= Solution
\b[Available <portfolio wealth> and the ruin boundary.] Put $h=\bar rD$, the constant interest payment on the fixed loan. The loan principal is already included in available <portfolio wealth>; it is not a growing portfolio holding. Therefore
$$
\boxed{dw=\sigma\theta\,dW+[rw+(\mu-r)\theta-h-c]\,dt,\qquad w_0=x_0+D.}
$$
In particular the interest outflow is $\bar rD$, not $(\bar r-r)D$. Writing $w=x+D$ would instead give net <portfolio wealth> drift $rx+(\mu-r)\theta-(\bar r-r)D-c$, which explains the distinction.
Let $\tau_0$ denote the ruin time, to avoid confusing it with a fixed terminal horizon. The objective stops at $\tau_0$; consequently \b[the absorbing boundary is $V(0)=0$], without an obligation to keep financing the loan after ruin. <Dynamic programming> gives, for $w>0$,
$$
\rho V=(rw-h)V'
+\sup_\theta\{(\mu-r)\theta V'+\tfrac12\sigma^2\theta^2V''\}
+\sup_{c\geq0}\{U(c)-cV'\}.
$$
For increasing strictly <concave> value, put $z=V'(w)>0$ and use <inverse marginal utility> $I$. The optimal controls and the optimized <HJB equation> are
$$
c^*=I(z),\qquad \theta^*=-\frac{\kappa}{\sigma}\frac{V'}{V''},\qquad
0=(rw-h)V'-\rho V-\frac{\kappa^2(V')^2}{2V''}+\widetilde U(V'),
$$
where $\widetilde U(z)=\sup_{c\geq0}[U(c)-zc]$. For <CRRA utility> with $0<R<1$, write $p=1-R$ and $q=1-1/R<0$. Then
$$
I(z)=z^{-1/R},\qquad \widetilde U(z)=\frac R p z^q.
$$
\b[Dualization and the printed constant.] Use the convex <wealth-variable Legendre dual>
$$
J(z)=\sup_{w\geq0}[V(w)-zw].
$$
At an interior maximizing <portfolio wealth>, $J'=-w$, $J''=-1/V''>0$ and $V=J-zJ'$. The dual <HJB equation> is the linear <Euler differential equation>
$$
\boxed{\tfrac12\kappa^2z^2J''+(\rho-r)zJ'-\rho J-hz+\frac R p z^q=0.}
$$
A trial power $z^t$ gives
$$
Q(t)=\tfrac12\kappa^2t(t-1)+(\rho-r)t-\rho.
$$
Direct substitution gives
$$
Q(q)=-\gamma_M,\qquad
\boxed{\gamma_M=\frac{\rho+(R-1)(r+\kappa^2/(2R))}{R}.}
$$
The PDF prints an additional factor $1/2$ before $\kappa^2/(2R)$ in its definition of $\gamma_M$. That printed definition is inconsistent with its own identity for $Q(q)$. The expression above is the one used here; assume this corrected $\gamma_M>0$.
\b[Solution when $\kappa\ne0$.] Assume positive discount $\rho$, nonzero $\sigma$, and $h>0$. Let $m>0$ and $n<0$ be the two roots of $Q$:
$$
m,n=\frac{-(\rho-r-\kappa^2/2)\ \pm\
\sqrt{(\rho-r-\kappa^2/2)^2+2\kappa^2\rho}}{\kappa^2}.
$$
Since $Q(q)<0$, $n<q<0<m$. Put $C=R/(p\gamma_M)$. For $r\ne0$, the general interior solution is
$$
J(z)=Az^m+Bz^n-\frac h r z+Cz^q.
$$
The appropriate large-wealth condition is the <Merton consumption-investment problem> bound
$$
0\leq V(w)\leq \frac{\gamma_M^{-R}}p\,w^p.
$$
Indeed any original control consumes $c+h$ in the debt-free comparison model until ruin, and $U(c+h)\geq U(c)$. Dualizing this bound gives $0\leq J(z)\leq Cz^q$. Because $n<q$, convexity and this upper bound force \b[$B=0$] as $z\downarrow0$.
At the other endpoint <portfolio wealth> reaches zero. If $z_*=V'(0+)$, the <dual ruin boundary with debt service> requires
$$
\boxed{J(z_*)=0,\qquad J'(z_*)=0,\qquad J(z)=0\quad(z\geq z_*).}
$$
Solving these two equations gives
$$
\boxed{
z_*=\left[\frac{Cr(m-q)}{h(m-1)}\right]^R,\qquad
A=\frac{C(1-q)}{m-1}\,z_*^{q-m}.}
$$
For $r>0$, $m>1$, and for $r<0$, $0<m<1$, so the quantity defining $z_*$ is positive in either case. These formulas determine the entire value. For each $w>0$, choose the unique $z\in(0,z_*)$ satisfying
$$
\boxed{w=\frac h r-Amz^{m-1}-Cqz^{q-1}.}
$$
Then
$$
\boxed{V(w)=A(1-m)z^m+C(1-q)z^q,\qquad
c^*=z^{-1/R},\qquad
\theta^*=\frac{\kappa z}{\sigma}
\left[Am(m-1)z^{m-2}+Cq(q-1)z^{q-2}\right].}
$$
Both terms in the bracket are positive, even when $r<0$ and $A<0$. Thus $J''>0$, $-J'$ decreases from infinity to zero as $z$ increases, and the <portfolio wealth> inversion really is unique. The extended dual is continuously differentiable and convex.
There is no additional condition $J''(z_*^-)=0$. In fact
$$
J''(z_*^-)=C(1-q)(m-q)z_*^{q-2}>0.
$$
Available <portfolio wealth> is killed at zero; the portfolio can have a nonzero limiting volatility immediately before ruin. Imposing a reflecting-boundary or zero-curvature condition would solve a different problem.
For $r=0$, the linear forcing resonates with the root $m=1$. Put $b=\rho+\kappa^2/2=Q'(1)$. The dual and boundary constants instead are
$$
\boxed{\begin{aligned}
J(z)&=Az+\frac h b z\log z+Cz^q &&(0<z<z_*),\\
z_*&=\left[\frac{Cb(1-q)}h\right]^R,\qquad
A=-\frac h b\log z_*-Cz_*^{q-1}.
\end{aligned}}
$$
Extend by zero for $z\geq z_*$. Here
$$
w=-A-\frac h b(\log z+1)-Cqz^{q-1},\qquad
V(w)=C(1-q)z^q-\frac h b z,
$$
and the controls remain $c=z^{-1/R}$ and $\theta=\kappa zJ''/\sigma$, with $J''=h/(bz)+Cq(q-1)z^{q-2}>0$.
\b[Verification and transversality.] The candidate is nonnegative, increasing, strictly <concave> and zero at ruin. Its <HJB equation> makes the discounted value plus accrued utility a <local supermartingale> for every admissible control, and a <local martingale> for the stated feedback. Localization at positive lower and finite upper <portfolio wealth> levels gives the finite-horizon comparison. The <investment value transversality condition> follows from the same debt-free bound: applying the <Itô formula> to $w^p$ and maximizing its risky term gives
$$
\mathbb E[e^{-\rho t}w_t^p\mathbf1_{\{t<\tau_0\}}]
\leq w_0^p e^{-[\rho-p(r+\kappa^2/(2R))]t}
=w_0^p e^{-R\gamma_M t}.
$$
The nonnegative <consumption> and debt-service drifts only decrease this bound. Thus the expected terminal candidate tends to zero. The feedback has at most linear growth, including a finite limit as <portfolio wealth> decreases to zero; stopping it at ruin gives an admissible policy. Letting localization levels and then the horizon tend to their limits proves that the candidate is the value, rather than just a formal dual solution. If $D=0$, the absorbing-debt boundary disappears and the ordinary <Merton consumption-investment problem> formula is recovered.
\b[Zero market price of risk.] With $\sigma\ne0$ and $\kappa=0$, the dual equation is first order. If $\rho>r$, the preceding formulas remain valid with $m=\rho/(\rho-r)$ for $r\ne0$, and the logarithmic formula with $b=\rho$ for $r=0$; there is no $B$ term and $\theta^*=0$. The same boundary and transversality argument verifies this deterministic <consumption> policy.
The remaining finite-value case has $r>0$ and $pr<\rho\leq r$. Put $\gamma_0=(\rho-pr)/R$, $w_c=h/(rR)$ and
$$
z_c=\left(\frac{Rr}{p\gamma_0h}\right)^R.
$$
The correct convex dual and its corresponding value are
$$
\boxed{
J(z)=\max\left\{\frac{R}{p\gamma_0}z^q-\frac h r z,\ 0\right\},\qquad
V(w)=
\begin{cases}
z_cw,&0\leq w\leq w_c,\\
\gamma_0^{-R}(w-h/r)^p/p,&w\geq w_c.
\end{cases}}
$$
The two value branches have the same value and derivative at $w_c$. Above $w_c$, hold no stock and consume $\gamma_0(w-h/r)$; the surplus over $h/r$ grows at rate $r-\gamma_0\geq0$. If $\rho=r$, the lower branch is attained by zero stock holding and constant <consumption> $ph/R$: <portfolio wealth> solves $\dot w=r(w-w_c)$ until ruin, and direct integration gives $z_cw$.
If $\rho<r$ and $0<w<w_c$, the lower branch is a supremum attained in a limit of increasingly rapid fair stock lotteries between zero and $w_c$, followed by the upper-branch policy on success. The success probability tends to $w/w_c$ and the fixed service cost during the lottery tends to zero. This is possible because unrestricted dollar holdings in the nonzero-volatility stock produce a fair <Brownian motion> exposure even when its excess drift is zero. The supporting linear branch has optimized waiting residual $(r-\rho)z_c(w-w_c)<0$; the fast lotteries, rather than a finite feedback optimizer, supply the missing control limit. The piecewise candidate is <concave>, has nonpositive waiting residual everywhere, and the preceding moment bound still supplies an upper-bound verification. \b[This degenerate case can have a supremum without an ordinary maximizing strategy.]
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