Question

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

$d[(S_t -K)^+ ]$ ??

I guess I need to use local times but how?

Answer 1

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

$(K-S_t)^+=(K-S_0)^+-\int_0^t1_{K\leq S_u}dS_u+L_t$

and the martingale part is

$(K-S_0)^+-\int_0^t1_{K\leq S_u}\sigma S_u dB_u$

Answer 2

Hi,

Simply use Itô-Tanaka's formula I guess this should give something like : $df(S_t)=D_-f(S_t)dS_t+\frac{1}{2}dL^s_tf''(ds)$

with $f(S)=(S-K)^+$ so $D_-f(S)=1_{]K,+\infty}(S)$ and $f''(ds)=\delta_K(ds)$

This gives if I am not mistaken :

$d(S_t-K)^+=df(S_t)=1_{]K,+\infty}(S_t)dS_t+\frac{1}{2}dL^K_t$

With $L^K_t$ being the local time of your geometric Brownian Motion $S$ around level $K$ at time $t$.

Regards

Edit NB:
-$D_-$ stands for the left derivative of $f$

-$f''(ds)$ stands for second derivatives in the distribution-sense.

-The use of Itô-Tanaka's formula allows to avoid the derivation of the Mollifiers-type argument for the direct proof of the result (which is quite cumbersome in my opinion). I should add that Ito-Tanaka's formula is applicable to every $f$ that is the differnce of two convex functions if I remember well, which is the case here with $f(x)=(X-K)^+$.