\problemdata      
        {?} % latex-friendly label for the prob.
        {Probability-Intro} % The topic(s) of the problem content.
        {?} % Source (if known)
        {F01, PS10} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).

\begin{problem}
\begin{theorem*}[Relative Probability]
Let $\mathcal{P} \eqdef (S, \pr{})$ be a probability space and $C$ be an
event with nonzero probability.  Then
\[
\mathcal{P}_C \eqdef (S, \prsub{C}{})
\]
is also a probability space, where for any event $A$,
\begin{equation}\label{defprc}
\prsub{C}{A} \eqdef \prcond{A}{C}.
\end{equation}
\end{theorem*}
We call $\prsub{C}{}$ the \emph{probability relative to $C$}, and
$\mathcal{P}_C$ the \emph{probability space relative to $C$}.

\bparts 

\ppart Prove the Relative Probability Theorem. (Hint: Review the
definition of a probability space in
\hyperref{http://theory.lcs.mit.edu/classes/6.042/fall01/lectures/l10.pdf}{l10}{probsp}{Notes 10, Definition 4.2}.

\solution{
A probability space is defined to be a sample space $S$ and a probability function $\pr{}$ which satisfies two properties:
\begin{enumerate}
\item $\pr{S} =1$, and

\item if $A$ is the disjoint union of a sequence of events
$A_0,A_1,\dots$, then $A$ is an event, $\sum_{i\in\naturals} \pr{A_i}$
converges and
\begin{equation}\tag{Sum Rule}
\pr{A} = \pr{\lgunion_{i\in\naturals}A_i} = \sum_{i\in\naturals} \pr{A_i}.
\end{equation}
\end{enumerate}

First, we see that the sample space for the new probability space has not 
changed.  We must then just show that the new probability function $\prsub{C}{}$
still satisfies the two conditions.

\begin{enumerate}
\item 
To show this:
\begin{align*}
\prsub{C}{S} &= \prcond{S}{C} & \text{definition~(\ref{defprc}) of $\prsub{C}{}$}\\
             &= \frac{\pr{S \cap C}}{\pr{C}} & \text{definition of $\prcond{S}{C}$}\\
             &= \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{C}{\lgunion_{i\in\naturals}A_i}
    &= \pr{\lgunion_{i\in\naturals}A_i \biggm| C} &
          \text{definition~(\ref{defprc}) of $\prsub{C}{}$}\\\\
    &= \frac{\pr{(\lgunion_{i\in\naturals}A_i) \cap C}}{\pr{C}} &
           \text{definition of $\prcond{\cdot}{C}$}\\
    &= \frac{\pr{\lgunion_{i\in\naturals}(A_i \cap C)}}{\pr{C}}&
             \text{distributive law}\\
    &= \sum_{i \in \naturals} \frac{\pr{A_i \cap C}}{\pr{C}} & \text{Sum Rule}\\
    &=  \sum_{i \in \naturals} {\prcond{A_i}{C}} &
           \text{definition of $\prcond{\cdot}{C}$}\\\\
    &= \sum_{i \in \naturals} {\prsub{C}{A_i}}&
          \text{definition~(\ref{defprc}) of $\prsub{C}{}$}
\end{align*}

\end{enumerate}
}

\ppart Explain why the two-set Inclusion-Exclusion Law for conditional
probabilities,
\[
\prcond{A\union B}{C} = \prcond{A}{C} + \prcond{B}{C} - \prcond{A
\intersect B}{C},
\]
now follows immediately from ordinary two-set Inclusion-Exclusion
(without having to mimic the ordinary proof using conditional
probabilities).

\solution{
Now that we have this nifty theorem about relative probability, we no
longer have to re-derive formulas and laws with conditional probabilities and
treat them differently. 
The new probability space we defined was different from the original in that
the probability function now represents probabilities that are conditioned on
event $C$. These are exactly the conditional probabilities used in the Inclusion-Exclusion Law for conditional probabilities.  Since in part a, we have shown that the original sample space together with the probability function representing these probabilities form a valid probability space, we are able to extend
any laws for probability spaces to conditional probabilities. 
}
\eparts      
\end{problem}
