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\title{Sets\footnote{ This slide show is an open-source document. See last slide for copyright information.}}
\subtitle{STA 256: Fall 2019}
\date{} % To suppress date

\begin{document}

\begin{frame}
  \titlepage
\end{frame}

\begin{frame}
\frametitle{Statistical Experiment}
\framesubtitle{This vocabulary is not in the text} 
A statistical experiment is a procedure whose outcome is not known in advance with certainty. \pause

\vspace{3mm}
Sample Space: set of outcomes $s \in S$ \pause

\begin{itemize}
    \item Sell 500 lottery tickets, pick the winning number.  \pause\\
            $S = \{1, 2, \ldots, 500 \}$ \pause
    \item Hold your breath as long as you can. \pause \\
            $S = \{t: t \geq 0 \}$ \pause
    \item Pick coin or die from jar, roll or toss. \pause \\
            $S = \{H,T,1,2,3,4,5,6 \}$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Event: Set of outcomes, $A \subset S$}
%\framesubtitle{} 
\pause
\begin{center}
\includegraphics[width=2in]{Venn}
\end{center} \pause
\begin{itemize}
    \item $A \cap B = \{s \in S:s \in A \mbox{ and } s \in B \}$ \pause
    \item $A \cup B = \{s \in S:s \in A \mbox{ or } s \in B \}$ \pause
    \item $A^c = \{s \in S: s \notin A \}$ \pause
    \item $A \cap B^c = \{s \in S: s \in A \mbox{ and } s \notin B \}$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Disjoint sets} 
%\framesubtitle{} 
\begin{itemize}
    \item $A$ and $B$ are said to be \emph{disjoint} if $A \cap B = \emptyset $ \pause
    \item The idea is that $A$ and $B$ have no elements in common; they do not overlap. \pause
\begin{center}
\includegraphics[width=2in]{Venn2}
\end{center} \pause
    \item However, recall that the null set is a subset of every set:  $\emptyset \subseteq A$. \pause
    \item So $\emptyset \cap A = \emptyset$. \pause
    \item And the null set is also disjoint from every set.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Set Laws}
\framesubtitle{No proofs, just Venn diagrams at most}  \pause
\begin{itemize}
    \item Commutative: $A \cup B = B \cup A$, $A \cap B = B \cap A$ \pause
    \item Associative
        \begin{itemize}
            \item  $(A \cup B) \cup C = A \cup (B \cup C)$, 
            \item  $(A \cap B) \cap C = A \cap (B \cap C)$
        \end{itemize}   \pause
    \item Distributive (like multiplication)
        \begin{itemize}
            \item  $A \cap (B \cup C) \pause = (A \cap B) \cup (A \cap C)$ \pause
            \item  $A \cup (B \cap C) \pause = (A \cup B) \cap (A \cup C)$
        \end{itemize}
\end{itemize} % The distributive laws will go on the formula sheet.
\end{frame}

\begin{frame}
\frametitle{De Morgan Laws}
\framesubtitle{Not in the text}  \pause
\begin{itemize}
    \item $(A \cap B)^c = A^c \cup B^c$ 
    \item $(A \cup B)^c = A^c \cap B^c$  
    \item Rule: complement and flip $\cup\cap$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Extend the notation to larger number of sets} 
\framesubtitle{Not in the text}  \pause
Distributive laws
\begin{itemize}
    \item $A \cap \left(\cup_{j=1}^n B_j\right) = \cup_{j=1}^n (A \cap B_j)$\pause, or even 
    \item $A \cap \left(\cup_{j=1}^\infty B_j\right) = \cup_{j=1}^\infty (A \cap B_j)$ \pause
    \item[] and \pause
    \item $A \cup \left(\cap_{j=1}^n B_j\right) = \cap_{j=1}^n (A \cup B_j)$
    \item $A \cup \left(\cap_{j=1}^\infty B_j\right) = \cap_{j=1}^\infty (A \cup B_j)$ 
\end{itemize} \pause
De Morgan Laws (complement and flip)
\begin{itemize}
    \item $(\cap_{j=1}^\infty A_j)^c = \cup_{j=1}^\infty A_j^c$ 
    \item $(\cup_{j=1}^\infty A_j)^c = \cap_{j=1}^\infty A_j^c$ 
\end{itemize}
\end{frame}

\begin{frame}
% \frametitle{{\color{red} WARNING} } \pause
%\framesubtitle{} 

\begin{center}
{\color{red} \LARGE WARNING!}
\end{center} \pause

\begin{itemize}
    \item Addition and subtraction apply only to numbers. They are not set operations. \pause
    \item For example, if $A$ and $B$ are sets, then $A+B$ is not defined. \pause
    \item $A+B$ is not the same as $A \cup B$. \pause
    \item Because what is $A+A-A$? \pause
    \item[] 
    \item Some very bad probability proofs use addition and subtraction of sets. \pause
    \item Such proofs will receive a {\color{red} zero}.
\end{itemize}
\end{frame}



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\begin{frame}
\frametitle{Copyright Information}

This slide show was prepared by  \href{http://www.utstat.toronto.edu/~brunner}{Jerry Brunner},
Department of Statistical Sciences, University of Toronto. It is licensed under a 
\href{http://creativecommons.org/licenses/by-sa/3.0/deed.en_US}
     {Creative Commons Attribution - ShareAlike 3.0 Unported License}. Use any part of it as you like and share the result freely. The \LaTeX~source code is available from the course website:

\vspace{5mm}
\href{http://www.utstat.toronto.edu/~brunner/oldclass/256f19} {\small\texttt{http://www.utstat.toronto.edu/$^\sim$brunner/oldclass/256f19}}

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$ = \{s \in S: \}$

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    \item  \pause
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