# Binary logistic regression: Interpreting odds ratio vs. comparing predictive probabilities

Consider the following logistic regression:

set.seed(1839)
n <- 200
x <- rbinom(n, 1, .5)
y <- cut(x + rnorm(n), 2)
mod <- glm(y ~ x, family = binomial())
summary(mod)$coef  Which produces:  Estimate Std. Error z value Pr(>|z|) (Intercept) -1.172720 0.2244242 -5.225463 1.737202e-07 x 2.490961 0.3421728 7.279833 3.342331e-13  During my time in academia, I would report this as an OR of 12.07 (because exp(coef(mod)[[2]])) and say that "people with an x score of 1 are 12 times as likely then people with an x score of 0 to have a positive outcome on y" (where we assume that 1 is a positive outcome for y). Now that I'm in industry and working with clients more, people like to see raw percentage point increases. Which we can approximate from comparing the predicted values for people where x = 0 and x = 1: > inv_logit(coef(mod)[[1]]) [1] 0.2363636 > inv_logit(sum(coef(mod))) [1] 0.7888889  So we could show predicted probabilities (usually shown as percentages for ease to the reader) as 78.9% when x = 1 and 23.6% when x = 0. This is a difference of 55.3 percentage points. However, when we take those two predicted probabilities and divide them by one another, we don't get "12 times as likely." Instead, we get: > inv_logit(sum(coef(mod))) / inv_logit(coef(mod)[[1]]) [1] 3.337607  Which is 3.34 times, not 12.07. Now, I know that what I'm doing is:$p(y = 1 | x = 1) \over p(y = 1 | x = 0)$Whereas the odds ratio is doing:$p(y = 1 | x = 1) \over 1 - p(y = 1 | x = 1)$Which aren't the same, because inv_logit(coef(mod)[[1]]) + inv_logit(sum(coef(mod))) is greater than 1 (as it usually is). I feel as if I am missing something basic here, but can't quite place it. Can anyone tell me where I'm going wrong in my interpretation here? Or perhaps I'm interpreting the odds ratio incorrectly? How do I reconcile these differences? • EXP(B) in logistic regression is by how many times the estimated odds grows when you switch x by one unit. So, it is the ratio of two odds. – ttnphns Jun 14 '18 at 15:36 • ...that is [p(event)/p(noevent) | x=1] / [p(event)/p(noevent) | x=0] – ttnphns Jun 14 '18 at 15:47 • – gung - Reinstate Monica Jun 14 '18 at 17:45 ## 1 Answer We often interpret the odds ratio like you did:$\newcommand{\Odds}{\text{Odds}}\newcommand{\event}{\text{event}}\newcommand{\noevent}{\text{no event}}\$

people with an x score of 1 are 12 times as likely then people with an x score of 0 to have a positive outcome on y

But likely is purposefully vague here so as not to confuse non-statisticians.

What we really mean is

people with an x score of 1 have an odds 12 times greater than people with an x score of 0 of having a positive outcome on y

As ttnphns showed in the comments, this is expressed as $$\frac{\Odds(\event|x=1)}{\Odds(\noevent|x=0)} = \frac{\frac{p(\event|x=1)}{p(\noevent|x=1)}}{\frac{p(\event|x=0)}{p(\noevent|x=0)}}$$

Basically, this is the difference between interpreting odds ratio and relative risk. (See here for more explanation)

• Bingo, I knew I was missing something simple! Funny how the shorthand explanations we get from introductory material can end up confusing us way down the road. The "what we really mean is" you gave is very clear. – Mark White Jun 15 '18 at 13:53