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# Thread Subject: Central Moment for a Gaussian Mixture model

 Subject: Central Moment for a Gaussian Mixture model From: kumar vishwajeet Date: 18 Jul, 2011 20:45:10 Message: 1 of 2 Hi, 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. Or, Should I calculate it as follows: \int{(x-mu)^3 P(x)dx} = \int{(x-mu)^3 \SUM{1}{50}{w_i N(mu_i, sigma_i)}dx where \SUM{1}{50}{w_i N(mu_i, sigma_i) indicates summation of the pdf of all the mixture components. I get different answers with each of them. But I don't see any error with either of these methods. What do you think???? Thanks.
 Subject: Central Moment for a Gaussian Mixture model From: Roger Stafford Date: 18 Jul, 2011 21:34:09 Message: 2 of 2 "kumar vishwajeet" wrote in message ... > 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