Re: Symmetry and moments counterexample

In article <1138311341.326200.181110@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<pamelafluente@xxxxxxxxx> wrote:

>> Suppose F and G are two different distributions on (0,
>> infinity) which have the same moments. Then if we choose X
>> and Y from F and G respectively,
>...
>But since X and Y have different distributions, the
>> distribution of Z will not be symmetric.


>If F and G have all the moments equal they "usually" the same
>distribution.
>In such a case, X and Y cannot "have different distributions" and the
>proof fails.

In a certain sense, this fails for "almost all"
distributions. If the tail is not fast enough, it may
fail; one cannot make a stronger statement, as changing
one moment of such a distribution will make it unique.

>So my question becomes:
>Can you make one example of 2 different distributions F and G with
>equal moments?

Here is a relatively simple example; I will indicate also how
one can get at it.

Consider the integral of x^{c-1)*exp(-x)*cos(x) dx on the
positive real axis. It is Gamma(c)*R((1+i)^(-c)). Now if
c is an integer of the form 4k+2, this integral is 0. That
means that the integral of x^{4k+1}*exp(-x)*cos(x) dx on
the positive reals is 0. Now set x = y^(1/4), and one gets
the integral of y^{k-.75}*exp(-y^.25)*(1+h*cos(y^.25) dy is
independent of h, which gives a family of such
distributions with the same k-th moments for all k.
--
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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