Why is glmnet ridge regression giving me a different answer than manual calculation?

Question

I'm using glmnet to calculate ridge regression estimates. I got some results that made me suspicious in that glmnet is really doing what I think it does. To check this I wrote a simple R script where I compare the result of ridge regression done by solve and the one in glmnet, the difference is significant:

n    <- 1000
p.   <-  100
X.   <- matrix(rnorm(n*p,0,1),n,p)
beta <- rnorm(p,0,1)
Y    <- X%*%beta+rnorm(n,0,0.5)

beta1 <- solve(t(X)%*%X+5*diag(p),t(X)%*%Y)
beta2 <- glmnet(X,Y, alpha=0, lambda=10, intercept=FALSE, standardize=FALSE, 
                family="gaussian")$beta@x
beta1-beta2

The norm of the difference is usually around 20 which cannot be due to numerically different algorithms, I must be doing something wrong. What are the settings I have to set in glmnet in order to obtain the same result as with ridge?

Answer 1

The difference you are observing is due to the additional division by the number of observations, N, that GLMNET uses in their objective function and implicit standardization of Y by its sample standard deviation as shown below.

$$ \frac{1}{2N}\left\|\frac{y}{s_y}-X\beta\right\|^2_{2}+\lambda\|\beta\|^2_{2}/2 $$

where we use $1/n$ in place of $1/(n-1)$ for $s_y$, $$ s_y=\frac{\sum_i(y_i-\bar{y})^2}{n} $$

By differentiating with respect to beta, setting the equation to zero,

$$ X^TX\beta-\frac{X^Ty}{s_y}+N\lambda\beta =0 $$

And solving for beta, we obtain the estimate,

$$ \tilde{\beta}_{GLMNET}= (X^TX+N\lambda I_p)^{-1}\frac{X^Ty}{s_y} $$

To recover the estimates (and their corresponding penalties) on the original metric of Y, GLMNET multiplies both the estimates and the lambdas by $s_y$ and returns these results to the user,

$$ \hat{\beta}_{GLMNET}=s_y\tilde{\beta}_{GLMNET}= (X^TX+N\lambda I_p)^{-1}X^Ty $$ $$ \lambda_{unstd.}=s_y\lambda $$

Compare this solution with the standard derivation of ridge regression.

$$ \hat{\beta}= (X^TX+\lambda I_p)^{-1}X^Ty $$

Notice that $\lambda$ is scaled by an extra factor of N. Additionally, when we use the predict() or coef() function, the penalty is going to be implicitly scaled by $1/s_y$. That is to say, when we use these functions to obtain the coefficient estimates for some $\lambda^*$, we are effectively obtaining estimates for$\lambda=\lambda^*/s_y$.

Based on these observations, the penalty used in GLMNET needs to be scaled by a factor of $s_y/N$.

set.seed(123)

n    <- 1000
p   <-  100
X   <- matrix(rnorm(n*p,0,1),n,p)
beta <- rnorm(p,0,1)
Y    <- X%*%beta+rnorm(n,0,0.5)

sd_y <- sqrt(var(Y)*(n-1)/n)[1,1]

beta1 <- solve(t(X)%*%X+10*diag(p),t(X)%*%(Y))[,1]

fit_glmnet <- glmnet(X,Y, alpha=0, standardize = F, intercept = FALSE, thresh = 1e-20)
beta2 <- as.vector(coef(fit_glmnet, s = sd_y*10/n, exact = TRUE))[-1]
cbind(beta1[1:10], beta2[1:10])

           [,1]        [,2]
[1,]  0.23793862  0.23793862
[2,]  1.81859695  1.81859695
[3,] -0.06000195 -0.06000195
[4,] -0.04958695 -0.04958695
[5,]  0.41870613  0.41870613
[6,]  1.30244151  1.30244151
[7,]  0.06566168  0.06566168
[8,]  0.44634038  0.44634038
[9,]  0.86477108  0.86477108
[10,] -2.47535340 -2.47535340

The results generalize to the inclusion of an intercept and standardized X variables. We modify a standardized X matrix to include a column of ones and the diagonal matrix to have an additional zero entry in the [1,1] position (i.e. do not penalize the intercept). You can then unstandardize the estimates by their respective sample standard deviations (again ensure you are using 1/n when computing standard deviation).

$$ \hat\beta_{j}=\frac{\tilde{\beta_j}}{s_{x_j}} $$

$$ \hat\beta_{0}=\tilde{\beta_0}-\bar{x}^T\hat{\beta} $$

mean_x <- colMeans(X)
sd_x <- sqrt(apply(X,2,var)*(n-1)/n)
X_scaled <- matrix(NA, nrow = n, ncol = p)
for(i in 1:p){
    X_scaled[,i] <- (X[,i] - mean_x[i])/sd_x[i] 
}
X_scaled_ones <- cbind(rep(1,n), X_scaled)

beta3 <- solve(t(X_scaled_ones)%*%X_scaled_ones+1000*diag(x = c(0, rep(1,p))),t(X_scaled_ones)%*%(Y))[,1]
beta3 <- c(beta3[1] - crossprod(mean_x,beta3[-1]/sd_x), beta3[-1]/sd_x)

fit_glmnet2 <- glmnet(X,Y, alpha=0, thresh = 1e-20)
beta4 <- as.vector(coef(fit_glmnet2, s = sd_y*1000/n, exact = TRUE))

cbind(beta3[1:10], beta4[1:10])
             [,1]        [,2]
 [1,]  0.24534485  0.24534485
 [2,]  0.17661130  0.17661130
 [3,]  0.86993230  0.86993230
 [4,] -0.12449217 -0.12449217
 [5,] -0.06410361 -0.06410361
 [6,]  0.17568987  0.17568987
 [7,]  0.59773230  0.59773230
 [8,]  0.06594704  0.06594704
 [9,]  0.22860655  0.22860655
[10,]  0.33254206  0.33254206

Added code to show standardized X with no intercept:

set.seed(123)

n <- 1000
p <-  100
X <- matrix(rnorm(n*p,0,1),n,p)
beta <- rnorm(p,0,1)
Y <- X%*%beta+rnorm(n,0,0.5)

sd_y <- sqrt(var(Y)*(n-1)/n)[1,1]

mean_x <- colMeans(X)
sd_x <- sqrt(apply(X,2,var)*(n-1)/n)

X_scaled <- matrix(NA, nrow = n, ncol = p)
for(i in 1:p){
    X_scaled[,i] <- (X[,i] - mean_x[i])/sd_x[i] 
}

beta1 <- solve(t(X_scaled)%*%X_scaled+10*diag(p),t(X_scaled)%*%(Y))[,1]

fit_glmnet <- glmnet(X_scaled,Y, alpha=0, standardize = F, intercept = 
FALSE, thresh = 1e-20)
beta2 <- as.vector(coef(fit_glmnet, s = sd_y*10/n, exact = TRUE))[-1]
cbind(beta1[1:10], beta2[1:10])

             [,1]        [,2]
 [1,]  0.23560948  0.23560948
 [2,]  1.83469846  1.83469846
 [3,] -0.05827086 -0.05827086
 [4,] -0.04927314 -0.04927314
 [5,]  0.41871870  0.41871870
 [6,]  1.28969361  1.28969361
 [7,]  0.06552927  0.06552927
 [8,]  0.44576008  0.44576008
 [9,]  0.90156795  0.90156795
[10,] -2.43163420 -2.43163420

Answer 2

According to https://web.stanford.edu/~hastie/glmnet/glmnet_alpha.html, when the family is gaussian, glmnet() should minimize $$\frac{1}{2n} \sum_{i=1}^n (y_i-\beta_0-x_i^T\beta)^2 +\lambda\sum_{j=1}^p(\alpha|\beta_j| +(1-\alpha)\beta_j^2/2). \tag{1}$$

When using glmnet(x, y, alpha=1) to fit the lasso with the columns in $x$ standardized, the solution for the reported penalty $\lambda$ is the solution for minimizing $$\frac{1}{2n} \sum_{i=1}^n (y_i-\beta_0-x_i^T\beta)^2 +\lambda \sum_{j=1}^p |\beta_j|.$$ However, at least in glmnet_2.0-13, when using glmnet(x, y, alpha=0) to fit ridge regression, the solution for a reported penalty $\lambda$ is the solution for minimizing $$\frac{1}{2n} \sum_{i=1}^n (y_i-\beta_0-x_i^T\beta)^2 +\lambda \frac{1}{2s_y} \sum_{j=1}^p \beta_j^2.$$ where $s_y$ is the standard deviation of $y$. Here, the penalty should have been reported as $\lambda/s_y$.

What might happen is that the function first standardizes $y$ to $y_0$ and then minimizes $$\frac{1}{2n} \sum_{i=1}^n (y_{0i}-x_i^T\gamma)^2 +\eta \sum_{j=1}^p(\alpha|\gamma_j| +(1-\alpha)\gamma_j^2/2), \tag{2}$$ which effectively is to minimize $$\frac{1}{2n s_y^2} \sum_{i=1}^n (y_i-\beta_0-x_i^T\beta)^2 +\eta \frac{\alpha}{s_y} \sum_{j=1}^p |\beta_j| +\eta \frac{1-\alpha}{2s_y^2} \sum_{j=1}^p \beta_j^2,$$ or equivalently, to minimize $$\frac{1}{2n} \sum_{i=1}^n (y_i-\beta_0-x_i^T\beta)^2 +\eta s_y \alpha \sum_{j=1}^p |\beta_j| +\eta (1-\alpha) \sum_{j=1}^p \beta_j^2/2.$$

For the lasso ($\alpha=1$), scaling $\eta$ back to report the penalty as $\eta s_y$ makes sense. Then for all $\alpha$, $\eta s_y$ has to be reported as the penalty to maintain continuity of the results across $\alpha$. This probably is the cause of the problem above. This is partly due to using (2) to solve (1). Only when $\alpha=0$ or $\alpha=1$ there is some equivalence between problems (1) and (2) (i.e., a correspondence between the $\lambda$ in (1) and the $\eta$ in (2)). For any other $\alpha\in(0,1)$, problems (1) and (2) are two different optimization problems, and there is no one-to-one correspondence between the $\lambda$ in (1) and the $\eta$ in (2).