 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-2/6e/d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

f_{n} = n^{2}

, the birth increment is

2 n + 1

, while

f_{n} - f_{n - 2} = n^{2} - (n - 2)^{2} = 4 n - 4.

With

D = 2 d

, the generator identity gives

\frac{d}{d t} ⟨ n^{2} ⟩ = b (2 μ + 1) - 4 d (⟨ n^{2} ⟩ - μ) - 2 d p_{1} .

At stationarity

μ = b / (2 d)

, and solving this equation for the variance yields

σ^{2} = ⟨ n^{2} ⟩ - μ^{2} = \frac{3 b}{4 d} - \frac{p _{1}}{2} .

For

b, d > 0

the chain is irreducible on the nonnegative integers, so its stationary distribution has

0 < p_{1} < 1

. Therefore

\frac{3 b}{4 d} - \frac{1}{2} < σ^{2} < \frac{3 b}{4 d} .

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