The precision of a plug-in survival probability

Question

Lifetimes X1, \dots, Xn are i.i.d. Exponential with rate \lambda, and \hat\lambda = 1/\bar X is the maximum likelihood estimator. You want the survival probability past a fixed time t, S = P(X t) = e^{-\lambda t}, estimated by the plug-in \hat S = e^{-\hat\lambda t}. Use the delta method to find the approximate variance of \hat S.

Accepted Answer

Start from the estimator's spread. The maximum likelihood rate is asymptotically normal with \operatorname{Var}(\hat\lambda) \approx \frac{\lambda^2}{n}, since its asymptotic variance factor is 1/I(\lambda) = \lambda^2. Apply the delta method. For a smooth function h(\lambda), the plug-in h(\hat\lambda) has approximate variance [h'(\lambda)]^2 \operatorname{Var}(\hat\lambda). Here h(\lambda) = e^{-\lambda t}, \qquad h'(\lambda) = -t\,e^{-\lambda t}, \qquad [h'(\lambda)]^2 = t^2 e^{-2\lambda t}. So \operatorname{Var}(\hat S) \approx t^2 e^{-2\lambda t}\cdot\frac{\lambda^2}{n} = \frac{t^2\lambd…

The precision of a plug-in survival probability

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