# Solving the Geometric Brownian Motion on a general interval.

I know that the Geometric Brownian Motion, with the expression $dX_t = v X_t dt + \sigma X_t dW_t$ has the next solution $$X_t = X_0 e^{\sigma W_t+ (v-\frac{\sigma ^2}{2})t}$$ on the interval $[0,T]$. But, what would be the solution on a general interval $[t_1,t_2]$?

Would it be $$X_t = X_{t_1} e^{\sigma W_t-W_{t_1}+ (v-\frac{\sigma ^2}{2})(t-t_1)}$$

Using Ito's Lemma we have

$$d\log X_t=\left(v-\frac12\sigma^2\right)dt+\sigma dW_t \tag 1$$

Integrating $(1)$ between $t_1$ and $t_2$ yields

\begin{align} \log(X_{t_2}/X_{t_1})&=\left(v-\frac12\sigma^2\right)\left(t_2-t_1\right)+\sigma\int_{t_1}^{t_2}dW_t\\\\ &=\left(v-\frac12\sigma^2\right)\left(t_2-t_1\right)+\sigma\left(W_{t_2}-W_{t_1}\right) \end{align}

from which we have

$$X_{t_2}=X_{t_1}e^{\left(v-\frac12\sigma^2\right)\left(t_2-t_1\right)+\sigma\left(W_{t_2}-W_{t_1}\right)}$$

• Thanks a lot, @Dr.MV. Why, when applying the Ito Formula, $\frac{\partial log(X_t)}{\partial t}=0$? Commented Jul 3, 2015 at 19:41
• You're welcome. My pleasure. Good question. The function $\log x$ is not an explicit function of time and therefore its partial derivative with respect to time is zero. Commented Jul 3, 2015 at 19:44
• Ah, okey! Since $Y_t=g(t,X_t)=\log(X_t)$ where $g(t,x)=log(x)$ and then: $$dY_t= (( \frac{\partial \log x}{\partial t}+\frac {\partial \log x}{\partial x}vx+\frac{1}{2}\frac{\partial^2 \log x}{\partial x^2}(\sigma x^2))( dt+ \frac{\partial \log x}{\partial x}\sigma xdW_t)_{x=X_t}$$ Correct? I think I filnally got to undertand. Moreover, we can apply your procedure and that $W_0=0$ to get the general formula: $X_t = X_0 e^{\sigma W_t+ (v-\frac{\sigma ^2}{2})t}$ OK? Commented Jul 3, 2015 at 20:02
• Well, the term $\frac{\partial \log x}{\partial t}=0$. And yes, by setting $W_0=0$ we recover the well-known result. Commented Jul 3, 2015 at 20:06
• You're welcome!! It was my pleasure. Glad I could help. Commented Jul 3, 2015 at 20:09