Exercise 2.25 (Independence of events)

Question

\begin{enumerate}
\item Show that two events $E,F$ are independent if and only if ${\bf P}(E \wedge F) = {\bf P}(E) {\bf P}(F)$.
\item If $E,F,G$ are events, show that the condition ${\bf P}(E \wedge F \wedge G) = {\bf P}(E) {\bf P}(F) {\bf P}(G)$ is necessary, but not sufficient, to ensure that $E,F,G$ are jointly independent.
\item Given an example of three events $E,F,G$ that are pairwise independent, but not jointly independent.
\end{enumerate}

Elu Thingol · Accepted Answer

\paragraph{(i) Mutually independent events.} 
This part is simply a matter of dissecting the definition and applying the identity $1_{E \cap F} = 1_{E} \cdot 1_{F}$ to it:
\begin{equation}
{\bf P}(E \wedge F) = {\bf P}(1_{E \wedge F} = 1) = {\bf P}\left(1_{E} \cdot 1_{F} = 1ight) = {\bf P}(1_{E} = 1 \wedge 1_{F} = 1) \stackrel{!}{=} {\bf P}(1_{E} = 1) \cdot {\bf P}(1_{F}=1) = {\bf P}(E){\bf P}(F).
\end{equation}

\paragraph{(ii) Jointly independent events.} 
\begin{itemize}
\item[$\impliedby$] Suppose that $E,F,G$ are jointly independent. By the previous part, 
\begin{equation}
{\bf P}([E \wedge F] \wedge G) = {\bf P}(E \wedge F) \cdot {\bf P}(F)  = {\bf P}(E) \cdot {\bf P}(F) \cdot {\bf P}(G) = {\bf P}(E) \cdot {\bf P}(F \wedge G) = {\bf P}(E \wedge [F \wedge G]).
\end{equation}

\item[$
ot\implies$] Consider the finite probability space $\Omega = \{1,2,3,4\}$ equipped with the $\sigma$-algebra $\mathcal{F} = 2^\Omega$. Let the probability measure be uniform, i.e., it is defined by the probabiliyt on atoms $P(\{i\}) = \frac{1}{4}$ for $i \in \Omega$. Define three events:
\begin{align*}
A &= \{1,2\}\
B &= \{1,3\}\
C &= \{2,3\}.
\end{align*}
Then ${\bf P}(A \wedge B) = 1/4$, ${\bf P}(A \wedge C) = 1/4$ and ${\bf P}(B \wedge C) = 1/4$ yet ${\bf P}(A \wedge B \wedge C) = 0$. In other words, $A$, $B$, and $C$ are pairwise independent despite the fact that they are not jointly independent.
\end{itemize}

Exercise 2.25 (Independence of events)

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Exercise 2.25 (Independence of events)

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