# Find the UMVUE of $6\theta^2$ given $f(x\mid\theta) = \frac{1}{2\theta^2} e^{\frac{-\sqrt{x}}{\theta}} I_{(0,\infty)}(x)$

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...

• 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 – Greenparker Apr 1 '17 at 18:51
• @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 – Greenparker Apr 1 '17 at 19:05
• @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. – Greenparker Apr 1 '17 at 19:51
• 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 – C Sutton Apr 12 '17 at 19:15
• "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.) $\qquad$ – Michael Hardy Jun 18 at 0:24
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