 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability/9/c/i/solution

Y = Z^{2} /2

is nonnegative, so its cumulative distribution function is zero for

y < 0

. If

y \geq 0

, then

F_{Y} (y) = P (Z^{2} /2 \leq y) = P (- 2 y \leq Z \leq 2 y) = Φ (2 y) - Φ (- 2 y) = 2Φ (2 y) - 1.

Φ (- z) = 1 - Φ (z)

. Hence

F_{Y} (y) = {2Φ (2 y) - 1, 0, y \geq 0, y < 0.

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