In logistic regression, we derive the odds ratio by exponentiating the coefficient (estimate). If one of my independent variables ($X_1$) was log transformed (natural log), how should I interpret the odds ratio in such cases?

For example:

  • $X_1$ has an OR of 0.46 (95% C.I: 0.24, 0.93) for a binary outcome of $Y$.
  • $X_1$ was log transformed (natural log)

In the above results, usually (if no log transformation was done) one would have interpreted as "for every one unit increase in $X_1$, the odds of outcome $Y$ would have increased by 0.46". But this variable was log transformed (natural log, base = 2.72) during the pre-processing stage of the model creation.

Any suggestions pls


Let's look at what's going on behind the scenes. A logistic regression model (binomial GLM) looks like this: $$ \log\left(\frac{\pi}{1-\pi}\right)=\beta_0+\beta_1X_1 $$ assuming we're only fitting an intercept and one parameter for your $X_1$ variable.

As you correctly state, exponentiating the coefficients gives us the odds: $$ \exp\left[\log\left(\frac{\pi}{1-\pi}\right)\right]=\exp(\beta_0+\beta_1X_1)\\ \implies \frac{\pi}{1-\pi} = \exp(\beta_0)\exp(\beta_1X_1) $$ For simplicity, let's focus on $\beta_1$ and assume that $\beta_0=0$. This makes $\exp(\beta_0)=\exp(0)=1$, and therefore the above expression reduces to: $$ \frac{\pi}{1-\pi} = \exp(\beta_1X_1) $$ Now, let's look at what happens when $X_1$ increases by one unit e.g. from $2$ to $3$. Let's also imagine $\beta_1=-0.74$: $$ (1) \space\space\frac{\pi}{1-\pi} = \exp(-0.74\times2)=\exp(-1.48)\approx0.23 \\ (2) \space\space\frac{\pi}{1-\pi} = \exp(-0.74\times3)=\exp(-2.22)\approx0.11 $$ A unit increase in $X_1$ has resulted in a decrease of $0.11/0.23=0.478$ in the odds. Note that this is just the odds ratio, and $\exp(\beta_1)=\exp(-0.74)=0.478$. So a unit increase in $X_1$ results in the odds of success being multiplied by the odds ratio, which is nothing but $\exp(\beta_1)$. Essentially, if $\beta_1>0$, an increase in $X_1$ will yield higher odds (because the odds ratio > 1), whereas if $\beta_1<0$, an increase in $X_1$ will yield lower odds (because the odds ratio < 1).

Now, you should be able to see for yourself what would happen if we were working with $\ln(X_1)$ rather than with $X_1$. Hint: $$ (3) \space\space\frac{\pi}{1-\pi} = \exp(-0.74\times\ln(2))=2^{-0.74}\approx0.60\\ (4) \space\space\frac{\pi}{1-\pi} = \exp(-0.74\times\ln(3))=3^{-0.74}\approx0.44 $$ That is, a decrease in the odds of $0.44/0.60=0.74$. However, the odds ratio is still $\exp(\beta_1)=\exp(-0.74)=0.478$, so what's going on here?

By taking the natural logarithm of $X_1$, you change the interpretation of its coefficient. Now, the odds ratio represents the change in the odds if you multiply $X_1$ by $e$. Therefore: $$ (5) \space\space\frac{\pi}{1-\pi} = \exp(-0.74\times\ln(2))=2^{-0.74}\approx0.599\\ (6) \space\space\frac{\pi}{1-\pi} = \exp(-0.74\times\ln(2\times e))=(2\times e)^{-0.74}\approx0.286 $$ We see that multiplying $X_1$ by $e$ results in the odds decreasing from $0.599$ to $0.286$. Here $0.286/0.599=0.478$ which is precisely the odds ratio i.e. $\exp(\beta_1)=\exp(-0.74)$.

Following the same logic, it's easy to see that:

  • If we had transformed $X_1$ to be $\log_2(X_1)$, the interpretation would be "by how much would the odds decrease after doubling $X_1$"
  • If we had transformed $X_1$ to be $\log_{10}(X_1)$, the interpretation would be "by how much would the odds decrease after multiplying $X_1$ by 10"
  • $\begingroup$ Luz thanks for the wonderful and detailed post. Apologies that still my confusion remains towards the end of the post...may be I am not mathematically sound. $\endgroup$
    – Venkat Pgi
    Sep 21 '21 at 10:54
  • $\begingroup$ Luz Thanks so much for the detailed reply My doubt is - from the last step - does this mean, I should exponentiate 1.584 further to get the odds ratio. Say something like 2.72^1.584 which will be 4.879 $\endgroup$
    – Venkat Pgi
    Sep 21 '21 at 10:59
  • $\begingroup$ @VenkatPgi The odds ratio is just $\exp(\beta_1)=\exp(0.46)=1.584$. $\endgroup$
    – Adrià Luz
    Sep 21 '21 at 11:07
  • $\begingroup$ Luz thanks once again...may be it was not clear from my initial question. The estimate which I got after running logistic regression was -0.743. After exponentiating, it was 0.46. If the original estimate is having a negative sign, the change in predictor would cause a change in the reverse direction in the outcome, right? In that case, should I do like this: exp(-0.743 x ln(3))/exp(-0.743 x ln(2))? $\endgroup$
    – Venkat Pgi
    Sep 21 '21 at 11:20
  • $\begingroup$ @VenkatPgi Ah I see, let me update with $\beta_1=-0.74$. $\endgroup$
    – Adrià Luz
    Sep 21 '21 at 11:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.