 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-resit/9/b/iii/solution

Since

X \geq 0

X \mapsto X^{2}

y \geq 0

F_{Y} (y) = P (X^{2} \leq y) = F_{X} (y) = 1 - e^{- y /2},

and

F_{Y} (y) = 0

for

y < 0

f_{Y} (y) = {\frac{1}{2} e^{- y /2}, 0, y \geq 0, y < 0.

The density value at the single point

y = 0

is immaterial. Hence

Y

1/2

. Direct integration by parts, or its standard expected value, yields

E [X^{2}] = E [Y] = \int_{0}^{\infty} \frac{y}{2} e^{- y /2} d y = 2.

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