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One of the predictors in my logistic model has been log transformed. How do you interpret the estimated coefficient of the log transformed predictor and how do you calculate the impact of that predictor on the odds ratio?

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If you exponentiate the estimated coefficient, you'll get an odds ratio associated with a $b$-fold increase in the predictor, where $b$ is the base of the logarithm you used when log-transforming the predictor.

I usually choose to take logarithms to base 2 in this situation, so I can interpet the exponentiated coefficient as an odds ratio associated with a doubling of the predictor.

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    $\begingroup$ Interesting. I always use natural logs because many of the coefficients tend to be close to zero and then can be interpreted as proportional (relative) differences. That's not possible in any other base of logarithm. I see some merit in using other bases, but I think you need to clarify your response, because prima facie your interpretation doesn't use the value of the coefficient at all! $\endgroup$ – whuber Mar 15 '11 at 15:12
  • $\begingroup$ @whuber sorry what does prima facie mean? First face?? $\endgroup$ – onestop Mar 16 '11 at 7:56
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    $\begingroup$ google.com/search?q=define+"prima+facie" $\endgroup$ – whuber Mar 16 '11 at 14:26
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@gung is completely correct, but, in case you do decide to keep it, you can interpret the coefficient has having an effect on each multiple of the IV, rather than each addition of the IV.

One IV that often should be transformed is income. If you included it untransformed, then each (say) \$1,000 increase in income would have an effect on the odds ratio as specified by the odds ratio. On the other hand, if you took log(10) of income, then each 10 fold increase in income would have the effect on the odds ratio specified in the odds ratio.

It makes sense to do this for income because, in many ways, an increase of \$1,000 in income is much bigger for someone who makes \$10,000 per year than someone who makes \$100,000.

One final note - although logistic regression makes no normality assumptions, even OLS regression doesn't make assumptions about the variables, it makes assumptions about the error, as estimated by the residuals.

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    $\begingroup$ +1, good points. I suppose I could have been more complete. In addition, I turned off the inadvertent mathjax by putting a backslash "\" immediately before the dollar signs. I hope you don't mind. $\endgroup$ – gung Jul 18 '12 at 1:05
  • $\begingroup$ What do you mean by 'logistic regression makes assumption about the errors' ? $\endgroup$ – user83346 Jan 16 '16 at 8:20
  • $\begingroup$ No, OLS regression makes assumptions about the errors. That's what I said. $\endgroup$ – Peter Flom Jan 16 '16 at 14:08
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This answer is adapted from The Statistical Sleuth by Fred L. Ramsey and Daniel W. Schafer.

If your model equation is:

$log(p/(1-p)) = \beta _{0} + \beta log(X)$

Then, each $k$-fold increase in $X$ is associated with a change in the odds by a multiplicative factor of $k^{\beta }$.

For example, I have the following model for presence of bed sores regressed on length of stay at a hospital.

$log(odds of bedsore)= -.44 + 0.45(length of stay)$

So my $\beta = 0.45$.

You can choose any $k$, based on what's works best for your model's interpretability.

I decide that $k=2$ and get the following:

$k^{\beta } = 2^{0.45} = 1.37$

Each doubling ($k=2$) of the length of stay is associated with a change in the odds of getting a bedsore by a factor of 1.37. Or if you double my length of stay, my odds of getting a bedsore will be 137% what they would have been otherwise.

Or if you decide $k=0.5$.

$k^{\beta } = 0.5^{0.45} = 0.73$

Each halving ($k=0.5$) of the length of stay is associated with a change in the odds of getting a bedsore by a factor of .73. Or if you cut my length of stay in half, my odds of getting a bedsore will only 73% of what they would have been otherwise.

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protected by whuber Jul 20 '12 at 13:22

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