= Solution
Write $\alpha=2r/\sigma^2\in(0,1)$. In a continuation interval, the <Euler differential equation> has power solutions $s$ and $s^{-\alpha}$: substitution of $s^\beta$ gives $(\beta-1)(\beta+\alpha)=0$. A decreasing bounded value uses the second solution.
Let $b$ be the exercise boundary. Value matching and <smooth fit> require $Ab^{-\alpha}=(1+b)^{-1}$ and $-\alpha Ab^{-\alpha-1}=-(1+b)^{-2}$. Dividing gives $b/(1+b)=\alpha$. Consequently
$$
\boxed{b=\frac\alpha{1-\alpha},\qquad
V(s)=\begin{cases}
(1+s)^{-1},&0<s\leq b,\\
\displaystyle\frac1{1+b}\left(\frac{s}{b}\right)^{-\alpha},&s>b.
\end{cases}}
$$
This is a <perpetual reciprocal-payoff American option>.
To verify the obstacle inequality, observe that $s^\alpha/(1+s)$ increases up to $b$ and decreases afterwards, because its logarithmic derivative is $\alpha/s-1/(1+s)$. Therefore $V(s)\geq(1+s)^{-1}$ for $s>b$. In the continuation region $(\mathcal L_r-r)V=0$. In the exercise region,
$$
(\mathcal L_r-r)g(s)
=\frac{\sigma^2}{(1+s)^3}
\left((1-\alpha)s^2-\frac{3\alpha}2s-\frac\alpha2\right).
$$
The quadratic is convex. At the two endpoints of $[0,b]$ it equals $-\alpha/2$ and $-\alpha/[2(1-\alpha)]$, respectively, so it is negative throughout that interval. Thus the <obstacle problem> is satisfied on both regions, with value matching and <smooth fit> at $b$. The second derivative has a jump at $b$; the equation is understood piecewise and in the generalized Itô sense described in part (a).
For optimality, use the <Risk-neutral measure for the Black-Scholes model>, under which $dS_t=rS_tdt+\sigma S_tdW_t^{\mathbb Q}$. The <Itô formula> makes $e^{-rt}V(S_t)$ a nonnegative <supermartingale>, so every exercise time $\tau$ satisfies
$$
\mathbb E_{\mathbb Q}[e^{-r\tau}g(S_\tau)1_{\{\tau<\infty\}}]\leq V(S_0).
$$
Before hitting the exercise region, its drift vanishes. Since $0\leq V\leq1$, the stopped process is a bounded <martingale>. Apply the <optional sampling theorem> at $\tau_*\wedge T$ and let $T\to\infty$. The contribution from $\{\tau_*>T\}$ is at most $e^{-rT}$, while the boundary value equals the payoff. Hence equality holds for
$$
\boxed{\tau_*=\inf\{t\geq0:S_t\leq b\}.}
$$
If $S_0\leq b$, exercise immediately; otherwise wait for the first down-crossing. If that time is infinite, the discounted payout is zero. This proves both the value and the <optimal stopping> policy, and part (a) supplies its <superhedge>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-40-perpetual-boundary.png]
{title=Perpetual reciprocal-payoff option value and exercise boundary for two interest-to-variance ratios}
{height=340}
Back to article page
