= Solution
For a starting price above the trigger, maximizing $V_L(S)$ amounts to maximizing $g(L)=(X-L)L^{-\beta_-}$ on $0<L<X$. Its derivative is
$$
g'(L)=L^{-\beta_--1}\bigl[(\beta_--1)L-\beta_-X\bigr].
$$
Because $\beta_-<0$, this is positive below $\beta_-X/(\beta_--1)$ and negative above it. Both endpoint limits of $g$ are zero. Thus the unique maximizing trigger is
$$
\boxed{S^*=\frac{\beta_-}{\beta_--1}X,\qquad 0<S^*<X.}
$$
Equivalently <smooth fit> requires $\beta_-(X-S^*)/S^*=-1$, matching the derivative of $X-S$ below the boundary. The optimized <perpetual put option> value is $X-S$ for $S\le S^*$ and $(X-S^*)(S/S^*)^{\beta_-}$ above it.
We can verify optimality among all stopping rules, not only constant triggers. The continuation branch is convex, with value and slope matching $X-S$ at $S^*$, so it dominates $X-S$ above the boundary; it is also positive, and hence dominates $(X-S)^+$ everywhere. Its discounted generator vanishes in continuation. In the exercise region, the generator of $X-S$ is $\delta S-rX$. The root equation gives
$$
\delta S^*/X=r+\tfrac12\sigma^2\beta_-<r,
$$
so for $\delta\ge0$ this generator is nonpositive throughout $S\le S^*$. <Smooth fit> eliminates a local-time contribution at the boundary. <Itô formula> and localization therefore make $e^{-rt}V(S_t)$ a nonnegative <supermartingale>, providing an upper bound for every discounted exercise payoff. The threshold strategy attains that bound by the <martingale> calculation in part (a). This proves the boundary is globally optimal.
The positive-interest assumption is substantive. If $r=0$ and $\delta\ge0$, the <stock> tends to zero almost surely, so the <supremum> of perpetual put payoffs has value $X$, approached by triggers $L\downarrow0$. No positive finite trigger attains that <supremum> for positive spot. The displayed finite-boundary formula belongs to the $r>0$ case. If $r<0$ and $\delta\ge0$, the <stock> still tends to zero, while the <discount factor> grows without bound. Even exercise at deterministic times tending to infinity makes the expected discounted put payoff unbounded, so there is no finite perpetual price in that extension.
Back to article page
