 Solution

ID: past-exam-of-the-martingales-and-financial-mathematics-course-of-the-university-of-oxford/2011/a/2/c/ii/solution

There are exactly

n

summands in

S_{n}

E \frac{S _{n}}{n} - 1 \leq \frac{1}{n} \sum_{i = 2}^{n + 1} E ∣ X_{i} - 1∣ = \frac{1}{n} \sum_{i = 2}^{n + 1} \frac{1}{l o g i} ⟶ 0.

The last step is convergence of arithmetic means of a sequence tending to zero. Thus

S_{n} / n \to 1

L^{1}

S_{n} / n ⟶ 1 in probability .

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