 Solution

ID: past-exam-of-the-probability-course-of-the-university-of-oxford/2021/2/c/iii/solution

P (Y = - 2) = \frac{2}{5}, P (Y = 1) = \frac{1}{2}, P (Y = 3) = \frac{1}{10} .

The probabilities are positive and sum to one. Direct calculation gives

E Y = - \frac{4}{5} + \frac{1}{2} + \frac{3}{10} = 0, E Y^{3} = - \frac{16}{5} + \frac{1}{2} + \frac{27}{10} = 0,

and

E Y^{4} = \frac{32}{5} + \frac{1}{2} + \frac{81}{10} = 15 < \infty.

Nevertheless it is not a symmetric distribution, since

P (Y = 3) = 1/10

while

P (Y = - 3) = 0

. Vanishing of just the first and third moments does not force all odd moments to vanish.

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