Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 2 / 18H (Markov Chains)

Table of contents
- a 18H
  - Solution a
- b 18H
  - Solution b
- c 18H
  - Solution c
- d 18H
  - Solution d
- e 18H
  - Solution e

a

Solution

If

S_{k} < n \leq S_{k + 1}

, the construction gives

X_{n} = S_{k + 1} - n

. This is exactly the stated maximum over the next renewal endpoint.

b

Solution

From state

i > 0

the chain moves deterministically to

i - 1

. From zero it jumps to

j

when the next lifetime is

j + 1

. Thus

p_{i, i - 1} = 1 (i > 0), p_{0, j} = q_{j + 1} (0 \leq j < N),

with all other probabilities zero.

c

Solution

The balance recurrence is

π_{i} = π_{i + 1} + π_{0} q_{i + 1} .

Working down from

N - 1

gives

π_{i} = π_{0} P (T_{1} > i)

. Since the sum of these tails is

E [T_{1}]

, the unique invariant law is

π_{i} = \frac{P ( T _{1} > i )}{E [ T _{1} ]} = \frac{\sum _{j = i + 1}^{N} q _{j}}{\sum _{j = 1}^{N} j q _{j}} .

d

Solution

Take

N \geq 2

and

q_{2} = 1

. The chain alternates deterministically between zero and one, so

P (X_{n} = 0)

alternates between one and zero and does not converge.

e

Solution

Adding

ε I

leaves every invariant distribution unchanged. On the unique closed communicating class it also supplies a self-loop, making the chain aperiodic; transient states do not affect the limit. Therefore

P (X_{n}^{(ε)} = 0) ⟶ π_{0} = \frac{1}{E [ T _{1} ]} = \frac{1}{\sum _{j = 1}^{N} j q _{j}},

independently of

ε \in (0, 1)

.

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