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I am looking for a simple example of two independent discrete random variables that one would not expect to be independent because one knows that these two quantities have a causal relationship in real life.

Salamander
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    If one knows that two variables have a causal relationship, they are not independent. I think you mean "believes" instead of knows. – AlgorithmsX Sep 16 '16 at 21:46
  • If $A$ causes $B$, then $B$ is not an independent event. I should have said that both cannot be independent events. – AlgorithmsX Sep 16 '16 at 21:52
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    Obligatory xkcd reference: cell phones. – dxiv Sep 16 '16 at 21:57
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    One example is if $X_1$ and $X_2$ are independent Bernoulli with $P[X_i=1]=1/2$ and $Y=(X_1+X_2) (mod 2)$. Then $X_1$ and $Y$ are independent. – Michael Sep 16 '16 at 22:04
  • @AlgorithmsX I wanted to clarify my response to what you said earlier. You said that if $A$ causes $B$, they cannot be independent. There may be a way to argue this, but I'm not convinced. Simpson's paradox shows that it's possible for $A$ to be independent from $B$ overall, and yet if we split the population into groups $1, 2, 3, \ldots, n$, then in each group $P(B | A)$ is much higher than $P(B | \lnot A)$. But, $P(B|A) = P(B|\lnot A)$ overall! – Caleb Stanford Sep 16 '16 at 22:14
  • So would it be correct to say $A$ causes $B$ in these Simpson's-paradox situations? I don't know, because I don't know the definition of causes. – Caleb Stanford Sep 16 '16 at 22:16
  • This question has an example that I like: http://math.stackexchange.com/questions/255769/why-is-that-the-events-sum-of-dice-roll-6-first-die-4-are-dependent-but-the/255799#255799 – Micah Sep 17 '16 at 05:37
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    It’s not an “in real life” example, so I guess not quite what you want, but the proof of Kolmogorov’s zero-one law is a lovely counter-intuitive use of independence. It shows that certain “tail events” on a sequence of random variables — roughly, events like “there will be infinitely many heads in the sequence of coin tosses” — must have probability 0 or 1, because any such event is independent of itself! – Peter LeFanu Lumsdaine Sep 17 '16 at 07:01

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A classic basic example of events that seem dependent, but aren't, is when flipping two coins:

  • $A$ is the event that coin 1 is heads.

  • $B$ is the event that coin 1 and coin 2 are the same.

If you don't think about it too hard, it seems like the value of coin 1 is important or relevant when determining if coins 1 and 2 are equal, and therefore it is intuitively tempting to say $A$ is relevant to $B$ or in other words they are not independent.

Caleb Stanford
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Along the lines of Michael's suggestion in the comments, we have the finger game of "odds and evens". Assuming each person's play is i.i.d.* at $50$-$50$ (by no means a guarantee, to be sure), the result is statistically independent of either person's play in isolation, but of course is completely determined by their joint play.

*i.i.d. = identically and independently distributed (thanks to Bakuriu for clarifying this for the OP in the comments)

Brian Tung
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Gambler's Fallacy is a pretty good example. People often expect past events to interfere with future independent events. For instance, if you flipped a coin ten times and got heads all ten times, you would expect to get tails the next time, even though the last ten flips have no effect on the next flip.

AlgorithmsX
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