 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-211/3/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

X_{j} = a^{j} x + b \sum_{ℓ = 0}^{j - 1} a^{ℓ} + \sum_{r = 1}^{j} a^{j - r} ξ_{r} .

For

q_{r} = \sum_{j = r}^{t} θ_{j} a^{j - r}

, regrouping the sum of the exponents gives

\sum_{j = 1}^{t} θ_{j} X_{j} = x \sum_{j = 1}^{t} θ_{j} a^{j} + b \sum_{r = 1}^{t} q_{r} + \sum_{r = 1}^{t} q_{r} ξ_{r} .

The independent innovations factorize the moment-generating function. Consequently

A_{t} = j = 1 \sum t θ_{j} a^{j}, B_{t} = r = 1 \sum t (b q_{r} + ψ (q_{r})), q_{r} = j = r \sum t θ_{j} a^{j - r} .

Finite sums avoid a division by

a - 1

, so these formulas include

a = 0

and

a = 1

; zeroth powers in these sums are

1

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