diff --git a/Advanced/Stopping Problems/parts/0 probability.tex b/Advanced/Stopping Problems/parts/0 probability.tex index 2ba3aec..6eb30bb 100644 --- a/Advanced/Stopping Problems/parts/0 probability.tex +++ b/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{} diff --git a/Advanced/Stopping Problems/parts/1 intro.tex b/Advanced/Stopping Problems/parts/1 intro.tex index 90b0250..940bb85 100644 --- a/Advanced/Stopping Problems/parts/1 intro.tex +++ b/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} diff --git a/Advanced/Stopping Problems/parts/2 secretary.tex b/Advanced/Stopping Problems/parts/2 secretary.tex index ac9b823..2cd7c89 100644 --- a/Advanced/Stopping Problems/parts/2 secretary.tex +++ b/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) diff --git a/Advanced/Stopping Problems/parts/3 orderstat.tex b/Advanced/Stopping Problems/parts/3 orderstat.tex index 6a376ca..82fc4e5 100644 --- a/Advanced/Stopping Problems/parts/3 orderstat.tex +++ b/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{} Say we have a random variable $\mathcal{X}$ uniformly distributed on $[0, 1]$, of which we take $5$ observations. \par diff --git a/Advanced/Stopping Problems/parts/4 again.tex b/Advanced/Stopping Problems/parts/4 again.tex deleted file mode 100644 index d3ab939..0000000 --- a/Advanced/Stopping Problems/parts/4 again.tex +++ /dev/null @@ -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 \ No newline at end of file