Perpetual American call option ($S_{\ast }=pK/(p-1)$)

An American call option with no expiry. For geometric Brownian motion with drift $r-y$ under a risk-neutral measure, positive dividend yield $y$, $r>0$, and volatility $\sigma >0$, let $p>1$ solve ${\sigma }^{2}p(p-1)/2+(r-y)p-r=0$. Its value is $(S_{\ast }-K)(S/S_{\ast }{)}^p$ below $S_{\ast }=pK/(p-1)$ and $S-K$ above. The power solves the stationary Black-Scholes equation with continuous stock dividends; value matching and smooth fit determine the threshold.

The candidate dominates the positive part $(S-K{)}^{+}$ because its continuation branch is convex and tangent to $S-K$ at $S_{\ast }$. In the exercise region its discounted-generator residual is $rK-yS\le 0$: the root equation gives $y{S}_{\ast }=K[r+{\sigma }^{2}p/2]>rK$. Applying Itô's formula to the discounted value, localized if necessary, gives an upper bound on every exercise policy. Stopping at the first hit of $S_{\ast }$ attains it; on paths not yet hitting, the bounded discounted continuation value tends to zero. This supplies the optimal stopping verification. As $y\downarrow 0$, $p\downarrow 1$, the threshold diverges and the value tends to $S$.

For the perpetual American call option model and a fixed exercise threshold $B>K$, this is the discounted expectation of $B-K$ at its first hit, with zero payoff if the hit never occurs. The stationary ordinary differential equation and boundary conditions $C_B(0)=0$, $C_B(B)=B-K$ prove the formula. At fixed $S<B$, differentiation gives ${\partial }_B{C}_B=S^p{B}^{-p-1}[pK-(p-1)B]$, so the unique maximizing threshold is $pK/(p-1)$. If a larger threshold is imposed, $C_B^{\prime }(B-)>1$ and its value lies below immediate exercise payoff just below $B$. This is allowed for a committed policy, but violates the exercise lower bound if quoted for an unrestricted American call option.