Re: Independent vs uncorrelated

Ben claimed:

In an earlier post you implied

Uncorrelated ------> rotational invariance ....(A)

I don't think that Jani Luhtanen did any such thing. He in
effect said what you have called (C)

Rotational invariance ---------> Uncorrelatedness ....(C) (opposite of
A)


The implication is *not* bidirectional; it is not true that

...... Rotational invariance <----------> Uncorrelatedness


Independence, spherical symmetry, rotational invariance etc.
are global properties of the distribution (pdf if you prefer).
For example, independence requires that for *every* choice
of real numbers a_1, a_2, .. a_n, it must be that
P(a_1, a_2, ... , a_n) = P(a_1)P(a_2)...P(a_n); the factorization
of the joint pdf into the product of the marginal pdfs holds
everywhere. In contrast, correlation is a pairwise property of
the random variables. For (zero-mean) random variables, a
positive correlation merely means that in a sort of **average**
sense, there is more probability mass in the first and third
quadrants than in the second and fourth; more specifcally

int_0^{infty} int_{0,infty} x*y*[f(x,y) + f(-x,-y) - f(x,-y) -f(-x,y)]
dx dy > 0.

This doesn't say anything f(x.y) itself. It doesn't even mean that
f(x,y) + f(-x,-y) > f(x,-y) + f(-x,y) for *all* x, y > 0, only that
f(x,y) + f(-x,-y) > f(x,-y) + f(-x,y) at enough points (x,y) that the
weighted average (the value of the above integral) is positive.
Similarly for negative correlation and zero correlation. For the
case of rotationally invariant distributions, f(x,y) is a function of
x^2 + y^2, and thus f(x,y) = f(-x, -y) = f(x, -y) = f(-x, y) and the
*integrand* is zero and hence so is the integral. But the integral
being zero does not mean rotational invariance of the pdf.

Hope this helps.

.

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