 Parity construction of pairwise independent bits

The parity construction of pairwise independent bits turns two independent fair bits into three pairwise independent random variables. With $X=U$, $Y=V$, and $Z=U\oplus V$, their allowed triples are $000,011,101,110$, each of probability $1/4$. Every coordinate has the Bernoulli distribution with parameter $1/2$. Every pair has the same joint probability mass function:All twelve pair cells equal the product of the corresponding probabilities in their marginal distributions. This establishes pairwise independent random variables, a requirement involving every pair of value-events, rather than merely one chosen event per coordinate. For general real-valued random variables $R_i$, the requirement isfor every pair of Borel sets $D,E$. For these binary random variables, checking all four cells of each pair suffices: any value-event is a union of the checked cells. In fact, for binary random variables, independence of their one-value events also gives the other cells by taking complements. Siegrist gives the all-set definition and its connection to indicator random variables in Random Services: independence of random variables.

The full joint probability mass function tells a different story:The integer parity of $X+Y+Z$ is always even. Any two bits determine the third: $Z=X\oplus Y$, $X=Y\oplus Z$, and $Y=X\oplus Z$. Each coordinate is independent of either other coordinate individually, but is dependent on the pair together. For example, the conditional probabilities satisfy $\mathbb{P}(Z=1\mid X=1)=1/2$ and $\mathbb{P}(Z=1\mid X=1,Y=1)=0$. See also pairwise independence does not imply joint independence for the general distinction.

The constraint changes the sum's probability distribution without changing its expected value or variance. Let $S=X+Y+Z$. Only $000$ gives $S=0$; each other allowed triple gives $S=2$. Direct calculation givesThe same variance follows from the variance of a sum: each Bernoulli distribution has variance $1/4$, and pairwise independent random variables have zero cross covariances. Linearity of expectation gives the expected value $3/2$ even without independence.

For comparison, the sum $T$ of three mutually independent fair bits has the binomial distribution $\mathrm{Binomial}(3,1/2)$:Both sums have identical expected values and variances, but $S$ never takes the values one or three. Even their third moments differ: $\mathbb{E}[S^3]=6$, whereas $\mathbb{E}[T^3]=27/4$. Pairwise independence preserves these first two sum summaries while allowing a different joint probability distribution and a different sum law. The calculation is an exact finite example.