 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iii/6/c/iii/solution

The probability density function must integrate to one, so

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

E [Y] = \frac{1}{2} E [Z^{2}] = \frac{1}{2} (var (Z) + (E [Z])^{2}) = \frac{1}{2} .

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

These are the values

Γ (1/2)

and

Γ (3/2)

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