I'm trying to come up with some real life examples of situations where you have unconditional dependence, but conditional independence. Here's one that I came up with, but I'm not sure if it's sound.

Consider $A, B, C$ to be the events that you win, lose, or tie in a game.

$A$ and $B$ are clearly dependent since if $A$ happens, $B$ can't happen.

Now say you're given $C$. Then I think knowing C happens would make $A$ and $B$ independent. But I am hesitant to believe this is sound because the definition of conditional independence is $P(A, B | C) = P(A|C) P(B|C)$. We know that $P(A, B | C) = 0$ since $A$ and $B$ are mutually exclusive regardless if we're conditioning on anything. We also know that $A,C$ and $B,C$ are mutually exclusive, so $P(A|C) = P(B|C) = 0$. But is $A|C$ and $B|C$ actually conditionally independent here, or is $P(A, B | C) = P(A|C) P(B|C)$ naturally satisfied due to the mutual exclusivity?


1 Answer 1


This is an example of conditional independence, since any probability-zero event (in this case $A\mid C$) is independent of any other event. However, it's rather a trivial example.

You can get examples where the independence is non-trivial. For example, roll a standard die, and set $A$ to be the event of getting an even number, $B$ to be the event of getting a prime number, and $C$ to be the event of getting at most $4$.

Now $P(A)=1/2$, $P(B)=1/2$ but these events are not independent since $P(A\cap B)=1/6$. However, $P(A\mid C)=P(B\mid C)=1/2$ and $P(A\cap B\mid C)=1/4$, so the conditional events are independent.

  • $\begingroup$ Is coming up with these kind of examples more art than science? Like what made you think of this die example and having $B$ be the event of getting a prime number. Did you do some guess and checking with the probabilities until you found one that worked? $\endgroup$
    – roulette01
    May 4, 2021 at 18:07
  • 2
    $\begingroup$ @roulette01 I was looking for conditional events of probability $1/2$. For conditional independence you need their intersection to have probability $1/4$, so have a conditional state space (i.e. event $C$) of size $4$. Then add some other states which correspond to $C^c$ and choose $A\cap C^c$, $B\cap C^c$ any way you like, just making sure that they aren't independent. It's easy to see such events will exist; getting them to have "natural" definitions is a bit more tricky and involves some trial and error. $\endgroup$ May 4, 2021 at 18:24

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