Past exam of the mathematical models of financial derivatives course of the University of Oxford / 2017 (A13374W1)

Source: released examination PDF, bundle pages 63–66.

Solutions to all three questions and all optional parts. Expectations in Question 1 are conditional on current information. The zero-drift variance is obtained by a limit. The exercise-policy calculation in Question 3(c) is distinguished from the market value of an unrestricted American option.

Apply Itô's formula to $f(u,s)=e^{-\alpha u}s$. Its second stock derivative is zero, and the two drift terms cancel, givingIntegrating and multiplying by $e^{\alpha T}$ solves the linear stochastic differential equation:The Itô integral has a deterministic integrand, so it is a centered Gaussian random variable independent of the current filtration ${\mathcal{F}}_t$. Its variance follows from the Itô isometry. With $\tau =T-t$,The second line is also the limit as $\alpha \to 0$. The normal law asserted here is conditional on ${\mathcal{F}}_t$; an arbitrary random initial value need not give an unconditional normal distribution.

Use a self-financing portfolio with ${\Delta }_t$ units of the traded asset and the remainder in the continuous-time bank account. If its value equals $V(S_t,t)$, its self-financing condition isMeanwhile Itô's formula givesMatching the Brownian motion coefficients gives the delta hedge ${\Delta }_t=V_S$. Matching drifts then cancels the physical drift $\alpha$, leaving the pricing equation for a linear asset with additive noise:The domain is the full real line: additive noise can take the asset through zero. The diffusion coefficient is $\sigma$, so the second-derivative term has no factor $S^2$. As usual, the replication argument assumes suitable differentiability and an admissible trading strategy.

The differential operator in part b is the infinitesimal generator of the auxiliary diffusion process $d{S}_u=r{S}_{u}du+\sigma d{W}_u$, with killing rate $r$. The Feynman-Kac formula therefore givesEquivalently, Itô's formula makes $e^{-r(u-t)}V(S_u,u)$ a local martingale under these dynamics; under appropriate growth and integrability assumptions it is a martingale, whose terminal conditional expectation yields the same formula. The expectation uses the drift-$r$ law, rather than the physical drift-$\alpha$ law in part a. This is risk-neutral pricing.

Apply part a with $\alpha =r$. For $\tau >0$, the transition density is the normal distribution with mean $m=S{e}^{r\tau }$ and positive variance $v={\sigma }^2(e^{2r\tau }-1)/(2r)$. Integrating the financial payoff against this probability density function givesThus $d_{\ast }=(x-S{e}^{r\tau })/\sqrt{v}$, as required. The variance is positive even if $r<0$, because numerator and denominator have the same sign. For $r=0$, replace $m$ by $S$ and $v$ by ${\sigma }^2\tau$; at expiry use the terminal value directly. The integral presumes an integrable financial payoff.

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.

A European call option has nonnegative financial payoff, so its price is nonnegative under no-arbitrage. A second arbitrage comparison buys one call and a zero-coupon bond paying $K$, and shorts one share. Its terminal value is $(S_T-K{)}^{+}+K-S_T=(K-S_T{)}^{+}\ge 0$, so its initial cost cannot be negative. ConsequentlyThe holder of an American call option can choose to wait until expiry, so $C_t^a\ge C_t^e$. Since $r>0$ and $t<T$, an in-the-money call has $C_t^a\ge S_t-K{e}^{-r(T-t)}>S_t-K$: immediate exercise sacrifices value. An out-of-the-money exercise pays zero and cannot improve the holder's position. This proves no early exercise of a call without dividends: waiting is always at least as good, and is strictly better for positive immediate exercise payoff. In a degenerate market with zero continuation value an out-of-the-money holder can be indifferent; the Black-Scholes model with positive volatility gives strictly positive continuation value before expiry.

For the perpetual American call option, substitute a power $S^q$ into the stationary Black-Scholes equation with continuous stock dividends. The characteristic equation isIts two roots have negative product $-2r/{\sigma }^2$, so one is negative and one positive. The condition at zero excludes the negative power. Since $F(1)=-y<0$ and $F(q)\to \infty$ as $q\to \infty$, the positive root satisfies $p>1$; explicitly,Value matching gives $C(S)=(\hat{S}-K)(S/\hat{S}{)}^p$. The smooth fit condition gives $p(\hat{S}-K)/\hat{S}=1$, henceAs $y\downarrow 0$, the positive root tends to $1$ from above, so $\boxed{p\to 1,\hat{S}\to \infty }$. For fixed $S$, the value tends to $S$. In the dividend-free perpetual limit no finite threshold attains that supremum; this agrees with the finite-maturity no early exercise of a call without dividends.

For a fixed exercise threshold $\bar{S}>K$, the continuation value solves the same stationary equation but is matched to $\bar{S}-K$ at that threshold. Boundedness at zero removes the negative power. Thus the fixed-threshold value of a perpetual call isIt is also $(\bar{S}-K){\mathbb{E}}_S^{\mathbb{Q}}[e^{-r{\tau }_{\bar{S}}};{\tau }_{\bar{S}}<\infty ]$, where ${\tau }_{\bar{S}}$ is the first up-crossing stopping time. Differentiating while keeping the current $S$ fixed givesThis is positive for $K<\bar{S}<\hat{S}=pK/(p-1)$, so increasing such an early threshold increases its continuation value. The comparison assumes $S<\bar{S}$; a policy already exercised has no remaining continuation value to vary.

If $\bar{S}>\hat{S}$, thenSince $\tilde{C}(\bar{S})=\bar{S}-K$, the derivative comparison gives $\tilde{C}(S)<S-K$ for $S$ just below $\bar{S}$. Therefore quoting this policy value as the price of an unrestricted American call permits arbitrage: buy the option below its immediate exercise value and exercise it at once. A contract that actually commits its holder to the threshold is a different claim; its lower value is a suboptimal policy value, not by itself an arbitrage. The perpetual American call option optimizes over all exercise policies and remains above its exercise payoff.