# Is independence indpendent on measure?

1. May 25, 2012

### tunaaa

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

Many thanks.

2. May 26, 2012

### chiro

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.

3. May 26, 2012

### micromass

Staff Emeritus
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 $P_2$.

4. May 26, 2012

Many thanks