Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 30K / b

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- Solution b

Solution

The variable

M_{n}

is

F_{n}

-measurable. Positivity, independence, and the mean-one assumption give

E [M_{n}] = \prod_{i = 1}^{n} E [Y_{i}] = 1,

so it is integrable. Since

Y_{n + 1}

is independent of

F_{n}

,

E [M_{n + 1} ∣ F_{n}] = E [M_{n} Y_{n + 1} ∣ F_{n}] = M_{n} E [Y_{n + 1}] = M_{n} .

Thus $(M_{n})$ is the product martingale from independent mean-one factors.

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