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

Table of contents
- Solution ii

Solution

Reaction 4 creates

X_{3}

, reaction 5 removes one

X_{3}

and one

X_{4}

, and reaction 6 creates

X_{4}

; reaction 7 leaves

X_{4}

unchanged. The exact first-moment equations are therefore

\frac{d}{d t} ⟨ x_{3} ⟩ = α_{4} - α_{5} ⟨ x_{3} x_{4} ⟩, \frac{d}{d t} ⟨ x_{4} ⟩ = α_{6} ⟨ x_{2} ⟩ - α_{5} ⟨ x_{3} x_{4} ⟩ .

At the assumed unique stationary distribution, both left-hand sides vanish. Eliminating the common mixed moment gives

α_{4} = α_{6} ⟨ x_{2}^{*} ⟩, ⟨ x_{2}^{*} ⟩ = \frac{α _{4}}{α _{6}} .

No moment closure or independence assumption is involved.

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