Is independence indpendent on measure?

[tunaaa]

Is it possible that 2 sigma-algebras could be independent under one measure but not independent under another?

Many thanks.

 

[chiro]

Is it possible that 2 sigma-algebras could be independent under one measure but not independent under another?

Many thanks.

Hey tunaaa and welcome to the forums.

I don't know much about measure theory so maybe you could give us the definition of independence for a measure. I've heard about decomposing measures into orthogonal parts but I don't think this is what you are asking about.

 

[micromass]

Hey tunaaa and welcome to the forums.

I don't know much about measure theory so maybe you could give us the definition of independence for a measure. I've heard about decomposing measures into orthogonal parts but I don't think this is what you are asking about.

He probably means independence wrt a probability measure. This is defined as follows: take a probability space (\Omega,\mathcal{F},P) and take \mathcal{B}_1 and \mathcal{B}_2 sigma-algebra's which are part of \mathcal{F}. They are independent if for every B_1\in \mathcal{B}_1 and B_2\in \mathcal{B}_2 holds that

P(B_1\cap B_2)=P(B_1)P(B_2)

Like the definition suggests, the measure P is critical here. If we have another measure on (\Omega,\mathcal{F}, then independence must not hold.

For example, look at (\Omega,\mathcal{F})=(\mathbb{N},\mathcal{P} (\mathbb{N})) and take P_1 uniquely defined by P_1(\{0\})=1. Further, take P_2 uniquely define by P_2(\{0\})=P_2(\{1\})=1/2.

Then {0} and {1} (which generate sigma-algebras) are independent for P_1, but dependent for P_2.

 

[tunaaa]

Many thanks