Free-Boundary Valuation of a Perpetual Option with Competing Put Payoffs

Summary

The document sets up a perpetual option whose payoff is the larger of two put-like payoffs, one with strike K1 and slope −1 and the other with strike K2 and slope −2. It begins with the standard perpetual-option ordinary differential equation under geometric Brownian motion. The general solution has a linear term and a power term; requiring the value to vanish as the asset price grows removes the linear term.

For a conventional American put, value matching and smooth pasting determine the exercise boundary and remaining coefficient. The question is whether those same conditions suffice when the payoff is the maximum of two lines. The document offers no answer or derivation for the altered payoff, so it establishes the valuation setup and identifies the free-boundary challenge without resolving whether there is one exercise region or multiple regions. Its formulation also assumes the stated perpetual model and does not discuss dividends or other market features.

Key ideas

• A perpetual option under the stated model satisfies a second-order ordinary differential equation.
• The boundary condition at high asset prices eliminates the linear component of the general solution.
• For a standard perpetual put, value matching and smooth pasting determine the exercise threshold and coefficient.
• The maximum of two put-like payoffs may complicate the exercise boundary, and the document leaves that problem open.

Tags

# Perpetual Option Paying Chooser Option


# Perpetual Option Paying Chooser Option












A perpetual option solves the ODE $$rSV_S+\frac{1}{2}\sigma^2S^2V_{SS}-rV=0$$ The general solution is $$V(S)=aS+bS^{\gamma}$$ where $\gamma=-\frac{2r}{\sigma^2}<0$.

For an American put option with payoff $K-S$, we find $a=0$ because we require $V(S)=0$ as $S\to\infty$. We find the free boundary (exercise point, $S^*$) and the remaining free parameter ($b$) by value-matching and smooth-pasting of $V(S)$ with the payoff $K-S$ at $S=S^*$, that is \begin{align} b(S^*)^{\gamma} &= K-S^* \\ b\gamma(S^*)^{\gamma-1} &= -1 \end{align} The option value is then \begin{align} V(S)=\begin{cases} K-S &if\; S<S^* \\ bS^{\gamma} &if\; S\geq S^* \end{cases} \end{align}

Question: What happens if the option pays $\max(K_1-S,K_2-2S)=K_2-2S+\max(S+K_1-K_2,0)$ instead of $K-S$? The payoff now resembles a chooser option (between two puts). We still require $a=0$ such that $V(S)=0$ as $S\to\infty$. But how to proceed? I don't think it's as simple as finding $b$ and $S^*$ by solving value-matching and smooth-pasting condition and setting \begin{align} V(S)=\begin{cases} \max(K_1-S,K_2-2S) &if\; S<S^* \\ bS^{\gamma} &if\; S\geq S^* \end{cases} \end{align}

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.