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Prof.~David Draper \\
Department of Applied Mathematics and Statistics \\
University of California, Santa Cruz

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\textbf{\large AMS 131: Quiz 4}

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Name: \underline{\hspace*{5.85in}}

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\begin{itemize}

\item

\textit{(Fact 1)} As a broad generalization (which you can verify empirically), statisticians tend to have shy personalities more often than economists do --- let's quantify this observation by assuming that 80\% of statisticians are shy but the corresponding percentage among economists is only 15\%. 

\item

\textit{(Fact 2)} Conferences on the topic of \textit{econometrics} are almost exclusively attended by economists and statisticians, with the majority of participants being economists --- let's quantify this fact by assuming that 90\% of the attendees are economists (and the rest statisticians). 

\end{itemize}

Suppose that you (a physicist, say) go to an econometrics conference --- you strike up a conversation with the first person you (haphazardly) meet, and find that this person is shy. The point of this problem is to show that the (conditional) probability $p$ that you're talking to a statistician is only about 37\%, which most people find surprisingly low, and to understand why this is the right answer. Let $St$ = (person is statistician), $E$ = (person is economist), and $Sh$ = (person is shy).

\begin{itemize}

\item[(a)]

Briefly explain what key assumption is necessary to validly bring probability into the solution of this problem.

\vspace*{0.75in}

\item[(b)]

Using the $St$, $E$ and $Sh$ notation, express the three numbers (80\%, 15\%, 90\%) above, and the probability we're solving for,
in unconditional and conditional probability terms.

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\item[(c)]

Briefly explain why calculating the desired probability is a good job for Bayes's Theorem.

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(over)

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\item[(d)]

Briefly explain why the following expression is a correct use of Bayes's Theorem in odds form in this problem. \vspace*{-0.1in}

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\Large

\[
\begin{array}{ccccc}
\left[ \frac{ P ( St \given Sh ) }{ P ( E \given Sh ) } \right] & = & \left[ \frac{ P ( St ) }{ P ( E ) } \right] & \cdot & \left[ \frac{ P ( Sh \given St ) }{ P ( Sh \given E ) } \right] \\
( 1 ) & = & ( 2 ) & \cdot & ( 3 )
\end{array}
\]

\normalsize

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\item[(d)]

Here are three terms that are relevant to the quantities in part (c) above:

\begin{itemize}

\item

(Prior odds in favor of $St$ over $E$)

\item

(Bayes factor in favor of $St$ over $E$)

\item

(Posterior odds in favor of $St$ over $E$)

\end{itemize}

Match these three terms with the numbers $( 1 ), ( 2 ), ( 3 )$ in the second line of the equation in part (c).

\vspace*{1.0in}

\item[(e)]

Compute the three odds values in part (d), briefly explaining your reasoning, thereby demonstrating that the posterior odds value $o$ in favor of $St$ over $E$ is $o = \frac{ 16 }{ 27 } \doteq 0.593$.

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\item[(f)]

Use the expression $p = \frac{ o }{ 1 + o }$ to show that the desired probability in this problem --- the conditional probability that you're talking to a statistician --- is $p = \frac{ 16 }{ 43 } \doteq 0.372$.

\vspace*{1.0in}

\item[(g)]

Someone says, ``That probability can't be right: 80\% of statisticians are shy, versus 15\% for economists, so your probability of talking to a statistician has to be over 50\%.'' Briefly explain why this line of reasoning is wrong, and why $p$ should indeed be less than 50\%.

\end{itemize}

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