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Geometric Brownian motion
(section)
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==Solving the SDE== For an arbitrary initial value ''S''<sub>0</sub> the above SDE has the analytic solution (under [[It么 calculus|It么's interpretation]]): : <math> S_t = S_0\exp\left( \left(\mu - \frac{\sigma^2}{2} \right)t + \sigma W_t\right).</math> The derivation requires the use of [[It么 calculus]]. Applying [[It么's formula]] leads to : <math>d(\ln S_t) = (\ln S_t)' d S_t + \frac{1}{2} (\ln S_t)'' \,dS_t \,dS_t = \frac{d S_t}{S_t} -\frac{1}{2} \,\frac{1}{S_t^2} \, dS_t \, dS_t </math> where <math> dS_t \, dS_t</math> is the [[quadratic variation]] of the SDE. :<math> d S_t \, d S_t \, = \, \sigma^2 \, S_t^2 \, d W_t^2 + 2 \sigma S_t^2 \mu \, d W_t \, d t + \mu^2 S_t^2 \, d t^2 </math> When <math> d t \to 0 </math>, <math> d t</math> converges to 0 faster than <math> d W_t</math>, since <math> d W_t^2 = O(d t) </math>. So the above infinitesimal can be simplified by :<math> d S_t \, d S_t \, = \, \sigma^2 \, S_t^2 \, dt </math> Plugging the value of <math>dS_t</math> in the above equation and simplifying we obtain : <math>\ln \frac{S_t}{S_0} = \left(\mu -\frac{\sigma^2}{2}\,\right) t + \sigma W_t\,.</math> Taking the exponential and multiplying both sides by <math>S_0</math> gives the solution claimed above.
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