 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/7/b/iv/solution

R_{k + 1}

r_{k + 1} = E [E [R_{k + 1} ∣ R_{k}]] = E [1 + (1 - \frac{2}{n}) R_{k}] = 1 + (1 - \frac{2}{n}) r_{k} .

This also holds for the first step from

r_{0} = n

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