= Solution
A <Bayes estimator> minimizes the integrated <loss function> under the prior and sampling distribution. For a nonnegative loss, <Tonelli's theorem> and the <posterior distribution> give the risk decomposition
$$
\int\!\int L(\theta,\delta(x))f(x\mid\theta)\pi(\theta)dxd\theta
=\int m(x)\left[\int L(\theta,\delta(x))\pi(\theta\mid x)d\theta\right]dx,
$$
where $m$ is the marginal density. Thus choosing, for each observed $x$, a measurable minimizer of the <posterior expected loss> minimizes the integrated risk. Observations of marginal probability zero do not affect the rule.
Here the likelihood is $\theta e^{-\theta x}$ and the prior is $\mu e^{-\mu\theta}$ on positive parameters. Put $b=\mu+x$. Their product is $\mu\theta e^{-b\theta}$, and $\int_0^\infty\theta e^{-b\theta}d\theta=b^{-2}$. The marginal density and posterior are therefore
$$
m(x)=\frac{\mu}{(\mu+x)^2},\qquad
\boxed{\pi(\theta\mid x)=b^2\theta e^{-b\theta},\quad\theta>0.}
$$
This <Gamma-exponential conjugacy> calculation gives a <Gamma distribution> with shape $2$ and rate $b$. It is the posterior, rather than the prior alone, which determines both requested estimators.
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