# Deriving Stochastic Differential Equations from Autocorrelation

Suppose I know my stationary stochastic process has the following autocorrelation function:

$$R(\tau) = \sigma^{2} e^{-\alpha |\tau|} \cos(\omega \tau)$$

How can I derive the stochastic differential equation (SDE) describes a process with this autocorrelation?

My attempt:

I will use the Wiener-Holmogorov whitening procedure to decompose the Laplace transform of $$R(\tau)$$ into the product of white noise and a system function:

$$\mathcal{L}_{R}(s) = W(s) H(-s) H(s)$$

Then letting $$h(t) = \mathcal{L}^{-1}(H(s))$$ and $$w(t) = \mathcal{L}^{-1}(W(s))$$ the SDE will be given by:

$$\frac{d h}{dt} = f(h,t) h(t) + w(t)$$

where $$f(h,t)$$ is a function that statistifes $$\frac{d h}{dt} = f(h,t) h(t)$$.

Solution:

The Laplace transform of $$R(\tau)$$ is given by:

$$\mathcal{L}_{R}(s) = \int_{-\infty}^{\infty} \sigma^{2} e^{-\alpha |\tau|} \cos(\omega \tau) e^{-s \tau} d\tau$$ $$\mathcal{L}_{R}(s) = \int_{-\infty}^{0} \sigma^{2} e^{\alpha \tau} \cos(\omega \tau) e^{-s \tau} d\tau + \int_{0}^{\infty} \sigma^{2} e^{-\alpha \tau} \cos(\omega \tau) e^{-s \tau} d\tau$$ $$\mathcal{L}_{R}(s) = \int_{-\infty}^{0} \sigma^2 \cos(\omega \tau) e^{-(s-\alpha) \tau} d\tau + \int_{0}^{\infty} \sigma^2 \cos(\omega \tau) e^{-(s + \alpha) \tau} d\tau$$ $$\mathcal{L}_{R}(s) = \int_{0}^{\infty} \sigma^{2} \cos(\omega \tau) e^{-(\alpha-s) \tau} d\tau + \int_{0}^{\infty} \sigma^{2} \cos(\omega \tau) e^{-(\alpha + s) \tau} d\tau$$ $$\mathcal{L}_{R}(s) = \sigma^{2} \left( \frac{\alpha - s}{ \omega^{2} + (\alpha -s)^{2}} + \frac{s + \alpha}{ \omega^{2} + (s + \alpha)^{2}} \right)$$ $$\mathcal{L}_{R}(s) = 2 \alpha \sigma^{2} \left( \frac{-s^{2} + \omega^{2} + \alpha^{2}}{(s^{2} - 2\alpha s + \alpha^{2} + \omega^{2}) (s^{2} + 2\alpha s + \alpha^{2} + \omega^{2})} \right)$$ $$\mathcal{L}_{R}(s) = 2 \alpha \sigma^{2} \left( \frac{\beta - s}{s^{2} - 2\alpha s + \beta^{2}} \right) \left( \frac{\beta + s}{s^{2} + 2\alpha s + \beta^{2}} \right)$$

where $$\beta = \sqrt{\omega^{2} + \alpha^{2}}$$. So we have:

$$H(s) = \frac{\beta + s}{s^{2} + 2\alpha s + \beta^{2}}$$ $$W(s) = 2 \alpha \sigma^{2}$$

To find $$h(t)$$ and $$w(t)$$ we need to take the Inverse Laplace Transforms. The noise term is trivial:

$$w(t) = \mathcal{L}^{-1} \left\{W(s)\right\} = 2 \alpha \sigma^{2} \delta(t)$$

The other term takes a bit of work and becomes rather complicated. I am not sure exactly if this is the right thing to do, so I am going to stop here.

Questions:

1. Is this the correct approach? If so, how can I derive this $$f(h,t)$$ term given that the result of the inverse Laplace becomes so complicated?
2. If not, what should I be doing instead?

Thanks!

Edit: A simpler example. Consider that instead $$H(s) = \frac{1}{s+\alpha}$$. Then $$h(t) = e^{-\alpha t}$$ and $$\frac{dh}{dt} = -\alpha h(t)$$. So our SDE would be:

$$\frac{dx}{dt} = -\alpha x(t) + 2 \alpha \sigma^{2} w(t)$$

Correct? Or should it be $$\frac{dx}{dt} = -\alpha x(t) + \sqrt{2 \alpha \sigma^{2}} w(t)$$