# $X,Y$ are independent iff conditional regular distribution of $X|Y$ is almost surely the same as the distribution of $X$.

I want to prove that $$X,Y$$ scalar random variables are independent iff conditional regular distribution of $$X|Y$$ is almost surely the same as the distribution of $$X$$. A simple result to prove for the... "regular" conditional distribution.

The definition employed for the regular distribution is the following. Let $$(\Omega,\mathcal F,P)$$ be a probability space, and $$(\Omega',\mathcal F')$$ a measurable space. Now $$\mu:\Omega\times\mathcal F'\rightarrow \mathbb R$$ is the regular conditional distribution if:

1) For every $$A\in \mathcal F'$$, $$\mu(.,A)=E(I_{X\in A}|\mathcal G)$$ almost surely. $$\mathcal G\subseteq \mathcal F$$, and in this case $$\mathcal G=\sigma (Y)$$, which is the pre-image of the borel sigma algebra.

2) For almost every $$\omega\in \Omega$$, $$\mu(\omega,.)$$ is a probability measure.

This definition I find very hard to use in order to prove almost anything.

## 1 Answer

Take $$A=(-\infty , x]$$. If $$X$$ and $$Y$$ are independent then $$\mu(\omega, A)=E(I_{(X \in A)} |Y)=P(X^{-1}(A))=P\{X\leq x\}$$ so the conditional distribution coincides with teh distribution of $$X$$. Conversely if this condition holds integrate both sides over $$\{Y\leq y\}$$ to get $$P\{X\leq x,Y\leq y\} =P\{X\leq x\}P\{Y\leq y\}$$.