Ridge regression degrees of freedom: limit according to sample size?

Question

I'm working on a high-ish dimensional logistic regression problem. I have 40 variables ($p=40$) and 900 samples ($n=900$), but only 30 of those samples are in the class I'm trying to predict.

I'm fitting a predictive model using penalized maximum likelihood (ridge regression, via glmnet). This requires setting a value for the penalty term $\lambda$. Typically, $\lambda$ is determined through cross-validation.

I've also been reading Frank Harrell, who suggests keeping a model's degrees of freedom below a fraction of the "limiting sample size", $m$. As a rule of thumb, he suggests $df < m/15$. In my case, where $m = 30$, this would mean keeping the degrees of freedom below 2.

The effective degrees of freedom in ridge regression can be calculated as a function of $\lambda$ , $$ df(\lambda) = \sum^{p}_{j=1} d_j^2 / (d_j^2 + \lambda) $$ where $d_j$'s are the singular values of the sample matrix (ESL, eq 3.50).

In ridge regression (or other penalized maximum likelihood techniques), is it ever advisable to choose $\lambda$ using the limiting sample size $m$, instead of choosing it through cross-validation? When might this be wise?

Answer 1

Harrell himself is a user here and might chime in. Failing that, and not having a copy of the book in question, I can only say that I see no reason to use degrees of freedom and $m$ to choose $λ$. The motivation for using cross-validation is that you want to choose the $λ$ that maximizes predictive accuracy, and cross-validation gives a good estimate of predictive accuracy. I don't see why using $m$ would get predictive accuracy that's any better.