Past exam of the martingales and financial mathematics course of the University of Oxford / 2007 / B / 6

Assume the SDE has the required solution and $f$ is sufficiently smooth. Itô's formula along that solution givesThe PDE cancels the first term. Integrating from $t$ to $T$ givesIf this integral is a true martingale, for example if its integrand has finite expected integrated square, its conditional expectation at time $t$ is zero. For the diffusion started at $x$ at time $t$, this yieldsThe notation in the question is the same Markov conditional expectation. The martingale or corresponding uniform-integrability growth hypothesis is part of the usual Feynman-Kac formula justification: a local martingale alone does not automatically have zero expected increment. Standard smooth solutions with the customary integrability bounds satisfy this requirement.

A stock holding receives both its price change and dividend, so its total gains are $dS+\delta Sdt$. The replicating portfolio therefore satisfiesItô's formula and the hedge $\Delta =(V_{\delta }{)}_S$ cancel the Brownian risk. Comparing drifts gives

To verify the claimed transformation, put $x=S{e}^{-\delta (T-t)}$ and set $U(t,S)=V_0(t,x)$. Then $x_t=\delta x$, $S{U}_S=x(V_0{)}_x$ and $S^2{U}_{SS}=x^2(V_0{)}_{xx}$. The dividend operator applied to $U$ becomesby the zero-dividend PDE. At maturity $x=S$, so the contingent claim payoff also agrees. Uniqueness in the pricing class proves

Differentiate the dividend PDE with respect to $S$. The diffusion term contributes ${\sigma }^{2}S{V}_{SS}+\frac{1}{2}{\sigma }^2{S}^2{V}_{SSS}$, and the stock-drift term contributes $(r-\delta )(V_S+S{V}_{SS})$. With $\Delta =V_S$, the coefficient of $\Delta$ is $(r-\delta )-r=-\delta$. ThereforeThe terminal condition is $\Delta (T,S)={\Psi }^{\prime }(S)$ wherever the contingent claim payoff is differentiable, with an appropriate weak or limiting interpretation at contingent claim payoff kinks. In the next part the squared positive-part contingent claim payoff is continuously differentiable, so its terminal option delta has no jump.

The contingent claim payoff derivative is ${\Psi }^{\prime }(s)=2(s-K{)}_{+}$. Differentiating the risk-neutral valuation is justified by the finite lognormal distribution, and $S_T$ is proportional to the starting stock value $s$. ThusFor a lognormal terminal stock, completing the square in its Gaussian exponent givesUse $p=2$ and $p=1$, and let $d_1=[\log (s/K)+(r+{\sigma }^2/2)\tau ]/(\sigma \sqrt{\tau })$. The result isAs $\tau \downarrow 0$ this tends to $2(s-K{)}_{+}$, including value zero at $s=K$, as required. The extra ${\sigma }^2$ in the option delta PDE's stock drift gives the same shifted Gaussian arguments. This is delta of a squared positive-part call.