= 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 is
$$
\boxed{\frac d{dt}\langle x\rangle=\lambda-2\beta\langle x^2\rangle
=\lambda-2\beta\big(\langle x\rangle^2+\operatorname{Var}x\big).}
$$
Neglecting fluctuations gives the <mean-field approximation> $\dot m=\lambda-2\beta m^2$, so the positive steady mean is
$$
\boxed{m=\langle x\rangle\simeq\sqrt{\lambda/(2\beta)}.}
$$
There 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 $2\beta E[x^2\mathbf1_{\{x\geq2\}}]$; 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.
Back to article page
