Re: Is there a natural phenomena which defines the Normal Curve?

In article <gv7ip11e5vjlb9ms3brs9gkohbaaj1mff2@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:
>On Thu, 08 Dec 2005 20:33:14 -0800, Ronald Bruck <bruck@xxxxxxxxxxxx>
>wrote:

>>In article <dnao4l$38o0@xxxxxxxxxxxxxxxxxxxx>, Herman Rubin
>><hrubin@xxxxxxxxxxxxxxxxxxxx> wrote:
>>> ... but
>>> for a sum of independent random variables to be normal,
>>> they both must be normal.

>>Is this obvious? It's sort of a unique factorization of N: if f*g =
>>N, then f = N and g = N. How do you do that?

>>--Ron Bruck

>Here's an intuitive explanation ...

Unfortunately, one step is drastically wrong.

>Suppose X,Y are independent and X+Y is normal.

>Let X_1, X_2, X_3, ..., X_n be n independent copies of X.

>Then X_k + Y is normal for all k, hence so is the sum,

>X_1 + X_2 + X_3 + ... + X_n + n*Y

How can you claim this? If you had X_k + Y_k, this
could be argued, but the X_k + Y are NOT independent.

>Dividing by n,

>(1/n)*(X_1 + X_2 + X_3 + ... + X_n) + Y

>is also normal.

>Letting n approach infinity, and applying the CLT,

>Z_1+ Y = Z_2 where Z_1, Z_2 are both normal.

>But then Y = Z_2 - Z_1 which implies Y is normal.

>By similar reasoning (or by subtraction), X is normal.

>quasi


--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@xxxxxxxxxxxxxxx Phone: (765)494-6054 FAX: (765)494-0558
.

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