Geometric brownian motion with more than one brownian motion term

Question

I know that if we have the following SDE:

$$dX_t = \alpha X_t dt + \sigma X_t dW_t$$ where $\alpha$ and $\sigma$ are constants and $X_0 = x_0$, then the solution is the geometric brownian motion: $$X(t) = x_0e^{(\alpha-\frac{1}{2}\sigma^2)t + \sigma W_t} $$

Now, what if we have more than one $dW$ term? Specifically, say we have the following SDE:

$$dX_t = \alpha X_t dt + \sigma_1 X_t dW^1_t+\sigma_2X_tdW^2_t$$ where $\alpha, \sigma_1, \sigma_2$ are all constants and $W^1_t$ and $W^2_t$ are independent standard Brownian motions, can we find a solution to $X(t)$? Can we say anything about the distribution of $X(t)$?

Answer 1

The respective solution would become $$ X_t=X_0\exp\left[\left(\alpha-\frac{1}{2}\sigma_1^2-\frac{1}{2}\sigma_2^2\right)t+\sigma_1W_t^1+\sigma_2W_t^2\right]. $$

Note that the SDE reads $$ {\rm d}X_t=X_t\left(\alpha{\rm d}t+\sigma_1{\rm d}W_t^1+\sigma_2{\rm d}W_t^2\right). $$ This gives the quadratic variation of $X_t$ by, intuitively, \begin{align} {\rm d}\left<X\right>_t&={\rm d}X_t{\rm d}X_t\\ &=X_t^2\left(\alpha{\rm d}t+\sigma_1{\rm d}W_t^1+\sigma_2{\rm d}W_t^2\right)^2\\ &=X_t^2\left(\alpha^2{\rm d}t^2+\sigma_1^2{\rm d}W_t^1{\rm d}W_t^1+\sigma_2^2{\rm d}W_t^2{\rm d}W_t^2+\right.\\ &\quad\quad\left.2\alpha\sigma_1{\rm d}t{\rm d}W_t^1+2\alpha\sigma_2{\rm d}t{\rm d}W_t^2+2\sigma_1\sigma_2{\rm d}W_t^1{\rm d}W_t^2\right)\\ &=X_t^2\left(\sigma_1^2{\rm d}t+\sigma_2^2{\rm d}t\right), \end{align} where the ${\rm d}W_t^1{\rm d}W_t^2$ term drops due to the independence of $W_t^1$ and $W_t^2$.

Thus thanks to Ito's lemma, \begin{align} {\rm d}\log X_t&=\frac{{\rm d}X_t}{X_t}-\frac{1}{2}\frac{{\rm d}\left<X\right>_t}{X_t^2}\\ &=\left(\alpha{\rm d}t+\sigma_1{\rm d}W_t^1+\sigma_2{\rm d}W_t^2\right)-\frac{1}{2}\left(\sigma_1^2+\sigma_2^2\right){\rm d}t\\ &=\left(\alpha-\frac{1}{2}\sigma_1^2-\frac{1}{2}\sigma_2^2\right){\rm d}t+\sigma_1{\rm d}W_t^1+\sigma_2{\rm d}W_t^2\\ &={\rm d}\left[\left(\alpha-\frac{1}{2}\sigma_1^2-\frac{1}{2}\sigma_2^2\right)t+\sigma_1W_t^1+\sigma_2W_t^2\right]. \end{align} This, together with the initial condition, eventually yields $$ X_t=X_0\exp\left[\left(\alpha-\frac{1}{2}\sigma_1^2-\frac{1}{2}\sigma_2^2\right)t+\sigma_1W_t^1+\sigma_2W_t^2\right]. $$