# Understanding conditional independence of two random variables given a third one

I am reading the Wikipedia article on conditional independence. There seems to be Two definitions for conditional independence of Two random variables $X$ and $Y$ given another one $Z$:

1. Two random variables $X$ and $Y$ are conditionally independent given a third random variable $Z$ if and only if they are independent in their conditional probability distribution given $Z$. That is, $X$ and $Y$ are conditionally independent given $Z$ if and only if, given any value of $Z$, the probability distribution of $X$ is the same for all values of $Y$ and the probability distribution of $Y$ is the same for all values of $X$.
2. Two random variables $X$ and $Y$ are conditionally independent given a random variable $Z$ if they are independent given $\sigma(Z)$: the $\sigma$-algebra generated by $Z$.

Two events $R$ and $B$ are conditionally independent given a $\sigma$-algebra $\Sigma$ if $$\Pr(R \cap B \mid \Sigma) = \Pr(R \mid \Sigma)\Pr(B \mid \Sigma)\ a.s.$$ where $\Pr(A \mid \Sigma)$ denotes the conditional expectation of the indicator function of the event $A$, given the sigma algebra $\Sigma$. That is,
$$\Pr(A \mid \Sigma) := \operatorname{E}[\chi_A\mid\Sigma].$$

Two random variables X and Y are conditionally independent given a $\sigma$-algebra $\Sigma$ if the above equation holds for all $R$ in $\sigma(X)$ and $B$ in $\sigma(Y)$.

I can understand the second definition, but my questions are:

1. What does the first definitions mean actually? I have tried several times on reading it, but fail to get what it means? Can someone rephrase it using rigorous and clean language, for example, by writing the definition in terms of some formulae?
2. Do the two definitions agree with each other? Why?
3. ADDED: I was wondering if the following is the correct way to understand the first definition. Notice that $P(\chi_A \mid Z)$ is defined as $E(\chi_A \mid Z)$ and therefore is a random variable. When the conditional probability $P(\cdot \mid Z)$ is "regular", i.e. when $P(\cdot \mid Z)(\omega)$ is a probability measure for each point $\omega$ in the underlying sample space $(\Omega, \mathcal{F}, P)$, does conditional independence between $X$ and $Y$ given $Z$ mean that $X$ and $Y$ are independent w.r.t. every probability measure defined by $P(\cdot \mid Z)(\omega), \forall \omega \in \Omega$? If yes, is the conditional probability $P(\cdot \mid Z)$ always guaranteed to be "regular"? So that there is no need to explicitly write this "regular" assumption?

Thanks and regards!

-
I just added my understanding of the first definition to my post as Part 3 of my questions. Hope someone can tell if it is correct. –  Tim Mar 10 '11 at 1:51

The first definition is the informal one, but at the same time seems rather convoluted to me.

I'd prefer: X and Y are conditionally independent with respect to a given Z iff

$P(X \; Y | Z) = P(X | Z ) P(Y | Z)$

Recall that conditioning one (or several) variables on the value of another, is (informally) the same as restricting the whole universe to a part of it. Then, if you are given the value of $Z$, you can think as if you are defining new variables that are the same as the unconditioned but that are restricted to our new (smaller universe) $X' \equiv X | Z$ $Y' \equiv Y | Z$ The above formula simply states that $X'$ and $Y'$ are independent.

The first definition says the same, but applying (in words) the property that two variables are independent iff their conditioned probabilities are the same as the unconditioned : $A$ indep $B$ iff $P(A | B ) = P (A)$

-
You should really replace your first equation by something like: "For every measurable sets $A$ and $B$, $P(X\in A,Y\in B|Z)=P(X\in A|Z)P(Y\in B|Z)$ almost surely". And get rid of these horrible $X'\equiv X|Z$ and $Y'\equiv Y|Z$. And finally, why not define conditional independence cleanly and fully rigorously through test functions... –  Did Mar 9 '11 at 23:30
@Didier: I added my understanding of the first definition to my post as Part 3 of my questions. Would you be able to see if it is correct? –  Tim Mar 10 '11 at 1:21
@Tim See the last sentence of my comment on the answer by @leonbloy. –  Did Mar 25 '11 at 13:56