diff --git a/Advanced/Stopping Problems/main.tex b/Advanced/Stopping Problems/main.tex deleted file mode 100755 index 3184e8a..0000000 --- a/Advanced/Stopping Problems/main.tex +++ /dev/null @@ -1,24 +0,0 @@ -% use [nosolutions] flag to hide solutions. -% use [solutions] flag to show solutions. -\documentclass[ - solutions, - singlenumbering, - unfinished -]{../../resources/ormc_handout} -\usepackage{../../resources/macros} - -\usepackage{units} - -\uptitlel{Advanced 2} -\uptitler{\smallurl{}} -\title{Stopping problems} -\subtitle{Prepared by Mark on \today{}} - -\begin{document} - - \maketitle - - \input{parts/0 intro.tex} - \input{parts/1 secretary.tex} - -\end{document} \ No newline at end of file diff --git a/Advanced/Stopping Problems/parts/0 intro.tex b/Advanced/Stopping Problems/parts/0 intro.tex deleted file mode 100644 index ddc278e..0000000 --- a/Advanced/Stopping Problems/parts/0 intro.tex +++ /dev/null @@ -1,25 +0,0 @@ -\section{Introduction} - -\generic{Setup:} -Suppose we toss a 6-sided die $n$ times. \par -It is easy to detect the first time we roll a 6. \par -What should we do if we want to annouce the \textit{last}? - -\problem{} -Given $l \leq n$, what is the probability that the last $l$ -tosses of this die contain exactly one six? - -\vfill - -\problem{} -For what value of $l$ is the probability in \ref{lastl} maximal? - -\vfill - -\problem{} -Finish your solution: \par -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 \ No newline at end of file diff --git a/Advanced/Stopping Problems/parts/1 secretary.tex b/Advanced/Stopping Problems/parts/1 secretary.tex deleted file mode 100644 index 09807f8..0000000 --- a/Advanced/Stopping Problems/parts/1 secretary.tex +++ /dev/null @@ -1,139 +0,0 @@ -\section{The Secretary} - -\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}. - -\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), \par -or select the applicant (which fills the position and ends the process). - -\vspace{2mm} - -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 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? - -\vfill - -\problem{} -If $n = 2$, what is the best hiring strategy, and what is the probability that we hire the best applicant? \par -Is this different than the probability of hiring the best applicant at random? - -\vfill -\pagebreak - - -\problem{} -If $n = 3$, what is the probability of hiring the best applicant at random? \par -Come up with a strategy that produces better odds. - -\vfill - -\problem{} -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.} - -\vfill - - -\remark{} -\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: -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 $k$. - -\pagebreak - -\problem{} -Consider the secretary problem with a given $n$. \par -What is the probability distribution of each of $I_1, I_2, ..., I_n$? \par -\note{That is, what are $\mathcal{P}(I_x = \texttt{0})$ and $\mathcal{P}(I_x = \texttt{1})$ for each $x$?} - -\vfill - -\problem{} -What is the probability that the $x^\text{th}$ applicant is \textit{not} the best-seen applicant? \par -In other words, what is the probability we reject the $x^\text{th}$ candidate? \par -\note{Assuming that $x > k$. Otherwise, the probability of rejection is 1!} - -\vfill - -\problem{} -Again, consider the secretary problem with a fixed $n$. \par -If we reject the first $k$ applicants and hire the first \say{best-yet} applicant we encounter, \par -what is the probability that we select the best candidate? \par -Call this function $\phi_n(k)$. - -\vfill - -\problem{} -Find $ - \underset{x \in \{0, ..., n\}}{\text{max}} - \Bigl(\phi_n(k)\Bigr) -$ for all $n$ in $\{1, 2, 3, 4, 5\}$. - -\vfill - -\problem{} -Find $ - \underset{x \in \mathbb{R}}{\text{max}} - \Bigl(~ - \underset{n \rightarrow \infty}{\text{lim}} - \phi_n(k) - ~\Bigr) -$ - - -\vfill \ No newline at end of file