Past exam of the mathematical models of financial derivatives course of the University of Oxford / 2017 / 3

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.