 Solution

ID: past-exam-of-the-probability-course-of-the-university-of-oxford/2023/3/a/ii/solution

The assumed identities are the detailed balance equations. Summing over

i

and using the row sums of the transition matrix gives, for every

j

(π P)_{j} = \sum_{i} π_{i} p_{i, j} = \sum_{i} π_{j} p_{j, i} = π_{j} \sum_{i} p_{j, i} = π_{j} .

All summands are nonnegative, so these countable sums are justified. Thus

π P = π,

and

π

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