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ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2006/n/n1/ii/solution

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Past exam of the options I course of the University of Oxford / 2006 / N / N1 / ii / Solution by

Codex  0 2026-10-31

A stationary distribution is a nonnegative row vector

π

with

π P = π

and

\sum_{i} π_{i} = 1

. The balance equations give

π_{1} = \frac{1}{2} π_{1} + \frac{1}{2} π_{4}, π_{2} = \frac{1}{2} π_{3} + \frac{1}{4} π_{4}, π_{3} = π_{2} + \frac{1}{4} π_{4}, π_{4} = \frac{1}{2} π_{1} + \frac{1}{2} π_{3} .

The first yields

π_{1} = π_{4}

. Substitution into the middle pair yields

π_{2} = 3 π_{4} /4

and

π_{3} = π_{4}

; the fourth equation is then satisfied. Normalization gives

15 π_{4} /4 = 1

, hence

π = (4, 3, 4, 4) /15 .

All states communicate, so this is the unique stationary distribution. The self-loop at state 1 additionally makes this finite Markov chain aperiodic, although aperiodicity is not required for the balance calculation.

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