 Solution

ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2009/m/1/c/solution

Part (b) gives

V_{i} \sim U (0, 1)

, so

E V_{i} = 1/2

and

Var (V_{i}) = 1/12

. Independence yields

E (S_{n} / n) = 1/2

and

Var (S_{n} / n) = 1/ (12 n)

. Apply part (a) to

Z = S_{n} / n - 1/2

P (∣ S_{n} / n - 1/2 ∣ \geq a) \leq \frac{1}{12 n a ^{2}} .

This is a direct Chebyshev inequality for the sample mean.

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