4  Independence

Definition 4.1 ($\mathbf{\text{Independent events}}$)
Given a probability space $(\mathrm{\Omega },\mathcal{B},\mathbb{P})$, two events $A,B\in \mathcal{B}$ are said to be independent if: $$\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$$ A finite sequence of events, namely $A_1,A_2,\dots ,A_n\in \mathcal{B}$, is said to be independent, if for all $2\le j\le n$ and $1\le k_1\le k_2\le \cdots \le k_j\le n$ we have: $$\mathbb{P}(A_{k_1}\cap A_{k_2}\cap \cdots \cap A_{k_n})=\prod _{j=1}^n\mathbb{P}(A_{k_j})$$

Events not pairwise independent.

Consider the probability space $(\mathrm{\Omega },\mathcal{P}(\mathrm{\Omega }),\mathbb{P})$, where $\mathrm{\Omega }=\{1,2,3,4,5,6\}$ and for every ${\omega }_i\in \mathrm{\Omega }$ we have a constant probability $\mathbb{P}({\omega }_i)=\frac{1}{6}\mathrm{\forall }i$. Consider the events $A_1=\{1,2,3,4\}$ and $A_2=A_3=\{4,5,6\}$, are these events independent? Note that the events have probabilities $\mathbb{P}(A_1)=\frac{2}{3}$, $\mathbb{P}(A_2)=\mathbb{P}(A_3)=\frac{1}{2}$. Consider all the events $A_1,A_2,A_3$, then the intersection of those sets gives $[A_1\cap A_2\cap A_3]=\{4\}$ that has probability $\mathbb{P}(\{4\})=\frac{1}{6}$. Then we can compute the product of the probabilities of the single events: $$\frac{1}{2}=\mathbb{P}([A_1\cap A_2\cap A_3])=\mathbb{P}(A_1)\mathbb{P}(A_2)\mathbb{P}(A_3)=\frac{2}{3}\frac{1}{2}\frac{1}{2}=\frac{1}{2}$$ Hence the events $A_1,A_2,A_3$ are $\mathbf{\text{pairwise independent}}$. Consider now only the events $A_2,A_3$, the probability of the joint set, namely $[A_2\cap A_3]=\{4,5,6\}$, is $\mathbb{P}(\{4,5,6\})=\frac{1}{2}$. However the product of the probabilities of the single events gives a different result: $$\frac{1}{2}=\mathbb{P}([A_2\cap A_3])\ne \mathbb{P}(A_2)\mathbb{P}(A_3)=\frac{1}{2}\frac{1}{2}=\frac{1}{4}$$ Hence the events $A_2,A_3$ are $\mathbf{\text{NOT pairwise independent}}$.

Proposition 4.1 ($\mathbf{\text{Independence and complementation}}$)
If two events $A$ and $B$ are independent, then also are $A$ and $B^c$, $B$ and $A^c$, $A^c$ and $B^c$ are independent.

Independence and complementation

Proof. If two events $A$ and $B$ are independent, then also are $A$ and $B^c$, $B$ and $A^c$, $A^c$ and $B^c$. In order to prove that $A$ and $B^c$ are independent let’s write the event $A$ as union of disjoint events (Equation 1.9). Then since $A$ and $B$ are assumed to be independent: $$\begin{array}{rl}\mathbb{P}(A)& =\mathbb{P}(A\cap B)+\mathbb{P}(A\cap B^c)=\\ & =\mathbb{P}(A)\mathbb{P}(B)+\mathbb{P}(A\cap B^c)\end{array}$$ Recovering $\mathbb{P}(A\cap B^c)$ one obtain: $$\begin{array}{rl}\mathbb{P}(A\cap B^c)& =\mathbb{P}(A)-\mathbb{P}(A)\mathbb{P}(B)=\\ & =\mathbb{P}(A)(1-\mathbb{P}(B))=\\ & =\mathbb{P}(A)\mathbb{P}(B^c)\end{array}$$ The same follows for $A^c$ and $B$. Now let’s consider the case of $A^c$ and $B^c$. Using the same trick done previously $A^c=[A^c\cap B]\bigsqcup [A^c\cap B^c]$. Since we have already proven that $B^c$ and $A$ are independent, we can write: $$\begin{array}{rl}\mathbb{P}(A^c)& =\mathbb{P}([A^c\cap B]\bigsqcup [A^c\cap B^c])=\\ & =\mathbb{P}(A^c\cap B)+\mathbb{P}(A^c\cap B^c)=\\ & =\mathbb{P}(A^c)\mathbb{P}(B)+\mathbb{P}(A^c\cap B^c)\end{array}$$ Recovering $\mathbb{P}(A^c\cap B^c)$ one obtain: $$\begin{array}{rl}\mathbb{P}(A^c\cap B^c)& =\mathbb{P}(A^c)-\mathbb{P}(A^c)\mathbb{P}(B)=\\ & =\mathbb{P}(A^c)(1-\mathbb{P}(B))=\\ & =\mathbb{P}(A^c)\mathbb{P}(B^c)\end{array}$$