In one paper describing results of survival analysis I have read a statement that implies that one can translate Hazard ratio (HR) into ratio of median survival times ($M_1$ and $M_2$) using the formula:
$HR = \frac{M_1}{M_2}$
I'm sure it doesn't hold when one cannot assume proportional hazard model (as nothing works if HR is not well-defined). But I suspect, that even then it wouldn't work for any survival distribution except exponential. Is my intuition right?
Best Answer
Your intuition is correct. The following relationship between survival functions holds: $$ S_1(t)=S_0(t)^r $$ where $r$ is the hazard ratio (see, e.g. the Wikipedia article Hazard ratio). From this we may show that your statement implies an exponential survival function.
Let us denote the medians by $M_r$, $M_1$ for two variables with hazard ratio $r$. Your statement implies $$ M_r = M_0/r $$ From the definition of the median, we get $$ S_r(M_0/r)=0.5 $$ Then, we substitute the relationship between survival functions $$ S_0(M_0/r)^r=0.5 \Rightarrow S_0(M_0/r) = 0.5^{1/r} $$ This holds for any $r$, hence $$ S_0(t) = 0.5^{t/M_0} = e^{t\frac{\log 0.5}{M_0}} $$ Hence, if the statement in your question holds for arbitrary HR, the survival distribution must be exponential.