 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-1/27k/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-05

Let

r_{i}

be the probability of returning to

i

after its first departure. Transience means

r_{i} < 1

; in particular

q_{i} > 0

N

of visits, including the initial visit, has the geometric distribution

P (N = n) = (1 - r_{i}) r_{i}^{n - 1}

n \geq 1

. The holding times of these visits are independent

Exp (q_{i})

variables and are independent of the jump-chain return decisions.

T_{i}

and

s \geq 0

, summing over

N

gives

E_{i} e^{- s T_{i}} = \sum_{n \geq 1} (1 - r_{i}) r_{i}^{n - 1} (\frac{q _{i}}{q _{i} + s})^{n} = \frac{q _{i} ( 1 - r _{i} )}{s + q _{i} ( 1 - r _{i} )} .

T_{i} \sim Exp (q_{i} (1 - r_{i})) .

Starting elsewhere adds an atom at zero if the state might never be hit; the purely exponential conclusion uses the specified initial state

i

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