Random variables that are triple-wise independent but quadruple-wise dependent

AI Thread Summary
The discussion revolves around finding four random variables that are triple-wise independent but not quadruple-wise independent, specifically taking values in {-1, 1}. The hint suggests using products of independent random variables, leading to the construction of events based on combinations of elementary events. A proposed solution involves using eight elementary events with equal probabilities and defining four events (A, B, C, D) through specific combinations of these events. Calculating the probabilities shows that while any three of the events are independent, the fourth event introduces dependence. This problem highlights the counterintuitive nature of independence in probability theory.
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Hi everyone, here's a probability problem that seems really counter-intuitive to me:

Find four random variables taking values in {-1, 1} such that any three are independent but all four are not. Hint: consider products of independent random variables.

My thoughts:
From a set perspective, there are no four sets such that any three are disjoint but all four overlap somewhere.

But according to the hint, let X_{i} be a random variable taking values 1 with probability p_{i} and -1 with probability 1-p_{i} for all i \in N. Then, I guess our four random variables will be products of these X_{i}'s but how should one construct them?
 
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Here's the solution for pairwise independent (3 events) but not all three (you can generalize to 4).

Four elementary events (j,k,l,m), each probability 1/4.
A=(j,k)
B=(j,l)
C=(j,m)
P(A)=P(B)=P(C)=1/2
P(A*B)=P(A*C)=P(B*C)=1/4=P(A)P(B)=P(A)P(C)=P(B)P(C) (independent)
P(A*B*C)=1/4, P(A)P(B)P(C)=1/8 (not independent)
 
For your problem work with 8 elementary events (G,H,I,J,K,L,M,N) each with Prob.=1/8 and 4 events of interest(A,B,C,D). Use the following table:
__G-H-I-J-K-L-M-N
A-1-1-1-1-0-0-0-0
B-1-1-0-0-1-1-0-0
C-1-0-1-0-1-0-1-0
D-1-0-0-1-0-1-1-0

1 means in the set, 0 means not in.

If you calculate the probabilities, you will see they are independent in 2's and 3's, but not all 4.
 
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