I feel like this is something we went over in class but it's not coming to me for some reason. I need to find the Bayesian estimator for $\tau(\theta)=e^{-\theta}$ under square error loss. I already found the Bayesian estimator for $\theta$ (idk if you'll need the context but the prior was Gamma($\alpha,\beta$) and the parameter of interest was that of iid Poisson variables). I don't know if I can just take the expectation of $e^{-\theta}$ and call it a day or what I need to do. Any help would be appreciated.
$\begingroup$
$\endgroup$
2
-
5$\begingroup$ No invariance property. Bayes estimator of $\theta$ under squared error loss is $E[\theta\mid X]$. Similarly Bayes estimator of $\tau(\theta)$ under squared error loss is $E[\tau(\theta)\mid X]$. $\endgroup$– StubbornAtomCommented Apr 12, 2020 at 7:43
-
3$\begingroup$ If $\exp\{-\theta\}$ is estimated under the squared error loss, the Bayes estimator is the posterior expectation. It is unclear what you mean by invariance in this context. $\endgroup$– Xi'anCommented Apr 12, 2020 at 9:43
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Thanks to the people in the comments, I can conclude that Bayesian estimators under square error loss do not have an invariance property, and the Bayes estimator of $\tau(\theta)$ under square error loss is $E(\tau(\theta)|X)$.