Independence of Events

Independence as a Product Rule

In a probability space, two events are independent when learning that one occurred does not change the probability assigned to the other, provided the conditioning event has positive probability. The defining product rule also makes sense when one of the events has probability zero, so it handles degenerate cases without a conditional probability.

Throughout, let \((\Omega,\mathcal F,\mathbb P)\) be a probability space: \(\mathcal F\) is a sigma-algebra of subsets of \(\Omega\), \(\mathbb P\) is a probability measure on \(\mathcal F\), and \(\mathbb P(\Omega)=1\). Thus the measure is finite. We do not assume that the probability space is complete. All events below belong to \(\mathcal F\), and all probability equalities are exact.

For two events \(A,B\), if \(\mathbb P(B)>0\), the definition is equivalent to \(\mathbb P(A\mid B)=\mathbb P(A)\), since \(\mathbb P(A\mid B)=\mathbb P(A\cap B)/\mathbb P(B)\). The product definition does not require \(\mathbb P(B)>0\): if \(\mathbb P(B)=0\), then \(\mathbb P(A\cap B)=0=\mathbb P(A)\mathbb P(B)\), so \(A\) and \(B\) are independent. Also, \(\varnothing\) and \(\Omega\) are each independent of every event.

Complements and Unions

Independence is preserved when either event is replaced by its complement. The probability of a union of two independent events then follows from the addition rule for probabilities. These conclusions depend on the product definition, not on an assumption that the events are disjoint.

Proof. Write \(p=\mathbb P(A)\) and \(q=\mathbb P(B)\). Since \(B=(A\cap B)\,\dot\cup\,(A^c\cap B)\), additivity on disjoint events gives \(\mathbb P(A^c\cap B)=q-\mathbb P(A\cap B)=q-pq=(1-p)q=\mathbb P(A^c)\mathbb P(B)\). Likewise, using \(A=(A\cap B)\,\dot\cup\,(A\cap B^c)\), we obtain \(\mathbb P(A\cap B^c)=p-pq=p(1-q)=\mathbb P(A)\mathbb P(B^c)\). Finally, \(\mathbb P(A^c\cap B^c)=1-\mathbb P(A\cup B)\). The addition rule gives \(\mathbb P(A\cup B)=p+q-pq\), so \(\mathbb P(A^c\cap B^c)=1-p-q+pq=(1-p)(1-q)=\mathbb P(A^c)\mathbb P(B^c)\). The same addition rule proves the stated union formula. \(\square\)

There is a corresponding formula for a finite mutually independent family. For events \(A_1,\ldots,A_n\), their complements have intersection \(\bigcap_{i=1}^n A_i^c=(\bigcup_{i=1}^n A_i)^c\). Mutual independence gives the probability of the left side as a product; taking its complement gives the union probability.

Proof. By the definition of mutual independence applied to the full index set, \(\mathbb P(\bigcap_{i=1}^n A_i^c)\) is not immediately given, since complements appear. The complement proposition extends to the family: for any choices of indices \(J,K\) that are disjoint, with \(A_j^c\) selected for \(j\in J\) and \(A_k\) selected for \(k\in K\), expand the indicators of the complements to obtain \(\mathbb P((\bigcap_{j\in J}A_j^c)\cap(\bigcap_{k\in K}A_k))=\sum_{L\subseteq J}(-1)^{|L|}\mathbb P(\bigcap_{i\in L\cup K}A_i)\). Each probability on the right is \(\prod_{i\in L\cup K}\mathbb P(A_i)\), by mutual independence. Factoring the finite sum gives \(\prod_{k\in K}\mathbb P(A_k)\prod_{j\in J}(1-\mathbb P(A_j))\). In particular, take \(J=\{1,\ldots,n\}\) and \(K=\varnothing\). Then \(\mathbb P(\bigcap_i A_i^c)=\prod_i(1-\mathbb P(A_i))\). Taking the complement proves the formula. \(\square\)

The same finite expansion shows that unions formed from disjoint groups of a mutually independent family are independent. In detail, let \(I,J\subseteq\{1,\ldots,n\}\) be disjoint, and put \(U=\bigcup_{i\in I}A_i\), \(V=\bigcup_{j\in J}A_j\). Inclusion-exclusion expands \(\mathbb P(U\cap V)\) as a sum over nonempty \(S\subseteq I\) and nonempty \(T\subseteq J\), with term \((-1)^{|S|+|T|+2}\mathbb P(\bigcap_{i\in S\cup T}A_i)\). Mutual independence factors that probability into \(\prod_{i\in S}\mathbb P(A_i)\prod_{j\in T}\mathbb P(A_j)\). The resulting double sum factors into the inclusion-exclusion expression for \(\mathbb P(U)\) times that for \(\mathbb P(V)\). Hence \(\mathbb P(U\cap V)=\mathbb P(U)\mathbb P(V)\). Empty groups give an empty union \(\varnothing\), and the equality holds there as well.

Testing Independence in Finite Models

In a finite probability model, each outcome has an assigned probability, the probabilities sum to \(1\), and the probability of an event is the sum of the probabilities of its outcomes. To test two events, compute their probabilities and the probability of their intersection, then compare the intersection probability with the product. A matching product is the criterion; visual separation or a plausible story is not.

A Common Mistake: Confusing Independence with Disjointness

Disjoint events cannot usually be independent. For a concrete check, take one fair die roll with \(\Omega=\{1,2,3,4,5,6\}\), uniform probability, and \(\mathcal F\) the power set. Let \(A=\{1\}\) and \(B=\{2\}\). These events are disjoint, so \(\mathbb P(A\cap B)=0\). But \(\mathbb P(A)\mathbb P(B)=(1/6)(1/6)=1/36\), not \(0\). Thus they are not independent. The incorrect calculation is to infer independence from \(A\cap B=\varnothing\); the correction is to test the defining equation directly.

More generally, if \(A\cap B=\varnothing\) and \(A,B\) are independent, then \(0=\mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B)\). Therefore at least one of the events must have probability zero. Disjointness and independence express different properties: disjointness rules out joint occurrence, whereas independence specifies the probability of joint occurrence as a product.