This is a hacky question. What I'm hoping for is answers that identify and explain the problems and possible solutions.

I'm dealing with some code that is already in production that uses a logistic regression model. The existing model was trained by treating the response variable as a boolean.

What I would like to try is modeling the same training data with a continuous response variable from -1 to 1 (conceptually the same response variable, just stop pretending it's boolean, which it isn't). I would train various linear regression models and try to evaluate whether or not there is any improvement over the existing logistic regression model.

My questions are:

  • Can I simply plug into the existing production logistic regression by substituting the linear regression coefficients and intercept? I'm wondering if the logit function will just be applied to the new linear regression equation, and the result will be a valid 0 or 1 classification.
  • How should I compare the results of the new linear regression to the old logistic regression? If the answer to my first question is affirmative, then I can just evaluate the precision and recall of the two "classification" models. Otherwise, I suppose I could run the linear regression model separately and then apply a threshold to its continuous response variable, and then treat that as a classification.
  • Is it reasonable to expect an improvement with this methodology?
  • 1
    $\begingroup$ How is the production code "pretending it's boolean"? Was the data preprocessed to be all $0, 1$ values prior to training, or was it trained on continuous values from $0$ to $1$? $\endgroup$ Nov 21, 2015 at 15:46
  • $\begingroup$ The data was preprocessed to be all 0 or 1 values prior to training. $\endgroup$
    – Tyro
    Nov 21, 2015 at 16:11

1 Answer 1

  1. No. The logistic regression equation describes, as a function of the various coefficients, {the log of the odds that the response variable will take a value of 1}. In contrast, a linear regression equation will describe, as a function of a different set of coefficients, {the actual value that the response variable will take}. This means that from one equation to the other the coefficients have a very different interpretation, and they are liable to be on very different scales.

(It is beneficial to an analyst to master the differences among three types of equations that are typically involved in logistic regression. These three have as their outcomes a) the log of the odds that the response variable will take a value of 1; b) the odds that the response variable will take a value of 1; and c) the probability that the response variable will take a value of 1.)

  1. You will likely meet with some highly divergent views on this. You could be in for a lot of reading--even on this site alone--to sort out the competing recommendations for comparisons that use R-squared; area under the ROC curve; -2LL; AIC; BIC; and pseudo-R-squareds of different stripes. But what unites people in many camps is their rejection of correct classification rate. One reason why this method is out of favor concerns the situation in which the probability of the response taking on a value of 1 is dichotomized into Predicted As Yes/Predicted As No based on a probability threshold of 0.5. In many applications of regression, few cases' probabilities will exceed this threshold, and so few or even no cases will be Predicted As Yes. This may give the erroneous impression that the regression has zero predictive value.

  2. It is reasonable to expect an improvement when your response variable is measured in a more fine-grained way, as long as those measurements are reliable and otherwise valid. You will have more information to work with in trying to match higher and higher values of Y to certain values of each X. Generally, more information yields better predictive accuracy.

  • $\begingroup$ But how will the Frankenstein model behave? The linear regression equation will evaluate to something between -1 and +1, which is then (incorrectly) interpretted as the log of the odds. So if my (incorrect) logistic regression uses a threshold of 0 for classifying, and if my linear regression continuous response variable similarly considers anything above the threshold of 0 as a positive result, won't the Frankenstein results still be valid? $\endgroup$
    – Tyro
    Nov 23, 2015 at 1:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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