 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/9/a/i/solution

0 < x < 2

2 x - x^{2} = x (2 - x) > 0

. A valid probability density function must integrate to one, so

1 = α \int_{0}^{2} (2 x - x^{2}) d x = α [x^{2} - \frac{x ^{3}}{3}]_{0}^{2} = \frac{4}{3} α .

Thus

α = 3/4

, which is positive and ensures nonnegativity. Its expected value is

E [X] = \frac{3}{4} \int_{0}^{2} (2 x^{2} - x^{3}) d x = \frac{3}{4} [\frac{2 x ^{3}}{3} - \frac{x ^{4}}{4}]_{0}^{2} = 1.

Therefore

Valid with α = \frac{3}{4}, E [X] = 1.

1

, providing a second check on the expected value.

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