Are components in a gaussian mixture independent?

[ConfusedCat]

Hello all,

I have used Expectation Maximization algorithm to approximate a probability density function (pdf) using a mixture of gaussians.

I need to square the pdf corresponding to the mixture of gaussians( it is a weighted sum of normal distributions with non-identical parameters). Rather than square a sum of gaussians, I thought that if the constituent normal pdfs of the mixture were independent, they could be summed resulting in a single gaussian, which I could then square.

I don't know how to prove (or disprove) the independence of the components. For a regular joint distribution f(x,y), independence can be proved if f(x,y)=f(x)f(y). Here, I have a series of real values for each of the normal components in the distribution. How do I find the joint distribution?

Any thoughts would be welcome.

Cheers

 

[I like Serena]

Hello all,

I have used Expectation Maximization algorithm to approximate a probability density function (pdf) using a mixture of gaussians.

I need to square the pdf corresponding to the mixture of gaussians( it is a weighted sum of normal distributions with non-identical parameters). Rather than square a sum of gaussians, I thought that if the constituent normal pdfs of the mixture were independent, they could be summed resulting in a single gaussian, which I could then square.

I don't know how to prove (or disprove) the independence of the components. For a regular joint distribution f(x,y), independence can be proved if f(x,y)=f(x)f(y). Here, I have a series of real values for each of the normal components in the distribution. How do I find the joint distribution?

Any thoughts would be welcome.

Cheers

Hi ConfusedCat! Welcome to MHB! ;)

We would need to test for independence of 2 continuous variables.
For instance this link gives a couple of possible approaches.

 

[ConfusedCat]

Hi ConfusedCat! Welcome to MHB! ;)

We would need to test for independence of 2 continuous variables.
For instance this link gives a couple of possible approaches.

Thank you for that link - it does look very useful.