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% use [nosolutions] flag to hide solutions.
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% use [solutions] flag to show solutions.
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\documentclass[
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solutions,
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singlenumbering,
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unfinished
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]{../../resources/ormc_handout}
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\usepackage{../../resources/macros}
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\usepackage{units}
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\uptitlel{Advanced 2}
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\uptitler{\smallurl{}}
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\title{Stopping problems}
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\subtitle{Prepared by Mark on \today{}}
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\begin{document}
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\maketitle
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\input{parts/0 intro.tex}
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\input{parts/1 secretary.tex}
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\end{document}
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\section{Introduction}
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\generic{Setup:}
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Suppose we toss a 6-sided die $n$ times. \par
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It is easy to detect the first time we roll a 6. \par
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What should we do if we want to annouce the \textit{last}?
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\problem{}<lastl>
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Given $l \leq n$, what is the probability that the last $l$
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tosses of this die contain exactly one six?
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\vfill
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\problem{}
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For what value of $l$ is the probability in \ref{lastl} maximal?
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\vfill
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\problem{}
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Finish your solution: \par
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In $n$ rolls of a six-sided die, when should we announce the last time we roll a 6? \par
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What is the probability of our guess being right?
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\vfill
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\pagebreak
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@ -1,139 +0,0 @@
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\section{The Secretary}
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\definition{The secretary problem}
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Say we need to hire a secretary. We have exactly one position to fill,
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and we must fill it with one of $n$ applicants. These $n$ applicants,
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if put together, can be ranked unambiguously from \say{best} to \say{worst}.
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\vspace{2mm}
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We interview applicants in a random order, one at a time. \par
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At the end of each interview, we either reject the applicant (and move on to the next one), \par
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or select the applicant (which fills the position and ends the process).
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\vspace{2mm}
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Each applicant is interviewed at most once---we cannot return to an applicant we've rejected. \par
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In addition, we cannot reject the final applicant, as doing so will leave us without a secretary.
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\vspace{2mm}
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For a given $n$, we would like to maximize our probability of selecting the best applicant. \par
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This is the only metric we care about---we do not try to maximize the rank of our applicant. \par
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Hiring the second-best applicant is no better than hiring the worst.
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\problem{}
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If $n = 1$, what is the best hiring strategy, and what is the probability that we hire the best applicant?
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\vfill
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\problem{}
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If $n = 2$, what is the best hiring strategy, and what is the probability that we hire the best applicant? \par
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Is this different than the probability of hiring the best applicant at random?
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\vfill
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\pagebreak
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\problem{}
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If $n = 3$, what is the probability of hiring the best applicant at random? \par
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Come up with a strategy that produces better odds.
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\vfill
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\problem{}<bestyet>
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Should we ever consider hiring a candidate that \textit{isn't} the best we've seen so far? \par
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Why or why not? \hint{Read the problem again.}
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\vfill
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\remark{}
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\ref{bestyet} implies that we should automatically reject any applicant that isn't
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the best we've seen. We can take advantage of this fact to restrict the types of
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strategies we consider.
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\vspace{2mm}
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We'll transform our sequence of $n$ secretaries into a sequence of $n$ random variables $I_1, I_2, ..., I_n$,
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each producing values in $\{\texttt{0}, \texttt{1}\}$. Each $I_x$ will produce \texttt{1} if the $x^\text{th}$
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secretary we interview is the best we've seen so far, and \texttt{0} otherwise.
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\problem{}
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What is the probability distribution of $I_1$? \par
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That is, what are $\mathcal{P}(I_1 = \texttt{0})$ and $\mathcal{P}(I_1 = \texttt{1})$?
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\vspace{1cm}
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\problem{}
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Convince yourself that the largest $x$ where $I_x = 1$ is the \par
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position of the best candidate in our list of applicants.
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\vspace{1cm}
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\remark{}
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Recall that we only know the \textit{relative} ranks of our applicants. \par
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We have no absolute metric by which to judge each candidate.
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\vspace{2mm}
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Thus, all $I_x$ defined above are independent:
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the outcome of any $I_a$ does not influence the probabilities of any other $I_b$.
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\vspace{2mm}
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We can therefore ignore any strategy that depends on the outcomes of
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previous $I_x$. Since all random variables in this sequence are independent,
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the results of past $I_x$ cannot possibly provide information about future $I_x$.
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\vspace{2mm}
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Given the above realizations, we are left with only one kind of strategy: \par
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We blindly reject the first $k$ applicants, and select the first \say{best-seen} applicant we encounter afterwards.
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All we need to do now is pick the optimal $k$.
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\pagebreak
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\problem{}
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Consider the secretary problem with a given $n$. \par
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What is the probability distribution of each of $I_1, I_2, ..., I_n$? \par
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\note{That is, what are $\mathcal{P}(I_x = \texttt{0})$ and $\mathcal{P}(I_x = \texttt{1})$ for each $x$?}
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\vfill
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\problem{}
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What is the probability that the $x^\text{th}$ applicant is \textit{not} the best-seen applicant? \par
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In other words, what is the probability we reject the $x^\text{th}$ candidate? \par
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\note{Assuming that $x > k$. Otherwise, the probability of rejection is 1!}
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\vfill
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\problem{}<phisubn>
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Again, consider the secretary problem with a fixed $n$. \par
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If we reject the first $k$ applicants and hire the first \say{best-yet} applicant we encounter, \par
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what is the probability that we select the best candidate? \par
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Call this function $\phi_n(k)$.
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\vfill
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\problem{}
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Find $
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\underset{x \in \{0, ..., n\}}{\text{max}}
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\Bigl(\phi_n(k)\Bigr)
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$ for all $n$ in $\{1, 2, 3, 4, 5\}$.
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\vfill
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\problem{}
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Find $
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\underset{x \in \mathbb{R}}{\text{max}}
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\Bigl(~
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\underset{n \rightarrow \infty}{\text{lim}}
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\phi_n(k)
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~\Bigr)
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$
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\vfill
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