Does $X ⊥ Y \leftrightarrow X ⊥ Y | Z$ implies $(X,Y) ⊥ Z$? Let $X, Y$ and $Z$ be random variables. Let


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*$p_1$ be the statement that $(X,Y) ⊥  Z$ (meaning $(X,Y)$ and $Z$  are independent), 

*$p_2$ be the statement that $X ⊥  Y$ (meaning $X$ and $Y$ are independent) 

*$p_3$ be the statement that $X ⊥  Y \mid Z$ (meaning $X$ and $Y$ are conditionally independent given $Z$) 
What I have known is that if $p_1$ is true, then $p_2$ and $p_3$ imply each other.
I wonder if the reverse is true. That is, if $p_2$ and $p_3$ imply each other, will $p_1$ be true? To disprove it, is there a counterexample? Thanks.
 A: No, the converse is not true.  Here are two counter-examples:
1) Consider the following triplets of values all with equal probability.  $p_2$ and $p_3$ are both true in that $X$ and $Y$ are unconditionally independent of each other and (given $Z$) are conditionally independent of each other, while $p_1$ is not true as the pair $(X,Y)$ is not independent of $Z$ 
 X    Y    Z 
 1    1    1
 1    2    1
 2    1    1 
 2    2    1
 1    3    2
 1    4    2
 2    3    2
 2    4    2 

2) Consider the following triplets of values all with equal probability.  Neither $p_2$ and $p_3$ are true in that $X$ and $Y$ are neither unconditionally independent of each other nor (given $Z$) conditionally independent of each other, while $p_1$ is not true as the pair $(X,Y)$ is not independent of $Z$ 
 X    Y    Z 
 1    1    1
 1    2    1
 2    1    1 
 1    3    2
 1    4    2
 2    3    2

A: If $p_1$false, then $p_3$ does not imply $p_2$ because assuming $p_1$ and $p_3$, we can't decide whether $X$ and $Y$ are dependent or independent.
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