Let $S_t$ and $Y_t$ be solutions to the following 2-dimensional SDE system
$$ \mathrm{d}S_t=S_t\mu(t,S_t,Y_t)\mathrm{d}t+S_t\sigma(Y_t)\left(\sqrt{1-\rho^2(t,S_t,Y_t)}\mathrm{d}W_t^1+\rho(t,S_t,Y_t)\mathrm{d}W_t^2\right)\\ \mathrm{d}Y_t=m(t,S_t,Y_t)\mathrm{d}t+v(t,S_t,Y_t)\mathrm{d}W_t^2$$
where $W^1_t$ and $W_t^2$ are two independent standard Brownian motions and $\mu, \sigma,m,v$ deterministic functions such that a unique solution to the SDE exists. In my notes it says that $\rho(t,S_t,Y_t)$ is the correlation process of $S_t$ and $Y_t$. Does that mean that $\mathrm{corr}(S_t,Y_t)=\rho(t,S_t,Y_t)$? And if it is so, how do I get this result since $S_t$ and $Y_t$ are only given implicitly.
Thanks in advance!
The quantity $\rho(t, S_t, Y_t)$ is the instantaneous, or local, correlation, which can be defined by $$d\langle S, Y\rangle_t/dt,$$ or $$corr(S_{t+\Delta}|{\mathscr{F}_t}, Y_{t+\Delta}|{\mathscr{F}_t}),$$ as $\Delta\rightarrow 0$.
On the other hand, the quantity $corr(S_t, Y_t)$ is the correlation for the $t$-term values $S_t$ and $Y_t$, which is usually called the term correlation.