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This question shows a counter-example to the statement

If $X_1,\dots,X_n,X_{n+1}$ are pairwise independent, then $\sum_{i=1}^n X_i$ is independent of $X_{n+1}.$

I wonder if the following statement is true (my professor used it in class but I was not convinced about it, especially after seeing the above linked question):

If $X_1,\dots,X_n,X_{n+1}$ are i.i.d. variables (note that besides from pairwise independency we are assuming that these variables are also identically distributed), then $\sum_{i=1}^n X_i$ is independent of $X_{n+1}$.

Any hint or full answer to my question is highly apreciatted. Thanks for any help in advance.

xyz
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  • The answer you cite has already shown that the claim is untrue for pairwise independent, identically distributed variables. – Aditya Dhawan Oct 16 '23 at 11:13
  • @AdityaDhawan Thanks for your comment. My probability theory is not on its best state, does the variable $X_3 = X_1 X_2$ from the post I linked also follows uniform distribution in $(-1,1)$ ? – xyz Oct 16 '23 at 11:18
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    i.i.d. includes independence, not just pairwise independence. – geetha290krm Oct 16 '23 at 11:21
  • The assertion that $\ X_1,X_2,\dots,X_n\ $ are i.i.d. implies that they are independent, a stronger condition than pairwise independence. Are you sure your professor made the parenthetical comment in precisely the same terms you've quoted here? – lonza leggiera Oct 16 '23 at 11:22
  • Thanks for all the comments. Yes, I am sure this is the reasoning the professor used. I will discuss this later today in the class I'll have with the professor and then I'll update you guys, before closing/deleting the question. – xyz Oct 16 '23 at 11:31
  • Consider the sigma algebras generated by said variables. – AlvinL Oct 16 '23 at 11:38
  • The assertion you've quoted is actually correct. Strictly speaking the parenthetical comment is also true, but oddly weaker than it needs to be to provide any assistance to anyone who might want to try proving the assertion. If $\ X_1,X_2,\dots,X_{n+1}\ $ are i.i.d. then they're not merely pairwise independent, which would not be sufficient to prove the assertion, but also independent, which is sufficient to do so. – lonza leggiera Oct 16 '23 at 13:53
  • @xyz Yes, $X_3$ is also uniform on ${ +1, -1}$. But as others have pointed out, i.i.d is very much reserved for independence and not pairwise independence. – Aditya Dhawan Oct 16 '23 at 18:52

1 Answers1

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In general, if $Z$ is independent of $(X_1, X_2, \cdots X_n)$ and $Y=g(X_1, X_2 \cdots X_n)$, then $(Y,Z)$ are independent.

This should be intuitively obvious. For formal proofs you can start here.

In our case, $Z=X_{n+1}$ and $g()=\sum()$.

Notice that it's not necessary that the variables are identically distributed, and neither that all $(X_1, X_2, \cdots X_n)$ are independent. All that's needed is that the set $(X_1, X_2, \cdots X_n)$ is independent from $X_{n+1}$. However, for this is not enough that $X_{n+1}$ is independent from each $X_i$ (as you noted).

leonbloy
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