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CH8.PCA.pdf

# CH8.PCA.pdf - Chapter 8 Principal Component Analysis(PCA 1...

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Chapter 8. Principal Component Analysis (PCA) 1. Objective Do we need two variables? “ˆ e 1 x 1 + ˆ e 2 x 2 ” is enough! Do we need two? “ x 1 , x 2 ?” or “ˆ e 11 x 1 + ˆ e 12 x 2 , ˆ e 21 x 1 + ˆ e 22 x 2 Last two : Easier to explain variability “Principal Components” 1

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X = ( X 1 , . . . , X p ) Two important parameters: mean and covariance matrix The mean has a direct interpretation. The covariance structure may be explained by a few linear combinations of th and often much of the total system variability of the p variables may be accounted for by these smaller number of linear combinations. The objective of PCA is to pick up a smaller number of principal components which explain the covariance structure and account for most of the total system variability of the p variables. One can replace the p variables by the k ( p ) principal components for further analysis of the data. 2
2. Population Principal Components Principal components of X = ( X 1 , · · · , X p ) (p.431) : X ( µ , Σ ) Σ = PΛP T , P = h e 1 e 2 · · · e p i , Λ = diag ( λ 1 , λ 2 , . . . , λ p ) X = < X , e 1 > e 1 + < X , e 2 > e 2 + · · · + < X , e p > e p . < X , e 1 > PC 1 , < X , e 2 > ± PC 2 ,... . formal . 1 1 st principal component : PC 1 a t 1 X = a 11 X 1 + a 12 X 2 + · · · + a 1 p X p a 1 arg max b { Var( b X ) : b = 1 } 2 r th principal component : Given 1 st , · · · , ( r 1) th principal components, PC r a t r X = a r 1 X 1 + a r 2 X 2 + · · · + a rp X p a r arg max b { Var( b X ) : b = 1 , Cov( b t X , a t j X ) = 0 ( j = 1 , · · · , r 1) } 3

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Mathematical formulation : 1 a 1 = arg max b { b Σb : b b = 1 } 2 Given a 1 , · · · , a r 1 , a r = arg max b { b Σb : b b = 1 , b Σa 1 = · · · = b Σa r 1 = 0 } Solution 1 maximize p i =1 λ i ( b t e i ) 2 subject to b 2 = 1 where Σ = p i =1 λ i e i e t i , e i : eigen vector, λ 1 ≥ · · · ≥ λ p 0. - maximizer a 1 = e 1 2 maximize p i =1 λ i ( b t e i ) 2 subject to b 2 = 1, b t e 1 = 0 - maximizer a 2 = e 2 3 maximize p i =1 λ i ( b t e i ) 2 subject to b 2 = 1, b t e 1 = 0 , · · · , b t e r 1 = 0 - maximizer a r = e r 4
Results (pp.432-433) (1) The r th principal component is PC r = e t r X ( r = 1 , · · · , p ) with Var(PC r ) = λ r , Cov(PC r , PC s ) = 0 ( r ̸ = s ) When r ̸ = s , Cov(PC r , PC s ) = Cov( e T r X , e T s X ) = e T r Σ e s = e T r p X k =1 λ k e k e k ! e s = 0 . (2) X = (PC 1 ) e 1 + · · · + (PC p ) e p , i.e., ( X ) [ e 1 , ··· , e p ] = (PC 1 , · · · , PC p ) (3) For Y (PC 1 , · · · , PC p ) , | Var( Y ) | = | Var( X ) | , tr(Var( Y )) = tr(Var( X )) Y = PC 1 PC 2 . . . PC p = e T 1 e T 2 . . . e T p X = P X .

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• Spring '16
• 행렬, choose principal components whose

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