Past exam of the mathematical models of financial derivatives course of the University of Oxford / 2017 / 2

For a positive starting value, Itô's formula applied to $\log S_u$ gives $d\log S_u=(\mu -{\sigma }^2/2)du+\sigma d{W}_u$. Hence the geometric Brownian motion isThe future Brownian increment $W_T-W_t$ is independent of ${\mathcal{F}}_t$ and has normal distribution $N(0,\tau )$. If ${\tilde{W}}_v=W_{t+v}-W_t$, then $\tilde{W}$ is a fresh Brownian motion and the increment is exactly ${\tilde{W}}_{\tau }$. This specifies the meaning of the paper's abbreviated $W_{\tau }$; the original process value $W_{T-t}$ is not generally that future increment. The stochastic exponential solution also verifies positivity for $S>0$.

Under the risk-neutral measure, write $Z\sim N(0,1)$, $a=\sigma \sqrt{\tau }$, and $m=\log (S/K)+(r-{\sigma }^2/2)\tau$. Then $\log (S_T/K)=m+aZ$ and $d_{-}=m/a$. By risk-neutral pricing, this log-call option has value $e^{-r\tau }\mathbb{E}[(m+aZ{)}^{+}]$.

Let $\varphi$ and $\Phi$ denote the standard normal density and standard normal distribution function. The Gaussian positive-part expectation follows by integration:because ${\varphi }^{\prime }(z)=-z\varphi (z)$. ThereforeFor $\tau =0$ take the continuous terminal limit $(\log (S/K){)}^{+}$.

The Black-Scholes reflection symmetry becomes a weighted reflection after taking a logarithm. Put $x=\log (S/B)$ and $v(x,t)=V(B{e}^x,t)$. The multivariable chain rule transforms the equation intoFor $w(x,t)=e^{\gamma x}v(-x,t)$, its left-hand side, after removing the factor $e^{\gamma x}$, isevaluated at $(-x,t)$. Taking $\gamma =-2b/{\sigma }^2$ makes the derivative coefficient $b$ and the additional zeroth-order term zero. Thussatisfies the same Black-Scholes equation.

Subtract the reflected solution from part c. On the alive side of the down-and-out barrier option,Both terms satisfy the Black-Scholes equation, and at $S=B$ they cancel. At expiry, $S\ge B$ gives $B^2/S\le B<K$, so the reflected log-call option has zero terminal financial payoff; the first term gives the required payoff. With the usual growth condition, uniqueness for the killed pricing problem identifies the expression as the risk-neutral price. After the barrier has been hit the contract value is zero. The algebraic image expression on $S<B$ is not its financial value.