1
$\begingroup$

The Brownian motion has a power spectral density (PSD) dependency on frequency like $\frac{1}{f^2}$. As far as I understand, power spectral density is defined only for wide sense stationary processes but the Brownian motion is not stationary.
Could anyone please explain why can we define PSD for the Brownian motion?

$\endgroup$
2
$\begingroup$

If you don't worry about technicality too much, there's a nice proof on Wikipedia. The gist of it is as follows: we know that white noise $\xi(t)$ has a flat power spectrum (just simulate some to see this!) and also that the Wiener process is the integral of white noise, so that we must have $$ S_{xx}^{(\xi)}(\omega) = \left| \mathcal{F}\left\{ \frac{d\mathcal{W}}{dt} \right\} \right|^2 \propto 1. $$ Recalling that $\mathcal{F}\left\{ \frac{d}{dt}f(t) \right\} \propto \omega\hat{f}(\omega)$, we see that we must have $\hat{f}(\omega) \propto \frac{1}{\omega}$. Then $S_{xx}^{(\mathcal{W})}(\omega) = \left| \hat{f}(\omega) \right|^2 \propto \frac{1}{\omega^2},$ completing the proof. Now, if you are concerned about some technicalities of convergence, there is no problem either. Here is what you can do. Redefine the Fourier transform as $$ \mathcal{F}\left\{f(t)\right\} = \frac{1}{\sqrt{T}}\int_0^T f(t) \exp(-i \omega t)\ dt $$ and note that this mean exists for all finite $T$. Then define the power spectral density as $$ S_{xx}(\omega) = \lim_{T \rightarrow \infty}\left \langle \left| \mathcal{F}\{x(t)\}\right|^2 \right\rangle, $$ where the mean is over realizations of the process $x(t)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.