Power Spectral Density of Brownian Motion despite non-stationary

Question

Note: I originally asked this on Physics Stack Exchange but haven't attracted any interest there so I'm asking here where it may be more relevant.

A white noise process, $\xi(t)$ with delta correlated two-correlation function $\langle \xi(t_1)\xi(t_2)\rangle = \delta(t_1-t_2)$, is clearly stationary and has a power spectral density which is the Fourier transform of the auto correlation function a la Wiener-Khintchine theorem.

$$ S_{\xi \xi}(f) = \int e^{i 2 \pi f t} \langle \xi(t)\xi(0) \rangle dt = 1 $$

We call such a process white noise because it has a flat power spectral density as indicated by the above equation.

I am concerned here with the power spectral density of a Brownian motion process or a Wiener process, $W(t)$ which is the integral (in some sense) of a white noise process.

$$ W(t) = \int_{t'=0}^t \xi(t) dt = \int_{t'=0}^t dW(t') $$

See the Wikipedia article on Brownian noise. In that article it is pointed out that since Brownian motion is the integral of white noise we have (using the Fourier transform convention above)

$$ \mathcal{FT}[W](f) = \frac{1}{-i 2\pi f} \mathcal{FT}[\xi](t) $$

Then the argument follows that the power spectral density is like the absolute value of the Fourier transform so we should have something like

$$ S_{WW}(f) = \frac{S_{\xi\xi}(f)}{|-i 2\pi f|^2} = \frac{1}{(2\pi)^2} \frac{1}{f^2} $$

That is, the Brownian process has a $\frac{1}{f^2}$ spectrum, or Red or Brown noise.

We can also perform a direct computation of the power spectral density from the definition of power spectral density as the limit of the magnitude of the windowed Fourier transform.

$$ S_{WW}(f) = \lim_{\tau \to \infty} \frac{1}{\tau} \int_{t_1=0}^{\tau} \int_{t_2=0}^{\tau} e^{i 2\pi f (t_1-t_2)} \langle W(t_1)W(t_2) \rangle dt_1 dt_2 $$

It can be shown for a Wiener process that the two-time correlation function is $\langle W(t_1) W(t_2) \rangle = \text{min}(t_1,t_2)$. We can write the integral as

$$ \frac{1}{\tau}\int_{t_1=0}^{\tau}\int_{t_2=0}^{t_1} e^{i 2\pi (t_1-t_2)} t_2 dt_1dt_2 + \frac{1}{\tau}\int_{t_1=0}^{\tau}\int_{t_2=t_1}^{\tau} e^{i 2\pi (t_1-t_2)} t_1 dt_1dt_2 $$

Mathematica (or a diligent calculus student) evaluates this integral to

$$ \frac{2\pi f\tau - \sin(2\pi f \tau)}{4 \pi^3 f^3 \tau} $$

In the limit of $\tau \to \infty$ the second term vanishes and we are left with

$$ S_{WW}(f) = \frac{1}{2 \pi^2} \frac{1}{f^2} $$

This is missing a factor of 2 compared to the formula above. I think the reason is related to the fact that I've only considered positive time for the windowed Fourier transform in the second derivation.

Here is my question I have given two sketches that show Brownian motion has power spectral density $\sim \frac{1}{f^2}$. However, I have also stated the two-time correlation function for Brownian motion is $\langle W(t_1)W(t_2) \rangle = \text{min}(t_1,t_2)$. This shows that Brownian motion is clearly not a stationary process. I know that the Wiener-Khintchine theorem tells us that if a process is stationary then the power spectral density is given by the Fourier transform of the two-time correlation function. But I also had the impression that non-stationary processes didn't have well-defined power spectral densities, or that you could only define a time-windowed power spectral density. However, Brownian motion seems to be an exception that rule. It seems to be an example of a process which is non-stationary (so you can't use the Wiener-Khintchine theorem, for example) yet still has a well-defined power spectral density. Am I correct in this characterization? Can anyone give me more information about non-stationary processes that still have power spectral densities?

Answer 1

Please see N.J. Kasdin, "Discrete Simulation of Colored Noise and Stochastic Processes and $1/f^\alpha$ Power Law Noise Generation", Proceedings of the IEEE, 83(5), (1995), 802-827. On page 806, Kasdin discusses the factor of 2 issue and states that "It is this author's belief that the factor of two in (24) is an error due to a broad-band bias introduced by the rectangular window." With a Hanning window (with 8/3 normalization), Kasdin gets an expression (equation 25) that converges to $1/\omega^2$ for large T. He also says "Since the formal spectral density of Brownian motion doesn't exist, the better question is which provides the most accurate information about the second order properties and possible linear models (see Section II-D) for Brownian motion?" Hope this helps a bit!

Answer 2

"The PSD is defined only for stationary processes" is a common misconception. The Wiener–Khinchin theorem states that for a wide-sense stationary process, the PSD is the Fourier transform of the autocorrelation function. This is sometimes (wrongly) taken as the "definition" of PSD. Actually, the PSD is defined as $$S_y(f)=\lim_{T\to\infty}\frac{1}{T}|X_T(f)|^2$$ where $X_T(f)$ is the spectrum of $x_T(t)$, which is the signal $x(t)$ truncated on a window of time $T$: $$x_T(t)=x(t)\cdot\sqcap(t/T)\,,\quad\text{with}\quad\sqcap(t)=\begin{cases}1&|t|\leq0.5\\0&|t|>0.5\end{cases}$$

Answer 3

We can't compute the PSD down to DC due to its divergence, but we can deduce the PSD everywhere else from knowing the PSD of a white noise process which is stationary (and the PSD is constant for all frequencies), and that the integration of white noise is a random walk process (Brownian), and the fact that the power response of integration goes as $1/f^2$ (which we see from the Laplace Transform of the time domain integration process is $1/s$).