Ludovic Duponchel wrote:
If x values have a normal distribution, is there a normal distribution
for x^2 ?
No. If the mean is 0, x^2 hasa chi-squared distribution with 1 DOF.
As the ratio mean/SD - infinity, the distribution of x^2 is
asymptotically normal.
-Robert Dawson
In article [EMAIL PROTECTED],
Ludovic Duponchel [EMAIL PROTECTED] wrote:
If x values have a normal distribution, is there a normal distribution
for x^2 ?
The only transformations one is likely to encounter which
preserve normality are linear.
--
This address is for information only. I do
If x values have a normal distribution, is there a normal distribution
for x^2 ?
Thanks a lot for your help.
Best regards.
Dr. Ludovic DUPONCHEL
UNIVERSITE DES SCIENCES DE LILLE
LASIR - Bât. C5
59655 Villeneuve d'Ascq
FRANCE.
Phone : 0033 3 20436661
Fax
On Thu, 29 Nov 2001 15:48:48 +0300, Ludovic Duponchel
[EMAIL PROTECTED] wrote:
If x values have a normal distribution, is there a normal distribution
for x^2 ?
If z is standard normal [ that is, mean 0, variance 1.0 ]
then z^2 is chi squared with 1 degree of freedom.
And the sum of S
Rich Ulrich wrote:
On Thu, 29 Nov 2001 15:48:48 +0300, Ludovic Duponchel
[EMAIL PROTECTED] wrote:
If x values have a normal distribution, is there a normal distribution
for x^2 ?
If z is standard normal [ that is, mean 0, variance 1.0 ]
then z^2 is chi squared with 1 degree of freedom
distribution, is there a normal distribution
for x^2 ?
If z is standard normal [ that is, mean 0, variance 1.0 ]
then z^2 is chi squared with 1 degree of freedom.
And the sum of S independent z variates
is chi squared with S degrees of freedom.
--
Richard Startz
On Thu, 29 Nov 2001 14:37:14 -0400, Gus Gassmann
[EMAIL PROTECTED] wrote:
Rich Ulrich wrote:
On Thu, 29 Nov 2001 15:48:48 +0300, Ludovic Duponchel
[EMAIL PROTECTED] wrote:
If x values have a normal distribution, is there a normal distribution
for x^2 ?
If z is standard normal
powerful intuitive use to them: The centre of the set is the mean
and 68% of values are in the interval [mean-SD to mean+SD], IF the set
have Normal Distribution. If the set distribution is NOT Normal, what
intuitive use have the values?
How about limiting distribution (CLT
cel v2.0.
If you did a search using google with search terms = simulation normal
distribution excel you should have found
http://phoenix.som.clarkson.edu/~cmosier/simulation/Week_6/norm_conv.html
and many others.
David Winsemius
janssen_w wrote:
Hi,
For some stats explaining I need
% of values are in the
interval [mean-SD to mean+SD], IF the set have Normal Distribution. If
we forecast the set distribution is Not Normal What intuitive use have
the values?
Other intuitive definition as that I see in RadioFrequency: The
bandwidth of one amplifier is between the frequencies where
:
The centre of the set is the mean and 68% of values are in the
interval [mean-SD to mean+SD], IF the set have Normal Distribution. If
we forecast the set distribution is Not Normal What intuitive use have
the values?
well, maybe the 68% values may not be totally relevant but, remember, the
SD
shold they) be
somehow transformed so that the resulting distribution looks and presumably
acts in the analyses) like a normal distribution.
Discriminant analysis, as usually done, is poor without
joint normality and linear comparison functions.
Marginal normality does not imply joint normality
distributions that are
not normal. My question is can these (and for that matter shold they) be
somehow transformed so that the resulting distribution looks and presumably
acts in the analyses) like a normal distribution.
TB
It depends. For some distributions it is easy to do
that the resulting distribution looks and presumably
acts in the analyses) like a normal distribution.
It depends. For some distributions it is easy to do the
transformations (e.g., log is often appropriate for +ve skew). An
alternative approach might be to consider logistic regression which
has several advantages
Gökhan wrote in message [EMAIL PROTECTED]...
Hi!
I wonder how the public is evaluating the normal distribution function
in realworld applications. I am implementing some methods where i have
to calculate different times probability functions relying on normal
distribution functions
of a multivariate normal
distribution
where I needed the to handle the operations on the covariance matrices .
Thanks anyway
Gökhan BakIr
Insitute of Robotics and Mechatronics
German National Research Institute for Aero and Space
82234 Oberpfaffenhofen
Tel
I presume that you want the density of a multivar normal distrib. You
don't calculate the inverse; you just need the quadratic form. I think
that Searle's matrix algebra book gives the computations. off hand, for
the quad form x'A-1x I'd get the cholesky factor of A = LL' and solve
for
Hi!
I wonder how the public is evaluating the normal distribution function
in realworld applications. I am implementing some methods where i have
to calculate different times probability functions relying on normal
distribution functions with steadily changing covariance matrix and mean
values
G?khan [EMAIL PROTECTED] wrote:
: Hi!
: I wonder how the public is evaluating the normal distribution function
I presume that you want the density of a multivar normal distrib. You
don't calculate the inverse; you just need the quadratic form. I think
that Searle's matrix algebra book gives
Hi all. I try to use the ratio between the sample averages of \mu and
\sigma to estimate the real ratio between \mu and \sigma. But I want to
know whether this estimator in any sense is optimum, and then is this one
the best estimator in Mean square estimation error sense?
Since the data are
Trying to use in finacial calcs. Hardcosed one to four decimals. Prefer more
precision.Thanks. [EMAIL PROTECTED]
===
This list is open to everyone. Occasionally, less thoughtful
people send inappropriate messages.
If you think you need more precision than given in the
usual tables or with a caculator, think again. You are
probably fooling yourself since no distribution in the real
world is _exactly_ normal.
Jon Cryer
At 03:55 PM 7/5/00 GMT, you wrote:
Trying to use in finacial calcs. Hardcosed one to
bet you can find something here ...
http://members.aol.com/johnp71/javastat.html
At 03:55 PM 7/5/00 +, MRFCLANCY wrote:
Trying to use in finacial calcs. Hardcosed one to four decimals. Prefer more
precision.Thanks. [EMAIL PROTECTED]
We offer six decimals at
http://www.stat.ucla.edu/calculators/cdf
but also the density, the quantile function, graphs of all these,
plus sets of random numbers emailed to you. And this for the most
common 20 distributions, including the noncentral ones.
At 14:05 -0400 07/05/2000, dennis
After I cited Stigler, to the effect that Quetelet never used the term
'normal' for the distribution,
on 14 Apr 2000 09:53:05 -0700, [EMAIL PROTECTED] (Alan Hutson)
wrote:
Kendall and Stuart have a footnote attributing the term to Galton
however there is no reference
I thought that Stigler
The normal distribution has often been called the Gaussian distribution,
although de Moivre and Laplace spoke of it well before Gauss.
The term "normal" had been used for the distribution by Galton (1877)
and Karl Person later recommended the routine use of that adjectiv
applies is because of the orthogonality properties of
the (multi)normal distribution.
If you take a simple random sample from a normal distribution, and represent
each Xi by a different axis, the axes will be mutually perpendicular.
Obviously there is more to it than this, but I can't remember
Hi,
I meet a problem to analysis a group data. The data consist of 2 or more
Normal distributions with different mean. I want to find the sigma and mu of
the distribution with the largest area. How can I seperate this normal
distribution from others? I would be appreciated if you can give me any
If you have attribute data that goes with the value data (such as batch #) this
can be sequenced etc. You can then perform an analysis that seperates them.
Be careful not to assume any distributions, but to let the distributions appear
from the data. After all, your data doesn't care what
seperate this normal distribution from others?
If you can do (2), then (1) becomes easy; and if (2) is what you really
want and need to do, that's what you need to focus on. But if all you
really need is (1), that's a different sort of technical problem, and
there probably are ways of estimating
Dear Donald:
Thank you so much for your help. You can find a group of data in the attached
file. Most value in this data locate arround 0.8. There is also some data
distribute arround 1. These data should be normal distribution. In these set
of data, most of data distribute arround 0.8. If I
Dear Members fo News Group,
I always appreciate that I could have received your help.
As I know, I can apply Kolmogorov-Smirnov goodness-of-fit test to
univariate sample. But, I don't know which method can be applied to
multivariate samples, especially, when I got the samples assumed to be
. That is, X1 and Y1 the non-negative truncations of X and
Y, respectively. Does anyone know whether in this case Z = X1 + Y1 is
still a truncated normal? Any reference on this? Thanks in advance!
It is not. An easy way to see this is to use the fact
that the truncated normal distribution has
.
The domain of random variable X and Y is -1 X, Y 1, which is points
in xy plane. The points is located clustring near origin (0,0), so I
try to approximate the its density to bivariate normal distribution.
Ah. That explains why (1 - sigma_max*sigma_min) would not be imaginary.
It is still unclear
estimates the median of Y if Y has a log-normal
distribution. Beware of non-robustness of geometric mean though.
The mean unlogged value is something like exp(mean unlogged + .5sigma2)
where sigma2=sd of logged values.
Did you mean "sigma2 = variance of logged values"? (Why else repr
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