 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-34/4/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Here

X_{1} \leq 1

, so for every

τ > 0

0 < e^{τ X_{1}} \leq e^{τ}, M (τ) = E e^{τ X_{1}} \leq e^{τ} < \infty.

M (τ) \geq e^{τ E X_{1}} = 1.

In fact the inequality is strict:

P (X_{1} = 1) > 0

and

E X_{1} = 0

exclude a constant random variable. In particular,

1 \leq M (τ) < \infty

as required.

For

Z_{n} = e^{τ S_{n}} / M (τ)^{n}

, independence and identical distribution give

E e^{τ S_{n}} = \prod_{i = 1}^{n} E e^{τ X_{i}} = M (τ)^{n} .

Thus

Z_{n}

is positive, integrable with mean one, and adapted to the natural filtration. Conditioning on the next increment gives

E [Z_{n + 1} ∣ F_{n}] = \frac{e ^{τ S_{n}}}{M ( τ ) ^{n + 1}} E [e^{τ X_{n + 1}} ∣ F_{n}] = \frac{e ^{τ S_{n}}}{M ( τ ) ^{n}} = Z_{n} .

Consequently

(Z_{n}) is a martingale .

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