Infinite Variance VS Zero Variance

cacosomoza

Hi there, this is caco, I´m a new user from planet Earth.

I´m studying Information Theory and lately It seems that I live in a fractal ocean of gaussian distributions...

I do know that zero variance processes relate to Dirac´s delta distribution, my favourite mathematical artifact ever. And then I considered the opposite case, a distribution in which its mean is all possible values at the same time. It would be somewhat like that its mean is its own domain (despite we are not talking of a function in the strict sense).

Anyway, I don´t wanna become too philosophical but I always loved the opposition between chaos and order and i think the presence of a mean is the presence of some intrinsic order within that process. We know the universe tends to set energy equiprobably around physical space, which would mean that the universe tends to an infinite variance distribution.

My question is, does it have any name such process, a totally chaotic and unpredictible process? Which is the opposite of the dirac delta function?

Thanks for your words

wofsy

If what you mean by opposite is Fourier transform then the Dirac delta transforms to the uniform distribution.

The universe will not tend to an infinite variance distribution because its volume is finite. Also it is not clear that it will ever reach a uniform energy distribution. It that were inevitable then how did it get wound up in the first place?

A distribution can have a mean and still have infinite variance.

A distribution without a mean is usually not studied and does not really relate to any theory I know except in degenerate cases such as particles of definite momentum or position in Quantum Mechanics.

A Brownian motion is totally unpredictable even though it has a definite mean.

The only continuous stochastic process with independent increments is a Brownian motion.

cacosomoza

[QUOTE=wofsy;2337441]
A distribution can have a mean and still have infinite variance.

[QUOTE]

Thanks for your answer. How is it possible that if all values are equally probable one may be selected as representing the most probable? Can somebody explain me this?

Thanks

wofsy

[QUOTE=cacosomoza;2337454][QUOTE=wofsy;2337441]
A distribution can have a mean and still have infinite variance.

? What do you mean - if all values have equal probability - how can one be most probable?

statdad

The mean doesn't have to be the most probable value.

I'm not quite getting what you mean by infinite variance: usually if a distribution has a moment that is not finite we say it doesn't exist. As an (artificial) example, consider the random variable who's density is

[tex]
f(x) = K \cdot \left(\frac 1 {x^3}\right), \quad x \ge 1
[/tex]

Here K is the constant needed to make

[tex]
\int f(x) \, dx = 1
[/tex]

This has a finite mean, since [tex] x f(x) [/tex] is of the order [tex] x^{-2} [/tex], and so the required integral converges. However, the variance doesn't exist, since [tex] x^2 f(x) [/tex] is of order [tex] x^{-1} [/tex], and the required integral does not converge.