\problemdata                                                             
        {Index variables} % latex-friendly label for the prob.
        {Proving Mutual Independence} % The topic(s) of the problem content.  
        {?} % Source (if known)                                          
        {S01 Tutorial 12, 5} % Usage (list ones you are aware of). 
        {Tina Wang, S02} % Last revision info (author, date).   
                                                                         
                                                                         
\begin{problem}           
Let $A,B,C$ be events over some probability space, and let
$I_A,I_B,I_C$ be the corresponding index variables, e.g., the event
$\set{I_A = 1}$ is the same as the event $A$ and $\set{I_A = 0}$ is the
same as $\bar{A}$.  Prove that $A,B,C$ are mutually independent iff the
random variables $I_A,I_B,I_C$ are mutually independent.

\solution{
($\Longrightarrow$):  
\begin{eqnarray*}
\lefteqn{\prob{I_A = 1 \wedge I_B= 0 \wedge I_C = 1}}\\
 & = & \prob{A \intersect \bar{B} \intersect C}\\
& = & \prob{(A \intersect C) - B}\\
& = & \prob{A \intersect C} - \prob{A \intersect C \intersect B}\\
& = & \prob{A} \cdot \prob{C} - \prob{A} \cdot \prob{C} \cdot \prob{B}
    \hspace{0.5in}\mbox{[because $A,B,C$ independent]}\\
& = & \prob{A} \cdot \prob{C} (1 -  \prob{B})\\
& = & \prob{A} \cdot \prob{C} \prob{\bar{B}}\\
& = & \prob{I_A = 1} \cdot \prob{I_C = 1} \prob{I_B = 0}\\
& = & \prob{I_A = 1} \cdot \prob{I_B = 0} \cdot \prob{I_C = 1}
\end{eqnarray*}
and similarly for any other three binary values in place of 101.

($\Longleftarrow$):  
\begin{eqnarray*}
\prob{A \intersect B \intersect C} & = & 
        \prob{I_A = 1 \wedge I_B = 1 \wedge I_C = 1}\\
& = & \prob{I_A = 1} \cdot \prob{I_B = 1 } \cdot \prob{I_C = 1}\\
& = & \prob{A} \cdot \prob{B} \cdot \prob{C}
\end{eqnarray*}
Also,
\begin{eqnarray*}
\prob{A \intersect B} & = & \prob{A \intersect B \intersect C}
                       + \prob{A \intersect B \intersect \bar{C}}\\
& = & \prob{I_A = 1 \wedge I_B = 1 \wedge I_C = 1} + \prob{I_A = 1 \wedge I_B = 1 \wedge I_C = 0}\\
& = & \prob{I_A = 1} \cdot \prob{I_B = 1} \cdot \prob{I_C = 1}
       + \prob{I_A = 1} \cdot \prob{I_B = 1} \cdot \prob{I_C = 0}\\
& = & \prob{A} \cdot \prob{B} \cdot \prob{C} + \prob{A} \cdot \prob{B}
                       \cdot \prob{\bar{C}}\\
& = & \prob{A} \cdot \prob{B} (\prob{C} + \prob{\bar{C}})\\
& = & \prob{A} \cdot \prob{B}
\end{eqnarray*}
and similarly for ${B \intersect C}$ and ${A \intersect C}$.
}

\end{problem}
                                 