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Intuitively, why may some of the slope coefficients in ridge regression increase in magnitude when the penalty parameter $\lambda$ is increased? Or in other words, why are the coefficient paths nonmonotonic?

enter image description here

Above is an example of ridge coefficient paths taken from James et al. "An Introduction to Statistical Learning" (Figure 6.4 on p. 216). As you can see, Income and Rating are nonmonotonic w.r.t. $\lambda$.

Algebraic and geometric arguments will be appreciated.

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My reasoning is somewhat similar to Cagdas', but I'd like to look at how the things develop as $\lambda$ goes into extremes.

If I did my algebra right, the derivative of the coefficients is given by:

$$\frac{\partial \hat{\beta}}{\partial\lambda} = -(X'X + \lambda I)^{-1}\hat{\beta}$$

Now, for $\lambda \rightarrow 0$, ridge regression approaches ordinary linear regression. The penalty term becomes negligible and you can approximate the derivative by:

$$\frac{\partial \hat{\beta}}{\partial\lambda} \approx -(X'X)^{-1}\hat{\beta}$$

That is, the gradient of $\hat{\beta}$ is mostly determined by the distribution of your data. The direction of change of $\hat{\beta}$ with increasing $\lambda$ can be positive or negative for any of the coefficients, depending both on the coefficients and on your data.

On the other hand, when $\lambda \rightarrow \infty$, $\lambda I \gg X'X$ and you can approximate the derivative by:

$$\frac{\partial \hat{\beta}}{\partial\lambda} \approx -(\lambda I)^{-1}\hat{\beta}$$

Here, the gradient is determined almost entirely by the value of $\hat{\beta}$, and ridge regression tries to force it to a null vector. The path of $\hat{\beta}$ for large and increasing $\lambda$'s is almost a straight line towards the origin.

Edit:

To illustrate this behaviour, here a simple data set with correlated variables:

set.seed(0)
tb1 = tibble(
  x1 = seq(-1, 1, by=.01),
  x2 = -2*x1 + rnorm(length(x1)),
  y  = -3*x1+x2 + rnorm(length(x1))
)

The path of the coefficients $\beta$ with an increasing $\lambda$ looks like this:

example path of the betas

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I think the key intuition to think of here is a situation where you are interested in modelling a single outcome of interest, $Y$, with three possible independent variables, $W$, $X$, $Z$. Imagine in a standard regression, $W$ and $X$ are found significant whereas the coefficient on $Z$ is found insignificant and close to zero, although $Z$ and $Y$ are correlated one-on-one.

Let's assume the reason that $Z$ is not significant in the full regression is because all the effect of $Z$ is subsumed in $W$ and $X$ (i.e. they jointly form a better predictor). Now in a ridge regression context, for some parameter values of $\lambda$ it might be the case that using $Z$ is preferred over having both $W$ and $X$ due to the inclusion of a parameter error term.

This example is also illustrative why ridge regression is used less within an inferential context, because even when controlling for potential confounders, they might still be 'the last coefficient standing'.

See below an example of an outcome $Y$ which is solely related to two variables, $Z$ and $X$. The variable $W$ is affected by both $Z$ and $X$ but does not affect $Y$. Therefore, the SSE (which is minimised in the regression contex) is much lower in a regression of the form $y_i = \beta_0 + \beta_x X + \beta_z Z + \beta_w W + u_i$ relative to one with only $W$: $y_i = \beta_0 + \beta_w W + u_i$.

This is illustrated in the R-squared of the former (at 0.97) relative to the latter (at 0.35). However, the L1-norm is strictly higher in the former case than in the latter (namely $L_1 = \sqrt{\beta_0^2 + \beta_1^2 + \beta_2^2}$). As you drive up $\lambda$, the model will prioritise minimising the $L_1$-norm over the SSE. At some point, the model will prefer having just a single non-zero $\beta$ driving the $L_1$-norm, even at the cost of a much poorer model fit. Therefore, you can see that the coefficients on $\beta_x$ and $\beta_z$ start to decrease up to the point that only $\beta_w$ is non-zero. Also note that further increasing $\lambda$ makes the $L_1$-norm important enough to completely lose interest in minimising any SSE (and therefore generating any model fit) and drives all coefficients to zero.

Results from plot(fit)

library(glmnet)

x <- rnorm(100, 10, 2)
z <- rnorm(100, 5, 2)
w <- 0.8*x + 0.8*z + rnorm(100, 0, 1)
u <- rnorm(100, 0, 1)

y <- 2*x + 2*z + u

# R-squared of 0.97
summary(lm(y ~ 1 + x + z + w))

# R-squared of 0.35
summary(lm(y ~ 1 + z))

fit <- glmnet(as.matrix(cbind(x, z, w)), as.matrix(y))
plot(fit)
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  • $\begingroup$ Another interesting question is whether the total error from the coefficient terms should always be monotonically decreasing. Intuitively they should if the 'normal' part of the loss function behaves normally, but i'm not 100% sure. $\endgroup$ Feb 12 '20 at 8:28
  • $\begingroup$ Thank you for your answer! I would appreciate some more algebraic details. In your example, the starting point is clear, but the end point is not obvious. It is not implausible, but I wish to see how that happens. Also, the question focuses on coefficient estimates (effect sizes), not their statistical significance. $\endgroup$ Feb 12 '20 at 8:51
  • $\begingroup$ Lets assume the DGP is as follows: y ~ 0.1 + 0.2X + 0.2W + u, that corr(X, Z) and corr(W,Z) are both 0.8 and corr(Z,y) is zero net of corr(X,y) and corr(W,y). In other words, standard regression would yield significant (and therefore non-zero) effect sizes for $\beta_x$ and $\beta_w$. y ~ Z would be an alternative specification with a less complete signal (due to the non-perfect correlation between actual effects X and W with Z). However, the additional error term due to coefficient size in the ridge structure might at some point prefer a single effect size for Z over two for W and X. $\endgroup$ Feb 12 '20 at 9:18
  • $\begingroup$ You could simulate the above evaluating the coefficient sizes $\beta_x$, $\beta_w$ and $\beta_z$ and evaluate the SSE and the joint size of the coefficients weighted by \lambda. At some point, the latter will dominate the former if still including two coefficients and prefer a less 'good' specification in terms of SSE but with a better score on the absolute coefficient sizes. $\endgroup$ Feb 12 '20 at 9:20
  • $\begingroup$ +1 for the effort, but I still do not get the intuition as well as I wish I did. Your last comment is too terse for me to understand. Working out the details step by step could help. Editing them into the answer could be a good idea. $\endgroup$ Feb 12 '20 at 9:37
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Ridge solution:

$$\hat{\beta_\lambda} = (X'X + \lambda I)^{-1}X'y$$

If my matrix algebra is right, derivative of Ridge with respect to $\lambda$:

$$\frac{\partial \hat{\beta_\lambda}}{\partial\lambda} = -(X'X + \lambda I)^{-2}X'y$$

which is:

$$\frac{\partial \hat{\beta_\lambda}}{\partial\lambda} = -(X'X + \lambda I)^{-1}\hat{\beta_\lambda} = -A_\lambda \hat{\beta_\lambda}$$

The derivative of each component of $\hat{\beta_\lambda}$ at each $\lambda$ depends on the values of other components of $\hat{\beta_\lambda}$. At this point it would be unreasonable to think that the derivative of a particular component of $\hat{\beta_\lambda}$ will not have a zero crossing when other components are changing.

Suppose that $X$s are not co-linear and are whitened before regression. In that case $\frac{X'X}{n}$ is $I$ where $n$ is the number of data points. Hence:

$$\frac{\partial \hat{\beta_\lambda}}{\partial\lambda} = -(n + \lambda)^{-1}\hat{\beta_\lambda} = -\frac{\hat{\beta_\lambda}}{n+\lambda} = 0$$

which will be $0$ when a component of $\hat{\beta_\lambda}$ becomes $0$. In this case we should have a monotonic scenario. Basically if we start with no correlation between regressors they will go to $0$ with increasing $\lambda$ without changing direction starting from their cross-covariance with $y$ as their initial values.

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    $\begingroup$ If I did my linear algebra right, you are missing an $I$ in your parentheses: $-(X'X + \lambda I)^{-1}$). $\endgroup$
    – Igor F.
    Feb 14 '20 at 14:10

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