 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2009/m/m1/c/solution

The preceding marginal calculation makes each

V_{i}

(0, 1)

1/2

1/12

E (S_{n} / n) = \frac{1}{2}, Var (S_{n} / n) = \frac{1}{12 n} .

Apply part (a) to the centred random variable

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

to obtain

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

This is Chebyshev's inequality in this setting.

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