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?