Let X belongs to Poisson distribution with parameter $\theta$ and ($x_1, x_2,\cdots,x_n$) is a random sample draw from the Poisson distribution. Let the prior of $\theta$ be the Gamma $\varGamma(1,1)$ with parameter all equals to 1. Also, let the loss function be the $\frac{e^{\theta}}{\theta}(\theta-d(x))^2$, where $d(x)$ be the decision rule.
How to prove that the decision $\bar x$ is admissable?
I have prove that the $\bar x$ is the bayes estimator, but the problem is that the Bayes risk is infinite in this case, so how can I prove it?
$\begin{split}r(\tau,d)=&\int_0^{\infty}E_{\theta}(\frac{e^{\theta}}{\theta}(\theta-\bar x)^2)\cdot \theta\cdot e^{-\theta}d\theta\\=&\int_0^{\infty}\frac{1}{\theta}\cdot\frac{\theta}{n}d\theta\\ =&\infty\end{split}$
If $X \overset{iid}{\sim}\textrm{Poisson}(\theta)$ and $\theta \sim \textrm{Gamma}(1,1)$; then $\theta | \bar{X} \sim \textrm{Gamma}(1 + n\bar{X},1+n)$.
Taking the expectation of the loss function with respect to the posterior gives us:
\begin{equation} \begin{split} E_{\theta|\bar{X}}[\frac{e^{\theta}}{\theta}(\theta-\delta)^2] & = \int_0^\infty \frac{e^{\theta}}{\theta}(\theta-\delta)^2 \theta^{n\bar{X}}e^{-(1+n)\theta}\\ &= \int_0^\infty (\theta-\delta)^2 \theta^{n\bar{X} - 1}e^{-n\theta} \end{split} \end{equation}
Which is the expectation of the squared error loss with respect to a $\textrm{Gamma}(n\bar{X},n)$ distribution.
The minimizer of the squared error loss is given by the expectation, so we have that $\delta = \frac{n\bar{X}}{n} = \bar{X}$.
Looking at the third property of admissibility for Bayes estimators here, we have that the loss function is continuous in $\theta$ for all $\delta$. Thus the Bayes estimator is admissible.