I am trying to solve the problem in the title, namely, and I would like to see whether it is correct or not.
Show that $\phi(t) = \exp(-|t|^\alpha)$ can be the characteristic function of a distribution with finite variance, if and only if $\alpha = 2$.
A similar question was asked here.
My thoughts so far on the implication: $\phi$ a characteristic function, hence $\alpha = 2$.
$\alpha$ needs to be positive, otherwise $\phi(0)$ wouldn't exist, hence $\alpha>0$.
If we calculate the second derivative of $\phi$ wrt $t$, we get the following:
$$ \phi^{\prime\prime}(t) = \begin{cases} \alpha^2t^{2(\alpha-1)}e^{-t^{\alpha}} - \alpha(\alpha-1)t^{\alpha-2}e^{-t^\alpha}, &\mbox{ }t>0\\\alpha^2t^{2(\alpha-1)}e^{-(-t)^{\alpha}} - \alpha(\alpha-1)(-t)^{\alpha-2}e^{-(-t)^\alpha}, &\mbox{ }t<0 \end{cases} $$
(This is the sentence where I think that a mistake may exist) In order for the left and right limits at $0$ to exist and be finite, we would need to have $\alpha\geq2$. If $\alpha>2$ then we would get $\phi^{\prime\prime}(0) = 0 = E[X^2]$. Given that $E[X^2]\geq 0$ it would mean that the pdf of $X$ would be constant, hence the characteristic function would be different than the one we considered, contradiction.
Hence $\alpha = 2$.
