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 in
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.
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 n
ote:
>
>> 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 independent z variates
>is chi squar
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 ]
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
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 2043666
these two values we can have one
> > 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
> > intui
plus ... many good REAL stat packages do this so easily
MTB > rand 5000 c1;
SUBC> norm 100 10. < mean and sigma
MTB > dotp c1
Dotplot: C1
..
.
.:.
ted with its features (and
> lack therof.) I, for one, cannot believe that histograms were not part of
> Excel 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/~cmo
itive 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
>we forecast the set distribution is Not Normal What intuitive use have
>the values?
well, maybe the 68% values may not
jim clark <[EMAIL PROTECTED]> wrote:
> "Chebyshev's Theorem: For any positive constant 'k', the
> probability that a random variable will take on a value within k
> standard deviations of the mean is at least 1 - 1/k2 ."
> This theorem holds for any distribution.
If you know that the distributi
itive 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
> we forecast the set distribution is Not Normal What intuitive use have
> the values?
Look up Chebyshev (or the many
o 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?
That of course depends on what the distribution actu
e 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?
Other intuitive definition as that I see in RadioFrequency: The
bandwidth of one amplifier is between the f
> >package. However, if you want a start, and feel that Excel is familiar
> >ground, here goes. The rand() function will generate random numbers from a
> >uniform distribution on the interval [0,1]. You can convert that to a
> >randomly distributed set of numbers using the inverse normal function,
u want with the data
you can generate normal like data for the unit normal distribution too ...
that is the default
mtb> rand 1 c1
that's it!
MTB > dotp c1
Dotplot: C1
Each dot repr
ere not part of
Excel 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
veral of the variables I am using show 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
n 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.
Discriminant analysis, as usually done, is poor without
joint normality and linear comparison functions.
Margina
> 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
>
alculate the density 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 Oberpfaf
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
>dis
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 mat
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
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 s
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 robert
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]
>
>
>=
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 fo
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. Pleas
ion is quite large for this data
(buyout funds are risky investments) negative numbers deviate significantly
from zero. A transformation from lognormal to normal seems not possible
therefore without "applying some tricks". Is this number large enough to
just use tests that are assumin
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 adjective to
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 Stigle
On Fri, 14 Apr 2000, Rich Ulrich wrote:
> > I believe that the term was at least popularized, if not
> > originated, by Quetelet, who called it the distribution of
> > the "normal person".
>
> Stephen Stigler, in his fine history, gives many pages to Quetelet and
> his fascination with the "aver
Kendall and Stuart have a footnote attributing the term to Galton
however there is no reference
Rich Ulrich wrote:
>
> On 13 Apr 2000 20:34:14 -0500, [EMAIL PROTECTED] (Herman
> Rubin) wrote, concerning the name of the "normal distribution" :
>
> >
> > I
On 13 Apr 2000 20:34:14 -0500, [EMAIL PROTECTED] (Herman
Rubin) wrote, concerning the name of the "normal distribution" :
>
> I believe that the term was at least popularized, if not
> originated, by Quetelet, who called it the distribution of
> the "normal person
Jan Souman wrote:
>
> Does anybody know why the normal distribution is called 'normal'? The most
> plausible explanations I've encountered so far are:
>
> 1. The value of a variable that has a normal distribution is determined by
> many different factors, each
exercises in which you
>>were asked
>>to find 'the equation to the normal to a curve', just after you were asked to
>>find the equation to the tangent.
>>
>>The reason why this name applies is because of the orthogonality
>>properties of
>>the (multi
In article <8d4fpl$em8$[EMAIL PROTECTED]>,
Jan Souman <[EMAIL PROTECTED]> wrote:
>Does anybody know why the normal distribution is called 'normal'? The most
>plausible explanations I've encountered so far are:
>1. The value of a variable that has a normal
ent.
The reason why this name 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
Does anybody know why the normal distribution is called 'normal'? The most
plausible explanations I've encountered so far are:
1. The value of a variable that has a normal distribution is determined by
many different factors, each contributing a small part of the total value.
Beca
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
can I 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 prob
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 yo
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
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
bivar
^2 + s_2^2)}.{PHI(x1) - PHI(x2)}
where s_1, s_2 are the two std.deviations,
the normalizing constant K is 4/sqrt{2.pi.(s_1^2 + s_2^2)},
PHI is the std. normal distribution function and the arguments
are:
x1 = (s_2/s_1).x / sqrt(s_1^2 + s_2^2)
x2 = (- s_1/s_2).x / sqrt(s_1^2 + s_2^2)
In the case of
My intuition tells me that although the sum of X1 and Y1 will not be exactly
a truncated normal, you will find that a truncated normal offers a good
approximation if both means are far from zero where the truncated part is
small. If you simulate this system, you will develop a better feeling for
i
f(X | X>=0) and g(Y|Y>=0),
>respectively. 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
Hi,
Suppose X, Y are independent random variables with normal distributions.
The means and variances are different. Assume X1 and Y1 are random
variables with the probability distributions f(X | X>=0) and g(Y|Y>=0),
respectively. That is, X1 and Y1 the non-negative truncations of X and
Y, respec
omments.
> 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 n
so I try to
>approximate the its density to bivariate normal distribution.
>To define normal distribution, I need to know three parameters.
>I could define the elliptical probability contour function by parameter
>sigma_Max, sigma_Min, and rotation angle Omega from reference axis to
>semi-axis.
s) for the benefit of those who have a low tolerance for ambiguity.)
Yes, the geometric mean 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
On Fri, 26 Nov 1999, Frank E Harrell Jr wrote:
> Beware - you can't just anti-log the mean and s.d. The median
> unlogged value is the antilog of the mean of the logged values.
That's interesting. The antilog of the mean of log(X) is the geometric
mean of X. Is the geometric mean necessari
thematician )
source
> >that explains the bivariate normal distribution or ( if there is any ) a
> >multi ( 3 variable )
> >equivalent. Namely, I would like find a method to sample a 2 or 3
variable
> >normal distribution
> >with a given mean and std. dev. like it is possible to
On Fri, 26 Nov 1999, Mr. SISAVATH Sourith wrote:
> Thanks for the advice.
> What I meant about the least square methods is as follows:
> If I calculate the mean and the variance of y=log(x)
> using the "standard" equations I mentioned in the previous mail
>mean value m = sum [ log x(i)*pr
g estimate: {A} nonparametric retransformation method},
journal = JASA,
volume = 78,
pages = {605-610},
annote = {smearing estimate;nonparametric;analysis of
cost;retransformation;empirical CDF;log-normal distribution}
}
"Donald F. Burrill" wrote:
> On Wed, 24
On Wed, 24 Nov 1999, Mr. SISAVATH Sourith wrote:
> I have a data sample of grains and the histogram of the
> grain size makes me think that the distribution is log-normal.
> Is it then reasonable to approximate the density function by a
> log-normal distribution, whose variance an
Hello
I have a data sample of grains and the histogram of the
grain size makes me think that the distribution is log-normal.
Is it then reasonable to approximate the density function by a
log-normal distribution, whose variance and mean value
has been calculated from the histogram, i.e.
mean
In article ,
Attila <[EMAIL PROTECTED]> wrote:
>Hello All!
>I am looking for a good ( and understandable to a non-mathematician ) source
>that explains the bivariate normal distribution or ( if there is any ) a
>multi ( 3 variable )
>equivalent. Namely, I would like find a
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