--- dasimov@earthlink.net wrote:

Suppose p: R -> R is a continuous mapping with p(x) > 0 everywhere, and integral{R} p(x) dx = 1.

Question: Can the set of Fourier transforms P(t) := (1/sqrt(2pi)) integral{R} p(x) exp(ixt) dx of such p's be characterized explicitly?

--Dan

I believe that the answer is "yes", but I can't remember the theorem. The Fourier transform of a probability density over the reals is called a "characteristic function". There is a trilogy of short monographs on this subject by Eugene Lukacs. Characteristic functions are very interesting. For example, there is the related subject of "stable distributions". The distribution is stable if sums of independent random variables possess the same distribution, with scaling and translation. The most obvious example is the Gaussian distribution. Another is the Cauchy distribution (a/pi)(1/(x^2+a^2). Since the sum distribution is a convolution, the characteristic functions multiply. For any a with 0<a<=2, exp(- w |t|^a + i c t) is the c.f. of a symmetric stable distribution. (Gaussian a=2, Cauchy a=1.) There are also nonsymmetric stable distributions. Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com