 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-2/10/a/solution

The support is

x \geq 1

, and its normalized Pareto distribution density integrates to

P (X \geq 1) = \int_{1}^{\infty} α x^{- α - 1} d x = [- x^{- α}]_{1}^{\infty} = 1 .

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