 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/discrete-mathematics-continuous-mathematics-and-probability/9/a/iv/solution

The logarithm is strictly increasing, so

P (Y \leq y) = P (X \leq e^{y})

. Since

X > 1

F_{Y} (y) = {0, 1 - e^{- cy}, y \leq 0, y > 0, f_{Y} (y) = {c e^{- cy}, 0, y > 0, y \leq 0.

c

. Its probability density function value at the single endpoint can be chosen arbitrarily without changing the law.

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