# Solution Geometric Brownian Brownian motion with no drift

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

The general solution is $$X_t=X_0 e^{(\mu-\frac{\sigma^2}{2})t+\sigma W_t}$$
If $\mu=0$, it is just $$X_t=X_0 e^{(\frac{\sigma^2}{2})t+\sigma W_t}$$
• I think that's buried in the general solution. Are you saying the closed-form general solution works for any $\mu$ other than zero? – eSurfsnake Mar 27 '18 at 5:36