Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 356 / 1 / e

The odd $X_1$ copy number decreases by two until it reaches one, so $⟨x_1(t)⟩\to 1$. The exact $X_2$ first-moment equation isBecause copy numbers are nonnegative integers, $x_2^2\ge x_2$, and the last mixed moment is nonnegative. HenceThe integrating factor $e^{{\alpha }_{3}t}$, together with $⟨x_1(t)⟩\to 1$, gives the comparison bound

Subtracting the exact $X_4$ first-moment equation from the $X_3$ equation cancels reaction 5:The assumed inequality is preciselyChoose a positive $\epsilon$ smaller than this gap divided by ${\alpha }_6$. The limit superior bound implies that, for all sufficiently large $t$,Thus $⟨x_3⟩-⟨x_4⟩$ grows at least linearly. Since $⟨x_4⟩\ge 0$,so the required species index is $\boxed{i=3}$.