1
$\begingroup$

Suppose I have a small data set $X$ that is $30\times6$. I am wondering if it makes sense to use ridge regression if I want to improve the predicting power of the model.

To my understanding, ridge regression can usually be used to solve the following problems:

  1. The data matrix $X$ is singular. In this case, OLS does not work.
  2. We have too many features. Ridge regression's objective function puts penalty on size of features, meaning $\sum \|\beta_i\|^2$.

However, I am wondering if the following reasoning makes sense:

OLS estimators are unbiased and have the least variance among all unbiased estimators. Since I have a very small data set, my OLS estimators have very big variance. Even though they are unbiased, since I do not have a big data set, the predicting power might still be low. By using ridge regression, I no longer have unbiased estimators, but a high value of $\lambda$ will give me estimators that have lower variance. As a result, it is possible that I end up with a model that has better predicting power.

$\endgroup$
  • $\begingroup$ I think "penalty on number of features" more describes LASSO, and is not really true for ridge regression. $\endgroup$ – GeoMatt22 Sep 7 '16 at 2:40
  • $\begingroup$ With 30 observations and 6 features, I think your best answer is actually applying both methods and testing which does a better job instead of guessing which theory might fit better. Ridge regression helps when there are a large number of multicollinear features which make it difficult to model small datasets. $\endgroup$ – Arun Jose Sep 7 '16 at 2:44
  • $\begingroup$ @GeoMatt22 Sorry. I meant to say "size of features" $\endgroup$ – 3x89g2 Sep 7 '16 at 5:20
  • 1
    $\begingroup$ It certainly could work. For any given sample size, the optimal amount of $L_2$ penalty is positive (so not zero). That is, there exists a positive $\lambda$ such that ridge will do better than OLS in terms of mean squared error. See Dave Giles blog post "A Regression "Estimator" that Minimizes MSE". $\endgroup$ – Richard Hardy Sep 7 '16 at 7:20
  • $\begingroup$ @RichardHardy This sounds like a really nice result. Knowing existence is always good. $\endgroup$ – 3x89g2 Sep 7 '16 at 7:26
0
$\begingroup$

Your reasoning does make sense.

Given a correctly specified linear model, the optimal amount of $L_2$ penalty is positive, for any given sample size. That is, there exists a positive $\lambda$ such that ridge will do better than OLS in terms of mean squared error of estimated parameters; see Dave Giles blog post "A Regression "Estimator" that Minimizes MSE" for details. If you are able to estimate $\lambda$ precisely enough, ridge will yield more precise estimates than OLS, which in turn will improve prediction accuracy.

Without the assumption of a correctly specified model, things are less straightforward, but ridge may still be a good alternative. You may test the performance of ridge vs. OLS on a test sample and see for yourself which one does better.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.