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

Table of contents
- i d
  - Solution i
- ii d
  - Solution ii

i

Solution

Every martingale is a supermartingale, with equality in its defining conditional expectation. Taking expectations and using the tower property of conditional expectation gives

E [X_{n + 1}] = E [X_{n}],

so induction yields

E [X_{n}] = E [X_{0}] for every n .

ii

Solution

Conversely, for a supermartingale define

Z_{n} = X_{n} - E [X_{n + 1} ∣ F_{n}] \geq 0.

The assumed equality of expectations gives

E [Z_{n}] = E [X_{n}] - E [X_{n + 1}] = 0.

A nonnegative random variable of expectation zero vanishes almost surely, so

E [X_{n + 1} ∣ F_{n}] = X_{n}

for every

n

. Hence

X

is a martingale. Together with part (i), this proves the constant-expectation supermartingale is a martingale criterion.

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