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The barrier-triggered put exercises at the particular stopping time ${\tau }_B\wedge T$. The American put option optimizes over all allowed exercise times, soIt can be strictly smaller, but can never exceed the American value. Equality is possible if immediate exercise is already triggered and optimal; a fixed barrier is not generally the optimal finite-maturity exercise rule.

The comparison with the European put option can go in either direction. By the early exercise premium for a fixed barrier,Thus an early payment gains interest but sacrifices remaining put time value; the expression inside braces need not have a fixed sign for arbitrary barriers.

For a concrete general construction take current spot $S\ge K$, $r>0$, $\sigma >0$, and $t<T$. Set $\gamma =2r/{\sigma }^2$ and $b_{\infty }=K\gamma /(1+\gamma )$, the perpetual put option exercise boundary. For $0<B<b_{\infty }$, the perpetual value at $B$ equals $K-B$ and strictly dominates the finite-maturity European value. The premium formula and positive probability of hitting $B$ therefore give $E_B>P$. Strictness follows from the perpetual verification argument: before a fixed positive expiry, its discounted value loses drift $rK$ whenever the stock is below $b_{\infty }$, with positive expected occupation time when started at $B<b_{\infty }$.

Conversely, for $B<K$ sufficiently close to $K$, every possible payment of the exploding put is at most $K-B$, whether paid at touch or on surviving to expiry. Hence $0\le E_B\le K-B\to 0$, while $P(S,t;K)>0$. This gives $E_B<P$ for such barriers, and also strict inequality below the American value.

These examples establish both possibilities in general, not a claim that every fixed initial state admits both. If $S<b_{\infty }$ and only alive barriers $B<S$ are allowed, all those barriers are in the same perpetual stopping region and the premium comparison is nonnegative.

For $S>b$, maximize the coefficient $H(b)=b^{\gamma }(K-b{)}^n$ of $S^{-\gamma }$. Its logarithmic derivative isIt changes sign once, at $b=\gamma K/(\gamma +n)$. The perpetual power put option therefore has candidate valueThe same boundary follows from smooth fit, $-\gamma (K-b{)}^n/b=-n(K-b{)}^{n-1}$. To establish optimality among all stopping times, rather than only fixed barriers, we verify the candidate. Let $\mathcal{L}=rS{\partial }_S+{\sigma }^2{S}^2{\partial }_{SS}/2$ be the infinitesimal generator. The maximum property of $H$ gives $V_n\ge g_n$ everywhere, and $(\mathcal{L}-r)V_n=0$ for $S>b$. For $S\le b$, set $y=S/(K-S)\le \gamma /n$. ThenThus the discounted candidate is a nonnegative supermartingale. The smooth fit removes the derivative-jump local time of a semimartingale term in the Itô formula. The optional stopping theorem and localization bound every exercise policy's discounted expected value by $V_n$. Immediate exercise attains it below $b$; the first hitting time of $b$ attains it above $b$. The terminal remainder vanishes since $0\le e^{-ru}{V}_n(S_u)\le K^n{e}^{-ru}\to 0$. This proves the displayed value.

For $n=1$ it reduces to the ordinary perpetual put option. Larger $n$ lowers the optimal trigger, because the more strongly curved financial payoff makes further downward moves more valuable. If $r=0$, under the risk-neutral measure the stock tends to zero almost surely, so the supremum is $K^n$, approached as the exercise time tends to infinity and not attained at a finite time when $S>0$. If $r<0$, the values at deterministic exercise times grow without bound; the perpetual value is infinite. These cases do not admit the positive-rate formula above.

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 isBecause ${\beta }_{-}<0$, this is positive below ${\beta }_{-}X/({\beta }_{-}-1)$ and negative above it. Both endpoint limits of $g$ are zero. Thus the unique maximizing trigger isEquivalently smooth fit requires ${\beta }_{-}(X-S^{\ast })/S^{\ast }=-1$, matching the derivative of $X-S$ below the boundary. The optimized perpetual put option value is $X-S$ for $S\le S^{\ast }$ and $(X-S^{\ast })(S/S^{\ast }{)}^{{\beta }_{-}}$ 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 $S^{\ast }$, 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 givesso for $\delta \ge 0$ this generator is nonpositive throughout $S\le S^{\ast }$. 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 $\delta \ge 0$, the stock tends to zero almost surely, so the supremum of perpetual put payoffs has value $X$, approached by triggers $L\downarrow 0$. 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 $\delta \ge 0$, 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.

For a dividend-free Black-Scholes model with $r>0$, $\sigma >0$, $K>0$ and positive integer $n$, the optimal stopping value of the power put option isIn the continuation region the stationary Black-Scholes equation has powers $S$ and $S^{-\gamma }$. Boundedness at infinity removes $S$. Value matching gives the coefficient $(K-b{)}^n{b}^{\gamma }$, maximized at the displayed $b$; equivalently this is the smooth fit boundary.

For a verification beyond constant-trigger policies, let $g(S)=(K-S{)}_{+}^n$. The maximum property of $b^{\gamma }(K-b{)}^n$ gives $V\ge g$. For the infinitesimal generator $\mathcal{L}=rS{\partial }_S+{\sigma }^2{S}^2{\partial }_{SS}/2$, $(\mathcal{L}-r)V=0$ above $b$. Below $b$, set $y=S/(K-S)\le \gamma /n$; thenThe candidate is continuously differentiable across $b$, so the generalized Itô formula has no derivative-jump local time of a semimartingale term. The discounted candidate is a nonnegative supermartingale, bounding every stopping reward by optional stopping. First hitting of $b$ attains the bound; the remainder at infinite time vanishes since $V\le K^n$ and $r>0$. For $n=1$ this recovers the perpetual put option. At $r=0$ the value is $K^n$ for $S>0$, approached but not attained at a finite exercise time. At $r<0$ the perpetual value is infinite, because exercise at deterministic times tending to infinity has unbounded discounted expected value.