Minor edits

Advanced/Stopping Problems/parts/0 probability.tex

@@ -117,7 +117,7 @@ Show that $\mathcal{E}(\mathcal{A} + \mathcal{B}) = \mathcal{E}(\mathcal{A}) + \
 
 \definition{}
 Let $A$ and $B$ be events on a sample space $\Omega$. \par
-We say that $A$ and $B$ are \textit{independent} if $\mathcal{P}(A \cap B) = \mathcal{P}(A) + \mathcal{P}(B)$. \par
+We say that $A$ and $B$ are \textit{independent} if $\mathcal{P}(A \cap B) = \mathcal{P}(A) \times \mathcal{P}(B)$. \par
 Intuitively, events $A$ and $B$ are independent if the outcome of one does not affect the other.
 
 \definition{}

Advanced/Stopping Problems/parts/1 intro.tex

@@ -11,42 +11,45 @@ tosses of this die contain exactly one six? \par
 \hint{Start with small $l$.}
 
 \begin{solution}
-	$\mathcal{P}(\text{last } l \text{ tosses have exactly one 6}) = (\nicefrac{1}{6})(\nicefrac{5}{6})^l \times l$
+	$\mathcal{P}(\text{last } l \text{ tosses have exactly one 6}) = (\nicefrac{1}{6})(\nicefrac{5}{6})^{l-1} \times l$
 \end{solution}
 
 \vfill
 
 \problem{}
 For what value of $l$ is the probability in \ref{lastl} maximal? \par
-The following table may help.
+The following table may help. \par
+\note{We only care about integer values of $l$.}
 
 \begin{center}
 	\begin{tabular}{|| c | c | c ||}
 		\hline
 		\rule{0pt}{3.5mm} % Bonus height for exponent
-		$l$ & $(\nicefrac{5}{6})^l$ & $(\nicefrac{1}{6})(\nicefrac{5}{6})^l$ \\
+		$l$ & $(\nicefrac{5}{6})^l$ & $(\nicefrac{1}{6})(\nicefrac{5}{6})^{l}$ \\
 		\hline\hline
-		1 & 0.83 & 0.133 \\
+		0 & 1.00 & 0.167 \\
 		\hline
-		2 & 0.69 & 0.115 \\
+		1 & 0.83 & 0.139 \\
 		\hline
-		3 & 0.57 & 0.095 \\
+		2 & 0.69 & 0.116 \\
 		\hline
-		4 & 0.48 & 0.089 \\
+		3 & 0.58 & 0.096 \\
+		\hline
+		4 & 0.48 & 0.080 \\
 		\hline
 		5 & 0.40 & 0.067 \\
 		\hline
-		6 & 0.33 & 0.055 \\
+		6 & 0.33 & 0.056 \\
 		\hline
-		7 & 0.27 & 0.045 \\
+		7 & 0.28 & 0.047 \\
 		\hline
-		8 & 0.23 & 0.038 \\
+		8 & 0.23 & 0.039 \\
 		\hline
 	\end{tabular}
 \end{center}
 
 \begin{solution}
-	$(\nicefrac{1}{6})(\nicefrac{5}{6})^l \times l$ is maximal at $x = 5.48$, so $l = 5$. \par
+	$(\nicefrac{1}{6})(\nicefrac{5}{6})^{l-1} \times l$ is maximal at $l = 5.48$, so $l = 5$. \par
 	$l = 6$ is close enough.
 \end{solution}
 

Advanced/Stopping Problems/parts/2 secretary.tex

@@ -69,7 +69,7 @@ Come up with a strategy that produces better odds.
 
 	The remark from the previous solution still holds: \par
 	When we're looking at the first applicant, we have no information; \par
-	when we're looking at the second, we have no choices.
+	when we're looking at the last, we have no choices.
 
 	\vspace{2mm}
 
@@ -178,7 +178,8 @@ if we use the \say{look-then-leap} strategy detailed above? \par
 
 	\vspace{2mm}
 
-	Unraveling our previous logic, we find that the probability we are interested in is also $\frac{k-1}{x-1}$.
+	Unraveling our previous logic, we find that the probability we are interested in is also $\frac{k-1}{x-1}$. \par
+	\note{Assuming that $x \geq k$. Of course, this probability is 0 otherwise.}
 \end{solution}
 
 \vfill
@@ -229,7 +230,7 @@ Let $r = \frac{k-1}{n}$, the fraction of applicants we reject. Show that
 \vfill
 
 \problem{}
-With a bit of faily unpleasant calculus, we can show that the following is true for large $n$:
+With a bit of fairly unpleasant calculus, we can show that the following is true for large $n$:
 \begin{equation*}
 	\sum_{x=k}^{n}\frac{1}{x-1}
 	~\approx~ \text{ln}\Bigl(\frac{n}{k}\Bigr)

Advanced/Stopping Problems/parts/3 orderstat.tex

@@ -88,7 +88,6 @@ Given some $y$, what is the probability that all five $\mathcal{X}_i$ are smalle
 Say we have a random variable $\mathcal{X}$ which we observe $n$ times. \note{(for example, we repeatedly roll a die)}
 We'll arrange these observations in increasing order, labeled $x_1 < x_2 < ... < x_n$. \par
 Under this definition, $x_i$ is called the \textit{$i^\text{th}$ order statistic}---the $i^\text{th}$ smallest sample of $\mathcal{X}$.
-a
 
 \problem{}<ostatone>
 Say we have a random variable $\mathcal{X}$ uniformly distributed on $[0, 1]$, of which we take $5$ observations. \par

Advanced/Stopping Problems/parts/4 again.tex

@@ -1,81 +0,0 @@
-\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