Problem 16.1: Dynamics of noise

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In the chapter, we worked out that the dynamics of the mean copy number of unregulated gene expression follow the same ODE we introduced at the very beginning,

\begin{align} \frac{\mathrm{d}\langle n \rangle}{\mathrm{d}t} = \beta - \gamma \langle n \rangle. \end{align}

We also worked out that the steady state distribution is Poisson;

\begin{align} P(n) = \frac{1}{n!}\left(\frac{\beta}{\gamma}\right)^n\mathrm{e}^{-\beta / \gamma}. \end{align}

Here, we will work out the dynamics of the variance in \(n\) and demonstrate that the steady state variance and noise are indeed what we get from a Poisson distribution.

a) Derive a differential equation for the variance \(\sigma^2 = \left\langle n^2 \right\rangle - \langle n \rangle ^2\). Your ODE should only have \(\langle n \rangle\) and \(\sigma^2\) as variables. Hint: First write down an expression for \(\mathrm{d}\left\langle n^2 \right\rangle / \mathrm{d}t\), and then use the fact that

\begin{align} \frac{\mathrm{d}\sigma^2}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\left\langle n^2 \right\rangle - \langle n \rangle^2\right) = \frac{\mathrm{d}\left\langle n^2 \right\rangle}{\mathrm{d}t} - \frac{\mathrm{d}\left\langle n \right\rangle^2}{\mathrm{d}t} = \frac{\mathrm{d}\left\langle n^2 \right\rangle}{\mathrm{d}t} - 2\langle n \rangle\,\frac{\mathrm{d}\left\langle n \right\rangle}{\mathrm{d}t}. \end{align}

b) Assuming we start at \(t = 0\) with no gene product (\(n = 0\)), solve for \(\langle n \rangle(t)\) and \(\sigma^2(t)\) to show that the mean and variance are equal for all time, and therefore the Fano factor is always unity. Hint: We have already solved for \(\langle n \rangle(t)\) in the notes. The resulting ODE for \(\sigma^2(t)\) may be solved by integrating factor.

c) Take the limit of \(t \to \infty\) of the above solutions to show that the steady state noise is \(\eta = \sqrt{\gamma/\beta}\), as we have shown in the chapter.