Lasso, Ridge, and Elastic Net

Transcription

1 Lasso, Ridge, and Elastic Net David Rosenberg New York University February 7, 2017 David Rosenberg (New York University) DS-GA 1003 February 7, / 29

2 Linearly Dependent Features Linearly Dependent Features David Rosenberg (New York University) DS-GA 1003 February 7, / 29

3 Linearly Dependent Features Suppose We Have 2 Equal Features Input features: x 1,x 2 R. Outcome: y R. Linear prediction functions f (x) = w 1 x 2 + w 2 x 2 Suppose x 1 = x 2. Then all functions with w 1 + w 2 = k are the same. give same predictions and have same empirical risk What function will we select if we do ERM with l 1 or l 2 constraint? David Rosenberg (New York University) DS-GA 1003 February 7, / 29

4 Linearly Dependent Features Equal Features, l 2 Constraint w 2 w 1 + w 2 = w 2 2 w 1 + w 2 = w 1 + w 2 = w 1 w 1 + w 2 = w 1 + w 2 = w 1 + w 2 = 2 2 Suppose the line w 1 + w 2 = corresponds to the empirical risk minimizers. Empirical risk increase as we move away from these parameter settings Intersection of w 1 + w 2 = 2 2 and the norm ball w 2 2 is ridge solution. Note that w 1 = w 2 at the solution David Rosenberg (New York University) DS-GA 1003 February 7, / 29

5 Linearly Dependent Features Equal Features, l 1 Constraint w2 w1 + w2 = w1 + w2 = 9 w 1 2 w1 + w2 = 7.25 w1 + w2 = 5.5 w1 w1 + w2 = 3.75 w1 + w2 = 2 Suppose the line w 1 + w 2 = 5.5 corresponds to the empirical risk minimizers. Intersection of w 1 + w 2 = 2 and the norm ball w 1 2 is lasso solution. Note that the solution set is {(w 1,w 2 ) : w 1 + w 2 = 2,w 1,w 2 0}. David Rosenberg (New York University) DS-GA 1003 February 7, / 29

6 Linearly Dependent Features Linearly Related Features Same setup, now suppose x 2 = 2x 1. Then all functions with w 1 + 2w 2 = k are the same. give same predictions and have same empirical risk What function will we select if we do ERM with l 1 or l 2 constraint? David Rosenberg (New York University) DS-GA 1003 February 7, / 29

7 Linearly Dependent Features Linearly Related Features, l 2 Constraint w 2 2 w 2 w 1 + 2w 2 = 10/ w 1 + 2w 2 = 10/ w 1 + 2w 2 = 10/ w 1 + 2w 2 = 10/ 5 w 1 w 1 + 2w 2 = 10/ corresponds to the empirical risk minimizers. Intersection of w 1 + 2w 2 = 10 5 and the norm ball w 2 2 is ridge solution. At solution, w 2 = 2w 1. David Rosenberg (New York University) DS-GA 1003 February 7, / 29

8 Linearly Dependent Features Linearly Related Features, l 1 Constraint w 2 w 1 + 2w 2 = 16 w 1 + 2w 2 = 12 w 1 2 w 1 + 2w 2 = 8 w 1 w 1 + 2w 2 = 4 Intersection of w 1 + 2w 2 = 4 and the norm ball w 1 2 is lasso solution. Solution is now a corner of the l 1 ball, corresonding to a sparse solution. David Rosenberg (New York University) DS-GA 1003 February 7, / 29

9 Linearly Dependent Features Linearly Dependent Features: Take Away For identical features l 1 regularization spreads weight arbitrarily (all weights same sign) l 2 regularization spreads weight evenly Linearly related features l 1 regularization chooses variable with larger scale, 0 weight to others l 2 prefers variables with larger scale spreads weight proportional to scale David Rosenberg (New York University) DS-GA 1003 February 7, / 29

10 Linearly Dependent Features Empirical Risk for Square Loss and Linear Predictors Recall our discussion of linear predictors f (x) = w T x and square loss. Sets of w giving same empirical risk (i.e. level sets) formed ellipsoids around the ERM. With x 1 and x 2 linearly related, we get a degenerate ellipse. That s why level sets were lines (actually pairs of lines, one on each side of ERM). KPM Fig David Rosenberg (New York University) DS-GA 1003 February 7, / 29

11 Linearly Dependent Features Correlated Features Same Scale Suppose x 1 and x 2 are highly correlated and the same scale. This is quite typical in real data, after normalizing data. Nothing degenerate here, so level sets are ellipsoids. But, the higher the correlation, the closer to degenerate we get. That is, ellipsoids keep stretching out, getting closer to two parallel lines. David Rosenberg (New York University) DS-GA 1003 February 7, / 29

12 Linearly Dependent Features Correlated Features, l 1 Regularization w2 w2 w 1 2 w 1 2 w1 w1 Intersection could be anywhere on the top right edge. Minor perturbations (in data) can drastically change intersection point very unstable solution. Makes division of weight among highly correlated features (of same scale) seem arbitrary. If x 1 2x 2, ellipse changes orientation and we probably hit a corner. David Rosenberg (New York University) DS-GA 1003 February 7, / 29

13 Linearly Dependent Features A Very Simple Model Suppose we have one feature x 1 R. Response variable y R. Got some data and ran least squares linear regression. The ERM is What happens if we get a new feature x 2, but we always have x 2 = x 1? ˆf (x 1 ) = 4x 1. David Rosenberg (New York University) DS-GA 1003 February 7, / 29

14 Linearly Dependent Features Duplicate Features New feature x 2 gives no new information. ERM is still ˆf (x 1,x 2 ) = 4x 1. Now there are some more ERMs: ˆf (x 1,x 2 ) = 2x 1 + 2x 2 ˆf (x 1,x 2 ) = x 1 + 3x 2 ˆf (x 1,x 2 ) = 4x 2 What if we introduce l 1 or l 2 regularization? David Rosenberg (New York University) DS-GA 1003 February 7, / 29

15 Linearly Dependent Features Duplicate Features: l 1 and l 2 norms ˆf (x 1,x 2 ) = w 1 x 1 + w 2 x 2 is an ERM iff w 1 + w 2 = 4. Consider the l 1 and l 2 norms of various solutions: w 1 w 2 w 1 w w 1 doesn t discriminate, as long as all have same sign w 2 2 minimized when weight is spread equally Picture proof: Level sets of loss are lines of the form w 1 + w 2 = c... David Rosenberg (New York University) DS-GA 1003 February 7, / 29

16 Correlated Features and the Grouping Issue Correlated Features and the Grouping Issue David Rosenberg (New York University) DS-GA 1003 February 7, / 29

17 Correlated Features and the Grouping Issue Example with highly correlated features Model in words: y is a linear combination of z 1 and z 2 But we don t observe z 1 and z 2 directly. We get 3 noisy observations of z 1. We get 3 noisy observations of z 2. We want to predict y from our noisy observations. Example from Section 4.2 in Hastie et al s Statistical Learning with Sparsity. David Rosenberg (New York University) DS-GA 1003 February 7, / 29

18 Correlated Features and the Grouping Issue Example with highly correlated features Suppose (x, y) generated as follows: z 1,z 2 N(0,1) (independent) ε 0,ε 1,...,ε 6 N(0,1) (independent) y = 1 1.5z 2 + 2ε 0 3z { x j = z 1 + ε j /5 for j = 1,2,3 z 2 + ε j /5 for j = 4,5,6 Generated a sample of (x,y) pairs of size 100. Correlations within the groups of x s were around Example from Section 4.2 in Hastie et al s Statistical Learning with Sparsity. David Rosenberg (New York University) DS-GA 1003 February 7, / 29

19 Correlated Features and the Grouping Issue Example with highly correlated features Lasso regularization paths: Lasso Paths Elastic-Net Paths coefficients coefficients λ λ 1 Lines with the same color correspond to features with essentially the same information Distribution of weight among them seems almost arbitrary David Rosenberg (New York University) DS-GA 1003 February 7, / 29

20 Correlated Features and the Grouping Issue Hedge Bets When Variables Highly Correlated When variables are highly correlated (and same scale, after normalization), we want to give them roughly the same weight. Why? Let their error cancel out How can we get the weight spread more evenly? David Rosenberg (New York University) DS-GA 1003 February 7, / 29

21 Correlated Features and the Grouping Issue Elastic Net The elastic net combines lasso and ridge penalties: 1 ŵ = arg min w R d n n { w T } 2 x i y i + λ1 w 1 + λ 2 w 2 2 We expect correlated random variables to have similar coefficients. i=1 David Rosenberg (New York University) DS-GA 1003 February 7, / 29

22 Correlated Features and the Grouping Issue Highly Correlated Features, Elastic Net Constraint w2.8 w w w1 Elastic net solution is closer to w 2 = w 1 line, despite high correlation. David Rosenberg (New York University) DS-GA 1003 February 7, / 29

23 Correlated Features and the Grouping Issue Elastic Net - Sparse Regions Suppose design matrix X is orthogonal, so X T X = I, and contours are circles (and features uncorrelated) Then OLS solution in green or red regions implies elastic-net constrained solution will be at corner Fig from Mairal et al. s Sparse Modeling for Image and Vision Processing Fig 1.9 David Rosenberg (New York University) DS-GA 1003 February 7, / 29

24 Correlated Features and the Grouping Issue Elastic Net A Theorem for Correlated Variables Theorem a Let ρ ij = ĉorr(x i,x j ). Suppose ŵ i and ŵ j are selected by elastic net, with y centered and predictors x 1,...,x d standardized. If ŵ i ŵ j > 0, then ŵ i ŵ j y 2 λ 2 1 ρij. a Theorem 1 in Regularization and variable selection via the elastic net : David Rosenberg (New York University) DS-GA 1003 February 7, / 29

25 Correlated Features and the Grouping Issue Elastic Net Results on Model Lasso Paths Elastic-Net Paths coefficients coefficients coefficients coefficients λ 1 λ λ 1 λ 1 Lasso on left; Elastic net on right. Ratio of l 2 to l 1 regularization roughly 2 : 1. David Rosenberg (New York University) DS-GA 1003 February 7, / 29

26 Extra Pictures Extra Pictures David Rosenberg (New York University) DS-GA 1003 February 7, / 29

27 Extra Pictures Elastic Net vs Lasso Norm Ball From Figure 4.2 of Hastie et al s Statistical Learning with Sparsity. David Rosenberg (New York University) DS-GA 1003 February 7, / 29

28 Extra Pictures The ( l q ) q Constraint Generalize to l q : ( w q ) q = w 1 q + w 2 q. Note: w q is a norm if q 1, but not for q (0,1) F = {f (x) = w 1 x 1 + w 2 x 2 }. Contours of w q q = w 1 q + w 2 q : David Rosenberg (New York University) DS-GA 1003 February 7, / 29

29 Extra Pictures l 1.2 vs Elastic Net From Hastie et al s Elements of Statistical Learning. David Rosenberg (New York University) DS-GA 1003 February 7, / 29