 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2008-printed-variant/m/m5/a/solution

Condition on

X_{m} = k

p_{ij} (m + n) = \sum_{k \in S} P_{i} (X_{m} = k) P (X_{m + n} = j ∣ X_{m} = k) = \sum_{k \in S} p_{ik} (m) p_{kj} (n) .

Thus the transition matrices satisfy

P (m + n) = P (m) P (n)

; since

P (1) = P

and

P (0) = I

, induction gives

P (n) = P^{n}

. Averaging over the initial state gives

P (X_{n} = i) = (μ P^{n})_{i} .

μ P = μ

, induction gives

μ P^{n} = μ

, so this probability equals

μ_{i}

for every

n

. This is stationarity of the initial distribution.

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