\problemdata                                                             
        {probability-conditional-4} % latex-friendly label for the prob.
        {Independence of events} % The topic(s) of the problem content.  
        {?} % Source (if known)                                          
        {S02 PS9} % Usage (list ones you are aware of). 
        {Tina Wang, S02} % Last revision info (author, date).   
                                                                         
                                                                         
\begin{problem}           
For any event, $A$, let $A^{(1)} \eqdef A$ and $A^{(-1)} \eqdef \bar{A}$.  Prove that events $A,B,C$ are mutually independent iff 

\begin{equation}
\pr{A^{(x)}}\pr{B^{(y)}}\pr{C^{(z)}} = \pr{A^{(x)} \cap B^{(y)} \cap C^{(z)}}
\end{equation}

for all $x,y,z \in {1,-1}$.

\emph{Note:} For this problem, do not use Theorem 12.3 of the Course
Notes without proof.  Feel free to prove the theorem and then apply
it.

\solution{Since we are asked to show an if and only if, we must prove this
in both directions.

First, we will show that if events $A,B,C$ are mutually independent, then
the above equation holds.  Let's consider all the possible combinations
for $(x,y,z)$ (there are 8 total).

\emph{(1,1,1)}:  This follows directly since $A,B,C$ are mutually independent.

\emph{(1,1,-1)}:  This can be shown as follows:
\begin{align*}
\pr{A^{(1)} \cap B^{(1)} \cap C^{(-1)}} 
&= \pr{A \cap B \cap \bar{C}} &
        \text{definition of $A^{(1)}$ and $A^{(-1)}$}\\
&= \pr{A \cap B} - \pr{A \cap B \cap C} &
        \text{see below}\\
&= \pr{A}\pr{B} - \pr{A}\pr{B}\pr{C} &
        \text{$A,B,C$ are mutually independent}\\
&= \pr{A}\pr{B}(1-\pr{C}) & \\
&= \pr{A}\pr{B}\pr{\bar{C}} &
        \text{complement rule}
\end{align*}
The event we are interested in is when events A and B happen, but C does
not.  This is the same as the event where A and B happen, but then we need
to subtract out the case when A, B, and C happen.  This will leave us with
the event that A and B happen, but C does not.  This can be easily
verified with a Venn Diagram.

\emph{(1,-1,1)}: similar to (1,1,-1) case.

\emph{(1,-1,-1)}: This can be shown as follows:
\begin{align*}
\lefteqn{\pr{A^{(1)} \cap B^{(-1)} \cap C^{(-1)}}}\\
&= \pr{A \cap \bar{B} \cap \bar{C}} &
        \text{definition of $A^{(1)}$ and $A^{(-1)}$}\\
&= \pr{A} - \pr{A \cap (B \cup C)} &
        \text{see below} \\
&= \pr{A} - (\pr{A \cap B} - \pr{A \cap C} + \pr{A \cap B \cap C}) & 
        \text{inclusion-exclusion} \\
&= \pr{A} - \pr{A}\pr{B} - \pr{A}\pr{C}\\
& \quad + \pr{A}\pr{B}\pr{C} &
        \text{$A,B,C$ are mutually independent}\\
&= \pr{A}(1-\pr{B}-\pr{C}+ \pr{B}\pr{C} &
        \text{algebra} \\
&= \pr{A}(1-\pr{B})(1-\pr{C})&
        \text{algebra} \\
&= \pr{A}\pr{\bar{B}}\pr{\bar{C}} &
        \text{complement rule}
\end{align*}
The event we are interested in is when A happens, but B and C do not.  We
can rewrite this as the event A but we need to subtract out the events
that also also in B or C.  To get this, we can use inclusion exclusion.
This is also easily verifiable via Venn Diagram.

\emph{(-1,1,1)}: similar to (1,1,-1) case. 

\emph{(-1,1,-1)}: similar to (1,-1,-1) case.

\emph{(-1,-1,1)}: similar to (1,-1,-1) case.

\emph{(-1,-1,-1)}: This can be shown as follows:
\[\begin{array}{rclr}
\lefteqn{\pr{A^{(-1)} \cap B^{(-1)} \cap C^{(-1)}} }\\
&=& \pr{\bar{A} \cap \bar{B} \cap \bar{C}} & 
        \text{definition of $A^{(\pm 1)}$}\\
&=& \pr{\bar{A \cup B \cup C}} & 
        \text{DeMorgan's} \\
&=& 1-\pr{A \cup B \cup C} &
        \text{complement rule} \\
&=& 1-(\pr{A}+\pr{B}+\pr{C} \\
        && - \pr{A \cap B}-\pr{A \cap C} \\
        && -\pr{B \cap C}+ \pr{A \cap B \cap C}) & 
        \text{inclusion exclusion} \\
&=& 1-\pr{A}-\pr{B}-\pr{C}\\
        && + \pr{A}\pr{B}+\pr{A}\pr{C} \\
        && +\pr{B}\pr{C} - \pr{A}\pr{B}\pr{C}) & 
        \text{mutual independence} \\
&=& (1-\pr{A})(1-\pr{B}-\pr{C} \\
        && +\pr{A}\pr{B}) &
        \text{algebra} \\
&=& (1-\pr{A})(1-\pr{B})(1-\pr{C}) &
        \text{algebra} \\
&=& \pr{\bar{A}}\pr{\bar{B}}\pr{\bar{C}} &
        \text{complement rule}
\end{array}
\]

Now, for the other direction, we must show that if we assume the equation
above is true for all $x,y,z \in {-1,1}$, then $A,B,C$ are mutally
independent.  The following must be shown for mutual independence:
\begin{eqnarray*}
\pr{A_1 \cap A_2} & = & \pr{A_1} \cdot \pr{A_2} \\
\pr{A_1 \cap A_3} & = & \pr{A_1} \cdot \pr{A_3} \\
\pr{A_2 \cap A_3} & = & \pr{A_2} \cdot \pr{A_3} \\
\pr{A_1 \cap A_2 \cap A_3} & = & \pr{A_1} \cdot \pr{A_2} \cdot \pr{A_3}
\end{eqnarray*}

Let's take $\pr{A_1 \cap A_2} = \pr{A_1} \cdot \pr{A_2}$.  We will rewrite the left side as some combination of intersections of events $A,B,C$ and their complements.   We can show that this is true as follows:

\begin{align*}
\pr{A \cap B} 
&= \pr{A \cap B \cap C} + \pr{A \cap B \cap \bar{C}} &
        \text{law of total probability} \\
&= \pr{A}\pr{B}\pr{C} +\pr{A}\pr{B}\pr{\bar{C}} &
        \text{(x,y,z)=(1,1,1) and (1,1,-1)} \\
&= \pr{A}\pr{B}(\pr{C}+\pr{\bar{C}}) &
        \text{algebra} \\
&= \pr{A}\pr{B} 
\end{align*}

The next two can be proven in symmetric ways.  What is left to prove is
the last statement.  This is obvious if we plug in $(x,y,z)=(1,1,1)$ into
the equation.

Now, while the above approach works, it is rather long and tedious.  Here
is a lemma that makes everything follow quite nicely:

\begin{lemma*}
$A_1,A_2,A_3,\dots,A_n$ are mutually independent iff
$\bar{A_1},A_2,A_3,\dots,A_n$ are mutually independent.
\end{lemma*}

\begin{proof}
$A_1,A_2,A_3,\dots,A_n$ are mutually independent

iff $A_1$ is independent of $\lgintersect_{j \in J} A_j$ for all $J
\subseteq \set{2,\dots,n}$ (by first def of Independence in Notes 9)

iff $\bar{A_1}$ is independent of $\lgintersect_{j \in J} A_j$
         for all $J \subseteq \set{2,\dots,n}$
        (from Class Problems 9-3, Problem 2 and Notes 9, Lemma 10.4)

iff $\bar{A_1},A_2,A_3,\dots,A_n$ are mutually independent
         (by first def of Independence in Notes 9).
\end{proof}

Now the whole Theorem follows straightforwardly by applying this Lemma
with $A_1^{(x_1)},\dots,A_n^{(x_n)}$ substituted for
$A_1,A_2,A_3,\dots,A_n$.
}
\end{problem}
                                 