Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 3 / 8H (Markov Chains)

Table of contents
- a 8H
  - Solution a
- b 8H
  - Solution b
- c 8H
  - Solution c

a

Solution

Let

π = (π_{A}, π_{B}, π_{C})

. Solving the stationary distribution equations

π P = π

together with

π_{A} + π_{B} + π_{C} = 1

gives

π = (\frac{1}{2}, \frac{1}{6}, \frac{1}{3}) .

b

Solution

Yes. The three nontrivial detailed balance identities are

π_{A} P_{A B} = \frac{1}{8} = π_{B} P_{B A},

π_{A} P_{A C} = \frac{1}{8} = π_{C} P_{C A},

and

π_{B} P_{BC} = \frac{1}{24} = π_{C} P_{CB} .

Thus the chain is a reversible Markov chain.

c

Solution

By stationarity,

P (X_{n - 2} = B ∣ X_{n} = B) = \frac{P ( X _{n - 2} = B , X _{n} = B )}{P ( X _{n} = B )} = \frac{π _{B} ( P ^{2} ) _{BB}}{π _{B}} = (P^{2})_{BB} .

The two-step return probability is

(P^{2})_{BB} = P_{B A} P_{A B} + P_{BB}^{2} + P_{BC} P_{CB} = \frac{3}{4} \frac{1}{4} + 0 + \frac{1}{4} \frac{1}{8} .

Therefore the requested probability is

\frac{7}{32} .

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