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Re: [Phys-l] Basic statistics



I have Feller in front of me and am too lazy to check Wiki, which is irrelevant here, anyhow.
Zeor mean is trivial. If the mean is nown, it can be subtracted.
Feller shows how to get around finite variance in some cases.. At any rate, in discussing n(0,1) the statement is meaningless until you specify
(I don't thing you can do it in words) what mathematical quantity it is that becomes n(0,1), and in what limit.

On Mon, 13 Nov 2006, Polvani, Donald G. wrote:

Jack Uretsky wrote:

I misremembered, and therefore mis-stated, what it is that the
Chi-square distribution does not do. You correctly
state that it becomes Gaussian in the large sample-size limit. What it
does not do is become
n(0,1) (0-mean, unit variance) as some other distributions do. The
zero-mean part is, of course, trivial because it >just invloves
subtraction of the the distribution means.
The correct statement of the CLT is that the means of samples
taken from finite variance distributions tend to
n(0,1) in the large sample-size limit. The are distributions for which
the CLT fails. See Feller, 2d Ed., Vol. II >(not an easy read).

I just checked with Wikipedia, see:

http://en.wikipedia.org/wiki/Central_limit_theorem

What I found at Wikipedia agrees with what I remember learning (but I do
not have an advanced probability/statistics text handy to check results
further, so I will rely on the collective wisdom of the list). Showing
convergence to n(0,1) (i.e. Gaussian with zero mean and unit variance)
is only an aid in simplifying the understanding and proof of the central
limit theorem. It says nothing about the applicability of the theorem.
The theorem states that ANY SUM of N independent identically distributed
random variables will tend to be approximately normally distributed
(with the approximation getting better as N gets larger) PROVIDED that
the SUM of the variables has a finite variance (I mistakenly left out
this requirement for finite variance in my original post on this
thread). The sums involved may have a non-zero mean and a non-unity
variance (provided the variance is finite); it's just mathematically
convenient to "normalize" (no pun intended, Jack) the sums so that they
have zero mean and unit variance by the standard technique). Presumably
(I have not tried), one could prove the central limit theorem without
working with zero mean, unit variance distributions for the SUMS of
independent variables involved.

Incidentally, the same Wikipedia article gives a remarkably simple three
line proof of the central limit theorem using characteristic functions.
It looked like a convincing proof to me, but I am the type of physicist
who took the applied option for my advanced math courses because I
disliked (and wasn't that good at) the pure math option courses with all
the delta and epsilon "rigorous" proofs.

Don Polvani
Northrop Grumman Corp.
Undersea Systems
Annapolis, MD 21404
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