# Problem 16.1: Dynamics of noise

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.