 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/iv/solution

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

For the standard normal distribution, the integrand in the second moment is even, so

E [Z^{2}] = \frac{1}{2 π} \int_{- \infty}^{\infty} z^{2} e^{- z^{2} /2} d z = \frac{2}{2 π} \int_{0}^{\infty} z^{2} e^{- z^{2} /2} d z .

Comparing these two integrals gives

E [X] = \frac{π}{2} E [Z^{2}], β = \frac{π}{2} .

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