 Solution

ID: past-exam-of-the-martingales-and-financial-mathematics-course-of-the-university-of-oxford/2008/a/2/c/solution

E A_{N}^{2} = \frac{1}{N ^{2}} \sum_{n, m \leq N} E (X_{n} X_{m}) = \frac{1}{N} .

Thus

A_{N} \to 0

L^{2}

P (∣ A_{N} ∣ \geq ϵ) \leq \frac{1}{N ϵ ^{2}} ⟶ 0.

Independence is unnecessary; the stated second-moment orthogonality is sufficient.

N_{j} = 2^{j}

m

\sum_{j} P (∣ A_{2^{j}} ∣ \geq 1/ m) \leq m^{2} \sum_{j} 2^{- j} < \infty.

The First Borel-Cantelli lemma makes only finitely many such exceedances occur. Intersecting the resulting probability-one events over

m

gives $A_{2^{j}} \to 0$ almost surely.

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