2
$\begingroup$

In my model I am considering the rate of hospital re-admission (outcome) and my covariates are non-log transformed while my main variable of interest - direct cost of home rehabilitation - is ln transformed (original values expressed in euros). My HR is 0.92 and I understand that increasing the cost of rehabilitation reduces the rate of re-admission. However, I would like to discuss the result in a easily understandable way (i.e. 10% increase of the direct cost of home rehabilitation associated with a xx reduction of the re-admission rate, or 1% increase, or percentage difference in rates for each 1 or 10 euros incremental). Could you help me converting it?

Thanks

$\endgroup$
1
$\begingroup$

There's an excellent discussion of how to interpret the coefficient of Cox regression models at How do I interpret Exp(B) in Cox regression?. In short, for coefficient $\hat\beta$, the interpretation is that a unit increase of the related variable, holding all other variables constant, is associated with the hazard being multiplied by $exp(\hat\beta)$. In your case, a unit increase in the log of the direct cost of home rehabilitation is associated with a 8% decrease in the hazard of re-admission.

All that remains is to convert "a unit increase in the log of the direct cost of home rehabilitation" into more interpretable terms. Assuming you are using the natural log, this unit increase means the cost is multiplied by $2.718\ldots$. As a result, the interpretation would be "a 2.718 times increase in the direct cost of home rehabilitation, holding all other variables constant, is associated with a 8% decrease in the hazard of re-admission."

You could also transform to an easier to interpret multiplier: "a 10% increase in the direct cost of home rehabilitation, holding all other variables constant, is associated with a 0.8% decrease in the hazard of re-admission." To obtain this statement, note that a 10% increase in direct cost of home rehabilitation is the same as a $log(1.1) = 0.095$ increase in your variable, so the hazard ratio is $exp(0.095\hat\beta) = 0.992$ in your case.

$\endgroup$
  • $\begingroup$ Hi, Thank you so much for your reply and link! I was trying to replicate your calculation and everything is clear but the last 0.992, how did you obtain it? $\endgroup$ – Vincent Jul 9 '16 at 20:59
  • $\begingroup$ @Vincent given that you reported a hazard ratio of 0.92, this means your coefficient in the Cox regression model is roughly $ln(0.92) = -0.083$, which I call $\hat\beta$. Given that you are interested in a 10% increase in this variable (aka it is multiplied by 1.1), you compute $ln(1.1) = 0.095$. The associated hazard ratio of your 10% increase in your variable is therefore $e$ raised to the power $-0.083\times0.095$, which comes out to be 0.992. Thus the 10% increase in the variable value is associated with a 0.8% decrease in hazard. $\endgroup$ – josliber Jul 9 '16 at 21:29
1
$\begingroup$

There is a very useful collection of general information about log-transformed variables in different combinations on this Cross Validated page.

For Cox models where you want to express a hazard ratio for some particular percentage change in a continuous predictor, it can be useful to make an appropriate change of base of the logarithm before you perform the regression.

For example, $\log_22$=1, so a doubling of cost would represent a 1-unit increase in the $\log_2$ scale. The hazard ratio associated with the coefficient for a 1-unit change in this $\log_2$ transformation of the cost variable would thus be the hazard ratio associated with a doubling of cost. If you want to express in terms of hazard ratio associated with a 10% change instead, then use $\log_{1.1}$. The option to change the logarithmic base might seem to be hidden if you want something other than natural log (base $e$), $\log_{10}$, or $\log_2$, which often have simple ways to specify. In R you can specify the desired base as log(x,base).

This is pretty much the approach suggested by @josliber already, but might save some hand calculation.

$\endgroup$

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.