# Show That $\sum_{K=1}^{n}\frac{X_k}{n^{\frac{1}{\alpha }}}$ If ${X_n}$ is $X_k$s Same Distribution

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?

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

(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).