 Solution

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.