For a project, I ran a logistic regression using continuous and dichotomous variables. How do I interpret the marginal effects of a dichotomous variable?

For example, one of our independent variables that has a binary outcome is "White", as in belonging to the Caucasian race. Our dependent variable also has a binary outcome (hence the use of the logit model) so our our outcomes are expressed in probabilities. So to interpret the marginal effect of being white on our outcome, would it be something like " a 1% increase in being white affect your probability of the dependent variable by x amount " ?

Any comments or suggestions welcome :)

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    $\begingroup$ How could you have a 1% increase in being white ?? That is the purpose of binary variables: you are white or you are not. $\endgroup$ – user89073 Nov 19 '15 at 15:19

It is easier to think about interpreting your dichotomous predictors by using the concept of the odds ratio.

Let me give you an example: Imagine you are trying to predict smoking status where our smoking variable is a 1 if you smoke and and 0 if you don't smoke (so a dichotomous outcome and so we can use logistic regression). Now, as in your case, imagine that you have a predictor variable called white where the variable is 1 if you are white or 0 if you are not white. In this example, you can fit a logistic regression model that looks something like this:

$$\text{logit}(p)=\beta_0+\beta_1\times \text{white}$$

And now, lets assume that you get an estimate of $\beta_1=-0.5108256$. Now, converting the estimate onto the odds ratio scale is as simple as exponentiating the parameter estimate, i.e, on the odds ratio scale we have $$e^{\beta_1}=e^{-0.5108256}=0.6$$. And so finally what this tells us is that if you are white, you are expected to be 60% less likely to be a smoker as compared to someone who is not white.

And so to answer your direct question, you wouldn't say that "a 1% increase in being white affect your probability of the dependent variable by x amount", but rather that, you are "y" times more likely to observe the dependent variable given that you are white as compared to not being white.

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    $\begingroup$ Thank you sooo much for your thorough explanation. This has really helped me in my interpretation of my results. I've been stuck on this for a while, so now I can move on. Thanks again :) !! $\endgroup$ – a_student Nov 20 '15 at 22:09
  • $\begingroup$ @a_student no worries. I am just glad I could be of help. $\endgroup$ – user95564 Nov 20 '15 at 22:23
  • $\begingroup$ -1 because, as discussed in @ChrisWright's answer, this answer mixes up the interpretation of odds and probabilities. "60% less likely" is understood by all to mean that the probability is 60% smaller, but in the example in this answer, .6 refers to the odds ratio, not a probability ratio. To get the reduction in probability you need to consider the intercept $\beta_0$ as well $\endgroup$ – Jake Westfall Apr 15 '20 at 15:11

The answers above are mostly correct, but not completely so.

In general exp(bi) is the estimate of odds(x+1)/odds(x). This can be shown with some simple arithmetic. As Alex noted, for the dummy in question, white is 1 and non-white is 0, so we can think about this as x+1 and x, when x = 0.

As such, if exp(bi) were 0.6, then:

odds(white)/odds(non-white) = 0.6


odds(white) = 0.6*odds(non-white)

which means the odds of being a smoker are 40% lower for white than non-white.

That does not mean that a white person is 40% less likely (that is ambiguous, but I assume they mean that the probability is multiplied by 0.6, not that 40% is subtracted off, but neither is true) to be a smoker. To see what it actually means, we need to know what the probability of a non-white person smoking is.

If, for example, it's 75%, then:

  • prob(smoke|non-white) = 0.75

  • odds(smoke|non-white) = 0.75/(1-0.75) = 3

  • odds(smoke|white) = 0.6*3 = 1.8

  • prob(smoke|white) = 1.8/(1.8 + 1) = 0.642

This means that the probability has been reduced by just 14.3% (0.642/0.75 = 0.857).

Whereas, if it's 50%, then:

  • prob(smoke|non-white) = 0.50
  • odds(smoke|non-white) = 0.50/(1-0.50) = 1
  • odds(smoke|white) = 0.6*1 = 0.6
  • prob(smoke|white) = 0.6/(0.6 + 1) = 0.375

This means that the probability has been reduced by 25% (0.375/0.5 = 0.75).

In fact, we only approach "40% less likely" in the limit as prob(smoke|non-white) goes to 0.

It's very common to use "odds" and "chance" interchangeably in conversation, but they are actually two very different things. It's one of the limits/challenges of logistic regression.

  • $\begingroup$ I wouldn't say that the marked answer is "mostly correct". It mixes up the most important part, i.e. odds != probabilities. Your answer points out the difference nicely! Also, it offers a correct interpretation of the results, which was the original question in the first place. $\endgroup$ – mc51 Jan 25 '20 at 19:34

"not being white" is the base or reference point. beta(not being white)=0 by definition, i.e. odds ratio(not being white)=1. If odds ratio(white)=0.6 then you are expected to be 100%-60%=40% less likely to be a smoker as compared to someone who is not white. Good Luck!


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