From Wikipedia:

The hazard ratio is simply the relationship between the instantaneous hazards in the two groups and represents, in a single number, the magnitude of distance between the Kaplan-Meier plots

As far as I know, a Kaplan-Meier plot is the estimate of the survival function under a level of the covariate. The hazard ratio is the ratio between the hazard function values at two different levels of the covariate. So I was wondering why the hazard ratio represents the magnitude of distance between the Kaplan-Meier plots?

  • $\begingroup$ The very next paragraph in the article you quote proceeds to answer this question. $\endgroup$
    – whuber
    Commented Aug 6, 2013 at 19:52
  • 1
    $\begingroup$ @whuber: Thanks. My understanding about that paragraph is limited. My question is rather "why" than "how". My confusion comes from that one is for survival function, and the other is for hazard function. $\endgroup$
    – Tim
    Commented Aug 6, 2013 at 19:56

3 Answers 3


The Cox proportional hazards model can be written in terms of the effect of predictor variables on the log relative hazard, which is also the effect on the log relative cumulative hazard scale. Log cumulative hazard is equal to the log of the -log of the cumulative survival function which Kaplan-Meier estimates. So you could say that the log hazard ratio (regression effect in the Cox model) estimates the average difference between two Kaplan-Meier estimates if you transform both of them by the log-log transformation.

  • $\begingroup$ I'd rather say: the log hazard ratio estimates the difference between the two "log minus log'' survival functions (this difference is constant by the proportional hazards assumption). Do you mean that the log hazard ratio is a kind of average between the two ``log minus log'' Kaplan-Meier estimates ? Is it the maximum likelihood estimate in the Cox model ? Actually I don't know how it is defined. $\endgroup$ Commented Aug 22, 2013 at 19:11

In the Cox proportional hazard models for two groups we assume that the hazard rate $\lambda_1$ in group 1 is related to the hazard rate $\lambda_2$ in group 2 by $\lambda_2(t)=k\lambda_1(t)$ for a certain proportionality constant $k>0$.

Then, denoting by $S_1$ and $S_2$ the survival functions in group 1 and group 2 respectively, the proportional hazard assumption is equivalently written $\log S_2(t) = k \log S_1(t)$ (by integration), and then $\boxed{\log(-\log S_2(t))=\log(k)+\log(-\log S_1(t))}$.

The boxed formula was already claimed in @FrankHarrell's answer.

I would add that this boxed formula provides a visual method to check the proportional hazards assumption: plot the two "$\log$ minus $\log$'' transformations of the Kaplan-Meier estimates $\hat S_1$ and $\hat S_2$ on the same graphic; under the proportional hazards assumption it is expected that the two curves are "parallel". With R, you get this graphic by typing:

fit <- coxph(Surv(time,status)~strata(group), mydataframe)
plot(survfit(fit), col=c("blue","red"), fun="cloglog") ### blue: group1, red: group2

The premise of the question is wrong. (So read Wikipedia with a critical eye. Most of the rest of that article appears correct, but that sentence is flawed as is the one immediately preceding it that mentions "odds".) The hazard ratio would be exp( log( H(t|covar=1)) - log( H(t|covar=2)) ) where the quantity inside the function that is difference in the log-cumulative-hazards estimated at values of covariates that differ by one unit. But that is not the ratio of KM(t) values. KM(t|covar) is 1-integral(h(t|covar)).

Differences on log transformed scales are ratios on the original scale. It is easy to see the error at the extremes of time = 0 (where S=1 by definition) and at time = infinity (where S=0) that the ratios of the KM plots will be 1 even with differing covariates.

If you want to look at instantaneous hazards, h(t) = d(H(t))/dt, then h(t|covar=2)/h(t|covar=1) will be an "instantaneous hazard ratio".


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