Cramér-Rao Lower Bound-Exponential distribution

Question

Let $X$ be a random draw from an Exponential($\theta$) distribution, which has a density $p_\theta(x)=\theta^{-1}e^{-x/\theta}$ with an unknown $\theta>0.$

a) Compute the Crámer-Rao lower bound for unbiased estimation of $g(\theta)=\theta^k, k\in N$.

b) Find the UMVUE of $g(\theta)=\theta^k, k\in N$.

c) Compute the variance of the UMVUE from b). For which $k\in N$, if any, does the UMVUE attain the CR-bound from a)?

My idea: for a), the Crámer-Rao lower bound would be (I think) $\frac{g'(\theta)^2}{I(\theta)} $ with $I(\theta)=-E_{\theta}[l''_{\theta}(X)]=1/\theta^2$ (here $l_\theta$ is MLE) so one would get $\frac{g'(\theta)^2}{I(\theta)} = \frac{(k \theta^{k-1})^2}{1/\theta^2}=k^2\theta^{2k} $. I'm not sure though if this is correct.

As for b) my first guess was to take a look at moment k of X and $E[X^k]=k!\theta^k$. I wanted to apply Lehmann-Scheffé Lemma, but I had no idea how to do that because of $k!$

Answer 1

a)

I'm not sure though if this is correct.

If you have a single observation yes it is! If you have a size-$n$ random sample you get

$$V(T)\geq \frac{k^2\theta^{2k}}{n}$$

b) Observe that $T=\overline{X}_n$ is unbiased estimator for $\theta$ and it is a function of $S=\Sigma_iX_i$, complete and sufficient statistics...thus T is UMVUE for $\theta$

Given this, one may suspect that $[\overline{X}_n]^k$ is UMVUE for $\theta^k$. To verify this it can be useful to calculate the expectation of

$$T_k=[\Sigma_iX_i]^k$$

This is easy because $\Sigma_iX_i$ has a known distribution (it's a gamma) and $T_k$ is function of $S$, complete and sufficient statistics.

Thus all you have to do, once you have calculated the expectation of $T_k$ is to correct, if necessary, its bias