# Given median survival and CI, and assuming an exponential distribution, can one predict survival and CI at a time point?

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?

• 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. May 21 at 20:34
• What is it you want the expected survival at 6 months for? Eg, is this for a power analysis? May 24 at 6:53
• @gung-ReinstateMonica kind of internal strategy, trying to decide if some scant preclinical data on our end can compete with big game-changer studies. May 26 at 18:06

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$$quantile semed <- ({sumfupper -sumf$$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'))