Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 2 / 6E (Mathematical Biology)

Apply the formal infinitesimal generator to the molecule count: its increment is $+1$ at rate $\lambda$ and $-2$ at rate $\beta x^2$. Thus the exact first-moment identity for the specified generator isNeglecting fluctuations gives the mean-field approximation $\dot{m}=\lambda -2\beta m^2$, so the positive steady mean isThere is a physical boundary qualification: the printed loss propensity is positive at $x=1$ and would produce a negative molecule count. Consequently it does not by itself define a molecule-count chain on all nonnegative integers. If losses are forbidden below two, the exact loss moment becomes $2\beta E[x^2{1}_{\{x\ge 2\}}]$; a mass-action two-molecule propensity instead uses $x(x-1)$. The calculation requested here uses the printed formal generator and its large-copy-number approximation, where such boundary corrections are negligible.

Let $J_{+}=\lambda$ and $J_{-}=2\beta m^2$ be the production and loss fluxes, counted in molecules per unit time. At the steady state, both equal $J=\lambda$. The restoring elasticity and turnover lifetime areFlux-weighting the magnitudes of the reaction steps, one for production and two for loss, givesThis is an average over molecular fluxes, as requested, not an average over reaction-event counts; those are different when a reaction removes two molecules at once.

For a stable one-species linear noise approximation, the normalized stationary fluctuation-dissipation relation for a reaction network becomesTo see the factors explicitly, the linearized restoring rate is $a=H/\tau =4\beta m$, while the chemical noise intensity is the rate-weighted squared step sizeStationarity of the variance gives $2a\mathrm{Var}x=B$. Using $\lambda =2\beta m^2$ therefore yieldsThis is the stationary fluctuation result within the stated negligible-fluctuation/linear-noise approximation, not an exact identity for the boundary-corrected discrete chemistry.