Kernel Methods and Support Vector Machines

Transcription

1 Kernel Methods and Support Vector Machines Oliver Schulte - CMPT 726 Bishop PRML Ch. 6

2 Support Vector Machines Defining Characteristics Like logistic regression, good for continuous input features, discrete target variable. Like nearest neighbor, a kernel method: classification is based on weighted similar instances. The kernel defines similarity measure. Sparsity: Tries to find a few important instances, the support vectors. Intuition: Netflix recommendation system.

3 SVMs: Pros and Cons Pros Very good classification performance, basically unbeatable. Fast and scaleable learning. Pretty fast inference. Cons No model is built, therefore black-box. Still need to specify kernel function (like specifying basis functions). Issues with multiple classes, can use probabilistic version. (Relevance Vector Machine).

4 Two Views of SVMs Theoretical View: linear separator SVM looks for linear separator but in new feature space. Uses a new criterion to choose a line separating classes: max-margin. User View: kernel-based classification User specifies a kernel function. SVM learns weights for instances. Classification is performed by taking average of the labels of other instances, weighted by a) similarity b) instance weight. Nice demo on web

5 Two Views of SVMs Theoretical View: linear separator SVM looks for linear separator but in new feature space. Uses a new criterion to choose a line separating classes: max-margin. User View: kernel-based classification User specifies a kernel function. SVM learns weights for instances. Classification is performed by taking average of the labels of other instances, weighted by a) similarity b) instance weight. Nice demo on web

6 Two Views of SVMs Theoretical View: linear separator SVM looks for linear separator but in new feature space. Uses a new criterion to choose a line separating classes: max-margin. User View: kernel-based classification User specifies a kernel function. SVM learns weights for instances. Classification is performed by taking average of the labels of other instances, weighted by a) similarity b) instance weight. Nice demo on web

7 Two Views of SVMs Theoretical View: linear separator SVM looks for linear separator but in new feature space. Uses a new criterion to choose a line separating classes: max-margin. User View: kernel-based classification User specifies a kernel function. SVM learns weights for instances. Classification is performed by taking average of the labels of other instances, weighted by a) similarity b) instance weight. Nice demo on web

8 Outline Theoretical View User View Building Kernels Computing the Max-Margin Classifier Non-Separable Data

9 Linear Classification Revisited Consider a two class classification problem Use a linear model y(x) = w x + b followed by a threshold function For now, let s assume training data are linearly separable (possibly after mapping to higher-dimensional space). Recall that the perceptron would converge to a perfect classifier for such data But there are many such perfect classifiers

10 Max Margin Classifiers y = 1 y = 0 y = 1 margin We can define the margin of a classifier as the minimum distance to any example In support vector machines the decision boundary which maximizes the margin is chosen. Intuitively, this is the line right in the middle between the two classes.

11 Support Vectors y = 1 y = 0 y = 1 The support vectors are the points at minimum distance to the decision boundary. The max-margin boundary depends only on the support vectors: other data points do not matter for classification and need not be stored.

12 Outline Theoretical View User View Building Kernels Computing the Max-Margin Classifier Non-Separable Data

13 Example: X-OR X-OR problem: class of (x 1, x 2 ) is positive iff x 1 x 2 > 0. Not linearly separable in input space, but linearly separable if we add extra dimensions. Use 6 basis functions φ(x 1, x 2 ) = (1, 2x 1, 2x 2, x 2 1, 2x 1 x 2, x 2 2 ). Simple classifier y(x 1, x 2 ) = φ 5 (x 1, x 2 ) = 2x 1 x 2. Linear in basis function space.

14 Example: X-OR X-OR problem: class of (x 1, x 2 ) is positive iff x 1 x 2 > 0. Not linearly separable in input space, but linearly separable if we add extra dimensions. Use 6 basis functions φ(x 1, x 2 ) = (1, 2x 1, 2x 2, x 2 1, 2x 1 x 2, x 2 2 ). Simple classifier y(x 1, x 2 ) = φ 5 (x 1, x 2 ) = 2x 1 x 2. Linear in basis function space.

15 Linear Separability Example x x 1 Decision Boundary x x x 1 x x x D mapping (x 2 1, x2 2, 2x 1 x 2 ).

16 Linear Separability Example x x 1 Decision Boundary x x x 1 x x x D mapping (x 2 1, x2 2, 2x 1 x 2 ).

17 The Kernel Trick There can be many extra dimensions, even infinite (see assignment). Don t want to compute basis function mapping φ(x). Key insight 1: Linear classification requires only the dot product. Key insight 2: The high-dimensional dot product φ(x) φ(z) can often be computed as a kernel function of the input vectors only: φ(x) φ(z) = k(x, z).

18 Kernel Trick Example Consider again the X-OR 6 basis functions φ(x 1, x 2 ) = (1, 2x 1, 2x 2, x 2 1, 2x 1 x 2, x 2 2 ). Exercise: find a closed form expression for φ(x) φ(z) Solution: Dot product φ(x) φ(z) = (1 + x z) 2 = k(x, z). A quadratic kernel.

19 Kernel Trick Example Consider again the X-OR 6 basis functions φ(x 1, x 2 ) = (1, 2x 1, 2x 2, x 2 1, 2x 1 x 2, x 2 2 ). Exercise: find a closed form expression for φ(x) φ(z) Solution: Dot product φ(x) φ(z) = (1 + x z) 2 = k(x, z). A quadratic kernel.

20 The Kernel Classification Formula Suppose we have a kernel function k and N labelled instances with weights a n 0, n = 1,..., N. As with the perceptron, the target labels +1 are for positive class, -1 for negative class. Then y(x) = a n t n k(x, x n ) + b n=1 x is classified as positive if y(x) > 0, negative otherwise. If a n > 0, then x n is a support vector. Don t need to store other vectors. a will be sparse - many zeros.

21 Example1 SVM with Gaussian kernel Support vectors circled. They are the closest to the other class. Note non-linear decision boundary in x space

22 the number the number of degrees of degrees of freedom of freedom is higher, is higher, for thefor linearly the linearly separable separable case (left case panel), (left panel), the the Theoretical View User View Building Kernels Computing the Max-Margin Classifier Non-Separable Data solution solution is roughly is roughly linear, linear, indicating indicating that the thatcapacity the capacity is being is being controlled; controlled; and that and the that the linearly linearly non-separable non-separable case (right case (right panel) panel) has become has become separable. separable. Example2 From Burges, A Tutorial on Support Vector Machines for Pattern Recognition (1998) Figure 9. Figure Degree 9. Degree 3 polynomial 3 polynomial kernel. kernel. The background The background colour colour shows the shows shape the of shape the of decision the decision surface. surface. Finally, Finally, note that notealthough that although the SVM the classifiers SVM classifiers described described above above are binary are binary classifiers, classifiers, they they are easily are easily combined combined to handle to handle the multiclass the multiclass case. case. A simple, A simple, effective effective combination combination trains trains SVM trained using cubic polynomial kernel k(x 1, x 2 ) = (x 1 x 2 + 1) 3 Left is linearly separable Note decision boundary is almost linear, even using cubic polynomial kernel Right is not linearly separable But is separable using polynomial kernel

23 Learning the Instance Weights The max-margin classifier is found by solving the following problem: Maximize wrt a L(a) = a n 1 2 n=1 subject to the constraints a n 0, n = 1,..., N N n=1 a nt n = 0 n=1 m=1 a n a m t n t m k(x n, x m ) It is quadratic, with linear constraints, convex in a Optimal a can be found With large datasets, local search strategies employed let s check SVM demo

24 Learning the Instance Weights The max-margin classifier is found by solving the following problem: Maximize wrt a L(a) = a n 1 2 n=1 subject to the constraints a n 0, n = 1,..., N N n=1 a nt n = 0 n=1 m=1 a n a m t n t m k(x n, x m ) It is quadratic, with linear constraints, convex in a Optimal a can be found With large datasets, local search strategies employed let s check SVM demo

25 Outline Theoretical View User View Building Kernels Computing the Max-Margin Classifier Non-Separable Data

26 Valid Kernels Valid kernels: if k(, ) satisfies: Symmetric; k(x i, x j ) = k(x j, x i ) Positive definite; for any x 1,..., x N, the Gram matrix K must be positive semi-definite: K = k(x 1, x 1 ) k(x 1, x 2 )... k(x 1, x N ) k(x N, x 1 ) k(x N, x 2 )... k(x N, x N ) Positive semi-definite means x Kx 0 for all x (like metric) then k(, ) corresponds to a dot product in some space φ a.k.a. Mercer kernel, admissible kernel, reproducing kernel Theorem of Mercer s 1909!

27 Examples of Kernels Some kernels: Linear kernel k(x 1, x 2 ) = x 1 x 2 Polynomial kernel k(x 1, x 2 ) = (1 + x 1 x 2 ) d Contains all polynomial terms up to degree d Gaussian kernel k(x 1, x 2 ) = exp( x 1 x 2 2 /2σ 2 ) Infinite dimension feature space

28 Constructing Kernels Can build new valid kernels from existing valid ones: k(x 1, x 2 ) = ck 1 (x 1, x 2 ), c > 0 k(x 1, x 2 ) = k 1 (x 1, x 2 ) + k 2 (x 1, x 2 ) k(x 1, x 2 ) = k 1 (x 1, x 2 )k 2 (x 1, x 2 ) k(x 1, x 2 ) = exp(k 1 (x 1, x 2 )) Table on p. 296 gives many such rules

29 More Kernels Stationary kernels are only a function of the difference between arguments: k(x 1, x 2 ) = k(x 1 x 2 ) Translation invariant in input space: k(x 1, x 2 ) = k(x 1 + c, x 2 + c) Homogeneous kernels, a. k. a. radial basis functions only a function of magnitude of difference: k(x 1, x 2 ) = k( x 1 x 2 ) Set subsets k(a 1, A 2 ) = 2 A 1 A 2, where A denotes number of elements in A Domain-specific: think hard about your problem, figure out what it means to be similar, define as k(, ), prove positive definite.

30 Outline Theoretical View User View Building Kernels Computing the Max-Margin Classifier Non-Separable Data

31 Marginal Geometry y > 0 y = 0 y < 0 R2 x2 R1 w x y(x) w x x1 w0 w See assignment. Projection of x in w dir. is w x w y(x) = 0 when w x = b, or w x w = b w So w x w b w = y(x) w is signed distance to decision boundary

32 Support Vectors y = 1 y = 0 y = 1 Assuming data are separated by the hyperplane, distance to decision boundary is tny(xn) w The maximum margin criterion chooses w, b by: { } 1 arg max w,b w min[t n(w x n + b)] n Points with this min value are known as support vectors

33 Canonical Representation This optimization problem is complex: { } 1 arg max w,b w min[t n(w x n ) + b)] n Exercise: Prove that rescaling w κw and b κb does not change distance tny(xn) w (many equiv. answers) So for x closest to surface, can set (how?): t (w x + b) = 1 All other points are at least this far away: n, t n (w x n + b) 1 Knowing that min distance = 1, the optimization becomes: arg max w,b 1 1 = arg min w w,b 2 w 2

34 Canonical Representation This optimization problem is complex: { } 1 arg max w,b w min[t n(w x n ) + b)] n Exercise: Prove that rescaling w κw and b κb does not change distance tny(xn) w (many equiv. answers) So for x closest to surface, can set (how?): t (w x + b) = 1 All other points are at least this far away: n, t n (w x n + b) 1 Knowing that min distance = 1, the optimization becomes: arg max w,b 1 1 = arg min w w,b 2 w 2

35 Canonical Representation This optimization problem is complex: { } 1 arg max w,b w min[t n(w x n ) + b)] n Exercise: Prove that rescaling w κw and b κb does not change distance tny(xn) w (many equiv. answers) So for x closest to surface, can set (how?): t (w x + b) = 1 All other points are at least this far away: n, t n (w x n + b) 1 Knowing that min distance = 1, the optimization becomes: arg max w,b 1 1 = arg min w w,b 2 w 2

36 Canonical Representation This optimization problem is complex: { } 1 arg max w,b w min[t n(w x n ) + b)] n Exercise: Prove that rescaling w κw and b κb does not change distance tny(xn) w (many equiv. answers) So for x closest to surface, can set (how?): t (w x + b) = 1 All other points are at least this far away: n, t n (w x n + b) 1 Knowing that min distance = 1, the optimization becomes: arg max w,b 1 1 = arg min w w,b 2 w 2

37 Canonical Representation So the optimization problem is now a constrained optimization problem: 1 arg min w,b 2 w 2 s.t. n, t n (w x n + b) 1 To solve this, we need to use Lagrange multipliers

38 Lagrange Multipliers - Inequality Constraints rf(x) Consider the problem: rg(x) xb xa min f (x) x s.t. g 1 (x) 0,..., g N (x) 0 g(x) > 0 g(x) = 0 Solutions are stationary points of the Lagrangian L(x, a) = f (x) a n g n (x) n=1 where for each n we have the KKT conditions, a n 0 g n (x) 0 a n g n (x) = 0 Therefore a n = 0 (inactive constraint) or g n (x) = 0 (active constraint).

39 Lagrange Multipliers - Inequality Constraints rf(x) Consider the problem: rg(x) xb xa min f (x) x s.t. g 1 (x) 0,..., g N (x) 0 g(x) > 0 g(x) = 0 Solutions are stationary points of the Lagrangian L(x, a) = f (x) a n g n (x) n=1 where for each n we have the KKT conditions, a n 0 g n (x) 0 a n g n (x) = 0 Therefore a n = 0 (inactive constraint) or g n (x) = 0 (active constraint).

40 Now Where Were We So the optimization problem is now a constrained optimization problem: w 2 arg min w,b 2 s.t. n, t n (w x n + b) 1 For this problem, the Lagrangian (with N multipliers a n ) is: L(w, b, a) = w 2 a n {t n (w x n + b) 1} 2 n=1 We can find the derivatives of L wrt w, b and set to 0: w = a n t n x n 0 = n=1 a n t n n=1

41 The Dual Formulation Recall the condition: w = a n t n x n n=1 Exercise: Show that w 2 2 = 1 2 N n=1 N m=1 a na m t n t m (x n x m ). Exercise: Show that n=1 m=1 a n {t n (w x n + b) 1} = n=1 a n a m t n t m (x n x m ) + b a n t n n=1 n=1 a n

42 Solving The Dual Formulation Combining the exercise results, we obtain the dual Lagrangian as a function of the Lagrange multipliers only: L(a) = a n 1 2 n=1 n=1 m=1 a n a m t n t m (x n x m ) The stationary points of L provide lower bounds on the original problem, so we want to maximize L (see http: //en.wikipedia.org/wiki/lagrange_duality). Apply the kernel trick to replace with kernel k, and remember the constraints on the Lagrange multipliers, to arrive at the following problem: Maximize L(a) = N n=1 a n 1 N N 2 n=1 m=1 a na m t n t m k(x n, x m ) subject to the constraints that a n 0, n = 1,..., N N n=1 a nt n = 0

43 From The Dual Solution a to a Classifier Given the solution a, we have formulas for w and for b (omitted). Then classify as follows: w = a n t n φ(x n ) n=1 y(x) = w φ(x) + b = a n t n k(x, x n ) + b n=1 Recall that every constraint is either inactive (a n = 0) or active (a n > 0 and t n y(x n ) = 1). If a n > 0, then x n is a support vector. a will be sparse - many zeros. Don t need to store x n for which a n = 0

44 Outline Theoretical View User View Building Kernels Computing the Max-Margin Classifier Non-Separable Data

45 Non-Separable Data > 1 y = 1 y = 0 y = 1 < 1 = 0 = 0 For most problems, data will not be linearly separable (even in feature space φ) Can relax the constraints from t n y(x n ) 1 to t n y(x n ) 1 ξ n The ξ n 0 are called slack variables ξ n = 0, satisfy original problem, so x n is on margin or correct side of margin 0 < ξ n < 1, inside margin, but still correctly classifed ξ n > 1, mis-classified

46 Loss Function For Non-separable Data > 1 y = 1 y = 0 y = 1 < 1 = 0 = 0 Non-zero slack variables are bad, penalize while maximizing the margin: min C ξ n w 2 n=1 Constant C > 0 controls importance of large margin versus incorrect (non-zero slack) Set using cross-validation Optimization is same quadratic, different constraints, convex

47 SVM Loss Function The SVM for the separable case solved the problem: 1 arg min w 2 w 2 s.t. n, t n y n 1 Can write this as: arg min w E (t n y n ) + λ w 2 n=1 where E (z) = 0 if z 1, otherwise Non-separable case relaxes this to be: arg min E SV (t n y n ) + λ w 2 w n=1 where E SV (z) = [z] + hinge loss [z] + = z if u 1, 0 otherwise

48 SVM Loss Function The SVM for the separable case solved the problem: 1 arg min w 2 w 2 s.t. n, t n y n 1 Can write this as: arg min w E (t n y n ) + λ w 2 n=1 where E (z) = 0 if z 1, otherwise Non-separable case relaxes this to be: arg min E SV (t n y n ) + λ w 2 w n=1 where E SV (z) = [z] + hinge loss [z] + = z if u 1, 0 otherwise

49 SVM Loss Function The SVM for the separable case solved the problem: 1 arg min w 2 w 2 s.t. n, t n y n 1 Can write this as: arg min w E (t n y n ) + λ w 2 n=1 where E (z) = 0 if z 1, otherwise Non-separable case relaxes this to be: arg min E SV (t n y n ) + λ w 2 w n=1 where E SV (z) = [z] + hinge loss [z] + = z if u 1, 0 otherwise

50 Loss Functions E(z) z Linear classifiers, compare loss function used for learning z = y n t n 0 iff there is an error (with t n {+1, 1}). z = y n t n 1 iff the point is on the right side of the margin boundary. Black is misclassification error Transformed simple linear classifier, squared error: (y n t n ) 2 Transformed logistic regression, cross-entropy error: t n ln y n SVM, hinge loss: positive only if the point is on the wrong side of the margin boundary. Sparse solutions.

51 Two Views of Learning as Optimization The original SVM goal was of the form: Find the simplest hypothesis that is consistent with the data, or Maximize simplicity, given a consistency constraint. This general idea appears in much scientific model building, in image processing, and other applications. Bayesian methods use a criterion of the form Find a trade-off between simplicity and data fit, or Maximize sum of the type (data fit - λ simplicity) e.g., ln(p(d M)) λln(p(m)) where the model prior M is higher for simpler models.

52 Pros and Cons of Learning Criteria The Bayesian approach has a solid probabilistic foundation in Bayes theorem. Seems to be especially suitable for noisy data. The constraint-based approach is often easy for users to understand. Often leads to sparser simpler models. Suitable for clean data.

53 Conclusion Readings: Ch (pp ) Non-linear features, or domain-specific similarity measurements are useful Dot products of non-linear features, or similarity measurements, can be written as kernel functions Validity by positive semi-definiteness of kernel function Can have algorithm work in non-linear feature space without actually mapping inputs to feature space Advantageous when feature space is high-dimensional