3.3.2 Generating a stationary Gaussian Markov sequence

Solve the stochastic differential equation

$$\dot{x}(t) = - \beta \, x(t) + s(t)$$

with a stochastic ``driving'' process $s(t)$, assumed to be uncorrelated Gaussian noise, i.e. Gauss distributed about $\langle s \rangle =0$, with $\langle s(0) \, s(t) \rangle = A\, \delta(t)$.

The general solution to this equation reads

$$x(t) = x(0)\,e^{-\beta \, t} + \int_{0}^{t} e^{-\beta (t-t')} \, s(t')\, dt'$$

Inserting $t=t_{n}$ and $t=t_{n+1} \equiv t_{n}+\Delta t$ one finds that

$$x(t_{n+1}) = x(t_{n})\, e^{-\beta \, \Delta t} + \int_{0}^{\Delta t} e^{-\beta (\Delta t-t')} \, s(t_{n}+t')\, dt'$$

At any time $t$, the values of $x(t)$ belong to a stationary Gauss distribution with $\langle x^{2} \rangle$$=A/2\beta$, and the process $\{ x(t_{n}) \}$ has the Markov property.

The integrals

$$z(t_{n}) \equiv \int_{0}^{\Delta t} e^{-\beta (\Delta t-t')} \, s(t_{n}+t')\, dt'$$

are elements of a random sequence, with $z$ Gauss distributed with zero mean and $\langle z(t_{n})\,z(t_{n+k}) \rangle = 0$ for $k \neq 0$. Their variance is

$$\langle z^{2} \rangle = \frac{A}{2\beta}\,(1-e^{-2\beta\,\Delta t})$$

Here is the resulting recipe for generating a stationary, Gaussian Markov sequence: $\Longrightarrow$





EXERCISE: Employ the above procedure to generate a Markov sequence $\{ x_{n}\}$ with a given $\beta$. Check if the sequence shows the expected autocorrelation.