 Noisy parity total correlation

Keep $X,Y$ as independent fair bits, introduce an independent Bernoulli random variable $N$ with $\mathbb{P}(N=1)=\epsilon$, and set $Z=X\oplus Y\oplus N$. The extra bit flips the integer parity rule with probability $\epsilon$. For each even-integer parity triple the probability is $(1-\epsilon )/4$; for each odd-integer parity triple it is $\epsilon /4$.

Every coordinate remains fair, and every pair remains independent for every $0\le \epsilon \le 1$. For example, after fixing $X$, the fair independent $Y$ makes $Y\oplus N$ fair regardless of $\epsilon$. A pair's joint probability distribution therefore does not change as noise destroys the three-way relation.

Knowing $X,Y$, uncertainty in $Z$ is exactly the uncertainty in $N$. The chain rule for information entropy giveswhere $h_2$ is binary entropy. HenceAt $\epsilon =1/4$, the result is approximately $0.188722$ bits, although all three pairwise mutual informations are zero. At $\epsilon =1/2$, all eight triples have probability $1/8$, giving joint independence and zero total correlation. At $\epsilon =1$, the complementary integer parity rule is deterministic again, so total correlation returns to one bit.

The example distinguishes "there is a dependence" from "a pair reveals it." Once $X,Y$ are known together, $Z$ is predictable at zero noise; knowing either one alone leaves a fair bit. Total correlation detects the joint departure from independence, but does not by itself specify which variables cause which others.