 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability/7/c/ii/solution

Each

X_{i}

- X_{i}

. By independence, the entire vector and its negative have the same joint law, so

S_{m}

has a symmetric distribution about zero. The two disjoint tail events consequently have equal probabilities:

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} .

Using only containment in the two-sided event would give

8/ (3 m)

; the symmetry supplies the necessary extra factor

1/2

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