Why can’t a subject die at time 0?
Why can’t a subject enter and die at the same time?
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Title
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Failure time, censoring time, and entry time in the Cox model
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Author
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William Gould, StataCorp
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Date
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January 1999; updated July 2011
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Consider a subject who enters at t0 and dies at t1.
Stata interprets the interval as [t0,t1)—closed on the left, open on
the right—or equivalently, as t such that t0 <= t < t1. By that
logic, t0=t1=0 makes no sense since it results in the interval
[0,0)—the interval would be 0 <= t < 0.
Mechanically, when events happen at the same time, Stata interprets them as
really occurring in the order of failures, censorings, and finally entries.
Thus, a subject who entered and died at the same time would first die and
then enter the sample.
Solving the problem of subjects who die at time 0
What you really mean by this is that subjects entered at time 0 and then,
almost instantly, died. You need to change your death times to 0+epsilon,
where epsilon is some small number.
Choose epsilon so that 0+epsilon is less than the time of the first death
after time 0. What value of epsilon you choose that meets that constraint
does not matter.
Say the earliest failure among those failing after time 0 is time 1. You
could
. replace time = .1 if time==0
. stset ...
. stcox ...
stcox would report the same results if we changed the time 0 deaths
to occur at time .2:
. replace time = .2 if time==0
. stset ...
. stcox ...
The Cox proportional hazards model is sensitive only to the ordering of the
failure events, so as long as we keep the earliest failure events occurring
first, the results will remain unchanged.
Solving the problem of subjects who enter and die at the same time
This problem deals with situations where you explicitly specify both the
entry and the exit times:
. stset t1, failure(outcome) enter(t0) ...
The solution is
. replace t0 = t0 - .1 if t0==t1
. stset t1, failure(outcome) enter(t0) ...
where .1 is like epsilon in the previous case; it is a small number that
does not change the ordering of events.
We shift the entry time back, not the failure time forward. To
understand why, let’s say
Subject A enters and dies at 5.
Subject B enters at 0 at dies at 5.
This means that subjects A and B died at the same time. Thus we must keep
them dying at the same time and so shift the entry time of subject A to be
just a little before time 5. If we instead shifted subject A’s death
time forward a little bit, we would be saying that subject A died after B.
Thinking carefully about censoring and failure times
Stata orders the events occurring at the same time as failures, then
censorings, then entries.
Think about the following:
Subject C: enters at 0, censored at 5
Subject D: enters at 0, fails at 5
Could subject C have died at time 5? That is, was subject C in the risk
pool when D died?
Answer: yes. Here is how it happened:
At time 0:
First, deaths (remove from risk pool): none.
Then, censorings (remove from risk pool): none.
Finally, entries (add to risk pool): C and D.
At time 5:
First, deaths (remove from risk pool): D
Then, censorings (remove from risk pool): C
Finally, entries (add to risk pool): none
Therefore, C was in the risk pool when D died.
Is that what we meant when we wrote that Subject C was censored at 5 and D
died at 5? If what you mean is Subject C could not have died at time 5, you
need to change Subject C’s censoring time to be 5 minus a little, to
make it, say, 4.9.
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