 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/ib/paper-4/9h/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

P (X_{2 n + 2} = j ∣ X_{2 n} = i, X_{2 n - 2}, \dots, X_{0}) = \sum_{k \in S} p_{ik} p_{kj} = (P^{2})_{ij} .

For each intermediate state, one first uses the next-step transition probability and then the following one; the sampled history adds no information once

X_{2 n}

is known. Thus

W

P^{2}

and initial law

λ

. In particular,

P (W_{1} = 0) = i \in S \sum λ_{i} j \in S \sum p_{ij} p_{j 0} .

If zero is not a state of

S

, this probability is zero.

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