= Solution
{c}
Take a <discrete probability distribution> on three points:
$$
\boxed{\mathbb P(Y=-2)=\frac25,\qquad\mathbb P(Y=1)=\frac12,\qquad\mathbb P(Y=3)=\frac1{10}.}
$$
The <probabilities> are positive and sum to one. Direct calculation gives
$$
\mathbb EY=-\frac45+\frac12+\frac3{10}=0,\qquad
\mathbb EY^3=-\frac{16}5+\frac12+\frac{27}{10}=0,
$$
and
$$
\boxed{\mathbb EY^4=\frac{32}5+\frac12+\frac{81}{10}=15<\infty.}
$$
Nevertheless it is not a <symmetric distribution>, since $\mathbb P(Y=3)=1/10$ while $\mathbb P(Y=-3)=0$. Vanishing of just the first and third <moments> does not force all odd <moments> to vanish.
Back to article page
