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I understand how to find 95% CI of Kaplan-Meier survival estimate using Greenwood's formula, and then using a log-log transformation so that the CI is bounded correctly. My question is, what are the corresponding 95% CI of 1 - S(t)? Is it simply (1 - upper bound of S(t), 1 - lower bound of S(t))?

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Yes, because 1-S(t) is monotonically decreasing function of S(t). Generally, if Y=f(X) is monotonically increasing function, the confidence interval of Y can be derived from CI of X by YL = f(XL) and YU=f(XU). If Y=f(X) is monotonically decreasing function, the confidence interval of Y can be derived from CI of X by YL = f(XU) and YU=f(XL). Function must be $monotonic$.

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