Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm preparing a lecture on stable distributions, and I'm trying to find a simple explanation of the following fact.

Suppose we are trying to come up with stable distributions. From the definition, it's clear that a distribution is stable iff its characteristic function $\phi$ satisfies $\phi(t)^n = e^{i t b_n} \phi(a_n t)$. The normal distribution, with chf $\phi(t) = e^{-t^2/2}$ clearly satisfies this with $b_n = 0$, $a_n = \sqrt{n}$. This suggests that we look for distributions with chfs of the form $\phi(t) = e^{-c |t|^\alpha}$. For $0 \le \alpha \le 2$, this is indeed a chf, and there is a nice proof in Durrett's book, constructing it as a weak limit using Lévy's continuity theorem. But:

For $\alpha > 2$, is there a simple reason why $\phi(t) = e^{-c |t|^\alpha}$ cannot be a chf?

Breiman's Probability proves a general formula for the chf of a stable distribution, using a representation formula for infinitely divisible distributions, but it's more work than I want to do for this.

share|cite|improve this question
up vote 7 down vote accepted

If $\phi$ is a characteristic function, then, for every real values of $s$ and $t$, $K(t,s)\geqslant0$ where $K(t,s)$ is the determinant $$ K(t,s)=\det\begin{pmatrix}\phi(0) & \phi(t) & \phi(t+s) \\ \phi(-t) & \phi(0) & \phi(s) \\ \phi(-t-s) & \phi(-s) & \phi(0)\end{pmatrix}. $$ Using $\phi_\alpha(t)=\mathrm e^{-c|t|^\alpha}$ for every $t$, one gets, for every fixed $x$, $K_\alpha(t,xt)=c^2|t|^{2\alpha}k_\alpha(x)+o(|t|^{2\alpha})$ when $t\to0$, with $$ k_\alpha(x)=2x^\alpha(1+x)^\alpha+2x^\alpha+2(1+x)^\alpha−x^{2\alpha}−(1+x)^{2\alpha}−1. $$ If $\alpha>2$, $k_\alpha(x)=−\alpha^2x^2+o(x^2)$ when $x\to0$ hence $k_\alpha(x)<0$ for some values of $x$ and $K_\alpha(t,tx)<0$ for some (small) values of $t$ and $x$. This proves that $\phi_\alpha$ is not a characteristic function.

First edit To prove that the condition that $K$ is nonnegative is necessary for $\phi$ to be a characteristic function, consider more generally the matrix $M=(M_{k,\ell})$ where $M_{k,\ell}=\mathrm E(\mathrm e^{\mathrm i(t_k-t_\ell)X})$ for some given real numbers $(t_k)$. Then, for every complex valued vector $v=(v_k)$, $$ v^*Mv=\sum\limits_{k,\ell}M_{k,\ell}v_k\bar v_\ell=\mathrm E\left(\sum\limits_{k,\ell}Z_k\bar Z_\ell v_k\bar v_\ell\right)=\mathrm E\left(\left|\sum\limits_{k}Z_kv_k\right|^2\right), $$ with $Z_k=\mathrm e^{\mathrm it_kX}$, hence $v^*Mv\ge0$ for every $v$. This means that $M$ represents a nonnegative form, and in particular, $\det M\geqslant0$.

Second edit Here is an alternative proof. I seem to remember that the second moment of $X$ with characteristic function $\phi$, be it finite or not, is $$ \mathrm E(X^2)=\lim\limits_{t\to0}\ t^{-2}(2-\mathrm E(\mathrm e^{\mathrm itX})-\mathrm E(\mathrm e^{-\mathrm itX})). $$ Assuming $\phi_\alpha$ is the characteristic function of $X_\alpha$ and using $\phi_\alpha(t)=1-c|t|^\alpha+o(|t|^\alpha)$ when $t\to0$, one gets $\mathrm E(X_\alpha^2)=0$ when $\alpha>2$, which is absurd.

share|cite|improve this answer
Thanks! I guess the fact that $K(t,s) \ge 0$ follows from Bochner's theorem, which unfortunately we have not covered. Is there another way to see it? – Nate Eldredge Oct 27 '11 at 15:47
Oh, never mind, that's the easy direction of Bochner's theorem. Ok, this could work. – Nate Eldredge Oct 27 '11 at 15:50
Nate, one can bypass Bôchner for the implication "if c.f. then det nonnegative", I am adding this now. – Did Oct 27 '11 at 15:58
@Didier Piau: The second edit works fine. Note that $t^{-2}(2-e^{itX}-e^{-itX})=\left(2t^{-1}\sin(tX/2)\right)^2$ is nonnegative, bounded by $X^2$, and tending to $X^2$. Fatou's lemma+dominated convergence implies that its expectation tends to $\mathbb{E}[X^2]$. – George Lowther Oct 27 '11 at 16:25
I like your second method the best. Essentially, the issue is that $\phi_\alpha$ is twice differentiable at zero but the second derivative vanishes, which can only happen when $X=0$. Actually, I found afterwards that this argument also appears in Durrett, in a different section from where I was looking. Thanks very much! – Nate Eldredge Oct 27 '11 at 19:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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