False positive/negative rate in ridge and lasso regressions

Question

I have a confusion matrix of true and estimated $\boldsymbol{\beta}$ vectors of lasso and ridge models from a replicate of a simulation study, say. The following tables illustrate the scenario.

$$\begin{array}{c|c|c|} \text{lasso} & \#\left(\beta_{j}=0\right) & \#\left(\beta_{j}\ne0\right)\\ \hline \#\left(\hat{\beta}_{j}=0\right) & 4 & 2\\ \hline \#\left(\hat{\beta}_{j}\ne0\right) & 1 & 5 \\\hline \end{array}   \begin{array}{c|c|c|} \text{ridge} & \#\left(\beta_{j}=0\right) & \#\left(\beta_{j}\ne0\right)\\ \hline \#\left(\hat{\beta}_{j}=0\right) & 0 & 0\\ \hline \#\left(\hat{\beta}_{j}\ne0\right) & 5 & 7 \\\hline \end{array}$$

The false positive rate from the lasso estimates is $2/(2+5)=2/7$.

Since the ridge can not perform variable selection, the false positive rate should be 0 all the time.

But when I study a paper, I come across a simulation result in which the median false positive rate from the ridge is reported as a value different from zero.

The paper is here (or here in arxiv).

Relevant part of the table:

Do I overlook something?

Answer 1

The definition of the false positive rate is $FPR=FP/(FP+TN)$. Here, the false positives are the coefficients which are wrongly detected to be different from zero, while the true negatives are the coefficients who are rightfully detected to be different from zero (true positives are the coefficients rightfully detected to be equal to zero).

Thus, in your example the false positive rate for lasso would be $1/(1+5)=1/6$ and for ridge $5/(5+7)=5/12=5/p$ if $p$ is the total number of coefficients.

Obviously, ridge regression always finds all coefficients to be different from 0, so the $FPR$ for ridge is always equal to $N/p$ where $N$ is the number of negatives (or coefficients equal to 0)...

In addition, the authors write down $0.62$ as $FPR$ for an example where the true coefficient vector is $\beta= (3, 1.5, 0, 0, 2, 0, 0, 0)$. Now, there's five zeros out of 8 coeffcients, and $5/8=0.625$. So the authors did round $0.625$ to $0.62$ which is weird.