# How to find the asymptotic variance of a UMVUE?

Suppose that $X_1, \ldots , X_n$ are iid observations from an exponential distribution with a pdf: $p(x) = \frac{1}{\theta}e^{-\frac{x}{\theta}}$. Suppose that my interest is in estimating the probability: $h(\theta) = P(X \geq t)$. I have found that my maximum likelihood estimator is: $\hat{h(\theta)} = exp(-t/\bar{X})$, and that the UMVUE of $h(\theta)$ is:

$$\left(1-\frac{t}{n\bar{x}}\right)^{(n-1)}$$

I am currently trying to find the asymptotic variance of the UMVUE (Uniformly Minimal Variance Unbiased Estimator). One method I know would be the the central limit theorem, but am not sure it it applies here. Could anyone give me an idea how to find the asymptotic distribution of the UMVUE?

Since $E(X_1) = \theta$ and $Var(X_1) = \theta^2$, by the CLT, $$\sqrt{n}(\bar{X} - \theta) \overset{d}{\to}N(0, \theta^2).$$
You can now apply the Delta Method using $$g(\theta) = \left(1 - \dfrac{t}{n \theta} \right)^{n-1}$$.
You can fill in the gaps, and you should get $$\sqrt{n}(g(\bar{X}) - g(\theta) \overset{d}{\to} N \left(0, \dfrac{t}{\theta^2}e^{-t/\theta^2} \right).$$