= Solution
Write $h=T-t$. Conditional on $\mathcal F_t^S$, <independent increments> of the <Brownian motion> give
$$
S_T=S_t\exp\{-\sigma^2h/2+\sigma\sqrt h\,Y\},
\qquad Y\sim N(0,1),
$$
where $Y$ has the <standard normal distribution> and is <independent> of $\mathcal F_t^S$. Factoring $S_t$ out of the positive part identifies the <normalized Black-Scholes call function>:
$$
\mathbb E[(S_T-K)^+\mid\mathcal F_t^S]
=S_t F(\sigma^2h,K/S_t)=C_t.
$$
The <European call option> payoff is <integrable>, because $0\leq(S_T-K)^+\leq S_T$ and $\mathbb E S_T=S_0$. The <tower property of conditional expectation> therefore proves that $C$ is a true <martingale>. At $t=T$, the boundary value $F(0,m)=(1-m)^+$ gives $C_T=(S_T-K)^+$. \b[The answer is the conditional payoff martingale]
$$
\boxed{C_t=\mathbb E[(S_T-K)^+\mid\mathcal F_t^S].}
$$
Back to article page
