 Solution

For constant relative risk aversion utility, $I(z)=z^{-1/R}$. Define the Merton consumption constantThe Gaussian exponential moment givesWhen ${\gamma }_M>0$, the budget is $w_0=y^{-1/R}/{\gamma }_M$. Conditional pricing then gives $c_t^{\ast }={\gamma }_M{w}_t^{\ast }$, and the Merton consumption-investment problem hasThe 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.

The problem has a finite value precisely when ${\gamma }_M>0$. For $0<R<1$ and ${\gamma }_M\le 0$, the supremum is $+\infty$. Indeed, use the displayed risky fraction and consumption $c_t=k{w}_t$, $k>0$. Its expected utility isFor ${\gamma }_M<0$, sufficiently small $k$ already gives infinite utility. For ${\gamma }_M=0$, the finite values tend to infinity as $k\downarrow 0$.

For $R>1$ and ${\gamma }_M\le 0$, every admissible stream has utility $-\infty$. To see this, put $q=(R-1)/R$ and apply Holder inequality on $[0,T]\times \Omega$: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.