2
$\begingroup$

The Cox proportional hazards model for survival data with covariate ${\bf z}$ is defined through the hazard function $h(t,{\bf z})$ by $$ h(t,{\bf z}) = h_0(t)~\cdot\theta~~,~~~\theta = \theta(\beta, {\bf z}).$$The proportional odds model is sometimes given by

$$ O(t,{\bf z}) = O_0(t)\cdot\theta $$ where

$O_0(t) = O(t,{\bf 0})$ and $$ O(t,{\bf z}) = \frac{1-S(t,{\bf z})}{S(t,{\bf z})}$$ is the odds of an event occurring in the time interval $(0,t)$ for an individual with covariate ${\bf z}.$

My question is this: Is there a way to define the proportional odds model via its hazard function?

$\endgroup$
  • $\begingroup$ Relevant: the complementary log-log link can be used to model discrete survival, cases where "time-to-event" is either aggregated over intervals (such as number of drug injections until infection from hepatitis C, or such). $\endgroup$ – AdamO Jan 10 '18 at 20:04
2
$\begingroup$

Yes. You can use the relationship between the survival and hazard function: $$\lambda(t) = \frac{d}{dt} -\ln S(t) \iff S(t) = e^{-\int_0^t \lambda(t)\,dt}.$$

This gives $$O(t) = \frac1{S(t)} - 1 = -1 + e^{\int_0^t \lambda(t)\,dt}$$ which you can simplify however you want.

$\endgroup$

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.