 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iii/6/b/i/solution

1 = \int_{0}^{b} a x d x = \frac{a b ^{2}}{2}, 1 = E [X] = \int_{0}^{b} a x^{2} d x = \frac{a b ^{3}}{3} .

Dividing the second by the first gives

2 b /3 = 1

. Consequently

b = \frac{3}{2}, a = \frac{2}{b ^{2}} = \frac{8}{9} .

Both are positive, as required for the stated probability density function.

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