Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 40 / 1

Put ${\mathcal{L}}_{r}V(s)=\frac{1}{2}{\sigma }^2{s}^2{V}^{\prime \prime }(s)+rs{V}^{\prime }(s)$ and define the nonnegative reserve rate $a(s)=rV(s)-{\mathcal{L}}_{r}V(s)$. The obstacle problem gives $V\ge g$ and $a\ge 0$. The American-option superhedge with a funded reserve invests the local surplus in the bond rather than consuming it.

For initial wealth $x\ge V(S_0)$, setand choose the stock and bond holdingsThus $D_t\ge 0$ and $F_t\ge V(S_t)\ge g(S_t)$, pathwise at every time. The Itô formula under the original drift givesAdding these equations yieldsThis is a self-financing strategy, and its nonnegative wealth makes it an admissible trading strategy. Continuity of the stock and local regularity of $V$ ensure local integrability of the holdings. The construction does not require $\mu =r$.

For a classical solution, the usual Itô formula applies directly. The smooth fit solution below is $C^1$ and piecewise $C^2$, with locally absolutely continuous first derivative. The generalized Itô formula applies with its almost-everywhere second derivative; the absence of a derivative jump means no boundary local time of a semimartingale term. This is the usual regularity interpretation of the perpetual American option obstacle equation.

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 $A{b}^{-\alpha }=(1+b{)}^{-1}$ and $-\alpha A{b}^{-\alpha -1}=-(1+b{)}^{-2}$. Dividing gives $b/(1+b)=\alpha$. ConsequentlyThis 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 $V(s)\ge (1+s{)}^{-1}$ for $s>b$. In the continuation region $({\mathcal{L}}_r-r)V=0$. In the exercise region,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 $d{S}_t=r{S}_{t}dt+\sigma S_{t}d{W}_t^{\mathbb{Q}}$. The Itô formula makes $e^{-rt}V(S_t)$ a nonnegative supermartingale, so every exercise time $\tau$ satisfiesBefore hitting the exercise region, its drift vanishes. Since $0\le V\le 1$, the stopped process is a bounded martingale. Apply the optional sampling theorem at ${\tau }_{\ast }\wedge T$ and let $T\to \infty$. The contribution from $\{{\tau }_{\ast }>T\}$ is at most $e^{-rT}$, while the boundary value equals the payoff. Hence equality holds forIf $S_0\le b$, 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.