Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 39 / 5

Use the Black-Scholes equation with the claim's terminal condition $V(T,s)=g(s)$:For $t<T$, the Itô formula under the physical measure givesChoose the delta hedge and its bank-account holding byTheir portfolio value is $V(t,S_t)$, andThus the portfolio is self-financing and, with initial capital $V(0,S_0)$, has terminal wealth $g(S_T)$.

Localizing to compact stock-price intervals and times below $T$ justifies these stochastic integrals from the classical smoothness of $V$. The extension to maturity also follows from boundedness. Under the Risk-neutral measure for the Black-Scholes model, discounted $V(t,S_t)$ is a bounded local martingale, hence a true square-integrable martingale. Its quadratic variation has finite expectation, so the stock-integral coefficient is square integrable through $T$. Equivalence of measures preserves its almost-sure finiteness, and the physical drift integral is finite by Cauchy-Schwarz. The terminal limit is $g(S_T)$.

Finally, $B_t=B_0{e}^{rt}$ is bounded away from zero on this finite horizon, and bounded $V$ makes $V(t,S_t)/B_t$ bounded below by a deterministic constant. The hedge is therefore an admissible trading strategy. The displayed holdings replicate the payoff admissibly, even if the bounded payoff can be negative.

Set $\theta =(\mu -r)/\sigma$. The positive densityhas mean one, and the Girsanov theorem makes $W_t^Q=W_t+\theta t$ a Brownian motion under the equivalent measure $Q$. HenceThe Black-Scholes equation makes $e^{-rt}V(t,S_t)$ a local martingale. Bounded $V$ on the finite horizon makes it a true martingale, so its terminal condition gives the risk-neutral valuationThe geometric Brownian motion solution under $Q$ isThe Brownian increment is independent of ${\mathcal{F}}_t$ and equals $\sqrt{T-t}Z$ in distribution, where $Z$ has the standard normal distribution. ThereforeThe expectation on the right is only an integral against the standard normal law; the physical drift $\mu$ has disappeared.

Let $L=\exp ((r-\frac{1}{2}{\sigma }^2)(T-t)+\sigma \sqrt{T-t}Z)$. Boundedness of $g^{\prime }$ and integrability of the lognormal distribution variable $L$ justify differentiation under the expectation, by dominated convergence:A differentiable increasing payoff has $g^{\prime }\ge 0$, and $L>0$. Thus the delta hedge hasMoreover $\mathbb{E}L=e^{r(T-t)}$, so $0\le {\pi }_t\le \Vert g^{\prime }{\Vert }_{\infty }$. At maturity the delta is $g^{\prime }(S_T)\ge 0$. The admissible trading strategy already constructed therefore uses nonnegative stock holdings throughout replication. After maturity, liquidate the stock position and retain the proceeds in the bank account, so ${\pi }_t=0$ for $t>T$ if holdings are to be defined for all times.