Averaging the slow propensities over the conditional Poisson distribution uses $\mathbb E[Y(Y-1)\mid X=x]=q(x)^2$. Thus the effective birth and death rates of $X$ are
$$
\boxed{\lambda_1(x)=\frac{\alpha_1}{V}q(x)^2
=\frac{\alpha_1\alpha_3^2V^3}{\alpha_4^2x^2},}
$$
$$
\boxed{\lambda_2(x)=\frac{\alpha_2}{V}x(x-1).}
$$
Consequently
$$
\partial_tp_0(x,t)
=([E_x^{-1}-1]\lambda_1(x)
+[E_x^{+1}-1]\lambda_2(x))p_0(x,t).
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 356 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-356/3/c/solution