\section{The Secretary, Again}

Now, let's solve the secretary problem as as a stopping rule problem. \par
The first thing we need to do is re-write it into the form we discussed in the previous section. \par
Namely, we need...
\begin{itemize}
	\item A sequence of random variables $\mathcal{X}_1, \mathcal{X}_2, ..., \mathcal{X}_t$
	\item A sequence of reward functions $y_0, y_1(\sigma_1), ..., y_t(\sigma_t)$.
\end{itemize}

\vspace{2mm}

For convenience, I've summarized the secretary problem below:
\begin{itemize}
	\item We have exactly one position to fill, and we must fill it with one of $n$ applicants.
	\item These $n$ applicants, if put together, can be ranked unambiguously from \say{best} to \say{worst}.
	\item We interview applicants in a random order, one at a time.
	\item After each interview, we reject the applicant and move on, \par
	or select the applicant and end the process.
	\item We cannot return to an applicant we've rejected.
	\item Our goal is to select the \textit{overall best} applicant.
\end{itemize}

\definition{}
First, we'll define a sequence of $\mathcal{X}_i$ that fits this problem. \par
Each $\mathcal{X}_i$ will gives us the \textit{relative rank} of each applicant. \par
For example, if $\mathcal{X}_i = 1$, the $i^\text{th}$ applicant is the best of the first $i$. \par
If $\mathcal{X}_i = 3$, two applicants better than $i$ came before $i$.

\problem{}
What values can $\mathcal{X}_1$ take, and what are their probabilities? \par
How about $\mathcal{X}_2$, $\mathcal{X}_3$, and $\mathcal{X}_4$?

\vfill

\remark{}
Now we need to define $y_n(\sigma_n)$. Intuitively, it may make sense to set $y_n = 1$ if the $n^\text{th}$
applicant is the best, and $y_n = 0$ otherwise---but this doesn't work.

\vspace{2mm}

As defined in the previous section, $y_n$ can only depend on $\sigma_n = [x_1, x_2, ..., x_n]$, the previous $n$ observations.
We cannot define $y_n$ as specified above because, having seen $\sigma_n$, we \textit{cannot} know whether or not the $n^\text{th}$
applicant is the best.

\vspace{2mm}

To work around this, we'll define our reward for selecting the $n^\text{th}$ applicant as the \textit{probability}
that this applicant is the best.

\problem{}
Define $y_n$.

\begin{solution}
	\begin{itemize}
		\item An applicant should only be selected if $\mathcal{X}_i = 1$
		\item if we accept an the $j^\text{th}$ applicant, the probability we select the absolute best is equal to \par
		the probability that the best of the first $j$ candidates is the best overall. \par

		\vspace{1mm}

		This is just the probability that the best candidate overall appears among the first $j$, \par
		and is thus $\nicefrac{j}{n}$.
	\end{itemize}

	So,
	\begin{equation*}
		y_j(\sigma_j) =
		\begin{cases}
			\nicefrac{j}{n} & x_j = 1 \\
			0 & \text{otherwise}
		\end{cases}
	\end{equation*}

	\vspace{2mm}
	Note that $y_0 = 0$, and that $y_n$ depends only on $x_n$.

\end{solution}

\vfill
\pagebreak