Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: Central Moment for a Gaussian Mixture model Date: Mon, 18 Jul 2011 21:34:09 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 16 Message-ID: <j028sh$bs1$1@newscl01ah.mathworks.com> References: <j0260l$3u7$1@newscl01ah.mathworks.com> Reply-To: <HIDDEN> NNTP-Posting-Host: www-06-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: newscl01ah.mathworks.com 1311024849 12161 172.30.248.38 (18 Jul 2011 21:34:09 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Mon, 18 Jul 2011 21:34:09 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1187260 Xref: news.mathworks.com comp.soft-sys.matlab:736662 "kumar vishwajeet" wrote in message <j0260l$3u7$1@newscl01ah.mathworks.com>... > I am trying to calculate the third central moment of a gaussian mixture model having 50 components. I want to know whether the following is a correct method to calculate its third central moment. > \int{(x-mu)^3 P(x)dx} = E[(x-mu)^3] > = E[x^3] - 3*mu*E[x^2] - mu^3 > Where mu is the resultant mean of the mixture. and \int indicates the integration. > ....... - - - - - - - - - Both of these amount to about the same computation, I would think. To carry out the integration needed in E{x^3}, E{x^2}, and E{x} you would presumably have to do separate integrations on each of the 50 components. However I disagree with your last step in the first method. I believe it should read: = E{x^3} - 3*mu*E{x^2} + 2*mu^3 since 3*mu^2*E{x} = 3*mu^3. Roger Stafford