 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iii/4/c/ii/solution

Each

X_{i}

- X_{i}

, and independence makes

S_{m}

and

- S_{m}

identically distributed. Thus their two strict tail probabilities are equal:

P (S_{m} > m /2) = P (S_{m} < - m /2) = \frac{1}{2} P (∣ S_{m} ∣ > m /2) .

P (S_{m} > m /2) \leq \frac{1}{2} \frac{2 m /3}{( m /2 ) ^{2}} = \frac{4}{3 m} .

The equal-tail argument is essential: applying Chebyshev's inequality directly to the absolute tail alone gives the weaker bound

8/ (3 m)

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