The wiki page on semi-martingales states that
Every Lévy process is a semimartingale.
and that
The quadratic variation exists for every semimartingale.
Let $X_t$ be a stable Levy process with $X_t$ distributed as $S(\alpha, \beta, \mu \, t, \sigma \, t^{\frac{1}{\alpha}} ; 0)$. It's increments $X_{t+ \delta t}-X_t$ follow $ \mu \, \delta t + \sigma \, {\delta t}^{\frac{1}{\alpha}} Y$, where $Y$ follows standard stable distribution $S(\alpha, \beta, 0, 1\, ; 0)$.
Now the quadratic variation of such a process:
$ [X]_t = \lim_{\vert\vert P \vert\vert \to 0} \sum_i (X_{t_{i+1}}- X_{t_i})^2 = \lim_{\vert\vert P \vert\vert \to 0} \sum_i (\mu \, \delta t_i + \sigma \, {\delta t_i}^{\frac{1}{\alpha}} Y_i)^2 $
To simplify things, assume that $\mu = 0$, and that $\delta t_i$ are the same, i.e. $\delta t_i = t/n$: $ [X]_t = \sigma^2 t^{\frac{2}{\alpha}} \lim_{n \to \infty} {n}^{-\frac{2}{\alpha}} \sum_{i=0}^{n-1} Y_i^2 $
For $\alpha =2$ the limiting random variable is that of $S/n$ where $S$ follows $\chi^2_{n}$ which converges to a degenerate distribution concentrated at $x=1$.
Added: The sum can not converge in probability to a degenerate distribution for $\alpha<2$, because $\mathbb{E}(Y_i^2) = \infty$ for $0<\alpha<2$, which implies the convergence in distribution should be to a distribution with infinite mean.
Simulation shows this is indeed the case:
How can I formally see that the above limit does converge in probability ? Was the limiting distribution studied, even if only in the symmetric case of $\beta=0$ ?
Thank you.