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I'm looking for the answer to the problem in the title. I know it comes from the Marcinkiewicz theorem, but I need formal proof of it.

GoodDeeds
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mwrooo
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  • It appears that Marcinkiewicz's theorem in this context says that if $P$ is a polynomial of degree more than $2$, then $\exp(P(x))$ is not a characteristic function. – Cameron L. Williams Mar 16 '16 at 15:27
  • Take two derivatives and set $t=0$ to get $EX^2=0$ - contradiction. – A.S. Mar 16 '16 at 15:45

3 Answers3

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Assume on the contrary that $e^{-t^4}=E(e^{itX})$ for some random variable $X$. Taking the fourth derivative in both sides with respect to $t$ and setting $t=0$ you get $-24=E(X^4)$, a contradiction, since $X^4\geq 0$ and therefore $E(X^4)\geq 0.$

Nikolaos Skout
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By Bochner's theorem, it suffices to show that $\exp(-t^4)$ is not positive-definite. Meaning that for some collection $x_1,\ldots,x_n$, we have that the matrix $A = (\exp(-(x_i-x_j)^4))_{i,j=1}^n$ is not positive-semidefinite. $A$ is automatically self-adjoint so that condition is fulfilled.

Let us take $x_1 = 0$, $x_2 = .01$ and $x_3 = .02$, we get

$$A = \left(\begin{array}{ccc} 1 & e^{-.01^4} & e^{-.02^4} \\ e^{-.01^4} & 1 & e^{-.01^4} \\ e^{-.02^4} & e^{-.01^4} & 1\end{array}\right).$$

The reason I picked small $x_i$ is that the values will all be fairly close to $1$ which means that there is a good shot at getting negative eigenvalues. If $x_2$ and $x_3$ were greater than $1$, then the diagonal terms would dominate by far, which would lead to positive eigenvalues. If you compute the eigenvalues of $A$ (by whatever means.. in my case, Mathematica) you get

$$ \lambda_1 = 3,\quad \lambda_2 = 1.6\cdot 10^{-7}, \quad \lambda_3 = -4\cdot 10^{-8}$$

and thus $A$ has a negative eigenvalue which means it is not positive-semidefinite. This in turn says that $\exp(-t^4)$ is not positive-definite which in turn says that it is not the characteristic function of a probability measure.

What is nice about this approach is that it kind of tells you that in order for something to be a characteristic function, it cannot be very flat around $0$; instead, it has to immediately (and somewhat rapidly) decay away from $0$. The reason being that if it were fairly flat, you could do something similar to the above and likely get negative eigenvalues.

Cameron L. Williams
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  • You basically know that $\varphi_X''(0)<0$ for non-degenerate $X$ which significantly limits the flatness. – A.S. Mar 16 '16 at 18:08
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Seems that the $4n^{th}$ moment of the measure is alternating, so it is not a probability measure.

James
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