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\input{macros-course}


\begin{document}

\tutn{18}{11/19/99}{November 17, 1999}

\begin{problems}

\problem In the O.J. Simpson trial, defense attorney Dershowitz observed
that the fraction of wife batterers that murder their wives is $1/2000$.
Since the probability that a wife batterer would murder his wife is so
low, Dershowitz argued that the fact that Simpson battered his wife,
Nicole Brown Simpson, should be excluded from the evidence.  We will
develop a counter-argument to Dershowitz', by conditioning the probability
on the event that Mrs. Simpson is dead.

We have the following statistical information (all for 1994):
\begin{center}
\begin{tabular}{|l|c|}
\hline
women & 125M\\
married women & 50M\\
wives battered by husband & 3M\\
women murdered & 6K\\
wives murdered by husband & 1.5K\\
\hline
\end{tabular}
\end{center}
The number of battered wives includes those who were murdered by their
husbands.  Define the events
\begin{eqnarray*}
H & ::= & \mbox{wife is murdered by husband} \\
M & ::= & \mbox{woman is murdered} \\
B & ::= & \mbox{wife is battered by husband}
\end{eqnarray*}

\bparts

\ppart Dershowitz' statistical observation can be stated technically as
\[\prob{H \mid B} = 1/2000.\]
Prove it.

\solution{
\[
\prob{H \mid B}
= \frac{\prob{H \intersect B}}{\prob{B}}
=\frac{\prob{H}}{\prob{B}}=\frac{1500/125M}{3M/125M}
=\frac{1500}{3M}=1/2000.
\]}

\ppart Calculate $\prob{H \mid M}$, i.e., the probability that a murdered
woman was murdered by her husband.

\solution{
\[\prob{H \mid M}
= \frac{\prob{H \intersect M}}{\prob{M}}
=\frac{\prob{H}}{\prob{M}}=\frac{1500/125M}{6000/125M}
= \frac{1500}{6000}=1/4.
\]}

\iffalse
\ppart Prove
\begin{equation}

\label{ABA-B}
\prob{C \mid A}  =  \prob{C \intersect D \mid A} + \prob{C - D \mid A}
\end{equation}
for arbitrary events $A, C, D$.

\solution{
\begin{eqnarray*}
\prob{C \intersect A} & = & \prob{(C \intersect A) \intersect D}
    + \prob{(C \intersect A) - D},\\
& = & \prob{(C \intersect D) \intersect A}
    + \prob{(C - D) \intersect A}.\\
\end{eqnarray*}
Now dividing both sides of this equation by $\prob{A}$ yields the desired
identity.}
\fi

\ppart Calculate $\prob{H \mid B \intersect M}$, i.e., the probability that
a battered, murdered woman was murdered by her husband.  Assume that the
the probability of a wife being murdered by someone \emph{other} than her
husband is not influenced by whether or not she was being battered.

\hint $\prob{C \mid A}  =  \prob{C \intersect D \mid A} + \prob{C - D \mid
A}$.

\solution{
Note that
\[
\prob{H \mid B \intersect M}
= \frac{\prob{H \intersect B \intersect M}}{\prob{B \intersect M}}
= \frac{\prob{H}}{\prob{M \mid B} \prob{B}}.
\]

To evaluate $\prob{M \mid B}$, we apply the identity from the hint:
\[
\prob{M \mid B} = \prob{M \intersect H \mid B} + \prob{M - H \mid B}.
\]
We know that being murdered by someone other than the husband is not
influenced by whether the victim was battered by her husband or not,
namely,
\[\prob{M - H \mid B} = \prob{M - H},\]
and $H \intersect M = H$, so
\[
\prob{M \mid B} =  \prob{H \mid B} + \prob{M - H} =
\frac{1}{2000} + \frac{6K - 1.5 K}{125M}.
\]

Now we have
\begin{eqnarray*}
\prob{H \mid B \intersect M} & = & \frac{1500 / 125M}{(1/2000 +
(6K - 1.5 K)/125M)3M/125M}\\
 & = & \frac{1500}{3M/2000 + 4500 \cdot 3/125}\\
 & = & \frac{1}{1 + 9/125}\ =\ \frac{125}{134}\  \approx\  0.93.
\end{eqnarray*}
}

\ppart How could you counter Dershowitz' argument that evidence that a
husband battered his wife should not be admitted as evidence?

\solution{Being battered increases the probability that a murdered woman
was murdered by her husband from 25\% to 93\%, so the fact of being
battered greatly influences the probability of a husband's guilt.
Assuming it is proper for a jury to consider such probabilities---this is
questionable by the way---then Dershowitz' argument that battery is
unlikely to be relevant does not hold up.}

\eparts

\problem 

A number $m$ between 0 and 100 is chosen, with all numbers equally likely.
Then $m$ white balls and $100-m$ black balls are placed in an urn.
We draw 20 successive balls from the 100 in the urn, leaving 80
behind.  We're blindfolded and the urn is shaken between draws, so we
assume that all draws of 20 of the 100 balls are equally likely.

Now it turns out that all the balls we draw are white.  What is the
probability that if we draw a 21st ball out of the urn, it will be white
as well?

Bayes' rule tells us that
\[
\prob{21 \mbox{ white} \mid 20 \mbox{ white}}
  = \frac{\prob{21 \mbox{ white}}}{\prob{20 \mbox{ white}}}.
\]
Therefore the problem reduces finding the probability that if we draw
20 balls out of the bin, we get all white balls; similarly for 21.

In general, suppose we have $n$ balls in the box, and that for any
$m=0,\ldots,n$, the probability that exactly $m$ of these $n$ are white
balls is $1/(n+1)$.  We draw $k$ balls out of the box, what is the
probability that they are all white?  Somewhat surprisingly, we will see
that the answer does not depend on $n$.

To make the computation, we first use the rule for conditional
probability, conditioned on how many white balls there are.  If there are
fewer than $k$ white balls, then the probability of drawing $k$ is 0.
Otherwise,
\[
\prob{k \mbox{ white}}
  = \sum_{m=k}^n \prob{k \mbox{ white} \mid
   m \mbox{ white in the box}}\prob{m \mbox{ white in the box}}.
\]

\bparts

\ppart  Use the previous equation to prove that
\begin {equation}\label{eq:binom}
\prob{k \mbox{ white}} = \sum_{m=k}^n \frac{\binom{m}{k}}{\binom{n}{k}} 
\cdot \frac{1}{n+1} 
\end {equation}

\solution{

To evaluate the conditional probability, we have to define the sample
space and identify the events we are interested in.  The sample space is
given by the $\binom{n}{k}$ ways of drawing $k$ balls out of the box.  The
events whose probability we want to measure are the $\binom{m}{k}$ choices
of $k$ balls out of the $m$ white ones.

}


\ppart The summation of the binomial coefficients can be evaluated
with the following result.
\begin{lemma*} For any $k$ and $n$ such that $n \geq k \geq 0$,
\[
\sum_{m=k}^n  \binom{m}{k} = \binom{n+1}{k+1}.
\]
\end {lemma*}

Give a combinatorial argument that proves the Lemma.  \hint Classify
subsets of $k+1$ elements from $\set{1,\dots, n+1}$ by their maximum
element.

\solution{

Combinatorial argument:

\begin{proof}

We know the righthand side of the equation is the number of ways to choose $k+1$
elements from $\set{1,\dots, n+1}$.

For any such subset, the maximum element $m+1$ must satisfy $k+1 \leq m+ 1
\leq n+1$.  For each $m \in \set{k,\dots,n}$, there are $\binom{m}{k}$
subsets of $k+1$ elements from $\set{1,\dots, m+1}$ whose
maximum element is $m+1$.  So the lefthand side of the equation also counts the
number of such subsets.

\end{proof}

For the record, here's an algebraic proof:

\begin {proof}
We prove the lemma by induction on $n$.

{\em Base Case:} $n=0$. Then also $k=0$ and the statement
of the lemma simplifies to
\[ \binom{0}{0} = \binom{1}{1} \]
which is true.

{\em Inductive Step:}  We distinguish two cases.
If $n=k$ then the statement of the lemma simplifies to
\[ \binom{n}{n} = \binom{n+1}{n+1} \]
which is true. Otherwise, we proceed as follows:

\[ \begin {array}{rcll}
 \binom{n+1}{k+1} & = & \binom{n}{k} + \binom{n}{k+1} & \mbox{By
Pascal's equality}\\
& = & \binom{n}{k} +  \sum_{m=k}^{n-1}  \binom{m}{k} & 
\mbox{By Inductive Hypothesis}\\
& = & \sum_{m=k}^{n}  \binom{m}{k} &  \mbox{QED}
\end {array} \]
\end {proof}

}

\ppart Conclude that
\[
\prob{\mbox{first 21 all white} \mid \mbox{first 20 all white}}
 = \frac{21}{22},
\]
and more generally, 
\[
\prob{\mbox{first $k+1$ all white} \mid \mbox{first $k$ all white}}
 = \frac{k+1}{k+2},
\]
regardless of the value of $n \geq k+1$.


\solution{
We can now simplify Equation~\ref{eq:binom}
\[
\prob{k \mbox{ white}} = \sum_{m=k}^n \frac {\binom{m}{k}}{\binom{n}{k}} 
\cdot \frac 1 {n+1}  = \frac{\binom{n+1}{k+1}}{\binom{n}{k}}
 \frac{1}{n+1} = \frac{1}{k+1}.
\]
Notice that the result is independent of $n$ as claimed.

\textbf{Comment}: The answer $(k+1)/(k+2)$ is so simple that it begs for a
simpler argument than we developed above.  However, we haven't been able
to invent one or find one in the literature.

}

\ppart Taking the T to lecture, Ben Bitdiddle has noticed the same person
20 days in a row, and he decides to offer odds of 21 to 1 that he will see
that person again tomorrow.  Is this a fair bet?  Explain?

\solution{

If the person Ben saw for 20 days also got off at Kendall station, carries
an MIT folder, and tomorrow is not a holiday, then we can expect the
chances of Ben seeing the person tomorrow to be higher than 21/22, and we
should reject Ben's bet.  But if it is the last day of lectures, then we
can expect chances to be lower, and we would confidently bet against Ben.
In general such a real-world situation hardly fits the model where all
possible numbers of consecutive successes are \emph{a priori} equally
likely.

Ben's bet is similar to the following argument originally made by LaPlace,
one of the founders of probability theory, who is alleged to have offered
odds of a million to one that the Sun would rise tomorrow based on the
following reasoning:
\begin{quotation}\noindent
The sun has certainly been rising every morning in the last 27 centuries
or so, since otherwise such a remarkable event as the sun not rising would
have been recorded in several places and we would know about that.  This
sums up to about a million days, therefore the chance that the sun is not
going to rise tomorrow is at most $10^{-6}$.
\end{quotation}

Anyone who finds this argument persuasive has a very shaky understanding
of probability, history, and astronomy.  LaPlace's argument supposes that
whether the sun will rise for exactly $k$ days occurs with equal
probability for all known $k$, an unjustified, and implausible assumption.
It also assumes that we know with certainty that the sun has risen for the
past million days, which given the nature of historical records, is no
more likely than the probability of tomorrow's sunrise that we are trying
to estimate in the first place.

}

\eparts

\end{problems}

\end{document}
        
