# How to back-calculate lost at follow up from kaplan meier survival esitmate

Given an event rate at 1 year= 7% and a Kaplan Meier survival estimate for the same event at 1 year = 8.5%, is it possible (and how) to back calculate n(%) of subjects censored before 1 year, e.g lost at follow-up/not completing study?

I doubt it is possible without some distributional assumptions. If you are willing to assume an exponential distribution for event times, then $1-\exp(-\hat{\lambda} )= 0.085$ so that $\hat{\lambda}=-log(0.915)\approx. 0.08883$, while at the same time $\hat{\lambda} = 0.07 / \text{avg. follow-up}$ so that the average follow-up to first event or censoring would have been about 0.788 years.
If you then assume that losses to follow-up follow an exponential distribution with parameter $\mu$, too, then you can calculate the expected duration of follow to 1 year, an event or censoring to be $\frac{e^{\lambda +\mu }-1}{(\lambda +\mu ) e^{\lambda +\mu }}$, insert $\hat{\lambda}\approx. 0.08883$ and try to find a value of $\mu$ that makes this 0.788. Then you could simulate and see how many losses to follow-up you get.