Outlines

• Ridge regression
• Lasso
• Elastic net

linear regression model

• \(Y = (Y_1, \ldots, Y_n)^\top \in \mathbb{R}^n\)
• \(X = (1_n, X_1, \ldots, X_p) \in \mathbb{R}^{n \times (p+1)}\)
  • \(1_n \in \mathbb{R}^n\),
  • \(X_j \in \mathbb{R}^n\) with \(1 \le j \le p\)
• \(\beta = (\beta_0, \beta_1, \ldots, \beta_p)^\top \in \mathbb{R}^{p+1}\)
• \(\varepsilon = (\varepsilon_1, \ldots, \varepsilon_n)^\top \in \mathbb{R}^n\)

\[Y = X\beta + \varepsilon\]

Solution to linear regression model

• least square estimator (LS): \[\hat{\beta} = \arg \min_\beta \frac{1}{2}\| Y - X\beta\|_2^2\]

  • \(\|a\|_2 = \sqrt{a_1^2 + \ldots + a_p^2}\).

• Maximum likelihood estimator (MLE): \[\hat{\beta} = \arg \max_\beta L(\beta; X, Y)\]

• For linear regression model, LS estimator is the same as MLE \[\hat{\beta} = (X^\top X)^{-1} X^\top Y\] Assuming \(X^\top X\) is invertable

Problem with linear regression

• When \(n>p\), number of subjects is larger than number of features (variables), linear model works fine.
• When \(n<p\), \(\hat{\beta} = (X^\top X)^{-1} X^\top Y\), \(X^\top X \in \mathbb{R}^{p \times p}\) is singular. Thus we cannot estimate \(\hat{\beta}\)
  • Solution: apply PCA to \(X\) and reduce to \(X' \in \mathbb{R}^{n \times r}\), where \(r<n\). (lack of interpretation)
  • model selection: using backward selection, forward selection with BIC or AIC. (Searching space is large)
• When \(n>p\), if two features are highly correlated, the resulting coefficients will have high variance and thus not stable.
  • Solution: PCA
  • Solution: use one one variable among all highly correlated variables as representative (using VIF to detect)

Collinearity

n <- 100
set.seed(32611)
x1 <- rnorm(n,3)
x2 <- rnorm(n,5)
x3 <- x2 + rnorm(n,sd=0.1)
cor(x2, x3)

## [1] 0.9957515

x <- data.frame(x1,x2,x3)
y <- 2*x1 + 3*x2 + 4*x3 + rnorm(n, sd = 3)
xyData <- cbind(y,x)
lmFit <- lm(y~x1 + x2 + x3, data=xyData)
summary(lmFit)

## 
## Call:
## lm(formula = y ~ x1 + x2 + x3, data = xyData)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -10.2838  -1.7470   0.0118   1.8627   6.9231 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  -1.5342     1.7695  -0.867   0.3881    
## x1            2.2746     0.3116   7.299 8.39e-11 ***
## x2            5.2712     3.1574   1.669   0.0983 .  
## x3            1.8343     3.1201   0.588   0.5580    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.9 on 96 degrees of freedom
## Multiple R-squared:  0.8679, Adjusted R-squared:  0.8638 
## F-statistic: 210.3 on 3 and 96 DF,  p-value: < 2.2e-16

How to test collinearity

• Remove variable with VIF > 10

library(car)

## Loading required package: carData

vif(lmFit)

##         x1         x2         x3 
##   1.014903 118.852908 119.013351

summary(lm(y~x1 + x2, data=xyData))

## 
## Call:
## lm(formula = y ~ x1 + x2, data = xyData)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -10.3139  -1.7410  -0.1042   1.8065   7.0280 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  -1.5612     1.7629  -0.886    0.378    
## x1            2.2572     0.3092   7.301 7.97e-11 ***
## x2            7.1196     0.2895  24.595  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.891 on 97 degrees of freedom
## Multiple R-squared:  0.8674, Adjusted R-squared:  0.8647 
## F-statistic: 317.4 on 2 and 97 DF,  p-value: < 2.2e-16

To solve these problems

Regularization methods provide better solutions to this problem

• Ridge regression.
• lasso.
• Elastic net

Ridge regression

\[\hat{\beta} = \arg \min_\beta \frac{1}{2}\| Y - X\beta\|_2^2 + \lambda \| \beta\|^2_2\] - \(\|a\|_2 = \sqrt{a_1^2 + \ldots + a_p^2}\).

• \(\lambda \ge 0\) is a tuning parameter, controling the strength of the penalty term.
  • \(\lambda =0\), Ridge regression reduces to linear regression \(\hat{\beta}^{Ridge} = \hat{\beta}^{LS}\).
  • \(\lambda = \infty\), we get \(\hat{\beta}^{Ridge} = 0\)
  • Given a positive \(\lambda\), we both fit a linear model and shrink the coefficients.

Ridge regression solution

\[\hat{\beta} = \arg \min_\beta \frac{1}{2}\| Y - X\beta\|_2^2 + \lambda \| \beta\|^2_2\]

• \(\hat{\beta} = (X^\top X + \lambda I)^{-1} X^\top Y\)
• As \(\lambda\) increases, the bias increases and the variance decreases.

library(MASS)
lm.ridge(y~x1 + x2 + x3, data=xyData, lambda = 10)

##                  x1        x2        x3 
## 0.9529628 2.0535563 3.4643002 3.2671640

Prostate cancer data

The data is from the book element of statistical learning

library(ElemStatLearn)
str(prostate)

## 'data.frame':    97 obs. of  10 variables:
##  $ lcavol : num  -0.58 -0.994 -0.511 -1.204 0.751 ...
##  $ lweight: num  2.77 3.32 2.69 3.28 3.43 ...
##  $ age    : int  50 58 74 58 62 50 64 58 47 63 ...
##  $ lbph   : num  -1.39 -1.39 -1.39 -1.39 -1.39 ...
##  $ svi    : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ lcp    : num  -1.39 -1.39 -1.39 -1.39 -1.39 ...
##  $ gleason: int  6 6 7 6 6 6 6 6 6 6 ...
##  $ pgg45  : int  0 0 20 0 0 0 0 0 0 0 ...
##  $ lpsa   : num  -0.431 -0.163 -0.163 -0.163 0.372 ...
##  $ train  : logi  TRUE TRUE TRUE TRUE TRUE TRUE ...

prostate$train <- NULL

Ridge regression Prostate cancer data example

library(MASS)
lm_ridge <- lm.ridge(lpsa ~ ., data=prostate, lambda=0); lm_ridge

##                    lcavol      lweight          age         lbph          svi 
##  0.181560862  0.564341279  0.622019787 -0.021248185  0.096712523  0.761673403 
##          lcp      gleason        pgg45 
## -0.106050939  0.049227933  0.004457512

lm(lpsa ~ ., data=prostate)

## 
## Call:
## lm(formula = lpsa ~ ., data = prostate)
## 
## Coefficients:
## (Intercept)       lcavol      lweight          age         lbph          svi  
##    0.181561     0.564341     0.622020    -0.021248     0.096713     0.761673  
##         lcp      gleason        pgg45  
##   -0.106051     0.049228     0.004458

Ridge regression Prostate cancer data example 2

lm.ridge(lpsa ~ ., data=prostate, lambda=10)

##                    lcavol      lweight          age         lbph          svi 
## -0.023822581  0.470383515  0.595477805 -0.015328827  0.082534382  0.663639476 
##          lcp      gleason        pgg45 
## -0.022092251  0.066864682  0.003190709

lm.ridge(lpsa ~ ., data=prostate, lambda=Inf)

##            lcavol  lweight      age     lbph      svi      lcp  gleason 
## 2.478387 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 
##    pgg45 
## 0.000000

Summary for Ridge regression

• Ridge regression will stabilize the varaince of the coefficient estiamtes.
• Ridge regression will increase bias but decrease variance.
• Problem with Ridge regression: coefficient won’t be shrinked to exact 0, unless \(\lambda = +\infty\).

lasso

• The full name of Lasso is least absolute shrinkage and selection operator (Tibshirani, 1996)
• Lasso, the \(l_1\) norm penalty will shrink some of the coefficients to exact zero. \[\hat{\beta}^{lasso} = \arg\min_{\beta \in \mathbb{R}^P} \frac{1}{2} \|y - X\beta\|_2^2 + \lambda \|\beta\|_1\]
• \(\|\beta\|_1 = \sum_{j=1}^p |\beta_j|\) is called \(l_1\) norm of \(\beta\).
• \(\lambda \ge 0\) is a tuning parameter, controling the strength of the penalty term.
  • \(\lambda =0\), we have original linear regression.
  • \(\lambda = \infty\), we get \(\hat{\beta}^{lasso} = 0\)
  • For \(\lambda\), we both fit a linear model and shrink some coefficients to exact zero.

lasso example (penalty form)

• lasso was implemented in R lars package.

library(lars)

## Loaded lars 1.2

x <- as.matrix(prostate[,1:8])
y <- prostate[,9]
lassoFit <- lars(x, y) ## lar for least angle regression
coef(lassoFit, s=2, mode="lambda") ## get beta estimate when lambda = 2. 

##    lcavol   lweight       age      lbph       svi       lcp   gleason     pgg45 
## 0.4506712 0.2910916 0.0000000 0.0000000 0.3475427 0.0000000 0.0000000 0.0000000

## Mode="lambda" uses s as the lasso regularization parameter for lambda

lasso, the penalty form and the constraint form

• lasso regression equivalent forms.
  • in penalty form: \[\hat{\beta}^{lasso} = \arg\min_{\beta \in \mathbb{R}^P} \frac{1}{2} \|y - X\beta\|_2^2 + \lambda \|\beta\|_1\]
  • in constraint form: \[\hat{\beta}^{lasso} = \arg\min_{\beta \in \mathbb{R}^P} \frac{1}{2} \|y - X\beta\|_2^2, s.t, \|\beta\|_1 \le \mu\]
  • When \(\mu > \|\hat{\beta}^{LS}\|_1\), the solution will always be \(\beta = \hat{\beta}^{LS}\), we could also rewrite as \[\hat{\beta}^{lasso} = \arg\min_{\beta \in \mathbb{R}^P} \frac{1}{2} \|y - X\beta\|_2^2, s.t, \|\beta\|_1 \le s\|\hat{\beta}^{LS}\|_1\] where \(0 \le s \le 1\)

lasso example (constraint form)

• lasso was implemented in R lars package.

library(lars)
x <- as.matrix(prostate[,1:8])
y <- prostate[,9]
lassoFit <- lars(x, y) ## lar for least angle regression
coef(lassoFit, s=0.5, mode="fraction") ## get beta estimate when mu = 0.5*||beta_LS||_1

##       lcavol      lweight          age         lbph          svi          lcp 
## 0.4746756454 0.4143707391 0.0000000000 0.0000000000 0.4499667198 0.0000000000 
##      gleason        pgg45 
## 0.0000000000 0.0001502949

## If mode="fraction", then s should be a number between 0 and 1, 
## and it refers to the ratio of the L1 norm of the coefficient vector, 
## relative to the norm at the full LS solution. 

Re-visit AIC and BIC

• AIC and BIC will achieve feature selection.
  • \(AIC = 2k - 2 \log ( {\hat{L}} )\)
  • \(BIC = log(n)k - 2 \log ( {\hat{L}} )\)
• equivalently:

\[\hat{\beta}^{IC} = \arg\min_{\beta \in \mathbb{R}^P} \frac{1}{2} \|y - X\beta\|_2^2 + \lambda \|\beta\|_0\]

• \(\|\beta\|_0\) equals to \(k\) where \(k\) is number of non-zero entries of \(\beta\).
• \(\|y - X\beta\|_2^2\) is equivalent to the likelihood function for Gaussian error.

visualize lasso path

• lasso solution (beta estimate is piecewise linear with respect to lambda)
  • \(\hat{\beta}^{lasso} = \arg\min_{\beta \in \mathbb{R}^P} \frac{1}{2} \|y - X\beta\|_2^2, s.t, \|\beta\|_1 \le \mu\)
  • When \(\mu > \|\hat{\beta}^{LS}\|_1\), \(\hat{\beta}^{lasso} = \hat{\beta}^{LS}\)

plot(lassoFit)

• x-axis: \(\frac{\mu}{\|\hat{\beta}^{LS}\|_1}\)

Intuition for lasso and Ridge regression

• lasso regression equivalent forms.
  • in penalty form: \[\hat{\beta}^{lasso} = \arg\min_{\beta \in \mathbb{R}^P} \frac{1}{2} \|y - X\beta\|_2^2 + \lambda \|\beta\|_1\]
  • in constraint form: \[\hat{\beta}^{lasso} = \arg\min_{\beta \in \mathbb{R}^P} \frac{1}{2} \|y - X\beta\|_2^2, s.t, \|\beta\|_1 \le \mu\]
• Ridge regression equivalent forms.
  • in penalty form: \[\hat{\beta}^{Ridge} = \arg\min_{\beta\in\mathbb{R}^p} \frac{1}{2} \|y - X\beta\|_2^2 + \lambda\|\beta\|_2^2\]
  • in constraint form: \[\hat{\beta}^{Ridge} = \arg\min_{\beta\in\mathbb{R}^p} \frac{1}{2} \|y - X\beta\|_2^2, s.t, \|\beta\|_2^2 \le \mu\]

Intuition for lasso and Ridge regression

• Lasso: \(\hat{\beta}^{lasso} = \arg\min_{\beta \in \mathbb{R}^P} \frac{1}{2} \|y - X\beta\|_2^2, s.t, \|\beta\|_1 \le \mu\)
• Ridge: \(\hat{\beta}^{Ridge} = \arg\min_{\beta\in\mathbb{R}^p} \frac{1}{2} \|y - X\beta\|_2^2, s.t, \|\beta\|_2^2 \le \mu\)

From book Element of Statistical Learning

How to choose Tuning parameter

• cross validation, we leave this for future lectures.

lasso path

plot(lassoFit)

• lasso solution is piecewise linear.
• The lars package only calculates the solution at the knots

Elastic net

\[\hat{\beta}^{elastic} = \arg\min_{\beta \in \mathbb{R}^P} \frac{1}{2} \|y - X\beta\|_2^2 + \lambda_2 \|\beta\|_2^2 + \lambda_1 \|\beta\|_1\]

• In their paper, they claim elastic net has smaller Mean squared error.

Elastic net

\[\hat{\beta}^{elastic} = \arg\min_{\beta \in \mathbb{R}^P} \frac{1}{2} \|y - X\beta\|_2^2 + \lambda ( (1 - \alpha ) \|\beta\|_2^2 + \alpha \|\beta\|_1)\]

library(glmnet)

## Loading required package: Matrix

## Loaded glmnet 4.1

fit.lasso <- glmnet(x, y, family="gaussian", alpha=1)
fit.ridge <- glmnet(x, y, family="gaussian", alpha=0)
fit.elnet <- glmnet(x, y, family="gaussian", alpha=.5)

Group lasso

\[\min_{\beta=(\beta_{(1)},\dots,\beta_{(G)}) \in \mathbb{R}^p} \frac{1}{2} ||y-X\beta||_2^2 + \lambda \sum_{i=1}^G \sqrt{p_{(i)}} ||\beta_{(i)}||_2,\]

• Where \(y \in \mathbb{R}^n\) is outcome and \(X \in \mathbb{R}^{n\times p}\) is the design matrix.
• The design matrix can be partitioned in to \(G\) groups \(X = [X_{(1)}, \ldots, X_{(G)}]\) where \(X_{(i)} \in \mathbb{R}^{n \times p_{(i)}}\).
• \(\beta \in \mathbb{R}^p\) is the predictor and it is partitioned in to \(G\) groups.

Fused lasso

• \(y\in \mathbb{R}^p\), (e.g., stock price change)
• \(\beta \in \mathbb{R}^p\), (e.g., underlying stock momentum)

\[\min_{\beta \in \mathbb{R}^p} \frac{1}{2} || y - \beta ||_2^2 + \lambda \sum_{i=1}^{p-1} |\beta_i - \beta_{i+1}|\]

Generalized lasso

Consider a general setting \[\min_{\beta \in \mathbb{R}^p} f(\beta) + \lambda ||D\beta||_1\] where \(f: \mathbb{R}^n \rightarrow \mathbb{R}\) is a smooth convex function. \(D \in \mathbb{R}^{m\times n}\) is a penalty matrix.

• When \(D=I\), the formulation will reduce to the lasso regression problem.
• When \[D= \left( \begin{array}{cccccc} -1 & 1 & 0 & \ldots & 0 & 0 \\ 1 & -1 & 1 & \ldots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & -1 & 1 \end{array} \right),\] The penalty will be the fussed lasso penalty.