 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/discrete-mathematics-probability-and-continuous-mathematics/8/a/solution

1/ n

. Using the given finite sums,

E [X] = \frac{1}{n} \sum_{r = 1}^{n} r = \frac{n + 1}{2}, E [X^{2}] = \frac{1}{n} \sum_{r = 1}^{n} r^{2} = \frac{( n + 1 ) ( 2 n + 1 )}{6} .

Subtracting the square of the expected value gives the variance:

var (X) = \frac{( n + 1 ) ( 2 n + 1 )}{6} - \frac{( n + 1 ) ^{2}}{4} = \frac{( n + 1 ) ( n - 1 )}{12} .

Hence

E [X] = \frac{n + 1}{2}, var (X) = \frac{n ^{2} - 1}{12} .

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