Model 2 correlated Geometric Brownian Motions

Question

Let's say we have 2 GBM processes:
$$S_1(t)=S_1(0)\cdot \exp{(\mu_1t-\frac{1}{2}\sigma_1^2t+\sigma_1\cdot W(t))}\text{ and } S_2(t)=S_2(0)\cdot \exp{(\mu_2t-\frac{1}{2}\sigma_2^2t+\sigma_2\cdot W(t))}$$
such that $\mathbb{E}[dW_1;dW_2]=\rho$.
How can I model them together? Formulas above won't take the correlation into account. Since $S(t)$ are meant to be stock prices, I can look at daily returns directly as:
$$\frac{S(t+1)-S(t)}{S(t)}=\exp{(\mu-\frac{1}{2}\sigma^2+\sigma \cdot N(0;1))}-1=\exp{(N(\mu-\frac{1}{2}\sigma^2;\sigma^2))}-1$$
which is close to normal (transposed lognormal in fact). If it was in fact normally distributed then I would use 2-D $N(M;\Sigma)$ with $M=(\mu_1-\frac{1}{2}\sigma_1^2;\mu_2-\frac{1}{2}\sigma_2^2)$ and $\Sigma$ as the covariance matrix. But I want to be precise with the model and not assume (incorectly) that returns are normal if prices are GBM. So I'm struggling on how to include the correlation/covariance in the direct formulas for $S_1,S_2$.

Answer 1

Let $B_1$ and $B_2$ be independent Brownian motions. Set $W_1 = B_1$ and $W_2 = \rho B_1 + \sqrt{1-\rho^2} B_2$. Then $W_1$ and $W_2$ are correlated Brownian motions with correlation coefficient $\rho$.

In your model for correlated geometric Brownian motion, you may let $S_1$ be driven by $W_1$ and $S_2$ be driven by $W_2$, to obtain:

$$\begin{align*} S_1(t) &= S_1(0) \exp \left( (\mu_1 - 0.5\sigma_1^2)t + \sigma_1 W_1(t) \right) \\ S_2(t) &= S_2(0)\exp \left( (\mu_2 - 0.5\sigma_2^2)t + \sigma_2 W_2(t) \right) \end{align*} \\ $$

Or, in terms of $(B_1, B_2)$, $$\begin{align*} S_1(t) &= S_1(0) \exp \left( (\mu_1 - 0.5\sigma_1^2)t + \sigma_1 B_1(t) \right) \\ S_2(t) &= S_2(0)\exp \left( (\mu_2 - 0.5\sigma_2^2)t + \sigma_2 (\rho B_1(t) + \sqrt{1-\rho^2} B_2(t)) \right) \end{align*} \\ $$