 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/27k/e/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Let

m

i \geq 1

gives

m_{i - 1} λ_{i - 1} = m_{i} λ_{i} (1 + ρ_{i}),

and hence, after choosing

m_{0} > 0

m_{i} = \frac{m _{0} λ _{0}}{λ _{i} \prod _{j = 1}^{i} ( 1 + ρ _{j} )} .

Balance at zero requires

m_{0} λ_{0} = \sum_{i \geq 1} m_{i} λ_{i} ρ_{i} = m_{0} λ_{0} \sum_{i \geq 1} \frac{ρ _{i}}{\prod _{j = 1}^{i} ( 1 + ρ _{j} )} .

\sum_{i = 1}^{n} \frac{ρ _{i}}{\prod _{j = 1}^{i} ( 1 + ρ _{j} )} = 1 - \frac{1}{\prod _{j = 1}^{n} ( 1 + ρ _{j} )} .

Consequently a nonzero invariant measure exists exactly when

j = 1 \prod \infty (1 + ρ_{j}) = \infty.

It can be normalized to an invariant probability distribution exactly when, in addition,

\sum_{i = 0}^{\infty} \frac{1}{λ _{i} \prod _{j = 1}^{i} ( 1 + ρ _{j} )} < \infty,

where the empty product for

i = 0

is one and an overall constant has been suppressed.

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