# Nonexistence of stable random variables

Is there an "elementary" proof that $\alpha$-stable random variables only exist for $0 < \alpha \le 2$? By elementary I mean without using Fourier transforms. I'd be happiest with either a direct probabilistic argument, or a geometric argument, e.g., referring to embeddings of $L_p$ spaces (as long as it doesn't depend on results which are themselves proved via Fourier analysis).

Note that what I'm interested in here is the fact that $\alpha$-stable random don't exist for $\alpha > 2$, not the fact that they do exist for $\alpha \le 2$.

Edit: To clarify, for the purposes of this question I define an $\alpha$-stable random variable as follows. Given $\alpha > 0$, $X$ is $\alpha$-stable if, whenever $X_1, \ldots, X_n$ are independent copies of $X$, there is a real number $d=d(n)$ such that $X_1 + \cdots + X_n$ has the same distribution as $n^{1/\alpha} X + d$. And I'm quite happy to simplify to the situation when $d=0$.

• Stable distributions are essentially defined by their characteristic function, so to request a proof that side-steps the cf (Fourier transform) seems a bit unreasonable. – wolfies Jul 2 '14 at 8:06
• @wolfies: Definition: $X$ has a stable distribution if, whenever $Y$ is an independent copy of $X$ and $a$ and $b$ are real numbers, there are real numbers $c$ and $d$ such that $aX + bY$ has the same distribution as $cX + d$. No characteristic functions in sight. – Mark Meckes Jul 2 '14 at 8:09
• I would say this is the "real" definition of a stable distribution. Versions given in terms of characteristic functions are only stated that way because the cf is the usual technical tool used to study them. – Mark Meckes Jul 2 '14 at 8:10
• Good question. If $\alpha$ stable variables exist and have finite $p$-th moment, then $\ell_\alpha$ embeds isometrically into $L_r$ for all $0<r\le p$. So $\alpha \le 2$ because $L_r$ has cotype $2$ for $r\le 2$ and $\ell_\alpha$ does not. But why must an $\alpha$ stable variable have finite $p$-th moment for some $p$? – Bill Johnson Jul 2 '14 at 10:53
• Here's a purely probabilistic proof that if nonzero $\alpha$-stable variables with finite second moment exist then $\alpha = 2$. Suppose that $X$ is $\alpha$-stable with finite second moment, and assume without loss of generality that $X$ is symmetric, so that $\mathbb{E} X = 0$. Then $$\mathbb{E} X^2 = \mathbb{E} \left(n^{-1/\alpha} \sum_{i=1}^n X_i \right)^2 = n^{-2/\alpha} \left(\sum_{i=1}^n \mathbb{E} X_i^2 + \sum_{i\neq j} \mathbb{E} X_i X_j \right) = n^{1-2/\alpha} \mathbb{E} X^2,$$ and so $\alpha = 2$. – Mark Meckes Jul 2 '14 at 11:18

The argument can be subdivided into two parts. The first part (which was already mentioned in the comments) states, that if an $\alpha$-stable random vector has a finite non-zero variance, then $\alpha=2$. The second part is showing that variance is finite when $\alpha>2$, which is done by subdividing the integral $E(X^2)$ into a suitable series. Thus, $Var(X)=0$ when $\alpha>2$, which completes the proof.