203 lines
6.1 KiB
TeX
203 lines
6.1 KiB
TeX
\section{Another Secretary Problem}
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As you may have already noticed, the secretary problem we discussed in the previous section
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is somewhat disconnected from reality. Under what circumstances would one only be satisfied
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with the \textit{absolute best} candidate? It may make more sense to maximize the average rank
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of the candidate we hire, rather than the probability of selecting the best. This is the problem
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we'll attempt to solve next.
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\definition{}
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The problem we're solving is summarized below.
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Note that this is nearly identical to the classical secretary problem in the previous
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section---the only thing that has changed is the goal.
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\begin{itemize}
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\item We have exactly one position to fill, and we must fill it with one of $n$ applicants.
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\item These $n$ applicants, if put together, can be ranked unambiguously from \say{best} to \say{worst}.
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\item We interview applicants in a random order, one at a time.
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\item After each interview, we either reject or select the applicant.
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\item We cannot return to an applicant we've rejected.
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\item The process ends once we select an applicant.
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\vspace{2mm}
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\item Our goal is to maximize the rank of the applicant we hire.
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\end{itemize}
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\definition{}<mod>
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Just like before, we need to restate this problem in the language of probability. \par
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To do this, we'll say that each candidate has a \textit{quality} rating in $[0, 1]$. \par
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\vspace{2mm}
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Our series of applicants then becomes a series of random variables $\mathcal{X}_1, \mathcal{X}_2, ..., \mathcal{X}_n$, \par
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where each $\mathcal{X}_i$ is drawn uniformly from $[0, 1]$.
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\problem{}<notsatisfy>
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The modification in \ref{mod} doesn't fully satisfy the constraints of the secretary problem. \par
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Why not?
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\begin{solution}
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If we observe $\mathcal{X}_i$ directly, we obtain \textit{absolute} scores. \par
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This is more information than the secretary problem allows us to have---we can know which of
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two candidates is better, but \textit{not by how much}.
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\end{solution}
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\vfill
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Ignore this issue for now. We'll return to it later.
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\problem{}
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Let $\mathcal{X}$ be a random variable uniformly distributed over $[0, 1]$. \par
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Given a real number $x$, what is the probability that $\mathcal{P}(\mathcal{X} \leq x)$?
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\begin{solution}
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\begin{equation*}
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\mathcal{P}(\mathcal{X} \leq x) =
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\begin{cases}
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0 & x \leq 0 \\
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x & 0 < x < 1 \\
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1 & \text{otherwise}
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\end{cases}
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\end{equation*}
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\end{solution}
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\vfill
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\problem{}
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Say we have five random variables $\mathcal{X}_1, \mathcal{X}_2, ..., \mathcal{X}_5$. \par
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Given some $y$, what is the probability that all five $\mathcal{X}_i$ are smaller than $y$?
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\begin{solution}
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Naturally, this is $\mathcal{P}(\mathcal{X} \leq y)^5$, which is $y^5$.
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\end{solution}
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\vfill
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\pagebreak
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%
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% MARK: Page
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%
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\definition{}
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Say we have a random variable $\mathcal{X}$ which we observe $n$ times. \note{(for example, we repeatedly roll a die)}
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We'll arrange these observations in increasing order, labeled $x_1 < x_2 < ... < x_n$. \par
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Under this definition, $x_i$ is called the \textit{$i^\text{th}$ order statistic}---the $i^\text{th}$ smallest sample of $\mathcal{X}$.
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\problem{}<ostatone>
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Say we have a random variable $\mathcal{X}$ uniformly distributed on $[0, 1]$, of which we take $5$ observations. \par
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Given some $y$, what is the probability that $x_5 < y$? How about $x_4 <y $?
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\begin{solution}
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$x_5 < y$: ~This is a restatement of the previous problem.
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\vspace{2mm}
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$x_4 < y$: ~We need 4 measurements to be smaller,
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and one to be larger. Accounting for permutations, we get
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$
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5\mathcal{P}(\mathcal{X} \leq y)^4
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\mathcal{P}(\mathcal{X} > y)
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+
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\mathcal{P}(\mathcal{X} \leq y)^5
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$, which is $5y^4(1-y) + y^5$.
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\end{solution}
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\vfill
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\problem{}
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Consider the same setup as \ref{ostatone}, but with $n$ measurements. \par
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What is the probability that $x_i < y$ for a given $y$?
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\begin{solution}
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\begin{equation*}
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\mathcal{P}(x_i < y)
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~=~
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\sum_{j=i}^{n}
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\binom{n}{j} \times
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y^j
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(1-y)^{n-j}
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\end{equation*}
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\end{solution}
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\vfill
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\remark{}
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The expected value of the $i^\text{th}$ order statistic on $n$ samples of the uniform distribution is below.
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\begin{equation*}
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\mathcal{E}(x_i) = \frac{i}{n+1}
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\end{equation*}
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We do not have the tools to derive this yet.
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\pagebreak
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%
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% MARK: Page
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%
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\definition{}
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Recall \ref{notsatisfy}. We need one more modification. \par
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In order to preserve the constraints of the problem, we will not be allowed to observe $\mathcal{X}_i$ directly. \par
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Instead, we'll be given an \say{indicator} $\mathcal{I}_i$ for each $\mathcal{X}_i$, which produces values in $\{0, 1\}$. \par
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If the value we observe when interviewing $\mathcal{X}_i$ is the best we've seen so far, $\mathcal{I}_i$ will produce $1$. \par
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If it isn't, $\mathcal{I}_i$ produces $0$.
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\problem{}
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Given a secretary problem with $n$ applicants, what is $\mathcal{E}(\mathcal{I}_i)$?
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\begin{solution}
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\begin{equation*}
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\mathcal{E}(\mathcal{I}_i) = \frac{1}{i}
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\end{equation*}
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\end{solution}
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\vfill
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\problem{}
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What is $\mathcal{E}(\mathcal{X}_i ~|~ \mathcal{I}_i = 1)$? \par
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In other words, what is the expected value of $\mathcal{X}_i$ given that \par
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we know this candidate is the best we've seen so far?
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\begin{solution}
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This is simply the expected value of the $i^\text{th}$ order statistic on $i$ samples:
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\begin{equation*}
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\mathcal{E}(\mathcal{X}_i ~|~ \mathcal{I}_i = 1) = \frac{i}{i+1}
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\end{equation*}
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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In the previous section, we found that the optimal strategy for the classical secretary problem is to
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reject the first $e^{-1} \times n$ candidates, and select the next \say{best-yet} candidate we see. \par
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\vspace{2mm}
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How effective is this strategy for the ranked secretary problem? \par
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Find the expected rank of the applicant we select using this strategy.
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\vfill
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\problem{}
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Assuming we use the same kind of strategy as before (reject $k$, select the next \say{best-yet} candidate), \par
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show that $k = \sqrt{n}$ optimizes the expected rank of the candidate we select.
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\begin{solution}
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This is a difficult bonus problem. see
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\texttt{Neil Bearden, J. (2006). A new secretary problem with rank-based selection and cardinal payoffs.}
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\end{solution}
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\vfill
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\pagebreak |