1

Let $X, Y$ and $Z$ be random variables. Let

  • $p_1$ 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.

Michael Hardy
  • 1
Tim
  • 47,382

2 Answers2

1

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
Henry
  • 157,058
-3

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.

I <3 all Tim's.

user
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