Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 30K / b

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- Solution b

Solution

The process is adapted to its natural filtration. At each fixed time it has finite support and is therefore integrable. Conditional on

X_{n - 1} = 0

,

E [X_{n} ∣ X_{n - 1} = 0] = \frac{1}{2 n} - \frac{1}{2 n} = 0 = X_{n - 1} .

Conditional on

X_{n - 1} \neq = 0

,

E [X_{n} ∣ X_{n - 1}] = \frac{1}{n} (n X_{n - 1}) + (1 - \frac{1}{n}) 0 = X_{n - 1} .

The Markov property now gives

E [X_{n} ∣ F_{n - 1}] = E [X_{n} ∣ X_{n - 1}] = X_{n - 1},

so

(X_{n})

is a martingale.

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