Let ${\{X_n}\}$ be a sequence of independent random variables and the stable distribution of order alpha $(0\le\alpha\le2)$.

Show that $$\sum_{k=1}^{n}\frac{X_k}{n^{\frac{1}{\alpha }}}$$ if ${X_n}$ is $X_k$s same distribution.

I can't find anywhere this theory. Can somebody help me?

  • 1
    $\begingroup$ Although it seems clear what you're trying to ask, your text is really garbled. Could you edit it to make sense? $\endgroup$ – whuber Jan 9 at 20:56

This is only true if the location parameter is zero.

(In which case you can use the characteristic function of the stable distribution $e^{-|ct|^\alpha (1-i\beta sign(t) tan(\pi\alpha/2))}$ to proof it for $\alpha\neq 1$)

As a counter example consider the normal distribution (for which $\alpha=2$)

$$Y = \frac{1}{\sqrt{n}}\sum_{i=1}^n X_i$$

In this case you have $\sigma_Y = \sigma_X$ but $\mu_Y \neq \mu_X$ and instead $\mu_Y = \sqrt{n} \mu_X$. So $Y$ is not similarly distributed as $X$ (it is the same family though, maybe that is what you meant).


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