 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/iii/solution

The source prints bare differentials

d

in both integrals; the integration variable is

t

, so we write them as

d t

Y

gives

1 = \int_{0}^{\infty} f_{Y} (t) d t = \frac{1}{π} \int_{0}^{\infty} t^{- 1/2} e^{- t} d t .

Therefore

\int_{0}^{\infty} t^{- 1/2} e^{- t} d t = π .

E [Z] = 0

and

var (Z) = 1

E [Z^{2}] = 1

. Since

Y = Z^{2} /2

E [Y] = 1/2

\frac{1}{2} = \int_{0}^{\infty} t f_{Y} (t) d t = \frac{1}{π} \int_{0}^{\infty} t^{1/2} e^{- t} d t .

Consequently

\int_{0}^{\infty} t^{1/2} e^{- t} d t = \frac{π}{2} .

These evaluate the gamma function at

1/2

and

3/2

, respectively, directly from a transformed normal random variable.

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