sphere and negative otherwise Support Vector Ranking Rationale Order examples by relevance (e.g. urgence, movie rating) Learn scoring function predicting quality of example Constraint function to score x i higher than x j if it is more relevant (pairwise comparisons for training) Easily formalized as a support vector classification task 9

10 Support Vector Ranking Optimization problem min w X,w 0 IR, ξ i,j IR 1 2 w 2 + C i,j ξ i,j subject to w T Φ(x i ) w T Φ(x j ) 1 ξ i,j ξ i,j 0 i, j : x i x j Note There is one constraint for each pair of examples having ordering information (x i x j means the former is comes first in the ranking) Examples should be correctly ordered with a large margin Support Vector Ranking Support vector classification on pairs min w X,w 0 IR, ξ i,j IR subject to 1 2 w 2 + C i,j ξ i,j y i,j w T (Φ(x i ) Φ(x j )) 1 ξ i,j Φ(x ij) ξ i,j 0 i, j : x i x j where labels are always positive y i,j = 1 Support Vector Ranking Decision function f(x) = w T Φ(x) Standard support vector classification function (unbiased) Represents score of example for ranking it 10