 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/discrete-mathematics-probability-and-continuous-mathematics/9/b/v/solution

Part (a) gives

P (X \in Z) = 0

⌈ X ⌉ = ⌊ X ⌋ + 1

almost surely. Both are integrable, since

0 \leq ⌊ X ⌋ \leq X

and

⌈ X ⌉ \leq X + 1

. Thus $E [⌈ X ⌉] = ζ (t - 1) + 1$ .

Now take expectations in

⌊ X ⌋ \leq X \leq ⌈ X ⌉

ζ (t - 1) \leq \frac{t - 1}{t - 2} \leq ζ (t - 1) + 1.

s = t - 1

, which ranges over every

s > 1

, and rearranging yields

\frac{1}{s - 1} \leq ζ (s) \leq \frac{s}{s - 1} (s > 1) .

These are the zeta bounds from rounded Pareto variables: rounding below and above converts one elementary expected value into bounds on an infinite series.

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