I am finishing my undergraduate thesis and I was asked this problem of negative constant in my examination. So I am trying to find out how to explain what to say in regard to the negative constant. In my model, I have 15 predictors and one response.

And in binary logistic, the response is either 1 or 0. So the examiner insisted that the response should only be either 1 or 0, not negative. So why is the constant negative if all the predictors excluded? What should I say here? What is the problem? Much appreciated on the help.

enter image description here

  • 1
    $\begingroup$ Welcome to our site! Please add the [self-study] tag & read its wiki. $\endgroup$ – Silverfish Jun 7 '16 at 3:49

I suspect your examiner was trying to trick you and/or test your knowledge of logistic regression. A negative intercept is relatively easy to interpreter as very low proportion of occurrences of the event of interest in the original sample in the absence of further influence from variables $X_1 \dots X_{15}$.

In general, with logistic regression you analyse the association of a binary outcome with a set of predictors. In particular you are modelling the $\log(\text{odds})$ of that particular outcome. The odds themselves are simply: $\text{odds} = \frac{\text{Pr(of Occurring)}}{\text{Pr(of Not Occurring)}} = \frac{\text{Pr(of Occurring)}}{1 - \text{Pr(of Occurring)}}$ where $\text{Pr}$ refers to proportions (or loosely speaking probability). From this later equality it follows that: $\text{Pr(of Occurring)} = \frac{\text{odds}}{1 + \text{odds}} $.

Now, particular to your case, a negative constant ($\beta_0$) simply means that the baseline proportion of your sample is quite low: $\exp(-4.587) / (1+ \exp(-4.587)) \approx 0.01008.$ This not catastrophic; maybe you have not centred your variable $X_i$ for example, this commonly leads to this phenomenon, but in case you need to be able to explain why your baseline is so low. If you truly have a very low occurrence of events in the original sample you may want/need to consider rare-event logistic regression (see King & Zeng's 2001 paper on Logistic regression in rare events data for a first taste).

As a quick step-through though to find the change in terms of the proportions that are modelled you need to:

  1. Get the $\log(\text{odds})$ estimate.
  2. Exponentiate it to get the $\text{odds}$.
  3. Get the new proportions as: $\text{Pr}_{\text{new}} = \frac{\text{odds}}{1 + \text{odds}} $.

As a final comment: the statistical significance of parameters included the model you present seems relatively low so giving a solid reason as to why you included them is crucial. I am against $p$-value hunting -which is a bad thing- but a 15-variable model with not a single very strongly ($p \leq1e^{-3}$) statistical significant variable seems a bit awkward at first glance.

The user @gung has given a very good answer on the matter too here.

  • $\begingroup$ i am not graduating in statistic so I am not an expert at this. But intuitively I have an inkling that the interpretation of the equation will most likely involve probability and odds, one that isn't necessarily explained by the common perception of how a simple equation should works. I have answered my examiner by indicating the character of probability and odds but he insists otherwise. I don't know if he's serious or just testing, but it seems the former to me. $\endgroup$ – Galang BS Jun 8 '16 at 5:41
  • $\begingroup$ anyway, i am looking for more confirmation whether by the textbook or explanation from the internet so I can back my own uninformed opinion, and your answer is particularly helpful, especially I am using the quick-step formula for my explanation. And to clarify the comment that my statistical significance is low, I would say that it is part of the research to see if the variables are crucial or no, to disprove the usual preconceived relationship between them that is assumed to be related significantly. Thanks for also pointing that out. $\endgroup$ – Galang BS Jun 8 '16 at 5:49
  • $\begingroup$ Cool, I am glad I could help. If you believe this answers your question you could consider accepting the answer. $\endgroup$ – usεr11852 says Reinstate Monic Jun 8 '16 at 7:21

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.