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I have access to a manuscript where the median survival and CI are reported. The shape of the curve approximately takes the shape expected for an exponential distributed failure time. I don't know the exact censoring distribution. Is there a useful approximation to convert the median survival (and CI) to the expected survival (and CI) at, say, 6 months? What assumptions about the censoring distribution need to be made for the approximation to work?

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    $\begingroup$ If you feel comfortable assuming it's really exponential, & that the censoring is independent / uninformative, then I don't see any problems. You'd want to take the uncertainty in the CI into account. Then I suppose you'd integrate over the results. I think that's what I would do. $\endgroup$ May 21 at 20:34
  • $\begingroup$ What is it you want the expected survival at 6 months for? Eg, is this for a power analysis? $\endgroup$ May 24 at 6:53
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    $\begingroup$ @gung-ReinstateMonica kind of internal strategy, trying to decide if some scant preclinical data on our end can compete with big game-changer studies. $\endgroup$
    – AdamO
    May 26 at 18:06
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Brookmeyer and Crowley (1982) provides general case solutions and exponential distribution solutions for median survival and calculation of confidence intervals.

Suppose $T$ is the distribution of event times, with $C$ the indicator of censor versus failure. The Kaplan Meier estimate $\hat{S}(t)$ of the overall survival is a consistent estimate of $1-F_{(t)}$ where $F$ is the cumulative distribution function for the failure process. Supposing $F(t)=1-\exp (-\lambda t)$, the median survival $F^{-1}(0.5) = M$ is given by $$M = \log 2 \sum (\text{observed survival times})/d$$ where $d$ is the sum of observed deaths. In fact, $\sum (\text{observed survival times})/d$ is the MLE of $\lambda$. The 6 months survival by $F(6) = \exp(-6 \lambda )$.

On the linear scale, the variance estimate of $\hat{M}$ is $\hat{M}/\sum P_i$ where $$P_i = 1-\exp\left(T_i \log(2) / \hat{M}\right).$$

A symmetric interval for the median with this variance approximation is called the Bartholomew interval. However, most reported intervals will be asymmetric based on the variance stabilized interval. As a pragmatic solution, I'd recommend taking the midpoint of these estimates to recover the linear variance approximation. With an estimate of the variance, the remaining details can be recovered via the $\delta$-method.

The mapping $f(x)$ defined by $\exp(-6 \log 2 / x)$ gives $F(6) = f(F^{-1}(0.5))$. And $$f^{\prime}(x) = 6 \log (2) \exp(-6 \log 2 / x) / x^2$$

So by the $\delta$-method, a consistent estimate of the variance of the 6 month survival would be given by:

$$ \text{SE}(\hat{S}_6) = f^\prime(x) \text{SE}(\hat{M})$$

The approximation seems to work well for $M$ and $S(6)$ not close to 1 or 0. R code below:

set.seed(123)
x <- rexp(1000, 1/10)
d <- rbinom(1000, 1, 0.5)
f <- survfit(Surv(x,d) ~ 1)
sumf <- quantile(f, 0.5)

## median and SE median
med <- sumf$quantile
semed <- ({sumf$upper -sumf$lower}/2)/1.96

sixmo <- exp(-6*log(2)/med)
sesixmo <- 6*log(2)/med^2*exp(-6*log(2)/med) * semed

plot(f, xlim=c(3, 9), ylim=c(0.6, 1), xlab='Time', ylab='Survival')
title('Midpoint variance approximation for median to 6 month survival')
points(6, sixmo, cex=3)
segments(6, sixmo - 1.96*sesixmo, 6, sixmo + 1.96*sesixmo)
legend('topright', pch=c(1,-1), lty=c(0,1), pt.cex=3, c('Estimate', '95% CI'))

enter image description here

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