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.

  • $\begingroup$ Thank you so much. I check the theorem you mentioned, but the thing is that all those theorem can only be applied in the situation that has finite Bayes risk. As for our example, the Bayes risk is infinite, right? or did I make some mistake? $\endgroup$ – Nicolas Mar 6 '18 at 19:18
  • $\begingroup$ The minimizer of the posterior expected loss is also a Bayes estimator. You can read that on the Wikipedia page for Bayes estimators: "Equivalently, the estimator which minimizes the posterior expected loss ... also minimizes the Bayes risk and therefore is a Bayes estimator." $\endgroup$ – Ryan Warnick Mar 7 '18 at 2:33
  • $\begingroup$ That's the derivation I gave. $\endgroup$ – Ryan Warnick Mar 7 '18 at 2:34
  • $\begingroup$ Yep, but how to show it is admissible? the theorem you mentioned in Wikipedia only applicable when the Bayes risk is finite, but in the case above, the Bayes risk is infinite. $\endgroup$ – Nicolas Mar 7 '18 at 7:13

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