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Logistic regression models the log odds of an event as some set of predictors. That is, $\log(p/(1-p))$ where p is the probability of some outcome. Thus, the interpretation of the raw logistic regression coefficients for some variable ($x$) has to be on the log odds scale. That is, if the coefficient for $x = 5$ then we know that a 1 unit change in x correspondents to 5 unit change on the log odds scale that an outcome will occur.

However, I often see people interpret exponentiated logistic regression coefficients as odds ratios. However, clearly $\exp(\log(p/(1-p))) = p/(1-p)$, which is an odds. As far as I understand it, an odds ratio is the odds of one event occurring (e.g., $p/(1-p)$ for event A) over the odds of another event occurring (e.g., $p/(1-p)$ for event B).

What am I missing here? Is seems like this common interpretation of exponentiated logistic regression coefficients is incorrect.

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  • $\begingroup$ The value of $exp(log(p/(1-p)))$ is not an exponentiated logistic regression coefficient, unless the model contains only an intercept. In general, the intercept $β_0$ is the log of the odds of having the outcome when all $Xs$ are zero. $\endgroup$ Commented May 3, 2022 at 20:44

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@Laconic's answer is great and complete, in my opinion. Something I wanted to add is that the original coefficients describe a difference in the log odds for two units who differ by 1 in the predictor. E.g., for a coefficient on $X$ of 5, we can say that the difference in log odds between two units who differ on $X$ by 1 is 5. Mathematically,

$$\beta = \log(\text{odds}(p|X=x_0+1))-\log(\text{odds}(p|X=x_0)) $$

When you exponentiate $\beta$, you get

$$\exp(\beta) = \exp(\log(\text{odds}(p|X=x_0+1))-\log(\text{odds}(p|X=x_0))) \\ = \frac{\exp(\log(\text{odds}(p|X=x_0+1)))}{\exp(\log(\text{odds}((p|X=x_0)))} \\ = \frac{\text{odds}(p|X=x_0+1)}{\text{odds}(p|X=x_0))}$$

which is a ratio of odds, an odds ratio.

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    $\begingroup$ This is extremely clear to me. My question is resolved. $\endgroup$ Commented Aug 9, 2018 at 18:51
  • $\begingroup$ @Noah I'm probably missing something, but why do you go from exp(something/something) to exp(something)/exp(something)? This does not seem to hold. $\endgroup$ Commented Sep 23, 2024 at 19:31
  • $\begingroup$ @SantiagoValdivieso Where do I do that? The first line is exp(something - something) which does equal exp(something)/exp(something). $\endgroup$ Commented Sep 23, 2024 at 20:36
  • $\begingroup$ Oh, right! Thank you! $\endgroup$ Commented Sep 23, 2024 at 23:27
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Consider two set of conditions, the first described by the vector of independent variables $X$, and the second described by the vector $X'$, which differs only in the ith variable $x_i$, and by one unit. Let $\beta$ be the vector of model parameters as usual.

According to the logistic regression model, the probability of the event occurring in the first case is $p_1 = \frac{1}{1 + \exp(-X \beta)}$, so that the odds of the event occurring is $\frac{p_1}{1-p_1} = \exp(X \beta)$.

The probability of the event occurring in the second case is $p_2 = \frac{1}{1 + \exp(-X' \beta)}$, so that the odds of the event occurring is $\frac{p_2}{1-p_2} = \exp(X' \beta) = \exp(X \beta + \beta_i)$.

The ratio of the odds in the second case to the odds in the first case is therefore $\exp(\beta_i)$. Hence the interpretation of the exponential of the parameter as an odds ratio.

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