Martingale part of the discontinuous put payoff - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T14:05:38Z http://mathoverflow.net/feeds/question/57524 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57524/martingale-part-of-the-discontinuous-put-payoff Martingale part of the discontinuous put payoff Samuel A 2011-03-06T02:07:24Z 2011-03-07T17:14:36Z <p>I need the martingale part of the put payoff (not $C^2$..). Where $S_t=exp(\sigma W_t -\frac{\sigma^2t}{2})$</p> <blockquote> <p>$d[(S_t -K)^+ ]$ ??</p> </blockquote> <p>I guess I need to use local times but how?</p> http://mathoverflow.net/questions/57524/martingale-part-of-the-discontinuous-put-payoff/57581#57581 Answer by The Bridge for Martingale part of the discontinuous put payoff The Bridge 2011-03-06T15:40:55Z 2011-03-07T17:14:36Z <p>Hi,</p> <p>Simply use Itô-Tanaka's formula I guess this should give something like : $df(S_t)=D_-f(S_t)dS_t+\frac{1}{2}dL^s_tf''(ds)$</p> <p>with $f(S)=(S-K)^+$ so $D_-f(S)=1_{]K,+\infty}(S)$ and $f''(ds)=\delta_K(ds)$ </p> <p>This gives if I am not mistaken :</p> <p>$d(S_t-K)^+=df(S_t)=1_{]K,+\infty}(S_t)dS_t+\frac{1}{2}dL^K_t$</p> <p>With $L^K_t$ being the local time of your geometric Brownian Motion $S$ around level $K$ at time $t$.</p> <p>Regards</p> <p>Edit NB:<br> -$D_-$ stands for the left derivative of $f$ </p> <p>-$f''(ds)$ stands for second derivatives in the distribution-sense. </p> <p>-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 $f(x)=(X-K)^+$. </p> http://mathoverflow.net/questions/57524/martingale-part-of-the-discontinuous-put-payoff/57659#57659 Answer by Samuel A for Martingale part of the discontinuous put payoff Samuel A 2011-03-07T13:07:03Z 2011-03-07T13:07:03Z <p>Thanks you all!</p> <p>(proof, for $\phi(t,S_t)=(K-S_t)^+$:</p> <p><strong>Step 1</strong> <em>smoothing</em> : $\phi_\epsilon(x)=1_{S_t\leq K+\epsilon}\cdot\phi(x)+1_{S_t\in]K-\epsilon,K]} \cdot \psi(x)$, where $\psi(x)=-\frac{1}{\epsilon^2}(K-x)^2(K-x-2\epsilon)$ .</p> <p>This function is $C^1$, and also $C^2$ excepting in a countable set.</p> <p><strong>Step 2</strong> 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 $[K-\epsilon,K]$</p> <p>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</p> <p><strong>Step 3</strong> 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)</p> <p><strong>Finally</strong> We get the formula 'à la bridge' namely</p> <blockquote> <p>$(K-S_t)^+=(K-S_0)^+-\int_0^t1_{K\leq S_u}dS_u+L_t$ </p> </blockquote> <p>and the martingale part is </p> <blockquote> <p>$(K-S_0)^+-\int_0^t1_{K\leq S_u}\sigma S_u dB_u$</p> </blockquote>