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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.

Riccardo.Alestra
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    As usual, merely requiring $\nu(t)\sim N(0,\sigma^2)$ for all $t\in[0,T]$ does not even guarantee integrability of $t\mapsto\nu(t,\omega)$. – user10354138 Apr 19 '21 at 10:24
  • @user10354138: I agree, but I can accept also numerical evaluation. What method can I apply in the case in which $a(t)$ is a continuous function: $a(t)\in C^{\infty}(\Delta t,\mathbb{R})$ where $\Delta t$ is the integration interval? – Riccardo.Alestra Apr 19 '21 at 10:32

1 Answers1

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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 $\frac13t^3$.

Lutz Lehmann
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