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?
 A: 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.
