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$$ \mathrm{d}S_t=μS_t\,\mathrm{d}t+σS_t\,\mathrm{d}B_t $$

The payoff function for a european call is: $$ f(S,T)=(S(T)-K)⁺. $$ When I graph this it is obvious it is continuous. How would I mathematically prove that it is actually continuous? Would i need to establish pathwise continuity?

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The function $x\mapsto (x-K)_+$ is continuous because it is $1$-Lipchitz. Indeed $$ |(x-K)_+-(y-K)_+| \le |x-y| $$ since $|u_+-v_+|\le |u-v|$ for any $(u,v)\in \mathbb{R}^2$. The solution of you SDE is pathwise continuous since the unique solution is $$ S_t = S_0 e^{(\mu-\sigma^2 /2)t + \sigma B_t}$$ Then $T \mapsto (S_T-K)_+$ is also pathwise continuous.

To be more precise if $(\Omega,\mathcal{F},\mathcal{P})$ is your probability space then it means that there exists $\Omega' \in \mathcal{F}$ with $P(\Omega')=1$ such that for all $\omega \in \Omega'$, the latter function $T \mapsto (S_T(\omega)-K)_+$ is continuous.

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ok thanks, what about the discrete Asian option payoff, it is not as straightforward [(1/n)∑S(t{i})-K]⁺ – Johnny Byr Sep 25 '12 at 9:16
Different problem? Then ask a different question! – Did Sep 25 '12 at 9:34
@JohnnyByr : for the discrete Asian what is the function you want to be continuous ? – vanna Sep 25 '12 at 9:49

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