# Brownian noise variance

I have a question on a Brownian noise mean square which I get from the exercise (10-4) reference [p493, Athanasios Papoulis and Unni Krishna Pillai, “Probability, Random Variables and Stochastic Processes”, 4th edition, TMH, 2002.]. In brief, the noise is described from the equation: $$mx'' + f x' = F(t)$$ the force $$F(t)$$ on the right is a stationary process and has white PSD $$S_{FF} = 2 kTf$$, the boundary conditions are $$x(0) = x'(0) = 0$$. The goal is to find a mean square of a process $$\underline{x}(t)$$ $$E\{x^2(t)\}$$, and from calculating impulse response function it is calculated as $$E\{x^2(t)\} = 2D^2(t + \text{other terms})$$ where $$D^2 = \frac{kT}{f}$$ and $$\alpha = \frac{f}{2m}$$

It is clear for me how it is calculated, and in physics it is also clear why it happens, for exemple a small particle will follow this process and is variance willl be so. However, it is not clear

1. how the variance can reach infinity where $$t \rightarrow \infty$$, how is it not contradictary with energy reaching infinity?
2. If $$\underline{F}$$ is a Gaussian process with a certain variance, is it normal that the output process has infiinite variance. From what I know if a system is linear a Gaussian noise should transform into a Gaussian noise, and it seems not to be case here

In the problem statement for the stochastic differential equation $$mx''+fx'=\mathbf F(t)$$ it is specified that the power spectral density of the process $$\mathbf F$$ be $$S_F(\omega)=2kTf$$ where $$f$$ is the friction coefficient and is a constant. The assumption that the spectral density is independent of the frequency $$\omega$$ implies that its variance, the total power it represents, is already infinite. So if the drive has infinite power then it should not be surprising that the response tot he drive have should infinite variance; one is unphysical as the other. If instead you want to make the problem a bit more realistic then assume that $$S_F(\omega)=2kTf$$ for $$|\omega| and $$S_F(\omega)=0$$ for $$\omega|>B,$$ and then solve the same equation.
As regards your second question, all real processes have finite bandwidth, and if for convenience we assume a non-physical flat spectrum, it will immediately becomes "physical" by having it pass through any real system. If a linear system has impulse response $$h(t)$$ whose Fourier transform is $$H(\omega),$$ then when a flat spectrum $$S(\omega)=s_0$$ non-physical process is its input, the output process will have spectral density $$|s_0|^2|H(\omega)|^2.$$
• Thank you for the answer. There is a problem however if the input process is band limited noise, for example a white noise filtered with a low-pass filter, so the input signal has a DC component. Also, if not considering a boundary condition, when calculating directly the output mean square (which is integrated output power spectral density) will give an integral that will not converge (integration of $\frac{1}{\omega}$). I guess it should be a limit of the model, might be a cutoff frequency? Commented Apr 9 at 7:26
• On "paper" maybe, but in practice the dc component is not a problem for a lowpass filter because there is always some resistance in a filter and thus $H(0)$ be finite at $u=0$. Even a capacitor has some internal resistance and is always driven by a non-zero internal resistance source. Regarding the convergence, a passive RLC circuit must have a square integrable $H(u)$/$h(t)$ transfer function from which it follows that if its input is square integrable so will be its output. An ideal white noise process is not square integrable in power but it is so when confined to a finite bandwidth. Commented Apr 9 at 10:32