I have the following stochastic process:
$$\ddot x(t)=a(t)+\nu(t)$$
where $a(t)$ is a deterministic function of the time and $\nu(t)$ is a random variable:
$$\nu(t)\in\mathcal{N}(0,\sigma^2)$$Obviuosly if $\nu(t)=0$, the solution would be:
$$x_0(t)=\int_1^td\eta\int_1^\zeta d\zeta a(\zeta)+c_1+c_2t$$I need to know the difference $d(t)=|x_0(t)-x(t)|$ where $x(t)$ is the solution of the differential equation when $\nu(t)\neq0$
Thanks.
$\endgroup$ 21 Answer
$\begingroup$You might want the SDE system\begin{align} dY_t&=a(t)\,dt+σ\,dB_t\\ dX_t&=Y_t\,dt \end{align}which indeed can also be integrated as\begin{align} Y_t&=c_2+\int_0^ta(s)\,ds+σ\,B_t\\ X_t&=c_1+c_2t+\int_0^t(t-s)a(s)\,ds+σ\int_0^tB_s\,ds\\ &=c_1+c_2t+\int_0^t(t-s)a(s)\,ds+σ\,tB_t-σ\int_0^ts\,dB_s \end{align}There is no bound on the Brownian motion, thus also not on the last two terms. Their combination, the Gaussian process $\int_0^t(t-s)\,dB_s$, has mean zero and variance $\frac13t^3$.
$\endgroup$
