In article <[EMAIL PROTECTED]>,
Maarten Speekenbrink <[EMAIL PROTECTED]> wrote:
>Oops, I've made a mistake in my former post. the corrected message is:

>Hi there,

>I would like to know the distribution of a linear combination of (not
>identical) logistically distributed variables, i.e.

>f(x) = exp( (x - alpha)/beta )/(1+exp( (x-alpha)/beta ))^2 .

>Does anybody now what this distribution would look like, or know a good
>reference? I've looked in the 'Handbook of the logistic distribution' by
>Balakrishnan, as well as in Johnson & Kotz, but couldn't find anything.
>Since the distribution is very similar to the normal, I would expect a
>similar result to that for the normal distributed (it is logistically
>distributed itself). Thanks for your help.

The question is rather vague.  One can compute the moment 
generating function of a logistic random variable as

        m(t) = exp(alpha*t)*pi*beta*t/sin(pi*beta*t)

for |beta*t| < 1, and the moment generating function of a sum
of independent random variables is the product of the moment
generating functions.For numerical calculation, I would suggest
using the complex integral for the density or the cdf, whichever
is desired, and using steepest descent on it.

As it is a convolution of log-concave distributions, the density
is log-concave.  It is also infinitely divisible, with an easily
computed Levy measure, although it is hard to see what one can do
with this.

It is very definitely NOT logistically distributed.
-- 
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
[EMAIL PROTECTED]         Phone: (765)494-6054   FAX: (765)494-0558
.
.
=================================================================
Instructions for joining and leaving this list, remarks about the
problem of INAPPROPRIATE MESSAGES, and archives are available at:
.                  http://jse.stat.ncsu.edu/                    .
=================================================================

Reply via email to