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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.

Consider irreversible optimal stopping of employment paying income $\epsilon >0$ and imposing disutility $\lambda >0$. Assume $r,\rho >0$, nonzero market price of risk $\kappa$, a positive Merton consumption constant, and solvency allowing borrowing against perpetual income up to $\epsilon /r$ before retirement. Let $m<0$ solve ${\kappa }^{2}m(m-1)/2+(\rho -r)m-\rho =0$. The retirement threshold is the displayed wealth level.

To derive it, apply the wealth-variable Legendre dual. The working dual equals the retired dual plus $h(z)=\epsilon z/r-\lambda /\rho +D{z}^m$. Excluding the growing homogeneous solution enforces the solvency asymptotic. Value matching and smooth fit at the retirement marginal value $z_{\ast }$ give $h(z_{\ast })=h^{\prime }(z_{\ast })=0$, hence $z_{\ast }=(\lambda r/(\epsilon \rho ))m/(m-1)$ and $D=-\epsilon z_{\ast }^{1-m}/(rm)>0$. The retired inverse marginal value gives $\overline{w}={\gamma }_M^{-1}{z}_{\ast }^{-1/R}$.

The root equation implies $z_{\ast }/(\lambda /\epsilon )=1+{\kappa }^{2}m/(2\rho )<1$. Thus the boundary exceeds the instantaneous gain-zero level ${\gamma }_M^{-1}(\epsilon /\lambda {)}^{1/R}$. A tighter borrowing constraint changes this free-boundary problem. Smooth fit matches the first derivative; matching the second derivative would incorrectly remove the income option.