Stopping edits

Advanced/Stopping Problems/parts/0 intro.tex

@@ -18,9 +18,8 @@ For what value of $l$ is the probability in \ref{lastl} maximal?
 
 \problem{}
 Finish your solution: \par
-In $n$ rolls of a six-sided die, how do we (most reliably) detect the last time we roll a 6?
+In $n$ rolls of a six-sided die, when should we announce the last time we roll a 6? \par
+What is the probability of our guess being right?
 
 \vfill
-
-
 \pagebreak

Advanced/Stopping Problems/parts/1 secretary.tex

@@ -1,6 +1,6 @@
-\section{The Secretary Problem}
+\section{The Secretary}
 
-\definition{}
+\definition{The secretary problem}
 Say we need to hire a secretary. We have exactly one position to fill,
 and we must fill it with one of $n$ applicants. These $n$ applicants,
 if put together, can be ranked unambiguously from \say{best} to \say{worst}.
@@ -8,19 +8,19 @@ if put together, can be ranked unambiguously from \say{best} to \say{worst}.
 \vspace{2mm}
 
 We interview applicants in a random order, one at a time. \par
-At the end of each interview, we either reject the applicant (and move on to the next one),
+At the end of each interview, we either reject the applicant (and move on to the next one), \par
 or select the applicant (which fills the position and ends the process).
 
 \vspace{2mm}
 
-Each applicant is interviewed exactly once---we cannot return to an applicant we've rejected. \par
+Each applicant is interviewed at most once---we cannot return to an applicant we've rejected. \par
 In addition, we cannot reject the final applicant, as doing so will leave us without a secretary.
 
 \vspace{2mm}
 
 For a given $n$, we would like to maximize our probability of selecting the best applicant. \par
-This is called the \textit{secretary problem}.
+This is the only metric we care about---we do not try to maximize the rank of our applicant. \par
-
+Hiring the second-best applicant is no better than hiring the worst.
 
 \problem{}
 If $n = 1$, what is the best hiring strategy, and what is the probability that we hire the best applicant?
@@ -36,22 +36,60 @@ Is this different than the probability of hiring the best applicant at random?
 
 
 \problem{}
-What happens if $n = 3$?
+If $n = 3$, what is the probability of hiring the best applicant at random? \par
-\begin{itemize}
+Come up with a strategy that produces better odds.
-	\item Find the probability of hiring the best applicant at random
-	\item Find the best possible hiring strategy. \par
-	What is its probability of success?
-\end{itemize}
-\hint{In this case, three is a fairly small number. \par
-It is easy to show that a strategy is optimal by considering all cases.}
 
 \vfill
 
-\problem{}
+\problem{}<bestyet>
 Should we ever consider hiring a candidate that \textit{isn't} the best we've seen so far? \par
-Why or why not? \hint{Read the problem again.da}
+Why or why not? \hint{Read the problem again.}
 
 \vfill
 
+
 \remark{}
-To find the optimal solution to the secretary problem, we'll restrict ourselves to
+\ref{bestyet} implies that we should automatically reject any applicant that isn't
+the best we've seen. We can take advantage of this fact to restrict the types of
+strategies we consider.
+
+\vspace{2mm}
+
+We'll transform our sequence of $n$ secretaries into a sequence of $n$ random variables $I_1, I_2, ..., I_n$,
+each producing values in $\{\texttt{0}, \texttt{1}\}$. Each $I_x$ will produce \texttt{1} if the $x^\text{th}$
+secretary we interview is the best we've seen so far, and \texttt{0} otherwise.
+
+\problem{}
+What is the probability distribution of $I_1$? \par
+That is, what are $\mathcal{P}(I_1 = \texttt{0})$ and $\mathcal{P}(I_1 = \texttt{1})$?
+
+\vspace{1cm}
+
+
+\problem{}
+Convince yourself that the largest $x$ where $I_x = 1$ is the \par
+position of the best candidate in our list of applicants.
+
+\vspace{1cm}
+
+
+\remark{}
+Recall that we only know the \textit{relative} ranks of our applicants. \par
+We have no absolute metric by which to judge each candidate.
+
+\vspace{2mm}
+
+Thus, all $I_x$ defined above are independent: \par
+the outcome of any $I_a$ does not influence the probabilities of any other $I_b$.
+
+\vspace{2mm}
+
+We can therefore ignore any strategy that depends on the outcomes of
+previous $I_x$. Since all random variables in this sequence are independent,
+the results of past $I_x$ cannot possibly provide information about future $I_x$.
+
+\vspace{2mm}
+
+Given the above realizations, we are left with only one kind of strategy: \par
+We blindly reject the first $k$ applicants, and select the first \say{best-seen} applicant we encounter afterwards.
+All we need to do now is pick the optimal $k$.