Suppose $X$, $Y$ are i.i.d and $\dfrac{X+Y} {2^{1/\alpha} } \stackrel{\text{d}}{=} X$.
- If $0<Var(X)<\infty$, then show that $\alpha =2$ and $X \sim N(0,\sigma^2)$ for some $\sigma^2 \ge 0 $.
- If $X$ has characteristic function $e^{-c|t|^{\alpha}}$ with $\alpha >0$, then deduce that $Var(X)< \infty$ and conclude that $X=0$ (i.e., Stable-$\alpha$ distributions do not exist for $\alpha>0$).
What I tried is Var($X$) = Var($\dfrac{X+Y} {2^{1/\alpha} } $)
$\implies$ Var($X$) =$\dfrac{2} {2^{2/\alpha}}$Var(X)
Since Var($X$) $\neq$ $0$ , $2^{1-2/\alpha}=1$ $\implies$ $\alpha =2$
But I'm unable to prove that $X \sim N(0,\sigma^2)$ for some $\sigma^2 \ge 0 $.I know I have to work with characteristic functions but I'm not getting anything. Also I don't know how to proceed for 2nd part. It would be helpful if someone can give hints. Thanks in advance.
(Part 1) As $X$, $Y$ are i.i.d. the OPs argument shows that $\alpha=2$ and $\mathbb{E}[X]=0$. Let $\sigma^2=\mathbb{E}[X^2]$. Using again the i.i.d. property of $X,Y$ we obtain that \begin{align} \phi^2(2^{-1/2}t)=\phi(t)\tag{1}\label{one} \end{align} where $\phi$ is the characteristic function of $X$. By induction \begin{align} \phi^{2^n}(2^{-n/2}t)=\phi(t)\tag{2}\label{two} \end{align} Notice that $\phi^{2^n}(2^{-n/2}t)$ is the characteristic function of the distribution of $$\frac{X_1+\ldots + X_{2^n}}{2^{n/2}}$$ where $(X_k:k\in\mathbb{N})$ is an i.i.d sequence. Since $\sigma^2<\infty$, the central limit theorem yields $$\frac{X_1+\ldots + X_{2^n}}{\sigma 2^{n/2}}\Longrightarrow N(0,1)$$ Hence $$\phi^{2^n}(2^{-n/2}t)\xrightarrow{n\rightarrow\infty}\psi_{\sigma^2}(t)$$ where $\psi_{\sigma^2}$ is the characteristic function of a normal distribution with mean $0$ and variance $\sigma^2$. Consequently, $X\sim N(0,\sigma^2)$.
(Part 2) Is not very clear. For example, the Cauchy distribution $$\mu(dx)=\frac{1}{\pi c}\frac{1}{1+\tfrac{x^2}{c^2}}\,dx$$ has characteristic function $\phi(t)=e^{-c|t|}$ ($\alpha=1$), satisfies $\frac{X+Y}{2}\stackrel{d}{=}X$ if $X\sim\mu$ but $\operatorname{var}(X)=\infty$.