$\exp(-|x|^\alpha),\alpha>2$ is not a Fourier transform of a probability distribution Problem: 

Show that $\exp(-|x|^\alpha),\alpha>2$ is not a Fourier transform of a probability distribution

Hint:

What would the second moment look like?

Edit 1:
Following our definition of the moment generating function, we have that
$M_\mu(ix)=\exp(-|x|^\alpha)=\hat{\mu}(x)$
Now we know that the second momement is given by
$\frac{d^2}{dx^2}M_\mu(0)$.
Thus I have to compute $\frac{d^2}{dx^2}M_\mu(x)$. We do this by using the fourier transform, namely $\hat{\mu}''(x)$.
Am I correct? I just did a quick computation with wolfram and following that result I would get that $\hat{\mu}(0)''=0$ for $\alpha>2$. Why would the result follow from this?
 A: Following the definition of the moment generating function, we have that
$M_\mu(ix)=\exp(-|x|^\alpha)=\hat{\mu}(x)$,
where $\hat{\mu}$ denotes the Fourier transform of the measure $\mu$.
Now we know that the second moment is given by
$\frac{d^2}{dx^2}M_\mu(0)=\hat{\mu}''(0)$.
Thus we have to compute $\hat{\mu}''(x)$:
$$\hat{\mu}'(x)=\alpha x(-e^{-|x|^\alpha})|x|^{\alpha-2}\implies\hat{\mu}'(0)=0$$
$$\hat{\mu}''(x)=\alpha^2x^2e^{-|x|^\alpha}|x|^{2\alpha-4}+(\alpha-2)\alpha x^2(-e^{-|x|^\alpha})|x|^{\alpha-4}-\alpha e^{-|x|^\alpha}|x|^{\alpha-2}\overset{\alpha>2}{\implies}\hat{\mu}''(0)=0$$
Thus for a random variable $X$ with $X\sim\mu$, we have that $\mathbb{E}(X^2)=0$ and $\mathbb{E}(X)=0$. It follows that $\operatorname{Var}(X)=\mathbb{E}(X^2)-[\mathbb{E}(X)]^2=0$. Now we know that $$\operatorname{Var}(X)=0\implies\mathbb{E}(X)=X \text{ a.s.}$$
Thus $X$ is a constant random variable. However, the probability measure of a constant random variable is given by the Dirac measure. We know that the Fourier transform of the latter is given by $\widehat{\delta_c}(x)=e^{i<x,c>}$ with constant $c$. Due to the uniqueness of the Fourier transform of a measure, it follows that $\exp(−|x|^\alpha)$ is not the Fourier transform of a probability distribution.
