Let $X_i$ be sequence of i.i.d. random variables. Suppose $0 < \alpha \leq 2$ and characteristic function is given by \begin{equation} \varphi(t) = e^{-|t|^\alpha}\quad \left[{}=\exp(-|t|^\alpha)\right] \end{equation} For what values of $\alpha$, Strong Law of Large Number hold?
Note
This is the last part of the question. If I am not mistaken, I need to argue that $E|X|<\infty$ for some $\alpha$.
First thing that I thought was to show $\varphi(t)$ is differentiable at $0$, however, this does not imply $E|X|<\infty$.
I wrote down my attempt below, but even if it is true, it seems too long for a rather short exam question.
I would appreciate a reference if I am missing something.
Attempt (not in great detail.)
First,
\begin{equation}
E|X| = \int_0^\infty P(|X|>t) \, dt = 2\int_0^\infty P(|X|>2/u)/u^2 \, du
\end{equation}
and note that
\begin{equation}
P(|X|>2/u) \leq \frac{1}{u} \int_{-u}^u (1-\varphi(t)) \, dt = \frac{2}{u} \int_0^u (1-e^{-t^\alpha}) \, dt
\end{equation}
To bound $E|X|$, note that $P(\cdots)\leq 1$ and $\int^{\infty} \frac{1}{u^2} \, du <\infty$. Hence it is sufficient to consider $u$ near $0$.
Without details,
\begin{equation}
\frac{1}{u}\int_0^u (1-e^{-t^\alpha}) \, dt \sim u^\alpha
\end{equation}
and hence
\begin{equation}
E|X| \leq \int_0 u^{\alpha-2}
\end{equation}
and if $\alpha > 1$ it is finite. Noting that characteristic function is not differentiable for $\alpha \leq 1$, $E|X|$ is not finite.