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 $d=0$.

  • $\begingroup$ 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. $\endgroup$ – wolfies Jul 2 '14 at 8:06
  • $\begingroup$ @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. $\endgroup$ – Mark Meckes Jul 2 '14 at 8:09
  • $\begingroup$ 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. $\endgroup$ – Mark Meckes Jul 2 '14 at 8:10
  • 1
    $\begingroup$ 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 $p$? $\endgroup$ – Bill Johnson Jul 2 '14 at 10:53
  • 1
    $\begingroup$ 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 $\alpha = 2$. $\endgroup$ – Mark Meckes Jul 2 '14 at 11:18

I believe there to be a proof avoiding Fourier analysis in the literature given by Feller (1971) as part of Theorem VI.1.1.

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.

| cite | improve this answer | |
  • $\begingroup$ I think I really should just read Feller straight through one of these days. $\endgroup$ – Mark Meckes Jul 3 '14 at 7:26

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.