In a number of statistical packages including SAS, SPSS and maybe more, there is an option to "suppress the intercept". Why would you want to do that?


3 Answers 3


If for some reason you know the intercept (particularly if it is zero), you can avoid wasting the variance in your data for estimating something you already know, and have more confidence in the values you do have to estimate.

A somewhat oversimplified example is if you already know (from domain knowledge) that one variable is (on average) a multiple of another, and you are trying to find that multiple.

  • $\begingroup$ I don't totally understand it, but in a model I am creating in R, I have something like lm (a ~ b / c - 1) which creates interactions between b and c, and by suppressing the intercept ("- 1" in R), I get more easily-interpretable answers that are essentially the same as if I don't suppress the intercept. Somehow, the interaction makes this possible. $\endgroup$
    – Wayne
    May 21, 2011 at 23:36
  • $\begingroup$ More easily interpretable answers that are essentially the same? That appears to be a contradiction. Maybe you should introduce this as a new question? $\endgroup$
    – Nick Sabbe
    May 22, 2011 at 8:38
  • $\begingroup$ If I look at the coefficients, with the intercept there is an (intercept) and a tempwarmer (one of my variables is temp which can be warmer or cooler). To interpret the coefficients, I have to know that (intercept) directly corresponds to tempcooler and tempwarmer + (intercept) is the directly-interpretable tempwarmer. If I suppress the intercept, I see tempcooler and tempwarmer directly. Perhaps a quirk of R's formulas and linear modeling, but... $\endgroup$
    – Wayne
    May 22, 2011 at 13:41

Consider the case of a 3-level categorical covariate. If one has an intercept, that would require 2 indicator variables. Using the usual coding for indicator variables, the coefficient for either indicator variable is the mean difference compared to the reference group. By suppressing the intercept, you would have 3 variables representing the categorical covariate, instead of just 2. A coefficient is then the mean estimate for that group. A more concrete example of where to do this is in political science where one may be studying the 50 states of the United States. Instead of having an intercept and 49 indicator variables for the states, it is often preferable to suppress the intercept and instead have 50 variables.

  • $\begingroup$ Much easier to interpret coefficient that way $\endgroup$ May 21, 2011 at 0:31
  • 2
    $\begingroup$ Yes, but it breaks down with two or more categorical variables! $\endgroup$ Feb 12, 2017 at 15:50

To illustrate @Nick Sabbe's point with a specific example.

I once saw a researcher present a model of the age of a tree as a function of its width. It can be assumed that when the tree is at age zero, it effectively has a width of zero. Thus, an intercept is not required.

  • 8
    $\begingroup$ The wisdom or lack thereof depends on the range of the dependent variable of interest. Consider car braking data where you have speeds and stopping distances. You could fit a quadratic model with or without intercept. Speeds of interest usually start around 50 km/hr and go up to, say, 130 km/hr. Fitting a quadratic with intercept in this case makes more sense, I think, as forcing the intercept to zero can incur (practically) significant lack-of-fit problems. The fact that the "braking distance" of a stopped car is zero is not particularly relevant for the modeling problem at hand. $\endgroup$
    – cardinal
    May 21, 2011 at 13:39
  • $\begingroup$ @cardinal yes I was wondering if I should make a similar point. I've found in some nonlinear regression modelling contexts there is greater interest in having a model that provides a theoretically plausible model that predicts accurately outside the range of the data (e.g., in learning curve data speed, models should not predict speeds below 0 seconds). In such cases constraining an intercept to zero may be more appropriate even if it results in a drop in prediction for the data. $\endgroup$ May 23, 2011 at 1:16
  • $\begingroup$ @cardinal I agree polynomial models rarely predict plausibly outside the range of the data, and thus constraining the intercept to 0 in such models is rarely a good idea. $\endgroup$ May 23, 2011 at 1:17
  • 1
    $\begingroup$ Thanks for your comments. My remark wasn't so much aimed at polynomial models. The choice of a quadratic was simply based on an actual physical motivation (i.e., classical mechanics). The point I was trying to articulate was that one should carefully consider the modeling problem of interest; sometimes doing something that is (or seems) "theoretically unjustified" is actually more appropriate statistically. $\endgroup$
    – cardinal
    May 23, 2011 at 3:18

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.