Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 211 / 3

The bond pays one at maturity. If $\tau =\inf \{t<T:P_t^T\le 0\}$ had positive probability of being finite, buy one bond at $\tau$. A negative price supplies immediate consumption and a positive terminal payoff; a zero price supplies a free positive terminal payoff. Trading only on the stopping event gives an arbitrage. Therefore $P_t^T>0$ almost surely for every $t<T$.

For $K_1<K_2$, the lower-strike payoff dominates:If $C_t^{T,K_1}<C_t^{T,K_2}$, buy the cheaper lower-strike call and sell the higher-strike call. This gives positive initial consumption and a nonnegative terminal payoff, an arbitrage. Hence the monotonicity of a European call price in strike gives $C_t^{T,K_1}\ge C_t^{T,K_2}$.

First, no arbitrage implies the lower boundotherwise buy the call and $K$ maturity-$(T+1)$ bonds and short one non-dividend-paying stock; the initial receipt is positive and the terminal payoff is nonnegative. At time $T$, the assumption $P_T^{T+1}\le 1$ therefore gives $C_T^{T+1,K}\ge (S_T-K{)}^{+}$.

If $C_t^{T,K}>C_t^{T+1,K}$, sell the shorter call and buy the longer one. At time $T$, the longer call's no-arbitrage value covers the shorter call's payoff, with a strictly positive initial receipt. This is impossible, so $T\mapsto C_t^{T,K}$ is nondecreasing.

Order the support as $K_1<\cdots <K_N$ and putOn the finite support,This follows by telescoping: at $S_T=K_j$, only terms through $j$ survive and reconstruct successive increments of $g$. The static replication on a finite terminal support therefore has no-arbitrage price