Hi all, I would like to rephrase my question to clear the confusion. If the convolution of f(t) and f(-t) is a Gaussian, then is there anything that we can say about f(t)! For one I can see that f(t) itself could be a Gaussian. Can it be any other known distribution? *********************************************** Research Assistant Computer Networking Research Laboratory Department of Electrical and Computer Eng. Colorado State University, Fort Collins, CO 80523 Voice: +1 970-491-7974 Fax: +1 970-491-2249 http://lamar.colostate.edu/~nischal http://www.engr.colostate.edu/ece/Research/cnrl ***********************************************
[EMAIL PROTECTED] (Herman Rubin) wrote in message news:<[EMAIL PROTECTED]>... > In article <[EMAIL PROTECTED]>, > Nischal Piratla <[EMAIL PROTECTED]> wrote: > >Hi all, > >I was analyzing some data and reached a stage where the distribution > >that I would like to solve for has a autocorrelation equal to Gaussian > >distribution. Is there a selected group of distributions that I could be > >looking at? Or stating the same thing in the other manner, which set of > >distributions have a autocorrelation that is Gausssian? > >(It looks like Gaussian itself belongs to this set of distributions. > >Correct me, if I am wrong.) > >Any kind of pointers are suggestions will be highly appreciated. > >Thank you, > >Nischal Piratla > > There is no way that a correlation, let alone an autocorrelation, > can have a normal distribution. . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
