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Could you help me to show the following:

"Consider the random variables $X,Y$ i.i.d. uniform on $[-1,1]$. Then $(X,Y,XY)$ are pairwise independent but mutually dependent. "

I am confused about pairwise independent: how can $X,XY$ be pairwise independent?


I would also appreciate some more general clarification on pairwise independence, mutual independence and higher order interactions: I am reading here, page 1:

"Tests of total (mutual) and pairwise independence are insufficient, however, since they do not rule out all third order factorizations of the joint distribution. An important class of high order interactions occurs when the simultaneous effect of two variables on a third may not be additive. In particular, it may be possible that $X\perp Y$ and $Y\perp Z$, whereas $(X,Y)\perp Z$ does not hold.''

  • the adjective "third" in the expression "third order factorizations" comes from the fact that I am considering 3 variables? So, if I have 4 variables then I have also fourth orderfactorizations? What is exactly intended as third (or higher) order factorizations?

  • What does it mean that the effect of a variable on another is additive? Is the example above reporting an additive effect? It seems multiplicative.

  • "in particular, it may be possible that $X\perp Y$ and $Y\perp Z$, whereas $(X,Y)\perp Z$ does not hold." Isn/t this case captured as well by the mutual dependence situations mentioned above?

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It seems your source is wrong or at least uses a very unfortunate example to get its point across. $XY$ and $X$ are independent if knowledge of $X$ does not influence the probability distribution of $XY$, but this breaks down in the special (probability zero) case that $X = 0$.

A better example is if $X$ and $Y$ are drawn independently from $\{-1, 1\}$ (so the 2 element set instead of the interval) where $1$ and $-1$ have equal probability for both $X$ and $Y$. Now a priori $XY$ has prob 1/2 of being -1 and prob 1/2 of being one, and after observing the value of $X$ (whatever it is) these probabilities are still the same. That is what it means for $XY$ to be independent of $X$. The converse independence works similarly: after observing the value of $XY$ you would still assign prob 1/2 to event X = 1 and prob 1/2 to event X = -1. So $X$ is independent of $XY$ as well.

However, knowing both $X$ and $Y$ the distribution of $XY$ is quite different (concentrated in one point) than it was without this knowledge, so $X, Y, Z$ are not mutually independent.

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    $\begingroup$ Thank you, that is clear. $\endgroup$ – STF Jul 5 '17 at 13:58
  • $\begingroup$ I expect you are correct in re-interpreting the question as discrete on $\{-1, 1\}$ rather than continuous on $[-1,1]$ $\endgroup$ – Henry Jul 6 '17 at 21:58

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