Pairwise Independence Checks for Finite Families

Why a Family Needs More Than Pairwise Checks

For two random variables, independence is a single relationship between the pair. For three or more, there are additional relationships involving triples and larger subcollections. Checking every pair examines only part of the joint distribution; it does not, by itself, establish how the entire family behaves together.

This distinction matters whenever an argument uses a collection of variables as though it were jointly independent. A calculation based only on pairwise independence may omit a higher-order dependence. The definitions make clear exactly which probability factorizations have been checked and which remain necessary.

Definitions for a Finite Family

Definition: Let $X_1,\ldots,X_m$ be real-valued random variables, where $m\geq 2$. They are pairwise independent if, for every two distinct indices $i,j$ and all Borel sets $A,B\subseteq\mathbb{R}$, $$ P(X_i\in A,\ X_j\in B)=P(X_i\in A)P(X_j\in B). $$ They are mutually independent if, for every nonempty index set $S\subseteq\{1,\ldots,m\}$ and every choice of Borel sets $A_i\subseteq\mathbb{R}$, $i\in S$, $$ P\left(\bigcap_{i\in S}\{X_i\in A_i\}\right)=\prod_{i\in S}P(X_i\in A_i). $$

The singleton cases in the mutual-independence definition are automatic: an event has the same probability as itself. Thus, for a family of three variables, mutual independence requires factorization for each pair and for the triple. For a larger family, it requires the corresponding condition for every subcollection, including subcollections of size three or more.

The same definitions apply to events. A finite family of events $E_1,\ldots,E_m$ is mutually independent when the probability of the intersection of the events in every subfamily equals the product of their probabilities. It is pairwise independent when this equality holds for every pair of distinct events. Event indicators provide a direct link: a collection of events is mutually independent exactly when their indicator random variables are mutually independent.

Factorization for Finite-Valued Variables

When the variables take values in finite sets, independence can be checked through their joint probability mass function. Write $p_i(x)=P(X_i=x)$, and let $p(x_1,\ldots,x_m)=P(X_1=x_1,\ldots,X_m=x_m)$.

Theorem: For finite-valued $X_1,\ldots,X_m$, mutual independence is equivalent to the cell-by-cell factorization $$ p(x_1,\ldots,x_m)=\prod_{i=1}^{m}p_i(x_i) $$ for every possible value tuple $(x_1,\ldots,x_m)$. Pairwise independence requires the corresponding factorization only for the joint PMF of each pair.

Proof. If the variables are mutually independent, apply the definition with $S=\{1,\ldots,m\}$ and $A_i=\{x_i\}$. This gives the stated factorization for each tuple.

Conversely, suppose the cell-by-cell factorization holds. For any subsets $B_i$ of the finite value set of $X_i$, sum over all value tuples with $x_i\in B_i$:

\begin{aligned} P(X_1\in B_1,\ldots,X_m\in B_m) &=\sum_{x_1\in B_1}\cdots\sum_{x_m\in B_m} \prod_{i=1}^{m}p_i(x_i)\\ &=\prod_{i=1}^{m}\left(\sum_{x_i\in B_i}p_i(x_i)\right)\\ &=\prod_{i=1}^{m}P(X_i\in B_i). \end{aligned}

The second equality follows by distributing the finite sums over the product. The same argument applies to any subcollection, so the variables are mutually independent. For a pair, the identical argument shows that its joint PMF factors exactly when the pair is independent. $\square$

This result gives a practical distinction: to establish mutual independence in a finite-valued model, a product factorization for all joint cells suffices. Factoring each two-variable marginal PMF establishes pairwise independence, but does not supply the required factorization for cells of the full joint distribution.

Worked Example: Checking Mutual Independence of Three Coin Tosses

Let $X_1,X_2,X_3$ indicate whether three independent, fair coin tosses show heads, with 1 for heads and 0 for tails. For any $x_1,x_2,x_3\in\{0,1\}$, each specified sequence of outcomes has probability $1/8$. Each marginal assigns probability $1/2$ to either value, so

P(X_1=x_1,X_2=x_2,X_3=x_3)=\frac18 =\frac12\cdot\frac12\cdot\frac12 =\prod_{i=1}^{3}P(X_i=x_i).

This holds for all eight value tuples. By the finite-valued factorization theorem, the three variables are mutually independent. In particular, every pair factors, and the triple factors too. Checking just one cell would not be enough to establish this conclusion; the factorization must hold for every possible tuple.

Worked Example: A Pair of Variables That Fails the Pairwise Check

Let $X$ be a fair Bernoulli random variable, set $Y=X$, and let $Z$ be an independent fair Bernoulli random variable. The marginal probabilities satisfy $P(X=1)=P(Y=1)=P(Z=1)=1/2$.

The pair $X,Y$ is not independent: because $Y=X$, $$ P(X=1,Y=1)=\frac12,\qquad P(X=1)P(Y=1)=\frac14. $$ The pair $X,Z$ is independent by the assumption on $Z$. The pair $Y,Z$ is also independent: $Y=X$, and $Z$ is independent of $X$. Thus, the family is not pairwise independent, because one of its pairs fails the required factorization.

This example illustrates the first step in a family check: test every pair. Passing only some pairwise checks is not evidence that the family is pairwise independent, let alone mutually independent.

Pairwise Independence Does Not Imply Mutual Independence

A family can pass every pairwise check and still fail a factorization involving three or more variables. The following continuous example makes the missing condition visible through explicit probabilities. It uses the fractional part $\{t\}=t-\lfloor t\rfloor$, which lies in $[0,1)$.

Worked Example: Three Pairwise Independent Uniform Variables That Are Not Mutually Independent

Let $U$ and $V$ be independent uniform random variables on $[0,1)$, and define $W=\{U+V\}$. First, $U$ and $V$ are independent by assumption. To check the pair $U,W$, fix $u\in[0,1)$. The map $v\mapsto\{u+v\}$ rotates the unit interval: it translates the part before the wraparound point and translates the part after it, preserving lengths. Therefore, for every Borel set $C\subseteq[0,1)$, $$ \lambda\bigl(\{v\in[0,1):\{u+v\}\in C\}\bigr)=\lambda(C), $$ where $\lambda$ denotes interval length. Integrating over $u$ in any Borel set $A\subseteq[0,1)$ gives $$ P(U\in A,W\in C) =\int_A \lambda(C)\,du =\lambda(A)\lambda(C) =P(U\in A)P(W\in C). $$ So $U$ and $W$ are independent. By the same argument with the roles of $U$ and $V$ exchanged, $V$ and $W$ are independent. All three variables are pairwise independent.

Now choose $A=B=C=[0,1/4)$. If $U,V\in[0,1/4)$, then $U+V<1/2$, so $W=U+V$ without wraparound. The event that all three variables fall in $[0,1/4)$ is the triangle $u+v<1/4$ inside the square $0\leq u,v<1/4$. Its area is $$ P(U<1/4,V<1/4,W<1/4)=\frac12\left(\frac14\right)^2=\frac1{32}. $$ If the three variables were mutually independent, this probability would instead be $$ P(U<1/4)P(V<1/4)P(W<1/4) =\frac14\cdot\frac14\cdot\frac14 =\frac1{64}. $$ Since $1/32\ne 1/64$, the triple factorization fails. The variables are pairwise independent but not mutually independent.

A Common Mistake and a Reliable Check

Common mistake: “Every pair in the family is independent, so the family is independent.” This reasoning checks only the conditions for subsets of size two. Mutual independence also requires the factorization for larger subcollections, as the uniform-variable example demonstrates.

For a family of $m$ variables, a disciplined check separates the two claims:

Check pairwise independence.
For each distinct pair $i,j$, verify the factorization for all Borel sets, or for all value pairs when the variables are finite-valued.

Check higher-order requirements.
For mutual independence, verify factorization for every subcollection of size at least three as well. For finite-valued variables, checking the full joint PMF factorization cell by cell is sufficient.

State the conclusion at the strength actually established.
If only pairs were checked, conclude pairwise independence, not mutual independence.

Pairwise independence is a useful property in its own right, but it is a weaker one. Keeping the distinction explicit prevents a pairwise calculation from being used to justify a joint factorization that has not been established.

Check Your Understanding

Use the definitions to answer each question.

For three random variables, which probability factorizations must hold for mutual independence beyond the three pairwise checks?
For finite-valued variables, why does factorization of every full joint-PMF cell imply factorization for arbitrary sets of values?
In the fractional-part example, what is the probability that all three variables are less than $1/4$, and what would it be under mutual independence?
Can a family be mutually independent if one pair is dependent? Explain using the definitions.
If every pair in a family passes an independence check, what conclusion is justified without further checks?