 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/i/solution

Since

X

is nonnegative,

F_{X} (x) = 0

for

x < 0

. For

x \geq 0

, use

\frac{d}{d t} e^{- t^{2} /2} = - t e^{- t^{2} /2}

to obtain

F_{X} (x) = \int_{0}^{x} t e^{- t^{2} /2} d t = 1 - e^{- x^{2} /2} .

F_{X} (x) = {0, 1 - e^{- x^{2} /2}, x < 0, x \geq 0.

In particular it tends to one as

x \to \infty

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