Pairwise and Mutual Independence

Two Levels of Independence

Checking that every two events behave independently does not necessarily show that a whole collection does. For three or more events, mutual independence imposes additional conditions on intersections involving three or more events. This distinction matters whenever a problem asks about a collection of events rather than just a pair.

Throughout, $(\Omega,\mathcal{F},\mathbb{P})$ is a probability space: $\mathcal{F}$ is a sigma-algebra of subsets of $\Omega$, and $\mathbb{P}$ is a probability measure with $\mathbb{P}(\Omega)=1$. Events are members of $\mathcal{F}$. All equalities below are equalities of probabilities, not statements about random variables.

Definition: Two events $A,B\in\mathcal{F}$ are independent if $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$. A family $(A_i)_{i\in I}$ of events is pairwise independent if this equality holds for every pair of distinct indices $i,j\in I$. The family is mutually independent if, for every finite nonempty set $J\subseteq I$, $$ \mathbb{P}\left(\bigcap_{j\in J}A_j\right)=\prod_{j\in J}\mathbb{P}(A_j). $$ For a singleton $J=\{j\}$, the condition is an identity. Thus a family with fewer than two events is mutually independent.

The word “finite” in the definition applies even when $I$ is infinite: mutual independence requires the product formula for every finite selection from the family. Pairwise independence checks only selections of size two. For a family of three events, mutual independence therefore requires the three pair conditions as well as the condition for the intersection of all three.

Proposition: In any probability space $(\Omega,\mathcal{F},\mathbb{P})$, every mutually independent family of events is pairwise independent.

Proof. Let $(A_i)_{i\in I}$ be mutually independent, and choose any distinct indices $i,j\in I$. The set $J=\{i,j\}$ is a finite nonempty subset of $I$. Applying the defining condition for mutual independence to this $J$ gives $\mathbb{P}(A_i\cap A_j)=\mathbb{P}(A_i)\mathbb{P}(A_j)$. This is precisely the condition for the chosen pair to be independent. Since the pair was arbitrary, every pair is independent. $\square$

The converse is false. The following example checks every pair explicitly and then shows that the three-event condition fails.

Worked Example: Pairwise Independent Events That Are Not Mutually Independent

Take $\Omega=\{000,011,101,110\}$, let $\mathcal{F}$ be the collection of all subsets of $\Omega$, and assign probability $1/4$ to each outcome. This is a probability space: the four individual probabilities sum to $4\cdot(1/4)=1$. Define $A$, $B$, and $C$ to be the events that the first, second, and third coordinate, respectively, is $1$.

Reading the coordinates in the order written, membership gives $A=\{101,110\}$, $B=\{011,110\}$, and $C=\{011,101\}$. Each event contains two of the four outcomes, so $\mathbb{P}(A)=\mathbb{P}(B)=\mathbb{P}(C)=2/4=1/2$.

For the pair $A,B$, the only outcome whose first and second coordinates are both $1$ is $110$. Thus $A\cap B=\{110\}$, and $\mathbb{P}(A\cap B)=1/4=(1/2)(1/2)=\mathbb{P}(A)\mathbb{P}(B)$. For $A,C$, the only outcome with first and third coordinates both $1$ is $101$. Hence $\mathbb{P}(A\cap C)=1/4=(1/2)(1/2)=\mathbb{P}(A)\mathbb{P}(C)$. For $B,C$, the only outcome with second and third coordinates both $1$ is $011$. Hence $\mathbb{P}(B\cap C)=1/4=(1/2)(1/2)=\mathbb{P}(B)\mathbb{P}(C)$. All three distinct pairs satisfy the independence condition, so the events are pairwise independent.

But no outcome in $\Omega$ has all three coordinates equal to $1$, so $A\cap B\cap C=\varnothing$ and $\mathbb{P}(A\cap B\cap C)=0$. Mutual independence would require $\mathbb{P}(A\cap B\cap C)=\mathbb{P}(A)\mathbb{P}(B)\mathbb{P}(C)=(1/2)(1/2)(1/2)=1/8$. Since $0\ne1/8$, the events are not mutually independent. Check: the listed outcomes each have an even number of coordinates equal to $1$; none is $111$, while each pair intersection contains exactly one listed outcome.

Checking Mutual Independence in a Finite Family

For three events, the definition requires three pair-intersection equations and one triple-intersection equation. More generally, for $n$ events one must check the product formula for each nonempty subcollection, not just for the whole collection. If a family has repeated events at different indices, the definition still applies to distinct indices; repetitions can make some required equations impossible unless the repeated event has probability $0$ or $1$.

Worked Example: Three Mutually Independent Events

Let $\Omega=\{0,1\}^3$, with $\mathcal{F}$ the set of all subsets and each of the eight outcomes assigned probability $1/8$. This is a probability space because $8\cdot(1/8)=1$. Let $A,B,C$ be the events that the first, second, and third coordinate, respectively, is $1$.

Each coordinate is $1$ in four outcomes, so each event has probability $4/8=1/2$. Each specified pair of coordinates is $1$ in two outcomes: for example, $A\cap B=\{110,111\}$, so $\mathbb{P}(A\cap B)=2/8=1/4=(1/2)(1/2)$. The other two pair intersections are $A\cap C=\{101,111\}$ and $B\cap C=\{011,111\}$; each has probability $2/8=1/4$, equal to the product of its two event probabilities.

The triple intersection is $A\cap B\cap C=\{111\}$, which has probability $1/8$. The product of the three individual probabilities is $(1/2)(1/2)(1/2)=1/8$, so the triple condition holds too. The singleton conditions are identities, and these are all the nonempty subcollections of $\{A,B,C\}$. Thus the family is mutually independent. Check: the full list has eight equally likely outcomes, the pair intersections each have two outcomes, and the triple intersection has one; the corresponding probabilities are $1$, $2$, and $1$ outcomes divided by $8$, respectively.

Complements Preserve Mutual Independence

In applications, one often needs conditions involving both events and their complements. For instance, the event that some selected outcomes occur and other selected outcomes do not occur is an intersection of events of the form $A_i$ or $A_i^c$. Mutual independence supplies a product formula for all such intersections as well.

Theorem: Let $(A_i)_{i\in I}$ be a mutually independent family in a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Choose any finite set $J\subseteq I$, and divide it into disjoint sets $S,T$ with $J=S\cup T$. Then $$ \mathbb{P}\left(\bigcap_{i\in S}A_i\cap\bigcap_{j\in T}A_j^c\right) =\prod_{i\in S}\mathbb{P}(A_i)\prod_{j\in T}\bigl(1-\mathbb{P}(A_j)\bigr). $$ An intersection over an empty index set means $\Omega$, and an empty product means $1$.

Proof. If $J=\varnothing$, the left side is $\mathbb{P}(\Omega)=1$, and the right side is the empty product $1$. Now suppose $J\ne\varnothing$. The event on the left is $\left(\bigcap_{i\in S}A_i\right)\cap\left(\bigcap_{j\in T}A_j^c\right)$. For each $j\in T$, write $\chi_{A_j^c}=1-\chi_{A_j}$, where $\chi_E$ denotes the indicator of an event $E$. Distributing the finite product of these differences, or equivalently applying finite inclusion-exclusion, gives the probability identity $$ \mathbb{P}\left(\bigcap_{i\in S}A_i\cap\bigcap_{j\in T}A_j^c\right) =\sum_{L\subseteq T}(-1)^{|L|} \mathbb{P}\left(\bigcap_{i\in S\cup L}A_i\right). $$ For each $L\subseteq T$, the set $S\cup L$ is a finite subset of $I$. If it is nonempty, mutual independence says that its intersection probability is $\prod_{k\in S\cup L}\mathbb{P}(A_k)$; if it is empty, the probability is $\mathbb{P}(\Omega)=1$, which is the same formula under the empty-product convention. Consequently the sum equals $$ \sum_{L\subseteq T}(-1)^{|L|} \prod_{i\in S}\mathbb{P}(A_i)\prod_{j\in L}\mathbb{P}(A_j) = \prod_{i\in S}\mathbb{P}(A_i) \prod_{j\in T}\bigl(1-\mathbb{P}(A_j)\bigr). $$ The last equality is the finite distributive expansion of the product over $T$. This proves the claimed formula, including the cases $S=\varnothing$ or $T=\varnothing$. $\square$

Worked Example: An Intersection Involving Complements

Use the mutually independent events $A,B,C$ on $\{0,1\}^3$ from the preceding example. Consider $A^c\cap B\cap C$. By direct inspection of the coordinates, the only outcome in this event is $011$, so its probability is $1/8$.

The complement formula gives the same value: $\mathbb{P}(A^c\cap B\cap C)=(1-\mathbb{P}(A))\mathbb{P}(B)\mathbb{P}(C)=(1-1/2)(1/2)(1/2)=1/8$. As a second check, the event $A^c\cap B^c\cap C^c$ contains just $000$, so its probability is $1/8$; the formula gives $(1-1/2)(1-1/2)(1-1/2)=1/8$. Both calculations agree with the count of outcomes in the eight-point sample space.

A Common Mistake: Stopping After the Pair Checks

A common error is to verify the pair-intersection equations and then conclude that a three-event family is mutually independent. In the parity example, the pair calculations all give $1/4=(1/2)(1/2)$. But using only those calculations to predict the triple intersection would give $\mathbb{P}(A)\mathbb{P}(B)\mathbb{P}(C)=(1/2)^3=1/8$, whereas the actual triple-intersection probability is $0$. The correction is to check every nonempty finite subcollection required by the definition. For three events, that means all three pairs and the triple, in addition to the automatic singleton identities.

Pairwise independence is therefore a weaker condition, not an approximation to mutual independence. Conversely, when mutual independence is known, the complement theorem gives the product rule for any finite pattern of occurrences and non-occurrences. In either direction, the relevant calculation is about probabilities of intersections, and it must be made at the size of subcollection the claim requires.

Check Your Understanding

For each question, distinguish the pair conditions from the conditions involving larger subcollections.

State the definition of pairwise independence for a family $(A_i)_{i\in I}$ of events.
For three events, list all the probability equations that must hold for mutual independence.
In the parity example, compute $\mathbb{P}(A\cap B\cap C)$ and compare it with $\mathbb{P}(A)\mathbb{P}(B)\mathbb{P}(C)$.
Explain why checking only pair intersections cannot establish mutual independence of three events.
If $A$ and $B$ are mutually independent, what does the complement formula give for $\mathbb{P}(A^c\cap B)$?

Pairwise and Mutual Independence

What You'll Learn