I have a logit model where the explanatory variable is the percent of total a total. As this is not a binary model I am unsure how to read the regression results.

I understand that the coefficient normally equates to the log odds which I can then convert to the probability using:

exp(coef) / (1 + exp(coef))

but I am still unsure how to interpret this in terms of the share of total explanatory.

for example: b_1 = 0.7166

so after converting to a probability: b_0 = exp(0.7166) / (1 + exp(0.7166)) = 0.6718

If I were interpreting this like an OLS (which I do not believe makes sense) I would say that a 1 point increase in the explanatory variable is associated with a 0.6718 percentage point increase in the dependent variable.

However, I do not know how to construct similar statement with the logit.

To make matters more complicated (at least to me) my explanatory variable is expressed as a decimal percent (i.e 10% = 0.10).

Should I first divide my coefficient by 100 so that initial part of my statement is correct "A one percentage point increase in.."?

Ultimately my goal is to be able to make a statement about how a change in the explanatory variable affects the share value of my dependent variable.


1 Answer 1


Let's look at some particular toy data in R which also gives a coefficient of about $0.71$:

datf <- data.frame(x = c(0.19, 0.21, 0.37, 0.40, 0.55, 0.56),
             success = c(  0,    1,    1,    1,    0,    1))
fit <- glm(success ~ x, data=datf, family="binomial")
#    Coefficients:
# (Intercept)            x  
#      0.4251       0.7102  

What this is saying is the maximum likelihood logistic curve has

  • the log-odds as $0.4251+0.7102x$
  • the odds as $e^{0.4251+0.7102x}= 1.52968 \times 2.0343^x$
  • the probability as $\frac{1.52968 \times 2.0343^x}{1+1.52968 \times 2.0343^x}$

So when $x=0$ you would get predicted log-odds of $0.4251$, odds of $1.5297$ and a probability of $0.6047$; when $x=1$ you would get predicted log-odds of $1.1352$ ($0.7102$ more), odds of $3.1118$ (multiplying by $e^{0.7102}=2.0343$) and a probability of $0.7568$

A $1$ percentage point increase in $x$ increases the predicted log-odds by $0.004251$ and multiplies the predicted odds by $2.0343^{0.01}=1.00713$. You cannot make such a simple statement about the predicted probabilities.

  • $\begingroup$ Thank you this is tremendously helpful!!! $\endgroup$ Sep 17, 2021 at 16:35

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