What is the Fourier Transform of a brownian motion?

Question

I looked into this article http://en.wikipedia.org/wiki/Brownian_noise and it says that:

If we have a brownian motion $W(t) = \int _{0}^{t} dW(s)$, then given that the spectral density of white noise is constant $S_0 = \left|\mathcal{F}\left[\frac{dW(t)}{dt}\right](\omega)\right|^2 = \text{const}$

Note that here $\mathcal{F}$ denotes the Fourier transform and $S_0$ is a constant. An important property of this transform is that the derivative of any distribution transforms as

$\mathcal{F}\left[\frac{dW(t)}{dt}\right](\omega) = i \omega \mathcal{F}[W(t)](\omega) $

from which we can conclude that the power spectrum of Brownian noise is

$S(\omega)= \left|\mathcal{F}[W(t)](\omega)\right|^2= \frac{S_0}{\omega^2}$

I don't understand this demonstration. Do you have a more detailed explanation or a link to a more detailed proof?

Thanks a lot for your help.

Answer 1

As mentioned above, the first equation about which you were confused is a property of the Fourier transform. Here is a very explicit derivation. First define the Fourier transform over a finite interval $(a,b)$ as $$ \mathcal{F}\left\{f(t)\right\} = \int_{(a,b)} f(t) e^{-i \omega t}\ dt. $$ With suitable technical considerations (if you care: that $f(t)$ is in the Sobolev space $W^{1,1}(a,b)$, which means that both $f$ and its derivative $f'$ are absolutely integrable over $(a,b)$) we can use our usual integration by parts formula: $\int u\ dv = uv|_{a}^b - \int v\ du$, where we will set $u = e^{-i \omega t}$ and $dv = f'(t) dt$. Then we have $$ \begin{aligned} \int e^{i \omega t}\frac{d}{dt} f(t)\ dt &= -i \omega e^{-i\omega t}f(t)\Big|_a^b - \int -i\omega e^{-i \omega t}f(t)\ dt \\ &= -i \omega \left(e^{-i \omega b}f(b) - e^{-i\omega a} f(a) \right) + i\omega \int f(t) e^{-i\omega t}\ dt\\ &= -i \omega \left(e^{-i \omega b}f(b) - e^{-i\omega a} f(a) \right) + i \omega \mathcal{F}\left\{ f(t) \right\}. \end{aligned} $$

If your function $f$ is well-behaved enough (if you somehow define a sequence of functions $f_n$ that converge to a limiting function and agree with $f$ on $(a,b)$, and if you can find a function $g$ so that, for any sequence of intervals $I_n$ converging to $\mathbb{R}$ you have $|f_n| \leq g$ for all $n$) then the constant term above cancels and you have the desired result.

All the formality seems a little contrived, and indeed from the point of view of a physicist it is a little bit---we just do this and don't worry about the formality. But if you want to get into the details, they are important. Suppose you were trying to do this with a random function: the Wiener process $\mathcal{W}(t)$, for example. All right, since $\mathcal{W}(t)$ is a.s. continuous everywhere then we can a.s. Riemann-Stieltjes integrate it as above. But its "derivative" is not well-defined in the traditional sense since $\mathcal{W}(t)$ is a.s. differentiable nowhere! Oops.

Everything does end up working out, and so even though the derivation you gave above is, technically-speaking, wrong, since $\frac{d}{dt}\mathcal{W}(t)$ is not defined, it is morally correct and, in physics, we use it all the time.

Answer 2

Sorry, I know this thread is old, but I feel like some statements are not very clear and/or misleading, and also I would like to add a more mathematically sound perspective on the matter.

As was already pointed out, a Brownian path is with probability 1 not differentiable anywhere, at least not in a usual sense. It is correct that $dW/dt$ is defined in a distributional sense, and as such its Fourier transform can be taken. But the result is a priori a distribution. The actual problem is the premise that "the spectral density of white noise is a constant". At that point, you're miles away from any rigorous definition. Moreover, although you can find this claim a lot, it't not correct, or at least it's misleading.

First of all, we should restrict ourselves to a Fourier decomposition on a finite interval, i.e. with discrete frequencies (because this is the only setting in which we can make meaningful statements). Let's say we're on $[0, 1]$. Then it's true that white noise has a flat spectral density in the sense that $$ E \left[ \left\vert \int_0^1 e^{-i \omega t} dW_t \right\vert ^2\right] = 1 , $$ which is independent of $\omega$. It is not true, however, that $\left\vert \int_0^1 e^{-i \omega t} dW_t \right\vert ^2$ is a constant. Rather, it follows a $\chi^2$ distribution.

However, $\int \phi(t) dW_t$ is a stochastic integral, it cannot be understood as $\int \phi(t) \frac{dW_t}{dt}dt$. Now, stochastic calculus tells us that, for any deterministic function $\phi(t)$, $\int_0^1 \phi(t)\,dW_t$ is a random variable with normal distribution with mean zero and variance $\sigma^2 = \int_0^t \vert \phi(t) \vert^2\, dt$.

How does this help? Well, we can write the Fourier transform as $$ \int_0^1 e^{-i \omega t} W_t dt = \int_0^1 e^{-i \omega t} \left( \int_0^s dW_s \right) dt = \int_0^1 \left( \int_s^1 e^{- i \omega t} d t \right) dW_s = \int_0^1 \phi(s)\, dW_s ,$$ with $$\phi(s) = \frac{i}{ \omega} (1 - e^{- i \omega s}).$$ And this function has $$\int_0^1 \vert \phi (s) \vert^2\, ds = 2 / \omega^2 .$$

This means that the $\omega$-th Fourier coefficient, $\omega \in 2 \pi \mathbb Z$, has distribution $\mathcal N (0, 2 / \omega^2 )$.

In fact, you can simulate a Brownian path by sampling independent standard normal variables $Z_n \sim \mathcal N(0, 1)$, for $n = 0, 1, 2, \dots$, and puting $$ W(t) := Z_0 t + \sum_{n = 1}^\infty \frac{\sqrt 2 Z_n}{\pi n} \sin (\pi n t). $$ But if you fix the absolute value of $Z_n$ and just choose the sign randomly, you will end up with something else.

If you really want to know what white noise is: It can be rigorously defined as a random tempered distribution (i.e. an extremely singular object!) following a certain law. Like a finite-dimensional random variable, this law can be defined via a characteristic function (existence is guaranteed by the Bochner-Minlos-Sazanov theorem). In infinite dimensions, this looks like this: The white noise distribution $T$ is distributed such, that for every Schwartz function (every smooth function that decays faster than any inverse polynomial) $\phi (t)$, $$ E [ e^{i (T, \phi )} ] = e^{- \frac 1 2 \int_{-\infty}^{\infty} \vert \phi(t) \vert^2 d t }$$ The field of mathematics that deals with this is called White Noise Analysis.