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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.

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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.

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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
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@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
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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

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