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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.