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Let $V=V(S_t,t)$ be the option price and \begin{align} V_t+\mu\,S\,V_S+\frac{1}{2}\sigma^2\,S^2\,V_{SS}=0\\ V(S_T,T)=\ln (S_T)^{2}. \end{align} My question: How can I obtain a closed form solution of $V=V(S_t,t)$. Please help me.

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  • $\begingroup$ @Richard,Hi Richard I am New member.Is my question wrong? $\endgroup$ – user16764 Jul 3 '15 at 12:01
  • $\begingroup$ @Richard ,Please guide me $\endgroup$ – user16764 Jul 3 '15 at 12:05
  • $\begingroup$ Everything fine .. I just thought that the correct expression is "closed form solution". So I edited the question. I hope this is fine with you. $\endgroup$ – Richard Jul 3 '15 at 12:14
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Feynman–Kac Theorem: Assume that $F$ is a solution to the boundary value problem \begin{align} &F_t+\mu(t,x)F_x+\frac{1}{2}\sigma^2(t,x)F_{xx}-rF=0\\ &F(T,x)=\Phi(x), \end{align} Assume furthermore that the process $e^{-r_s}\sigma(s,X_s)F_s$ is in $\mathcal L^2$ where \begin{align} dX_s=\mu(s,x)ds+\sigma(s,x)dW_s, \end{align} then $F$ has the representation. \begin{align} F(t,x)=e^{-r(T-t)}E^Q_{t,x}[\Phi(X_T)] \end{align} Now, let \begin{align} dS_t=\mu S_tdt+\sigma S_tdW_t \end{align} by application Ito lemma,we have \begin{align} \ln S_T=\ln S_t\,\,+(\mu-\frac{1}{2}\sigma^2)(T-t)+\sigma\,(W_T-W_t) \end{align} then \begin{align} &V(S_t,t)=e^{-0\times(T-t)}E^Q_{t,s}[\ln(S_T)^2]=E^Q_{t,s}[2\ln(S_T)]\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=E^Q_{t,s}[2\ln(S_t)+2(\mu-\frac{1}{2}\sigma^2)(T-t)+2\sigma\,(W_T-W_t)]\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\ln(S_t)+2(\mu-\frac{1}{2}\sigma^2)(T-t)+2\sigma\,E^Q_{t,s}[(W_T-W_t)]\\ \end{align} The process $W_t$ has independent increments,therefor \begin{align} V(S_t,t)=2\ln(S_t)+2(\mu-\frac{1}{2}\sigma^2)(T-t) \end{align}

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  • $\begingroup$ It would be nice to have a solution without using any SDE for $S$, that is, only use the PDE and the boundary condition. $\endgroup$ – Gordon Jul 3 '15 at 14:11
  • $\begingroup$ Wasn't the payoff $\ln^2(S_T)$? $\endgroup$ – AFK Jul 3 '15 at 18:32

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