Correlation in geometric brownian motion

Question

Let $S_t$ the solution of \begin{equation} \frac{dS_t}{S_t}=Y_tdt+\sigma dB_t \end{equation} where $B$ is a brownian motion and $Y$ is a Ornstein-Uhlenbeck process, with $d<Y_t,B_t>=dY_tdB_t=\rho dt$. Is it true that the solution remains \begin{equation} S_t=S_0\exp\left(\int_0^tY_sds-\frac{1}{2}\sigma t+B_t\right) \end{equation} ?

Answer 1

There are typos in the exponential: $$ S_t=S_0\exp\Big(\int_0^t Y_s\,ds-\frac{1}{2}\sigma^{\color{red}2}t+\color{red}{\sigma }B_t\Big)\,. $$ Writing $A_t=\exp(\int_0^t Y_s\,ds)$ we get a process of finite variation that has zero covariation with the Brownian motion. Then $$ S_t=A_tM_t $$ where $M_t$ is an exponential martingale that satisfies $$ dM_t=\sigma M_t\,dB_t\,,\quad M_0=S_0\,. $$ Noting that $dA_t=A_tY_t\,dt$ the integration by parts formula yields $$ dS_t=M_t\,dA_t+A_t\,dM_t=M_tA_tY_t\,dt+\sigma A_tM_t\,dB_t=S_t\,Y_t\,dt+\sigma S_t\,dB_t\,, $$ in other words, $S_t$ is the solution of $$ \frac{dS_t}{S_t}=Y_t\,dt+\sigma B_t\,. $$ When $S$ is a stock not paying dividends and $Y_t$ is the risk-less short rate this and similar models are very popular in financial mathematics.