[R] [Re: Does a survival probability(the probability not, experiencing an event) have to be non-increasing?

Andrews, Chris chrisaa at med.umich.edu
Sun Mar 23 00:40:49 CET 2014


The survival function, S(t), gives you the probability of surviving beyond time t starting from time 0.  If you want to know the probability of surviving beyond time t *given* that you survived to get a heart surgery at time u>0, that is a different function.  It might be S(t)/S(u) depending on what you are willing to assume.


________________________________________
From: Zhiyuan Sun [sam.d.sun at gmail.com]
Sent: Friday, March 21, 2014 10:48 PM
To: Therneau, Terry M., Ph.D.
Cc: r-help at r-project.org
Subject: Re: [R] [Re: Does a survival probability(the probability not, experiencing an event) have to be non-increasing?

Thanks Terry. Your explanation is right on the point. You solved my
question about time-dependent covariate. When calculating survival at
time t, you have to consider the exposure history before ant at time
t.   It is reasonable when we assume cumulative hazard before time t
can be carried over to the next time, i.e., hazard is cumulative, no
mater what happened.

However,I still have a question. Is it possible that some
time-dependent covairate(like heart surgery) that at some point,
completely eliminated previous risk, so that cumulative hazard before
that point can no longer be assumed to be carried over to next time
point? For example, when a bad part of a machine was replaced, the
machine got a new life. Or, when  a surgery was done to a patient, the
patient return to a healthy status.  Intuitively, previous hazard
should not be accumulated anymore after such major events happened to
a person or a machine. They almost get new "lives". It's like a
renewal/recharge process.  So, intuitively, survival should be
recalculated then, instead of keep "non-increasing".   I hope I
explained my question clearly.

Thanks if you can provide any further clarification on my question.

Thanks,
Zhiyuan



On Thu, Mar 20, 2014 at 8:46 AM, Therneau, Terry M., Ph.D.
<therneau at mayo.edu> wrote:
>
>
> On 03/20/2014 06:00 AM, r-help-request at r-project.org wrote:
>>
>> My question is related to a cox model with time-dependent variable.
>> When I think about it more, I get a little confused about
>> non-increasing assumption for survival probability for an individual.
>> For example, for a time-dependent ,say x, assuming increasing x
>> increases the risk of event. Assume,time t1 < t2.  If at x at t1<< x
>> at t2, obviously, hazard at t1 will less than hazard at t2, assuming
>> no other covariaates. But is it possible that s(t2|x at t2) > s(t1|x
>> at t1), since at t2, an individual is at greater risk.  This is kind
>> of confusing to me.
>>
>> Thanks for any helpful insights!
>
>
> Time dependent covariates and survival curves are confusing to a lot of
> people.
> The Cox model is a hazard model
>         h(t, x) = h_0(t) exp(x beta)
> which means it is a model of the moment-by-moment risk.  A time dependent
> model
> replaces x with x(t) which is the moment-by-moment value of x.
>
> After the model is fit, one can compute the
> time dependent cumulative hazard as
>
>       H(t,x) = \integral_0^t  h_0(s) exp(x(s) beta) ds
> and the survival is S = exp(-H).
>
> Since everthing inside the integral is positive H(t) has to be an increasing
> function of t, and thus S a decreasing one.  The key thing to note is that H
> or S depend on the entire covariate history for a subject.  If you have a
> subject whose value of "x" changes from 1 to 2 at time 10, when computing
> their survival at time 15 you cannot just use a value of "2" all the way
> from 0 to 15 in the formula.
>
>  Many Cox model programs (e.g.SAS) allow for time dependent covariates when
> computing the Cox fit, but then only allow for fixed covariates when
> computing a curve.  You can only do predictions for people whose covariates
> never change.  (For some diseases I work with such people do not exist, e.g.
> in PBC your bilirubin WILL rise with time.  So such a curve is useless).
> This adds to the confusion.
>
> Terry T.


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