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Consider the following model: $X_1,...,X_n$ are iid $N(\theta,1)$ and $\theta \sim N(0, \tau^2)$ for some known $\tau^2$. Determine the Bayes estimator of $\theta^2$ under the squared error loss.

I know the posterior distribution of $\theta$ given the observed sample $X_1,...,X_n$ is $N(\frac{\tau^2}{\frac 1n + \tau^2} \bar{X}, (\frac{1}{\tau^2} +n)^{-1})$.

So if the question asks me the Bayes estimator of $\theta$ under the squared error loss, I know that is $\frac{\tau^2}{\frac 1n + \tau^2} \bar{X}$. But now the question is about the Bayes estimator of $\theta^2$ .

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  • $\begingroup$ Have you tried using the König-Huygens formula? $\endgroup$
    – statmerkur
    Nov 26, 2023 at 7:04
  • $\begingroup$ Bayes estimator of any function $g(\theta)$ under squared error loss is similarly the posterior mean of $g(\theta)$ given the sample. $\endgroup$ Nov 26, 2023 at 10:26
  • $\begingroup$ @StubbornAtom so the Bayes estimator of $\theta^2$ is the expectation of $\theta^2$ under the posterior distribution of $\theta$? $\endgroup$ Nov 26, 2023 at 14:43
  • $\begingroup$ Prove it. Don't take my word for it. $\endgroup$ Nov 26, 2023 at 17:30

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