Justify an unbiased estimator is UMVUE Suppose $X_1,\ldots,X_n$ are iid $N(\theta,\theta)$, with $\theta\in(0,\infty)$. Is $\bar{X}$ the UMVUE (beta unbiased estimator) of $\theta$?
I find the complete sufficient statistic is $T=\sum_{i=1}^{n}X_i^2$. So $\bar{X}$ is not a function $T$. Then we cannot justify it is UMVUE or not. Can someone help me here? 
How to get complete sufficient statistis?
$\frac{f(x\mid\theta)}{f(y\mid\theta)}=\exp(\frac{1}{2\theta}\sum_{i=1}^n (y_i^2-x_i^2)+\sum_{i=1}^n (x_i-y_i))$. Let $\sum_{i=1}^n y_i^2=\sum_{i=1}^n x_i^2$.
My work
I got $\log L(x\mid\theta) = -\frac{n}{2}\frac{1}{\theta} + \frac{\sum_{i=1}^n x_i^2}{2} \frac{1}{\theta^2}-\frac{n}{2}$. Then, I let $\frac{\partial \log (x\mid\theta)}{\partial \theta}=0$. Then, I have $-\frac{n}{2}\theta^2-\frac{n}{2}\theta+\frac{1}{2}\sum_{i=1}^n x_i^2=0$. Then, I find the solution is weird. Am I wrong?
 A: $\bar X$ is the UMVUE for $\theta$. In fact, you haven't expressed the minimal sufficient statistic correctly.
The following result will make this clear:
Theorem: Given simple random sample from a population $X$ with density of the form
$$f_\theta(x)=c(\theta)h(x)e^{\sum_{j=1}^{p}q_j(\theta)T_j(x)}$$
where the functions $q_j(\theta)$ and $T_j(x)$ are such that there are no linear combinations of them so that it could be possible to simplify the exponent, then the statistic
$$\bigg(\sum_{i=1}^{n}T_1(X_i), \sum_{i=1}^{n}T_2(X_i),...,\sum_{i=1}^{n}T_p(X_i)\bigg)$$
Is a minimal sufficient statistic for the given population
If in this case we have $X\sim N(\theta,\theta)$, the density of the population is
$$f_\theta(x)=\frac{1}{\theta\sqrt{2\pi}}e^{-\frac{(x-\theta)^2}{2\theta^2}}\quad,\quad x\in \mathbb{R},\ \theta>0$$
So we can give the following factorization
$$f_\theta(x)=\frac{1}{\theta\sqrt{2\pi}}e^{-\frac{1}{2}}e^{-\frac{x^2}{2\theta^2}+\frac{x}{\theta}}\quad,\quad x\in\mathbb{R},\ \theta>0$$
making the follwing identifications:
$$c(\theta)=\frac{1}{\theta\sqrt{2\pi}}e^{-\frac{1}{2}} \ , \ h(x)=1 \ , \ q_1(\theta)=-\frac{1}{2\theta^2} \ , \ q_2(\theta)=\frac{1}{\theta}, T_1(x)=x^2 \ , \ T_2(x)=x$$
And applying the previous theorem, gives us that the minimal sufficient statistic for this distribution is
$$T=\bigg(\sum_{i=1}^{n}X_i^2,\sum_{i=1}^{n}X_i\bigg)$$
Of course, we can't simplify the exponent further, as the vectors from $T$ are linearly independent.
We would like to check as well that this is a complete statistic, but there is no need to verify that is satisfies the definition, as the following result states:
Theorem: In the same conditions of the previous theorem, the statistic
$$\bigg(\sum_{i=1}^{n}T_1(X_i), \sum_{i=1}^{n}T_2(X_i),...,\sum_{i=1}^{n}T_p(X_i)\bigg)$$
is complete if the image of the function $(q_1(\theta),q_2(\theta),...,q_n(\theta))$ contains an open subset of $\mathbb{R}^p$
(see E.L. Lehmann "Testing Statistical Hypothesis", Springer Texts in Statistics, 1986)
It is clear that the function $q(\theta)=(q_1(\theta),q_2(\theta))=\big(\frac{-1}{2\theta^2},\frac{1}{\theta}\big)$ contains some open subset of $\mathbb{R}^2$ for $\theta>0$, so we conclude from the previous theorem that the statistic $T$ is minimal, sufficient and complete.
To conclude that $\bar X$ is the UMVUE for $\theta$, note that
$$T^\star=\big(\bar X, s^2\big)$$
Is another minimal sufficient and complete statistic, and also
$$E_\theta[\bar X]=\theta\qquad,\quad\theta>0 $$
So we conclude from the Lehmann-Scheffé theorem that $\bar X$ is the UMVUE for $\theta$ as it is the only unbiased estimator for $\theta$ that is function of $T^\star$
A: We have a random sample $(X_1,X_2,\cdots,X_n)$ drawn from $\mathcal N(\theta,\theta)$ population where $\theta>0$. 
Joint density of $(X_1,X_2,\cdots,X_n)$ is 
\begin{align}
f_{\theta}(x_1,x_2,\cdots,x_n)&=\prod_{i=1}^n\frac{1}{\sqrt{2\theta\pi}}\exp\left[-\frac{(x_i-\theta)^2}{2\theta}\right]
\\&=\frac{1}{(\sqrt{2\pi})^n{\theta}^{\,n/2}}\exp\left[-\frac{1}{2\theta}\sum_{i=1}^n(x_i-\theta)^2\right]
\\&=\frac{1}{(\sqrt{2\pi})^n{\theta}^{\,n/2}}\exp\left[-\frac{1}{2\theta}\sum_{i=1}^nx_i^2+\sum_{i=1}^nx_i-\frac{n\theta}{2}\right]
\\&=\exp\left[-n\ln \sqrt{2\pi}-\frac{n}{2}\theta-\frac{1}{2\theta}\sum_{i=1}^nx_i^2+\sum_{i=1}^nx_i-\frac{n\theta}{2}\right]
\\&=\exp\left[-\frac{1}{2\theta}\sum_{i=1}^nx_i^2-n(\theta+\ln\sqrt{2\pi})+\sum_{i=1}^nx_i\right],\quad(x_1,\cdots,x_n)\in\mathbb R^n\,,\,\theta>0
\end{align}
As such, $\mathcal N(\theta,\theta)\in\text{One parameter exponential family}$ and hence $\displaystyle T(\mathbf X)=\sum_{i=1}^nX_i^2$ is a complete sufficient statistic for $\theta$. The OP found this correctly.
Since $E(X_1)=\theta$, UMVUE of $\theta$ would be $E\left(X_1\mid\sum_{i=1}^nX_i^2\right)$ by the Lehmann-Scheffe theorem. Or we can say that $E(\bar X\mid \sum_{i=1}^nX_i^2)$ is the UMVUE as $\bar X$ is also unbiased for $\theta$. Both would give the same UMVUE as UMVUE is unique whenever it exists.

So we can be certain that $\bar X$ is not the UMVUE of $\theta$ as we do not expect the conditional expectation $E(\bar X\mid T)$ or $E(X_1\mid T)$ to simply equal $\bar X$. Moreover, in this case, $\bar X$ is not a function of the complete sufficient statistic $T$, so we can rule out the possibility of it being the UMVUE.


For the actual UMVUE, one can refer to this paper by S. Nadarajah. He had simplified the expression of the UMVUE found previously by Mukhopadhyay/Cicconetti in their paper.
According to Nadarajah, UMVUE of $\theta$ is 
$$T^*(\mathbf X)=\sqrt{\frac{\sum_{i=1}^nX_i^2}{n}}\frac{I_{n/2}\left(\sqrt{n\sum_{i=1}^nX_i^2}\right)}{I_{n/2-1}\left(\sqrt{n\sum_{i=1}^nX_i^2}\right)}$$
,where $I_m(\cdot)$ is the modified Bessel function of the first kind of order $m$. 
Other equivalent expressions for the UMVUE are also presented by the author.
