# Correlated Brownian Motion expectation

Say I have the following two stochastic differential equations: $$\frac{dS_1}{S_1} = 0.03dt + 0.1 dW_1 + 0.4dW_2$$ $$\frac{dS_2}{S_2} = 0.03dt + 0.5 dW_1 + 0.1dW_2$$ where the instantaneous correlation between the two shocks are $cov(dW_1, dW_2)=0.5dt$. Let $S_1(0) = 100$ and $S_2(0) = 80$. Calculate $\mathbb{E}[1(S_1(1)<50)]$.

My attempt: The expectation is just simply the probability $\mathbb{P}(S_1(1)<50)$. However, I am confused with the extra $dW_2$ term in the stochastic differential equation for $S_1$. If the $dW_2$ was not there, for example, then we have a Geometric Brownian Motion (GBM) and we can use the properties of a GBM to calculate the required probability, but what do I do when there are two (correlated) Wiener process terms?

Generally, the respective solution to $${\rm d}S_t=S_t\left(\mu{\rm d}t+\sigma_1{\rm d}W_t^1+\sigma_2{\rm d}W_t^2\right)$$ with $${\rm d}W_t^1{\rm d}W_t^2=\rho{\rm d}t$$ would be $$S_t=S_0\exp\left[\left(\mu-\frac{1}{2}\sigma_1^2-\frac{1}{2}\sigma_2^2-\rho\sigma_1\sigma_2\right)t+\sigma_1W_t^1+\sigma_2W_t^2\right].$$

According to the SDE, the quadratic variation of $S_t$ is, intuitively, \begin{align} {\rm d}\left<S\right>_t&={\rm d}S_t{\rm d}S_t\\ &=S_t^2\left(\mu{\rm d}t+\sigma_1{\rm d}W_t^1+\sigma_2{\rm d}W_t^2\right)^2\\ &=S_t^2\left(\mu^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\mu\sigma_1{\rm d}t{\rm d}W_t^1+2\mu\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)\\ &=S_t^2\left(\sigma_1^2{\rm d}t+\sigma_2^2{\rm d}t+2\sigma_1\sigma_2\rho{\rm d}t\right). \end{align}

Thus thanks to Ito's lemma, \begin{align} {\rm d}\log S_t&=\frac{{\rm d}S_t}{S_t}-\frac{1}{2}\frac{{\rm d}\left<S\right>_t}{S_t^2}\\ &=\left(\mu{\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+2\rho\sigma_1\sigma_2\right){\rm d}t\\ &=\left(\mu-\frac{1}{2}\sigma_1^2-\frac{1}{2}\sigma_2^2-\rho\sigma_1\sigma_2\right){\rm d}t+\sigma_1{\rm d}W_t^1+\sigma_2{\rm d}W_t^2\\ &={\rm d}\left[\left(\mu-\frac{1}{2}\sigma_1^2-\frac{1}{2}\sigma_2^2-\rho\sigma_1\sigma_2\right)t+\sigma_1W_t^1+\sigma_2W_t^2\right]. \end{align} This eventually yields $$S_t=S_0\exp\left[\left(\mu-\frac{1}{2}\sigma_1^2-\frac{1}{2}\sigma_2^2-\rho\sigma_1\sigma_2\right)t+\sigma_1W_t^1+\sigma_2W_t^2\right].$$

With this solution, as well as the given correlation between $W_t^1$ and $W_t^2$, it is no longer hard to figure out the distribution of $S_t$, and the probability would follow straightforwardly.

• Thank you! This is exactly what I was looking for. Could you take a look at another question of mine? math.stackexchange.com/questions/2758487/… – elbarto Apr 29 '18 at 8:06
• @elbarto: My pleasure. I will have a look at this. – hypernova Apr 29 '18 at 8:30