# Solving SDE system to reobtain geometric Brownian motion

I have peacefully derived geometric Brownian motion by applying Ito's formula to the process $Y(t) = e^{\alpha B(t) - \frac{t}{2}}$ and letting $\alpha = 1$.

The differential form of $dY$ is clearly equal to

$$dY = \alpha YdB + \left(\frac{\alpha^2}{2} - \frac{1}{2}\right)Ydt \qquad (1)$$

and I understand this to be the SDE that governs a geometric Brownian motion with drift.

I'm now trying to look at the issue from another perspective.

Say I have the following system of SDEs

$$\left\{\begin{matrix} dX(t) = Y(t)dB^{(1)}(t)\\ dY(t) = Y(t)dB^{(2)}(t) \end{matrix}\right.$$

where $B^{(1)}(t),t>0$ and $B^{(2)}(t),t>0$ are two standard and independent Brownian motions. I also have the following initial conditions

$$X(0) = 0 \qquad Y(0) = 1$$

I would like to obtain $Y(t)$ as a solution to this system and eventually find out also $X(t)$, but I'm having a hard time.

I can see that the second equation is exactly equal to $(1)$ for $\alpha =1$, however I would like to actually see how the initial conditions play a role.

My attempt was the following.

I take the second equation and treat as follows

$$dY(t) = Y(t)dB^{(2)}(t) \iff \\ \frac{dY(t)}{Y(t)} = dB^{(2)}(t) \iff \\\ln(Y(t)) = \int_0^tdB^{(2)}(t) = B^{(2)}(t) - B^{(2)}(0) \iff \\ Y(t) = e^{B^{(2)}(t)}$$

since $B^{(2)}(t)$ is a standard Brownian motion. Nowhere near the wanted result. And applying the initial condition only gives me a triviality.

What am I missing? I clearly need to reobtain $Y(t)$ to then solve $X(t)$...

You can show that $$d(\ln Y_t) = \frac{dY_t}{Y_t} - \frac{1}{2}dt$$ So we can write: $$dY_t = Y_t\, dB_t \;\;\;\implies\;\;\; \frac{dY_t}{Y_t} = dB_t \;\;\;\implies\;\;\; d(\ln Y_t) = dB_t - \frac{1}{2}dt$$ Thus, we get: $$\ln Y_t = B_t - \frac{1}{2}t\;\;\;\implies\;\;\; Y_t = \exp\left( B_t - \frac{1}{2}t \right)$$ which is the solution to geometric Brownian motion with $\mu=0,\sigma=1,Y_0=1$.