class05

# class05 - Support Vector Machines For Classication 9.520...

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Support Vector Machines For Classification 9.520 Class 05, 22 February 2006 Ryan Rifkin

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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 .

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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

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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 non-differentiable (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

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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 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

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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

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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 .
The Dual Program Substituting in our expression for c , we are left with the following “dual” program: n max i =1 α i 4 1 λ α T α IR n n subject to : = 0 i =1 y i α i 0 α i 1 i = 1 , . . . , n n Here, Q is the matrix

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