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I attempted to treat SDE as ODE to solve the Geometric Brownian Equation and obviously got the wrong answer. My question is, where did it go wrong?

$dX_t=\mu X_tdt+\sigma X_tdW_t$ as given.

$\dfrac{dX_t}{X_t}=\mu dt+\sigma d W_t$ dividing $X_t$ on both sides.

$\int\limits_0^t\dfrac{dX_s}{X_s}=\int\limits_0^t\mu ds+\int\limits_0^t\sigma d W_s$ integrating on both sides.

$\ln\left(\frac{X_t}{X_0}\right)=\mu t+\sigma W_t$

$X_t=X_0 e^{\mu t+\sigma W_t}$

Jesse P Francis
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Chris L
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    "where did it go wrong?" When you decided that $\int dX/X=C+\log X$. Actually, if $Y=\log X$ then Itô says that $dY=dX/X-d\langle X\rangle/(2X^2)$, not $dY=dX/X$. – Did Oct 09 '15 at 09:52
  • @ChrisL, Welcome to math.SE! For now, I have converted the image to math notation in this question. For some basic information about writing math at this site see e.g. here, here, here and here. – Jesse P Francis Oct 09 '15 at 10:17
  • @Did thank you for the response. You are right. I am fundamentally wrong to apply ODE to SDE because of the very nature that X(t) is stochastic, not deterministic. As such, any attempt to apply ODE formula (which means ignoring the Ito correction of 1/2*f'' ) will not be rigorous. – Chris L Oct 11 '15 at 03:33

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