Interpretation of Odds Ratio When Independent Variable is Log Transformed

data transformationgeneralized linear modelinterpretationlogisticmultiple regression

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

Best Answer

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"
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