= Solution
For observed data $x$, let $F_x$ be the <posterior> <cumulative distribution function>. Assuming finite <posterior> first moment, the <posterior> risk under <asymmetric absolute-error loss> is
$$
R_x(a)=\gamma\int_a^\infty(\theta-a)\pi(\theta\mid x)d\theta
+\delta\int_{-\infty}^a(a-\theta)\pi(\theta\mid x)d\theta.
$$
At points where the <posterior> has a density, differentiation gives $R_x'(a)=(\gamma+\delta)F_x(a)-\gamma$. Risk is convex, so a global minimum is a <posterior> <quantile>:
$$
\boxed{a(x)=F_x^{-1}\left(\frac\gamma{\gamma+\delta}\right).}
$$
More generally any $a$ with $F_x(a^-)\leq\gamma/(\gamma+\delta)\leq F_x(a)$ minimizes risk. This is the <Bayes quantile under asymmetric absolute-error loss>; a larger cost for underestimation moves the chosen <quantile> upward.
For the uniform sampling model set $M=\max_iX_i$. For a valid positive sample the <likelihood> is $\theta^{-n}\mathbf1_{\{\theta>M\}}$. Multiplying by the <prior> cancels its power of $\theta$ and gives <posterior> density proportional to $e^{-\theta}\mathbf1_{\{\theta>M\}}$. Normalization yields
$$
\pi(\theta\mid x)=e^{-(\theta-M)}\mathbf1_{\{\theta>M\}},\qquad
F_x(a)=1-e^{-(a-M)}\quad(a\geq M).
$$
Solving the <quantile> equation gives the explicit Bayes estimate
$$
\boxed{a(X_1,\ldots,X_n)=M+\log\left(\frac{\gamma+\delta}{\delta}\right).}
$$
The <posterior> is the sample maximum plus a unit-rate <exponential distribution> displacement; it has finite first moment, so the preceding risk minimization is justified here.
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