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This question has been asked before in here Geometric Brownian motion without drift but I can't find what I want in the answers so ask again differently: for $\mu=0$ $$ dX_t =\mu X_t dt + \sigma X_t dW_t = \sigma X_t dW_t $$ Does it become: $$ (1) x_T = e^{\sigma(W_T-W_t)} $$ or $$ (2) x_T = e^{0.5\sigma^2(T-t)+\sigma(W_T-W_t)} $$ (2) seems unlikely for me because the process is clearly a local Martingale but (2) is not

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The general solution is $$X_t=X_0 e^{(\mu-\frac{\sigma^2}{2})t+\sigma W_t} $$

If $\mu=0$, it is just $$X_t=X_0 e^{(\frac{\sigma^2}{2})t+\sigma W_t} $$

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  • $\begingroup$ How will you relate this to the Martingale theory? Shoudn't X be a martingale $\endgroup$ – Kim Mar 26 '18 at 4:50
  • $\begingroup$ I think that's buried in the general solution. Are you saying the closed-form general solution works for any $\mu$ other than zero? $\endgroup$ – eSurfsnake Mar 27 '18 at 5:36

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