 Solution

This relative absolute-error loss weights each parameter by $1/\theta$. Its posterior expected loss isThe effective density for the Bayes estimator under weighted absolute loss is proportional to $\pi (\theta \mid x)/\theta$, hence an exponential distribution of rate $b$. Its median solves $1-e^{-ba}=1/2$.

More explicitly, for $a\ge 0$, differentiating the split integrals at $\theta =a$ gives $R_x^{\prime }(a)=b(1-2e^{-ba})$ and $R_x^{\prime \prime }(a)=2b^2{e}^{-ba}>0$. For $a<0$, the loss has derivative $-b$, so its minimum cannot lie there. The unique global minimum is thereforeThis is the weighted posterior median; the ordinary median of the Gamma posterior would minimize unweighted absolute loss and would give a different answer.