 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-2/7h/solution

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

p_{n} = P (X_{n} = 1 ∣ X_{0} = 1) .

Conditioning on

X_{n}

p_{n + 1} = (1 - α) p_{n} + β (1 - p_{n}) = β + (1 - α - β) p_{n} .

π_{1} = \frac{β}{α + β} .

Thus

p_{n} - π_{1} = (1 - α - β)^{n} (p_{0} - π_{1})

, and

p_{0} = 1

gives

P (X_{n} = 1 ∣ X_{0} = 1) = \frac{β}{α + β} + \frac{α}{α + β} (1 - α - β)^{n} .

Starting instead from state two gives the same recurrence with initial value zero. Therefore

P (X_{n} = 1 ∣ X_{0} = 2) = \frac{β}{α + β} [1 - (1 - α - β)^{n}] .

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