= Solution
For <constant relative risk aversion utility>, $I(z)=z^{-1/R}$. Define the <Merton consumption constant>
$$
\gamma_M=\frac{\rho-(1-R)(r+\kappa^2/(2R))}{R}.
$$
The Gaussian exponential moment gives
$$
\mathbb E[\xi_t(ye^{\rho t}\xi_t)^{-1/R}]=y^{-1/R}e^{-\gamma_Mt}.
$$
When $\gamma_M>0$, the budget is $w_0=y^{-1/R}/\gamma_M$. Conditional pricing then gives $c_t^*=\gamma_Mw_t^*$, and the <Merton consumption-investment problem> has
$$
\boxed{V(w)=\frac{\gamma_M^{-R}w^{1-R}}{1-R},\qquad c_t^*=\gamma_Mw_t^*,\qquad\theta_t^*=\frac{\mu-r}{R\sigma^2}w_t^*.}
$$
The resulting wealth is a <geometric Brownian motion>, so positive initial wealth remains positive. Substitution in the <Hamilton-Jacobi-Bellman equation> verifies the first-order conditions and the value; the budget argument in part (a) supplies global optimality.
\b[The problem has a finite value precisely when $\gamma_M>0$.] For $0<R<1$ and $\gamma_M\leq0$, the supremum is $+\infty$. Indeed, use the displayed risky fraction and consumption $c_t=kw_t$, $k>0$. Its expected utility is
$$
\frac{w^{1-R}k^{1-R}}{1-R}\int_0^\infty e^{-[R\gamma_M+(1-R)k]t}\,dt.
$$
For $\gamma_M<0$, sufficiently small $k$ already gives infinite utility. For $\gamma_M=0$, the finite values tend to infinity as $k\downarrow0$.
For $R>1$ and $\gamma_M\leq0$, every admissible stream has utility $-\infty$. To see this, put $q=(R-1)/R$ and apply <Holder inequality> on $[0,T]\times\Omega$:
$$
\int_0^T e^{-\gamma_Mt}\,dt
=\mathbb E\int_0^T(e^{-\rho t}c_t^{1-R})^{1/R}(\xi_tc_t)^{(R-1)/R}\,dt
\leq\left(\mathbb E\int_0^T e^{-\rho t}c_t^{1-R}dt\right)^{1/R}w_0^{(R-1)/R}.
$$
The left side diverges as $T\to\infty$, forcing the positive utility-cost integral to diverge. Thus there is no finite-value optimization problem in either ill-posed case.
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