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

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SUMMARY

This discussion addresses the construction of four random variables that are triple-wise independent but quadruple-wise dependent. The solution involves using products of independent random variables, specifically defined as X_{i} taking values of 1 with probability p_{i} and -1 with probability 1-p_{i}. The example provided utilizes eight elementary events (G, H, I, J, K, L, M, N) with equal probabilities of 1/8 to demonstrate the independence of events A, B, C, and D in pairs and triples, while showing their dependence when all four are considered.

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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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