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\begin{document}

{\bf Tutorial 10 solutions}

\date{}

\begin{enumerate}
\item Let $A_1, \ldots , A_n$ be a collection of disjoint sets such that their 
union is equal to the sample space.  Prove by induction that for any event $B$
$$Pr(B) = Pr(B \cap A_1)+ \ldots +P(B \cap A_n)$$
And note the corollary:
$$Pr(B) = Pr(B \cap A) + Pr(B \cap \bar{A})$$


By induction on $n$.

I.H. $P(n): Pr(B) = Pr(B \cap A_1)+ \ldots +P(B \cap A_n)$

Base: $n=1:  Pr(B) = Pr(B \cap A_1)$ since $A_1$ is the sample space
and hence $B \cap A_1 = B$.

Ind. Step: Assume $P(n-1)$, to prove $P(n)$,
consider the collection of disjoint events $A_1, A_2,\ldots,A_n$ whose
union is the sample space. 
We want to show that
$$Pr(B) = Pr(B \cap A_1)+ \ldots +P(B \cap A_n)$$
For this define $A$ to be the event $A_1 \cup A_2$.
Then apply $P(n)$ to the collection $A,A_3,\ldots,A_n$ 
to get that 
$$Pr(B) = Pr(B \cap A)+ Pr(B \cap A_3) + \ldots +P(B \cap A_n)$$
But $$Pr(B \cap A) = Pr(B \cap (A_1 \cup A_2))$$ 
$$= Pr((B \cap A_1) \cup (B \cap A_2))$$
$$ = Pr(B\cap A_1) + Pr(B \cap A_2) - Pr((B \cap A_1) \cap (B\cap A_2))$$
$$ = Pr(B\cap A_1) + Pr(B \cap A_2)$$
since $A_1$ and $A_2$ are disjoint.
From this, $P(n)$ follows. 

To see the corollary, use $P(2)$ and notice that $A_2 = \bar{A_1}$
since they cover the sample space.

\item Use part 1 to show that if $A_1, \ldots ,A_n$ is a disjoint collection
that covers the sample space, then for any events $B,C$ such that $P(C) \neq 0$
$$Pr(B \mid C) = Pr(B \cap A_1 \mid C) + \ldots +Pr(B \cap A_n \mid C)$$
And note the corollary:
$$Pr(B \mid C) = Pr(B \cap A \mid C) + Pr(B \cap \bar{A} \mid C)$$

Using Part (1) we have
$$Pr(B \cap C) = Pr(B \cap C \cap A_1) + \ldots + Pr(B \cap C \cap A_n)$$
Divide throughout by $Pr(C)$ and
notice that $Pr( X \mid C) = {Pr(X \cap C) \over Pr(C)}$ where
$X = B$ or $X = B \cap A_i$.

The corollary follows from the statement for $n=2$.

\item
In the OJ Simpson trial it was claimed by the defense that the fact that
Simpson battered his wife should be excluded from the evidence as the fraction
of wife batterers that murder their wives is $1/2000$ (Dershowitz Lemma).
In this problem we will condition this probability on the event that Nicol is 
dead and see that we obtain a somewhat different result.

We have the following statistical information (all for 1994):
\begin{itemize}

\item number of women in 1994 = 125,000,000
\item number of women murdered in 1994 = 6000
\item number of women murdered by spouse in 1994 = 1500
\item number of women with spouse in 1994 = 50,000,000
\item number of women battered by spouse in 1994 = 3,000,000
\end{itemize}

Being murdered counts as being battered. Also, we assume that 
the event of being murdered by
somebody else is not influenced by whether the woman was being battered or not.

Define the events

$S$ = Woman is murdered by spouse  \\
$M$ = Woman is murdered \\
$B$ = Woman is battered by spouse \\

Use the given information to evaluate the following

\begin{enumerate}

\item  $Pr(S \mid B)$, i.e., the probability that a battered woman
is murdered by her spouse.

$$Pr( S \mid B) = {Pr (S \cap B) \over Pr(B)} = {Pr(S) \over Pr(B)}$$
This is because being battered and murdered is the same as being murdered,
$S \cap B = S$.
$$Pr(S \mid B) = {1500/ 125,000,000 \over 3,000,000/125,000,000} = {1 \over 2000}$$


\item  $Pr(S \mid M)$, i.e., probability that a murdered woman 
was done in by her spouse.

$$Pr(S \mid M) = { Pr(S \cap M) \over Pr(M)} = {Pr(S) \over Pr(M)}$$
$$ = {1500/125,000,000 \over 6000/125,000,000} = {1 \over 4}$$

\item  $Pr(S \mid B \cap M)$, i.e., probability that a battered, murdered
woman is done in by her spouse. 

First note that
$$Pr(S \mid B \cap M) = {Pr(S \cap B \cap M) \over Pr(B \cap M)} = 
{Pr(S) \over Pr(M \mid B) Pr(B)}$$

Then to evaluate $Pr(M \mid B)$, use the corollary from part (2):
$$Pr(M \mid B) = Pr(M \cap S \mid B) + Pr(M \cap \bar{S} \mid B)$$
$$ = Pr(S \mid B) + Pr(M \cap \bar{S})$$
In simplifying the second term we take note of the fact that being
murdered by someone other than the spouse 
is not influenced by whether the victim
was battered by her spouse or not.

Now we have
$$Pr(S \mid B \cap M) = { 1500 /125,000,000 \over (\frac{1}{2000} + \frac{4500}
{125,000,000})\frac{3,000,000}{125,000,000}}$$
$$ = {1500 \times 125 \over 67000 \times 3} = {125 \over 134} = 0.93$$ 
\end{enumerate}

\end{enumerate}

\end{document}


