I need the martingale part of the put payoff (not $C^2$..). Where $S\_t=exp(\sigma W\_t -\frac{\sigma^2t}{2})$



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> $d[(S\_t -K)^+ ]$ ??
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I guess I need to use local times but how?



Martingale part of the discontinuous put payoff
Thanks you all!


(proof, for $\phi(t,S\_t)=(K-S\_t)^+$:


**Step 1** *smoothing* : 
$\phi\_\epsilon(x)=1\_{S\_t\leq K+\epsilon}\cdot\phi(x)+1\_{S\_t\in]K-\epsilon,K]} \cdot \psi(x)$, where $\psi(x)=-\frac{1}{\epsilon^2}(K-x)^2(K-x-2\epsilon)$ .


This function is $C^1$, and also $C^2$ excepting in a countable set.


**Step 2** Itô on 
$\phi\_\epsilon(S\_t)= \phi\_\epsilon(S\_0)+\int^t\_0\phi\_\epsilon^{'}(S\_t)dS\_t+\frac{1}{2}\int^t\_0 1\_{S\_t\in[K-\epsilon,K]}\phi\_\epsilon^{''}(S\_t)d\langle S\rangle \_t$
because $\phi \_\epsilon=0$ out of $[K-\epsilon,K]$


Let's denote by $L\_t=lim\_{\epsilon \to0}{\frac{1}{2\epsilon^2}\*\int{\_{K-\epsilon}^K(3S\_t+4\epsilon-3K)dS\_t}}$ it's a finite variation process since it is increasing


**Step 3** We have that $\phi\_\epsilon(S\_t)-\phi\_\epsilon(S\_0)-\int^t\_0\phi\_\epsilon^{'}(S\_t)dS\_t \space \xrightarrow {L^2} \space\phi(S\_t) -\phi(S\_0) -\int^t\_0\phi^{'}(S\_t)dS\_t $ (because $\int^t\_0\phi\_\epsilon^{'}(S\_t) 1\_{S\_u\in[0,K-\epsilon]}dS\_u \xrightarrow {L^2}\int\_0^t\phi^{'}(S\_u)du$ by using Itô isometry)


**Finally** We get the formula 'à la bridge' namely



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> $(K-S\_t)^+=(K-S\_0)^+-\int\_0^t1\_{K\leq S\_u}dS\_u+L\_t$ 
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and the martingale part is 



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> $(K-S\_0)^+-\int\_0^t1\_{K\leq S\_u}\sigma S\_u dB\_u$
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