 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-1/28k/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

\sum_{n = 0}^{\infty} \frac{1}{λ _{n}} < \infty,

then

E T_{\infty} = \sum_{n = 0}^{\infty} E X_{n} = \sum_{n = 0}^{\infty} \frac{1}{λ _{n}} < \infty.

A nonnegative random variable with finite expectation is finite almost surely, so the process explodes.

Conversely, suppose the series diverges. For

s > 0

, independence and the Laplace transform of an exponential variable give

E e^{- s T_{\infty}} = \prod_{n = 0}^{\infty} \frac{λ _{n}}{λ _{n} + s} = exp [- \sum_{n = 0}^{\infty} lo g (1 + \frac{s}{λ _{n}})] .

The sum in the exponent diverges. If infinitely many

λ_{n} \leq s

, infinitely many terms are at least

lo g 2

; otherwise, eventually

s / λ_{n} \leq 1

and

lo g (1 + s / λ_{n}) \geq \frac{s}{2 λ _{n}} .

Thus

E e^{- s T_{\infty}} = 0

. Since

e^{- s T_{\infty}}

is strictly positive on

{T_{\infty} < \infty}

T_{\infty} = \infty

N is nonexplosive ⟺ n = 0 \sum \infty λ_{n}^{- 1} = \infty .

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