# A problem with the derivation of Poisson noise statistics from Langevin equations

A paper was passed around by our professor. The paper has a derivation of noise in gene expression from Langevin equations. The actual context is not so important, it's just that I think there might be a simple error in the derivation. Here is the part of the derivation:

$$\frac{d \delta r}{dt} + \gamma_R \delta r = \eta_R \, .$$

Fourier transforming these equations by setting $x(t)=\int x(\omega) \exp(i \omega t) d\omega/2\pi$,

$$\frac{\delta r(\omega)}{\eta_R(\omega)} = \frac{1}{\gamma_R + i \omega} \, \quad \left \langle |\eta_R |^2 \right \rangle = q_R \, ,$$

so that the steady state value of the fluctuation is

$$\langle \delta r \rangle = \int \frac{d \omega}{2\pi} \frac{1}{\gamma_R^2 + \omega^2} q_R = \frac{q_R}{2 \gamma_R}$$

$\eta$ is the random noise term. I think there is a square root missing in the derivation, or I just do not see how it went from $\eta$ to $\langle | \eta^2 | \rangle$

There are a couple of mistakes.

Though

$$\delta r(\omega) = \frac{\eta(\omega)}{\gamma + i \omega}$$

is correct. However, if we assume the noise is $\delta$-correlated in the time domain,

$$\langle \eta(t)\eta(t') \rangle = q\delta(t-t')$$

this leads to similarly $\delta$-correlated noise in the frequency domain

$$\langle \eta(\omega)\eta(\omega')\rangle = 2 \pi q \delta(\omega+\omega')$$

$$\begin{split} \langle (\delta r(t))^2\rangle &= \frac{1}{4 \pi^2}\int_{-\infty}^\infty d \omega \int_{-\infty}^\infty d \omega' \,\mathrm{e}^{i (\omega+\omega')t} \langle \delta r(\omega) \delta r(\omega') \rangle \\& = \frac{1}{4 \pi^2}\int_{-\infty}^\infty d \omega \int_{-\infty}^\infty d \omega' \, \mathrm{e}^{i (\omega+\omega')t}\,\frac{\langle \eta(\omega) \eta(\omega') \rangle}{(\gamma+i \omega)(\gamma+i \omega')} \\ &= \frac{1}{2 \pi}\int_{-\infty}^\infty d \omega \frac{q}{\gamma^2+\omega^2} \\ & = \frac{q}{2 \gamma} \end{split}$$
• I might be lacking in knowledge about correlation function, but why does delta function correlation in time domain give $2\pi q\delta(\omega+\omega')$ correlation in frequency domain? – Ilya Lapan Dec 20 '16 at 11:01
• The Fourier transform is pretty much defined by the fact that the Fourier transform of a single $\delta$ function is a plane wave and vice versa. Given this, one way to see the result you are asking about is as follows: You are Fourier transforming in both coordinates, $t$ and $t'$. One Fourier transform turns a $\delta$-function into a plane wave, and the second transform turns that plane wave back into a $\delta$-function. – ComptonScattering Dec 20 '16 at 12:03
There clearly is a square missing. You can see that dimensionally. Moreover you expect $\langle \delta r \rangle$ to be vanishing and only its squared value to be finite.