1
$\begingroup$

Given a survival curve, to find the median one can visualize a line at 0.5 and then find the time at which the survival curve intersects with the line. But what if the Kaplan-Meier curve is horizontal at S(t)=0.5?

I´ve tried to come up with the solution by thinking of S(M)=0.5 and solving that equation for M. but it does not make sense to me. Is it possible to find an exact time point for the median survival, or will I get an interval?

$\endgroup$
3
  • $\begingroup$ Please don't apologize for asking a question. This is a perfectly good question (+1). Also, please don't add 'thanks' to your posts, on this site there's no need to say "thank you" at the end of your post - it might seem rude at first, but it's part of the philosophy of this site to "Ask questions, get answers, no distractions", and it means future readers of your question don't need to read through the pleasantries. $\endgroup$ Nov 12, 2020 at 14:53
  • $\begingroup$ Can you post a small example dataset for people to work with? $\endgroup$ Nov 12, 2020 at 14:53
  • $\begingroup$ I´m not working with data, its more or less a conceptual question. I've tried to find a K-M-Curve showing this behavior, here's the one I found on the internet which resembles my problem, (the green line) miro.medium.com/max/700/1*ATbW_NU5GMtTaCCPpfkzEA.png $\endgroup$
    – Lillys
    Nov 12, 2020 at 15:13

1 Answer 1

1
$\begingroup$

The median survival time is generally may be defined1 as the shortest time at which the proportion surviving is $\le .5$ (for a more comprehensive overview, see: Finding median survival time from survival function). You need the Kaplan-Meier estimator to account for censoring correctly, but when you have that, you can read the median off the curve. Here is an example, coded in R:

library(survival)  # we'll need this package
times  = c( 1,  2,  3,  5,  7,  9, 12, 15, 19, 22, 
           25, 26, 27, 28, 29, 30, 31, 32, 33, 34 )
events = c( 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,
            0,  0,  0,  0,  0,  0,  0,  0,  0,  0 )

windows()  # the KM plot
  plot(Surv(times, events), conf.int=FALSE, mark.time=TRUE, 
       yaxp=c(0,1,4), ylab="proportion surviving", xlab="days")
  segments(x0=-2, x1=22, y0=.5,   col="gray")
  segments(x0=34, x1=36, y0=.5,   col="gray")
  arrows(  x0=22, y0=.5, y1=-.04, col="gray", length=.1)

survival curve that plateaus at .5

50% survive at least 22 days, so that's the median survival time. If you like, you can get R to compute it for you. Notice that the survival time is computed, as is the lower bound of a 95% confidence interval on the median survival time, but the upper bound cannot be computed in this case.

survfit(Surv(times, events)~1)
# Call: survfit(formula = Surv(times, events) ~ 1)
# 
#       n  events  median 0.95LCL 0.95UCL 
#      20      10      22      12      NA 

Although it's a much more advanced topic, you may wonder why the survival times are all censored after a certain point. It could be there there are two different processes at work, such that there is one distribution of survival times for some patients, and the other proportion may well die of something someday, but will never die of the cause under study. This phenomenon is the focus of cure rate models.

1. Another possibility is to define it as the time of the midpoint between the two middle events, if an even number.

$\endgroup$
3
  • $\begingroup$ Thank you so much!! That is really helpful! $\endgroup$
    – Lillys
    Nov 14, 2020 at 6:34
  • $\begingroup$ You're welcome, @Lillys. $\endgroup$ Nov 14, 2020 at 6:51
  • $\begingroup$ If one changes for example the event at time 26 from 0 to 1, the median becomes 24. I guess that is due to having $S(t)=0.5$ on the valid interval $[22;26]$ instead of $[22;\infty]$. Could you maybe extend your answer to this case? I would appreciate it :) $\endgroup$
    – LuckyPal
    Jan 17 at 19:56

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.