Solution
= Solution
The given <expected value> and <variance> imply $M_X(0)=1$, $M_X'(0)=0$ and $M_X''(0)=1$. The <Taylor expansion> therefore gives
$$
M_X(t)=1+\frac{t^2}{2}+o(t^2).
$$
Since $\log(1+u)=u+o(u)$ as $u\to0$, it follows that
$$
\boxed{\log M_X(t)=\frac{t^2}{2}+o(t^2),\qquad\alpha=\frac12.}
$$
This is the <quadratic expansion of a cumulant-generating function> for a centered variable of variance one.