Find the MRE for b of E[0,b] $X_1,\dots , X_n$ i.i.d. from the E(0,b) distribution. Find the MRE (Minimum Risk Equivalent estimator) for b under the scale transformation group with the standardized square loss $(,\delta)=(\delta−)^2/^2$.
My work:
Since the complete sufficient natural statistic of E(0,b) is $\delta_0=\sum_{i=0}^{n}Xi$ (we have proved in the previous work).
So, I plug it in the formula of the MRE is $$\delta^* = \frac{\delta_0(\underline{X})E[\delta_0(\underline{X})]}{E[(\delta_0(\underline{X}))^2]}=\frac{\sum_{i=0}^{n}Xi\times E[\sum_{i=0}^{n}Xi]}{E[(\sum_{i=0}^{n}Xi)^2]}$$
What should I do next? I do not know how to calculate $E[\sum_{i=0}^{n}Xi]$ and $E[(\sum_{i=0}^{n}Xi)^2]$
Please give me a little hint, thanks!
 A: I try to answer my question. If I am wrong, please help me out. I appreciate it!
$$\delta^* = \frac{\delta_0(\underline{X})E[\delta_0(\underline{X})]}{E[(\delta_0(\underline{X}))^2]}=\frac{\sum_{i=1}^{n}Xi\times E[\sum_{i=1}^{n}Xi]}{E[(\sum_{i=0}^{n}Xi)^2]}=\frac{\sum_{i=1}^{n}X_i\times E[\sum_{i=1}^{n}]}{Var[\sum_{i=1}^{n}X_i]+(E[\sum_{i=1}^{n}X_i])^2}=\frac{n\bar{X}\times \sum_{i=1}^{n}E[X_i]}{\sum_{i=1}^{n}Var(X_i)+(\sum_{i=1}^{n}E(X_i))^2}=\frac{nb\times nb}{nb^2+n^2b^2}=\frac{n}{1+n}$$
A: The formula you use to find your desired estimator is not quite correct.
I assume the $X_i$'s are i.i.d exponential with mean $b$. So
$\boldsymbol X=(X_1,\ldots,X_n)\sim f_b\in\mathcal P$, where $\mathcal P=\left\{\frac1{b^n}f\left(\frac{x_1}b,\ldots,\frac{x_n}b\right):b>0\right\}$ is invariant under the scale transformation group. An equivariant estimator $\delta$ of $b$ then satisfies $\delta(c\boldsymbol X)=c\,\delta(\boldsymbol X),\, c>0$.
In an invariant scale problem, we define $\boldsymbol Z=(Z_1,\ldots,Z_n)$ where $Z_i=X_i/X_n$ for $i=1,\ldots,n-1$ and $Z_n=X_n/|X_n|$. If $\delta_0(\boldsymbol X)$ is any equivariant estimator of $b$ with finite risk, the uniformly minimum risk equivariant estimator (UMREE) of $b$ under the scaled squared error loss is given by
$$\delta(\boldsymbol X)=\frac{\delta_0(\boldsymbol X)E_1[\delta_0(\boldsymbol X)\mid \boldsymbol Z]}{E_1[(\delta_0(\boldsymbol X))^2\mid \boldsymbol Z]}$$
Notice that the expectation is taken under $b=1$, hence the '$E_1$'.
[See for example, Theory of Point Estimation (2nd ed.)]
A natural choice of $\delta_0$ here is $\delta_0(\boldsymbol X)=\sum\limits_{i=1}^n X_i$, which as you say is also a complete sufficient statistic. This makes $\delta_0$ independent of $\boldsymbol Z$ by Basu's theorem (since $\boldsymbol Z$ is ancillary), which leads to
$$\delta(\boldsymbol X)=\frac{\delta_0(\boldsymbol X) E_1(\delta_0(\boldsymbol X))}{\operatorname{Var}_1(\delta_0(\boldsymbol X))+(E_1(\delta_0(\boldsymbol X)))^2}=\frac1{n+1}\sum_{i=1}^n X_i$$
