I have a question regarding the Ornstein -Uhlenbeck process. We have a simplified version with Stochastic Integral Equation: $X_t=-a\int^t_0 X_s\,ds +B_t$. B is the Brownian motion.
And its analytic solution is $X_t=e^{-at}\int^t_0 e^{as}\,dB_s$.
How do I prove that this is the case? I know that $e^{-at}=-a\int^t_0 e^{-as}\,ds$ and that I'm somehow supposed to make use of this result but I am still unable to get from the analytic solution to the integral equation.
Thanks!
Best Answer
The Ito formula, for $\mathrm{d} X_t = - a X_t \mathrm{d} t + \mathrm{d} B_t$, you need is: $$ \mathrm{d}\left( f(t, X_t) \right) = \left( \partial_t f(t,X_t) - a X_t \partial_x f(t,X_t) + \frac{1}{2} \partial_{xx} f(t,X_t) \right) \mathrm{d} t + \left( \partial_x f(t,X_t) \right) \mathrm{d} B_t $$
HINT: use $f(t,X_t) = \mathrm{e}^{a t} X_t$.
Edit 9/8/2018: Added $X_t$ to formula