Re: Moments

Andor wrote:

Jani Huhtanen wrote:
Andor wrote:


Jani Huhtanen wrote:
Andor wrote:


Jani Huhtanen wrote:
Andor wrote:

Jani Huhtanen wrote:
Andor wrote:

John Herman wrote:
I don't think that skewnew is the name for the third moment
or kurtosis for
the fourth. Skew and the kurtosis are the names for
parameters of distributions relative to the Gaussian.

All those "moments" with a name aren't really moments in
anycase, since they are centered. For example, for the Cauchy
distribution there exists no variance, but a second moment
(which is infinite).


So you're saying that a second moment exists although the
integral diverges? In fact any moments mu_k, where k>=1, diverge
for Cauchy distribution and are quite often declared to be
undefined.

The difference lies in the ability to specify a limit, even if it
is infinity. For example

limit_{x->inf} x = inf.

However,

limit_{x->inf} x sin(x)

does not exist. Check out the section "Why the second moment of
the Cauchy distribution is infinite" in

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


Interesting. I see the difference. Mathworld, however, defined all
the moments (except 0th) undefined. I begin to wonder, why is the
mean so hard to define? The distribution is symmetric, thus doesn't
it follow automatically that the mean is at the point of symmetry.
In this case, the maxima of the distribution. In fact, parameter x0
seems to define the place of the maxima.

Yet another pretender to the "mean" could be the 50% percentile
(median) of the distribution. I can easily think of an asymmetric
distribution where the mean, the maximum of the density and the
median are all diffierent.


Of course. But isn't Cauchy symmetric?

Yes - I was just giving examples where definitions such as point of
symmetry, maximum of density, median, etc. do _not_ have the same
answer as the mean (and are thus not universally applicable as
surrogate for the mean).


Are there any cases where the mean is defined for a symmetric
distribution and it actually is not at the point of symmetry?

Probably not (at least I can't think of any, but then again, this has
been shown to not carry any weight :-). Do you propose the define the
mean for symmetric densities as the point of symmetry in case it is not
defined by the integral?

Not really. It is more like

(symm. distribution and the mean at x0) => (symmetry point at x0),

and not

(symm. distribution and the mean at x0) <= (symmetry point at x0).

It is just counter-intuitive to say that the mean is undefined even though
rules of thumb place it intuitively to the maximum of the distribution. I'm
just trying to intuitively grasp that mean is not defined (i.e., not
through the definition).

Probably the intuition comes from trying to estimate the mean from the
population. I suppose the variance of the estimate will be infinite when
the population size goes to infinity?

--
Jani Huhtanen
Tampere University of Technology, Pori
.

Relevant Pages

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    ... The distribution is symmetric, thus doesn't it follow ... automatically that the mean is at the point of symmetry. ... the maxima of the distribution. ... int_{0}^{infinity} x fdx = infinity. ...
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  • Re: Moments
    ... All those "moments" with a name aren't really moments in anycase, ... For example, for the Cauchy distribution ... distribution and are quite often declared to be undefined. ... automatically that the mean is at the point of symmetry. ...
    (comp.dsp)
  • Re: Moments
    ... All those "moments" with a name aren't really moments in anycase, ... For example, for the Cauchy distribution ... distribution and are quite often declared to be undefined. ... automatically that the mean is at the point of symmetry. ...
    (comp.dsp)
  • Re: Moments
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