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Use a stochastic representation result (feynman kac theorem) to solve the following boundary value problem :

$\frac{\partial V}{\partial t}+\mu x\frac{\partial V}{\partial x}+\frac{1}{2}\sigma^2x^2\frac{\partial^2 V}{\partial x^2}=0$

with $V(T,x)= \log(x^2)$

By Feynman-kac the solution should be $E[\log(X_T^2)|\mathscr{F}_t)]$ where we use the Ito process:$dX_t=\mu(X_t,t)dt+\sigma(X_t,t)dW_t$

and initial $X_t=x$.

Can I take $dX_t=\mu x dt+\sigma x dW_t$ here?

Which gives $X_T=x+\mu (T-t)+\sigma(W_T-W_t)$

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I guess you got a typo a the start. Anyways, use $\mathrm{d} X_s = \mu X_s\mathrm{d}s + \sigma X_s\mathrm{d}W_s $ to get $$\log \frac{X_T}{X_t} = \left( \mu - \frac{1}{2}\sigma^2\right)(T-t) + \sigma (W_T - W_t)$$ Then $$ V(t,x) = \mathbb{E}[2\log X_T | \mathcal{F}_t] = 2\log x + (2\mu - \sigma^2)(T-t) $$

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  • $\begingroup$ Where does the log come from in front of $\frac{X_T}{X_t}$ $\endgroup$ – Eran Mar 28 '17 at 9:52
  • $\begingroup$ And I don't see where the $2\log x$ comes from, in my calculation it is $x(2\mu-\sigma^2)(T-t)$ $\endgroup$ – Eran Mar 28 '17 at 10:49
  • $\begingroup$ Note that $$\mathrm{d}\log X_s = \frac{1}{X_s}\mathrm{d} X_s + \frac{1}{2}\left( -\frac{1}{X_s^2}\right)\mathrm{d}[X]_s = \frac{1}{X_s} (\mu X_s \mathrm{d}s + \sigma X_s \mathrm{d}W_s) + \frac{1}{2}\left( -\frac{1}{X_s^2}\right)(\sigma^2 X_s^2 \mathrm{d}s) = \left( \mu - \frac{1}{2} \sigma^2\right) \mathrm{d}s + \sigma \mathrm{d}W_s$$ $\endgroup$ – ChargeShivers Mar 28 '17 at 15:11
  • $\begingroup$ And the $2 \log x$? Why isn't it $x(2\mu-\sigma^2)(T-t)$? $\endgroup$ – Eran Mar 28 '17 at 18:09
  • $\begingroup$ Should the $X_t$ be there, at $X_T/X_t$? or shouldn't it be $ \log X_T= \log X_t +\left( \mu - \frac{1}{2}\sigma^2\right)(T-t) + \sigma (W_T - W_t)$ $\endgroup$ – Eran Mar 28 '17 at 18:20

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