3
$\begingroup$

I'm working through some old qualifying exam problems and am wondering what I'm being asked to do in the following question: We have a random sample $\textbf{X} = (X_1, \dots, X_n)$ from a binomial$(m,\theta)$ distribution with prior distribution $\Theta \sim \text{Beta}(\alpha, \beta)$, where $m \in \mathbb{Z}^+, \alpha>0$, and $\beta>0$ are all known. Find the Bayes estimator for $\theta$ and $\theta^2$.

Here is where my confusion begins. Usually I am asked to find the Bayes estimator with respect to a loss function. There is no loss function given here. Hence, I am not sure how to proceed.

I have worked out that the posterior distribution is $\Theta | T \sim \text{Beta}(T + \alpha, \beta + nm - T)$ where $T$ is the sample sum.

$\endgroup$
  • 2
    $\begingroup$ I think you are correct and you should have a loss function. Without it, I might arbitrarily choose $\mathbb E (\Theta \mid T)$ and $\mathbb E (\Theta^2 \mid T)$, implicitly using a squared error loss function in each case $\endgroup$ – Henry Dec 26 '17 at 10:59
  • $\begingroup$ Yea, it seems like they just want you to find the first two raw moments. Someone may have forgotten to write "under squared error loss." $\endgroup$ – Ryan Warnick Dec 30 '17 at 7:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.