 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-3/8h/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The moving sums

(K_{n})

are not necessarily Markov. For a counterexample, let the

X_{n}

be independent Bernoulli variables with parameter

1/2

. On the event

K_{n} = 1, K_{n - 1} = 0,

we must have

X_{n - 1} = 0

and

X_{n} = 1

, so

P (K_{n + 1} = 2 ∣ K_{n} = 1, K_{n - 1} = 0) = \frac{1}{2} .

On the other hand,

K_{n} = 1

and

K_{n - 1} = 2

X_{n - 1} = 1

and

X_{n} = 0

, whence

P (K_{n + 1} = 2 ∣ K_{n} = 1, K_{n - 1} = 2) = 0.

Both conditioning events have positive probability. Knowledge of

K_{n}

alone therefore does not determine the next-step law. The overlapping moving sum need not be Markov.

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