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# Thread Subject: Central Moment for a Gaussian Mixture model

Subject: Central Moment for a Gaussian Mixture model

From: kumar vishwajeet

### kumar vishwajeet (view profile)

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

### Roger Stafford (view profile)

Date: 18 Jul, 2011 21:34:09

Message: 2 of 2

"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