Given $X_1, X_2,\ldots, X_n$ are i.i.d rvs with pdf $f(x\mid\theta) = \frac{1}{2\theta^2} e^{\frac{-\sqrt{x}}{\theta}} I_{(0,\infty)}(x)$ for $\ \theta > 0$. Find the UMVUE of $\ 6\theta^2$, and compute its variance.

My thought: We see that since $E(\overline{X}) = 6\theta^2$, $\overline{X}$ is an unbiased estimator of $\ 6\theta^2$. Now, since these random variables belong to an exponential family, and $w(\theta) = \frac{-1}{\theta}$ whose range is an open set in $\mathbb R^1$, it's a well-known fact that $\sum_{i=1}^n \sqrt{X_i}$ is a complete statistic, and it's also a sufficient statistics (by Factorization theorem). But unfortunately, $\overline{X}$ is not a function of this complete sufficient statistics $\sum_{i=1}^{n} \sqrt{X_i}$, so I cannot apply Lehmann - Scheffe theorem to conclude that $\overline{X}$ is a UMVUE of $\theta$. Or am I completely on the wrong track, since $\overline{X}$ is NOT a UMVUE of $\theta$ ??

I also tried with the unbiased estimator $\ \frac{\sum_{i=1}^n \sqrt{X_i}}{2n}$ but this one does not have the variance equal to Cramer-Rao Lower Bound $=\frac{72\theta^4}{n},$ so it's not the UMVUE of $6\theta^2$ either.

My question: Can anyone please help me with this problem? I'm getting stuck on it for a while despite trying various unbiased estimator...

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    $\begingroup$ If $\sum_{i=1}^{n} \sqrt{X_i}/2n$ is complete and sufficient and unbiased, then it is a UMVUE. Sometimes UMVUEs don't attain the Cramer-Rao Lower Bound $\endgroup$ – Greenparker Apr 1 '17 at 18:51
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    $\begingroup$ @user177196 Ok, $\sum_{i=1}^{n} \sqrt{X_i}$ is complete and sufficient, and $\sum_{i=1}^{n} \sqrt{X_i}/2n$ is a function of a complete and sufficient statistic, and its unbiased. Thus by Lehman-Scheffe's theorem, it is UMVUE. See the statement of the theorem $\endgroup$ – Greenparker Apr 1 '17 at 19:05
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    $\begingroup$ @user177196 I haven't worked out the math (this is self-study), so I just gave you the hint, that the CR lower bound need not be attained and the Lehmann-Scheffe's theorem applies still. The technicalities are for you to figure out. $\endgroup$ – Greenparker Apr 1 '17 at 19:51
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    $\begingroup$ I will provide the solutions in class tonight. I do not like that my students are using this site to get unauthorized help on homework problems. Why not ask me for a hint? C. Sutton $\endgroup$ – C Sutton Apr 12 '17 at 19:15
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    $\begingroup$ "this one does not have the variance equal to Cramer-Rao Lower Bound [ . . . . . ], so it's not the UMVUE" $$\S$$ Are you sure the UMVUE always achieves the Cramér–Rao lower bound? Nothing in the Cramér–Rao theorem says that in every instance it is the GREATEST lower bound. (Right now I don't know a counterexample, but I'll look around. Maybe you've got one right here.) $\qquad$ $\endgroup$ – Michael Hardy Jun 18 '19 at 0:24

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