I am taking a course in logistic regression, and currently my class is about to finish our discussion about simple logistic regression. My professor said that the following statement is correct:

For one unit increase in x, the odds(event) is increased (decreased) by the factor $\exp(β_1)/(1-\exp(β_1))$ when $β_1$ is positive (negative).

I understand all of this except for the $(1-\exp(\beta_1))$ part. Why wouldn't the $\rm odds(event)$ just decrease by $\exp(\beta_1)$?

Where $\rm odds(event) = P(event)/(1-P(event))$ and $x$ is a continuous predictor.

  • $\begingroup$ I took the liberty of editing your post to utilize the markup features CV supports. Please ensure it still says what you want it to. $\endgroup$ Commented Feb 15, 2015 at 16:17

3 Answers 3


For inference in logistic regression, it would be easier to think in terms of log odds instead of odds. A simple logistic regression model is a generalized linear model with the form $$ \newcommand{\logit}{\rm logit} \newcommand{\odds}{\rm odds} \newcommand{\expit}{\rm expit} \logit(\pi_i) = \beta_0 + \beta_1X_{i1}, $$ where $\logit(\pi_i) = \log(\frac{\pi_i}{1-\pi_i}) = \log(\odds(\pi_i))$. Notice that it has the exact same form as linear regression with a Bernoulli-distributed transformed dependent variable. This has the more intuitive interpretation that for every increase in $X_{i1}$, the expected log odds in favor of $X_{i1}$ increase by $\beta_1$.

If you want to interpret the model in terms of odds, you just have to exponentiate the logit, which is what you initially assumed. This gives you $\odds(\pi_i) = \exp(\beta_0 + \beta_1X_{i1})$, and not $\odds(\pi_i) = \frac{\exp(\beta_0 + \beta_1X_{i1})}{1 + \exp(\beta_0 + \beta_1X_{i1})}$. The interpretation in this case is that for every increase in $X_{i1}$, the expected odds in favor of $Y_i = 1$ increase by $\exp(\beta_1)$. I think either you or your professor got odds and probability mixed up in your professor's explanation.

If you want to draw inference on probabilities, you need to invert the $\logit$ transformation by taking the $\expit$ of both sides, where $\expit(x) = \frac{\exp(x)}{1 + \exp(x)}$. In that case, we have $\pi_i = \frac{\exp(\beta_0 + \beta_1X_{i1})}{1 + \exp(\beta_0 + \beta_1X_{i1})}$ as our expected probability that $Y_i=1$. In this case, it is difficult to draw inference in terms of probabilities, which is why we transform back to probabilities by taking the expit of our expected log odds.


If your professor said that, they were wrong. The estimated betas ($\hat\beta_1$) in logistic regression are on the scale of the linear predictor. That is, they are changes in log odds. Exponentiating them (i.e., $\exp(\hat\beta_1)$) converts them from additive changes in log odds to multiplicative changes in odds. In other words, you are right, the odds would decrease (increase) by a factor of $\exp(\hat\beta_1)$. This follows from the definition of logarithms and exponentiation. When the odds are divided by $1+\rm odds$ (not $1-\rm odds$), then you will get a predicted probability.

In addition, your professor's expressions seem to be missing the intercept, $\hat\beta_0$, unless it was suppressed (a bad idea). That is, the predicted probability of 'success' ($Y=1$) at a particular point, $X=x_i$, is $\exp(\beta_0 + \beta_ix_i)/(1+\exp(\beta_0 + \beta_ix_i))$. This seems to be confused on multiple levels; that may be a bad sign.


The manner in which I asked this question may have been misleading:

Let B1 = -1 for a dichotomous variable X.

Note that Odds(X=1) = exp(B1)*Odds(X=0)

If B1 < 0 then the odds decreases by the factor 1 - exp(B1)

Ex: Let B1 = -1 then Odds(X=1) = exp(-1)*Odds(X=0) = .367*Odds(X=0)

So, in fact, we have a 1- e^(-1) = 1-.367 = .633 => 63.3% decrease.

This is what I meant. Sorry for any confusion.


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