I'm having trouble approaching this question with the differences I noted. In an exponential distribution with mean $\frac{1}{\theta}$, I would go about finding the UMVUE like so: \begin{align*} L(\theta\vert x) &= \prod_i \frac{1}{\theta} e^{-\frac{x}{\theta}}\\ &= \frac{1}{\theta^n} e^{-\frac{\sum x_i}{\theta}} \end{align*}
$$\ell = -n\ ln(\theta) - \frac{\sum x_i}{\theta} $$
\begin{align*} \frac{\partial \ell}{\partial \theta} &= -\frac{n}{\theta} + \frac{\sum x_i}{\theta^2}\\ &= \frac{\sum x_i - n\theta}{\theta^2}\\ &= n \frac{\bar{x} - \theta}{\theta^2} \end{align*}
This tells me that $\bar{X}$ is the the UMVUE because Casella and Berger notes: "Any function of $\sum x_i$ that is an unbiased estimator of $\theta$ is the best unbiased estimator."
Now in this case, my mean is $\theta$ which is raising a few questions from me. First, how would you go about finding the UMVUE? Would it involve the Lehmann–Scheffé theorem? My professor sort of glazed over the topic so I'm not really sure how it works. Also, I'm told this UMVUE will have a variance strictly larger than its CRLB, what's the reasoning behind this?
