How do I prove the following:
$X,Y,Z$ are independent, then any pair is independent given the remaining
random variable.
I know that $E(X|Z)=E(X) $ and $E(Y|Z)=E(Y)$
so $E(XY)=E(X)E(Y)=E(X|Z)E(Y|Z)$
does it imply that $E(X|Z)E(Y|Z)=E(XY|Z)$
On the other hand, if $X,Y,Z$ are pairwise independent, how do I disprove that any pair is independent given the remaining random variable?
Best Answer
$X,Y,Z$ are independent, hence, $P(X,Y,Z) = P(X) P(Y) P(Z)$. Then,
$$P(X,Y \mid Z) = \frac{P(X,Y,Z)}{P(Z)} = \frac{P(X) P(Y) P(Z)}{P(Z)}$$ This proves that $X,Y \mid Z$ are independent, since $P(X,Y \mid Z) = P(X)P(Y)$. The same proof goes to the other cases.
For the case you want to disprove, just use the following counter-example:
You can easily check that they are pairwise independent. But their collection is not independent. Just doing some calculations:
$$ P(X,Y \mid Z) = \frac{P(Z \mid Y,X)P(X,Y)}{P(Z)}= \frac{P(Z \mid Y,X)P(X)P(Y)}{P(Z)} $$
But $P(Z=1 \mid X=1, Y=1)=0 \neq P(Z=1)=0.5$. So we conclude that $P(X,Y \mid Z) \neq P(X)P(Y)$ which is what we wanted to disprove.