\problemdata                                                             
        {probability-conditional-3} % latex-friendly label for the prob.
        {Conditional Probability, Sample Spaces} % The topic(s) of the problem content.  
        {?} % Source (if known)                                          
        {S02 PS9; F01 PS10} % Usage (list ones you are aware of). 
        {Tina Wang, S02} % Last revision info (author, date).   
                                                                         
                                                                         
\begin{problem}           
All our probability identities continue to hold when all probabilities are
conditioned on the same event.  For example,
\begin{equation}\tag{Inclusion-Exclusion}
\pr{A \union B} = \pr{A} + \pr{B} - \pr{A \intersect B}
\end{equation}
becomes
\begin{equation}\tag{Conditional Inclusion-Exclusion}
\prcond{A \union B}{C} = \prcond{A}{C} + \prcond{B}{C} - \prcond{A \intersect B}{C}.
\end{equation}
Such identities carry over because for any event $C$, we can define a new
probability measure, $\prsub{}{C}$ on the same sample space by the rule
that
\begin{definition}
\label{defprs}
\[
\prsub{A}{C} \eqdef \prcond{A}{C}.
\]
\end{definition}
Now the conditional-probability version of an identity is just an instance
of the original identity using the new probability measure.  Prove that
for any probability space, $\sspace$, and event $C \subseteq
\sspace$, the function $\prsub{}{C}$ is another probability measure on
$\sspace$.  \hint Review the definition of a probability space in
\hyperref{http://theory.lcs.mit.edu/classes/6.042/spring02/handouts/lectures/ln9.pdf}
{ln9}{probsp}{Notes 9, Definition 5.2}.
\end{problem}

\solution{The sample space for the new probability space has not changed.
We must just show that the new probability function $\prsub{C}{}$
still satisfies the two conditions required of a probability measure.

\begin{enumerate}
\item 
To show this:
\begin{align*}
\prsub{S}{C} &= \prcond{S}{C} & \text{Definition~\ref{defprs} of $\prsub{}{C}$}\\
             &= \frac{\pr{S \cap C}}{\pr{C}} & \text{Definition of $\prcond{C}{S}$}\\
             &= \frac{\pr{C}}{\pr{C}} & \text{since $C \subseteq S$}\\
             &= 1 & \text{since $\pr{C} \neq 0$.}
\end{align*}

\item
The Sum Rule can be verified as follows:
\begin{align*}
\prsub{\lgunion_{i\in\naturals}A_i}{C}
    &= \prcond{\lgunion_{i}A_i}{C} &
          \text{Definition~\ref{defprs} of $\prsub{}{C}$}\\\\
    &= \frac{\pr{(\lgunion_{i}A_i) \cap C}}{\pr{C}} &
           \text{Definition of $\prcond{\cdot}{C}$}\\
    &= \frac{\pr{\lgunion_{i}(A_i \cap C)}}{\pr{C}}&
             \text{distributive law}\\
    &= \sum_{i} \frac{\pr{A_i \cap C}}{\pr{C}} & \text{Sum Rule}\\
    &=  \sum_{i} {\prcond{A_i}{C}} &
           \text{Definition of $\prcond{\cdot}{C}$}\\\\
    &= \sum_{i} {\prsub{A_i}{C}}.&
          \text{Definition~\ref{defprs} of $\prsub{}{C}$}
\end{align*}

\end{enumerate}
}


                                 