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 $f(x)=(X-K)^+$.

3 added 15 characters in body

Hi,

Simply use Itô-Tanaka's formula I guess this should give something like : $df(X_t)=D_-f(X_t)dX_t+\frac{1}{2}dL^x_tf''(dx)$df(S_t)=D_-f(S_t)dS_t+\frac{1}{2}dL^s_tf''(ds)$with$f(X)=(X-K)^+$f(S)=(S-K)^+$ so $D_-f(X)=1_{]K,+\infty}(X)$ D_-f(S)=1_{]K,+\infty}(S)$and$f''(dx)=\delta_K(dx)$f''(ds)=\delta_K(ds)$

This gives if I am not mistaken :

$df(X_t)=1_{]K,+\infty}(X_t)dX_t+\frac{1}{2}dL^K_t$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$. Regards 2 added 8 characters in body Hi, Simply use Itô-Tanaka's formula I guess this should give something like :$df(X_t)=D_-f(X_t)dX_t+\frac{1}{2}dL^x_tf''(dx)$with$f(X)=(X-K)^+$so$D_-f(X)=1_{]K,+\infty}(X)$and$\delta_K(dx)$f''(dx)=\delta_K(dx)$

This gives if I am not mistaken :

$df(X_t)=1_{]K,+\infty}(X_t)dX_t+\frac{1}{2}dL^K_t$

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

Regards

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