 Solution

The protein fluctuation equation gives $\widetilde{\delta P}=(k_P\widetilde{\delta R}+{\tilde{\eta }}_P)/({\gamma }_P-i\omega )$. With the stated independent noises,Partial fractions and the supplied integral give the general varianceThe specific count-noise formula in the question additionally uses the usual steady birth-and-death noise strengths $q_R=2{\gamma }_R{R}_{\ast }$ and $q_P=2{\gamma }_P{P}_{\ast }$. Then $P_{\ast }=k_P{R}_{\ast }/{\gamma }_P$ reduces it to $P_{\ast }[1+k_P/({\gamma }_R+{\gamma }_P)]$. Arbitrary unspecified $q_R,q_P$ do not by themselves imply this particular Fano factor.

The requested rewriting isHere $b$ is the translational burst size, the expected number of proteins produced during one RNA's mean lifetime. The ratio $\varphi$ compares removal rates, equivalently RNA to protein mean lifetimes. Poisson molecule-number statistics have variance equal to mean, hence $F=1$.

When RNA is much less stable than protein, $\varphi \ll 1$ and $F\simeq 1+b$. At fixed degradation rates it rises linearly with the rate of biological translation, with exact slope $1/({\gamma }_R+{\gamma }_P)$. The formal intercept at zero rate is the limit from positive protein mean; at exactly zero production both mean and variance vanish and their ratio is undefined. A larger Fano factor need not imply a larger relative variance $\mathrm{var}P/P_{\ast }^2$, because the mean also increases.