Determine the Bayes estimator of $\theta^2$ under the conjugate prior for the normal distribution.

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 $$\theta^2$$ .

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