For $\alpha>0$, variation of constants gives the stationary <Ornstein-Uhlenbeck process>
$$
f(t)=c\int_{-\infty}^t
\Lambda_f(s)e^{-\alpha(t-s)}\,ds.
$$
For $t\geq t'$ its covariance is
$$
\begin{aligned}
\langle f(t)f(t')\rangle
&=c^2\int_{-\infty}^{t'}
e^{-\alpha(t-s)}e^{-\alpha(t'-s)}\,ds\\
&=\frac{c^2}{2\alpha}e^{-\alpha(t-t')}.
\end{aligned}
$$
Symmetry in $t,t'$ therefore gives
$$
\boxed{
\langle f(t)f(t')\rangle
=f_0^2e^{-\alpha|t-t'|},
\qquad
f_0^2=\frac{c^2}{2\alpha}
}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 344 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-344/4/a/solution