# Correlation in geometric brownian motion

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

• No, if $Y_t$ depended only on $t$, the answer would be yes, but it is not the case here. And even if there is no correlation, the answer is still no.
– NN2
Apr 27, 2023 at 11:04
• @NN2 . The answer is yes. See my answer. May 9, 2023 at 8:33

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.