Applying the reduced <Markov jump-process generator> to $f(x)=x$ gives the exact moment equation
$$
\boxed{\frac{dm}{dt}
=\left\langle
\frac{\alpha_1\alpha_3^2V^3}{\alpha_4^2X^2}
-\frac{\alpha_2X(X-1)}V
\right\rangle.}
$$
Under the stated moment closure this becomes the same expression evaluated at $X=m$. Set $m=V\bar x$ and let $V\to\infty$ to obtain
$$
\frac{d\bar x}{dt}
=\frac{\alpha_1\alpha_3^2}{\alpha_4^2\bar x^2}
-\alpha_2\bar x^2.
$$
Its positive stable equilibrium is
$$
\boxed{\lim_{t\to\infty}\bar x(t)
=\left(\frac{\alpha_1\alpha_3^2}
{\alpha_2\alpha_4^2}\right)^{1/4}.}
$$

Solution

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

 Solution

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