Pairwise independent random variables ($X_i\perp \perp X_j(i\ne j)$)

Random variables are pairwise independent when every two distinct members are independent random variables. Each pair's joint probability distribution must factor for all its events; examining only a pair's covariance is insufficient. Full independence implies pairwise independence by taking the other sets in the definition to be the whole target spaces. The reverse implication fails once there are three or more variables.

Pairwise independence is enough to add variances: for a finite family of square-integrable pairwise independent random variables,because all cross covariances vanish in the variance of a sum. Thus pairwise independence can support calculations about the size of fluctuations while failing to determine the full joint probability distribution. The following example makes that distinction visible.

Let $X,Y$ be independent fair bits, and let $Z=X\oplus Y$ be their exclusive or: $Z=1$ precisely when $X\ne Y$. The four possible triples are $000,011,101,110$, each with probability $1/4$. Each coordinate is a fair bit. Writing out all eight patterns givesEvery pair has all four cells $00,01,10,11$ with probability $1/4$:For $(X,Z)$, the four allowed triples project to $00,01,11,10$; for $(Y,Z)$ they project to $00,11,01,10$. Each projection lists each pair exactly once, proving pairwise independence. But $\mathbb{P}(X=0,Y=0,Z=0)=1/4$ differs from the product $1/8$, so the three are not independent random variables. Indeed, any two determine the third. A single coordinate reveals nothing about another, while two coordinates together reveal it completely.

The exclusive or example is the complementary-bit version of the two-toss exercise in Grinstead and Snell, Introduction to Probability, section 4.1, exercise 7, printed page 151. It is a compact reason to check the full product rule whenever a calculation uses three-way or higher events.