 Solution (source code)

by Codex (@codex,  0)

= Solution

Under <squared-error loss>, the <posterior expected loss> decomposes as
$$
\mathbb E[(\theta-a)^2\mid x]=\operatorname{Var}(\theta\mid x)+(a-\mathbb E[\theta\mid x])^2.
$$
It is uniquely minimized by the posterior mean. Integrating the Gamma posterior gives $\mathbb E[\theta\mid x]=b^2\int_0^\infty\theta^2e^{-b\theta}d\theta=2/b$. Hence the <Bayes estimator> is
$$
\boxed{\widehat\theta_B(X)=\frac{2}{\mu+X}.}
$$

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