2
$\begingroup$

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}

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.