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I've been studying the Cox Proportional Hazards model, and this question is glossed over in most texts.

Cox proposed fitting the coefficients of the Hazard function using a partial likelihood method, but why not just fit the coefficients of a parametric Survival function using the maximum likelihood method and a linear model?

In any cases where you have censored data, you could just find the area under the curve. For example, if your estimate is 380 with standard deviation of 80, and a sample is censored >300, then there is an 84% probability for that sample in the likelihood calculation assuming normal error.

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  • $\begingroup$ As much as I like to have actuarial science questions here, I have to say this question is probably going to get a better response in the statistics site, Cross Validated. You can request a moderator migrate it. $\endgroup$ – Graphth Jul 18 '13 at 14:44
  • $\begingroup$ Alright, didn't realize that existed. Not sure how to request a migration. Please migrate? $\endgroup$ – user1956609 Jul 18 '13 at 15:00
  • $\begingroup$ @Graphth, I also didn't realize there was one... I didn't find it in the "all sites" list, could you link to it here? Thanks $\endgroup$ – jameselmore Jul 18 '13 at 15:06
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If you know the parametric distribution that your data follows then using a maximum likelihood approach and the distribution makes sense. The real advantage of Cox Proportional Hazards regression is that you can still fit survival models without knowing (or assuming) the distribution. You give an example using the normal distribution, but most survival times (and other types of data that Cox PH regression is used for) do not come close to following a normal distribution. Some may follow a log-normal, or a Weibull, or other parametric distribution, and if you are willing to make that assumption then the maximum likelihood parametric approach is great. But in many real world cases we do not know what the appropriate distribution is (or even a close enough approximation). With censoring and covariates we cannot do a simple histogram and say "that looks like a ... distribution to me". So it is very useful to have a technique that works well without needing a specific distribution.

Why use the hazard instead of the distribution function? Consider the following statement: "People in group A are twice as likely to die at age 80 as people in group B". Now that could be true because people in group B tend to live longer than those in group A, or it could be because people in group B tend to live shorter lives and most of them are dead long before age 80, giving a very small probability of them dying at 80 while enough people in group A live to 80 that a fair number of them will die at that age giving a much higher probability of death at that age. So the same statement could mean being in group A is better or worse than being in group B. What makes more sense is to say, of those people (in each group) that lived to 80, what proportion will die before they turn 81. That is the hazard (and the hazard is a function of the distribution function/survival function/etc.). The hazard is easier to work with in the semi-parametric model and can then give you information about the distribution.

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    $\begingroup$ Nice answer. What is unique about time is that it passes in one direction, and once we have withstood a high-risk period we are mainly interested in risks now in effect. That is what the hazard function tells us. $\endgroup$ – Frank Harrell Jul 18 '13 at 19:20
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    $\begingroup$ Another point that is worth adding is that with censored data, inspecting distributional assumptions can be very difficult. For example, suppose only 20% of your subjects observe an event. Trying to determine whether the tails of the distribution follow a Weibull distribution is clearly not going to be possible! A Cox-PH model somewhat sidesteps the issue (but you have to be very wary of the proportional hazards assumption if you want to extrapolate to the areas of times that were highly censored) $\endgroup$ – Cliff AB Dec 21 '17 at 21:08
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"We" don't necessarily. The range of survival analysis tools ranges from the fully non-parametric, like the Kaplan-Meier method, to fully parametric models where you specify the distribution of the underlying hazard. Each has their advantages and disadvantages.

Semi-parametric methods, like the Cox proportional hazards model, let you get away with not specifying the underlying hazard function. This can be helpful, as we don't always know the underlying hazard function and in many cases also don't care. For example, many epidemiology studies want to know "Does exposure X decrease the time until event Y?" That they care about is the difference in patients who have X and who do not have X. In that case, the underlying hazard doesn't really matter, and the risk of misspecifying it is worse than the consequences of not knowing it.

There are times however when this also isn't true. I've done work with fully parametric models because the underlying hazard was of interest.

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    $\begingroup$ "...and the risk of misspecifying it is worse than the consequences of not knowing it." This was very helpful, thank you. $\endgroup$ – user44845 Apr 6 '16 at 0:29
  • $\begingroup$ Could you give an example in when would the underlying hazard be of interest ? $\endgroup$ – Dan Chaltiel Mar 21 at 11:08
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    $\begingroup$ @DanChaltiel Any estimate that's intended to go into a mathematical model or the like would be an example - the underlying hazard function there is of particular interest. $\endgroup$ – Fomite Apr 2 at 19:46

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