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Unformatted text preview: Support Vector Machines For Classification 9.520 Class 05, 22 February 2006 Ryan Rifkin Plan • Regularization derivation of SVMs • Geometric derivation of SVMs • Optimality, Duality and Large Scale SVMs • SVMs and RLSC: Compare and Contrast The Regularization Setting (Again) We are given n examples ( x 1 , y 1 ) , . . . , ( x n , y n ), with x i ∈ IR d and y i ∈ {− 1 , 1 } for all i . As mentioned last class, we find a classification function by solving a regularization: n 1 summationdisplay min V ( y i , f ( x i )) + λ  f  2 K . f ∈H n i =1 In this class we specifically consider binary classification . The Hinge Loss The classical SVM arises by considering the specific loss function V ( f ( x ) , y ) ≡ (1 − yf ( x )) + , where ( k ) + ≡ max( k, 0) . The Hinge Loss 0 0.5 1 1.5 2 2.5 3 3.5 4 Hinge Loss −3 −2 −1 0 1 2 y * f(x) 3 Substituting In The Hinge Loss With the hinge loss, our regularization problem becomes n 1 summationdisplay min (1 − y i f ( x i )) + + λ  f  2 K . f ∈H n i =1 Slack Variables This problem is nondifferentiable (because of the “kink” in V ), so we introduce slack variables ξ i , to make the problem easier to work with: min f ∈H 1 n ∑ n i =1 ξ i + λ  f  2 K subject to : y i f ( x i ) ≥ 1 − ξ i i = 1 , . . . , n ξ i ≥ 0 i = 1 , . . . , n summationdisplay ∑ ∑ Applying The Representer Theorem Substituting in: n ∗ f ( x ) = c i K ( x , x i ) , i =1 we arrive at a constrained quadratic programming problem: min 1 i n =1 ξ i + λ c T K c c ∈ IR n n n subject to : y i j =1 c j K ( x i , x j ) ≥ 1 − ξ i i = 1 , . . . , n ξ i ≥ 0 i = 1 , . . . , n ∑ ∑ Adding A Bias Term If we add an unregularized bias term b , we arrive at the “primal” SVM: min 1 n n i =1 ξ i + λ c T K c c ∈ IR n ,ξ ∈ IR n n subject to : y i ( j =1 c j K ( x i , x j ) + b ) ≥ 1 − ξ i i = 1 , . . . , n ξ i ≥ 0 i = 1 , . . . , n summationdisplay summationdisplay summationdisplay Forming the Lagrangian We derive the Wolfe dual quadratic program using La grange multiplier techniques: n 1 summationdisplay L ( c , ξ, b, α, ζ ) = ξ i + λ c T K c n i =1 n n − α i y i c j K ( x i , x j ) + b − 1 + ξ i i =1 j =1 n − ζ i ξ i i =1 We want to minimize L with respect to c , b , and ξ , and maximize L with respect to α and ζ , subject to the con straints of the primal problem and nonnegativity constraints on α and ζ . summationdisplay summationdisplay Eliminating b and ξ n ∂L summationdisplay = 0 = ⇒ α i y i = 0 ∂b i =1 ∂L 1 = 0 = ⇒ − α i − ζ i = 0 ∂ξ i n 1 = ⇒ 0 ≤ α i ≤ n We write a reduced Lagrangian in terms of the remaining variables: n n T L R ( c , α ) = λ c K c − α i ( y i c j K ( x i , x j ) − 1) i =1 j =1 Eliminating c Assuming the K matrix is invertible, ∂L R = 0 = ⇒ 2 λK c − KY α = 0 ∂ c α i y i = ⇒ c i = 2 λ Where Y is a diagonal matrix whose i ’th diagonal element is y i ; Y α is a vector whose i ’th element is α i y i . ∑...
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This note was uploaded on 11/11/2011 for the course BIO 9.07 taught by Professor Ruthrosenholtz during the Spring '04 term at MIT.
 Spring '04
 RuthRosenholtz

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