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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:

enter image description here


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.

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up vote 1 down vote accepted

I think I figured it out. When $0<\alpha<2$, $Y_i^2$ has a long tailed distribution. Specifically:

$$ Pr(Y_i^2 > z) \approx \frac{2}{\pi} z^{-\frac{\alpha }{2}} \Gamma(\alpha) \sin \frac{\alpha \pi}{2} $$

Therefore $n^{-\frac{2}{\alpha}} \sum_{i=0}^{n-1} Y_i^2$ is in the basin of attraction of the stable distribution and should converge to positive stable distribution with exponent $\alpha^\prime = \frac{\alpha}{2}$. One has to determine only the scale. To that end, the generalized CLT should be used which states that if a variable $X$ has asymptotic $F_X(x) \approx 1- c x^{-\alpha}$ for large $x$, and $F_X(-x) \approx d (-x)^{-\alpha}$ for large $x$, then $n^{-\alpha} \sum_{i=1}^n x_i$ converges in probability to $S\left(\alpha, \frac{c-d}{c+d}, 0, \left( \frac{\pi (c+d)}{2 \Gamma(\alpha) \sin \frac{\pi \alpha}{2}}\right)^{\frac{1}{\alpha}} ; 1\right)$. In the case at hand $d=0$ because $Y_i^2$ are non-negative.

Hence it seems that $n^{-\frac{2}{\alpha}} \sum_{i=0}^{n-1} Y_i^2$ converges in probability to $S( \frac{\alpha}{2}, 1, 0, 4 \pi^{-\frac{1}{\alpha}} \left( \cos \frac{\pi \alpha}{4} \Gamma( \frac{1+\alpha}{2}) \right)^{\frac{2}{\alpha}} ;1 )$.

Notice that the scale parameter becomes small as $\alpha \to 2^-$, which corresponds to the degenerate limit for normal variates.

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