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Add FAIKR3 Bayesian net reasoning
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src/fundamentals-of-ai-and-kr/module3/sections/_bayesian_net.tex
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src/fundamentals-of-ai-and-kr/module3/sections/_bayesian_net.tex
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\chapter{Bayesian networks}
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\section{Bayes' rule}
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\begin{description}
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\item[Bayes' rule] \marginnote{Bayes' rule}
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\[ \prob{a \,\vert\, b} = \frac{\prob{b \,\vert\, a} \prob{a}}{\prob{b}} \]
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\item[Bayes' rule and conditional independence]
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Given the random variables $\texttt{Cause}$ and\\
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$\texttt{Effect}_1, \dots, \texttt{Effect}_n$, with $\texttt{Effect}_i$ independent from each other,
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we can compute $\textbf{P}(\texttt{Cause}, \texttt{Effect}_1, \dots, \texttt{Effect}_n)$ as follows:
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\[
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\textbf{P}(\texttt{Cause}, \texttt{Effect}_1, \dots, \texttt{Effect}_n) =
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\left(\prod_i \textbf{P}(\texttt{Effect}_i \,\vert\, \texttt{Cause})\right) \textbf{P}(\texttt{Cause})
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\]
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The number of parameters is linear.
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\begin{example}
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Knowing that $\textbf{P} \models (\texttt{Catch} \perp \texttt{Toothache} \vert \texttt{Cavity})$:
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\[
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\begin{split}
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\textbf{P}&(\texttt{Cavity} \,\vert\, \texttt{toothache} \land \texttt{catch}) \\
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&= \alpha\textbf{P}(\texttt{toothache} \land \texttt{catch} \,\vert\, \texttt{Cavity})\textbf{P}(\texttt{Cavity}) \\
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&= \alpha\textbf{P}(\texttt{toothache} \,\vert\, \texttt{Cavity})
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\textbf{P}(\texttt{catch} \,\vert\, \texttt{Cavity})\textbf{P}(\texttt{Cavity}) \\
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\end{split}
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\]
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\end{example}
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\end{description}
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\section{Bayesian network reasoning}
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\begin{description}
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\item[Bayesian network] \marginnote{Bayesian network}
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Graph for conditional independence assertions and a compact specification of full joint distributions.
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\begin{itemize}
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\item Directed acyclic graph.
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\item Nodes represent variables.
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\item The conditional distribution of a node is given by its parents
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\[ \textbf{P}(X_i \,\vert\, \texttt{parents}(X_i)) \]
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In other words, if there is an edge from $A$ to $B$, then $A$ (cause) influences $B$ (effect).
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\end{itemize}
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\begin{description}
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\item[Conditional probability table (CPT)] \marginnote{Conditional probability table (CPT)}
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In the case of boolean variables, the conditional distribution of a node can be represented using
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a table by considering all the combinations of the parents.
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\begin{example}
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Given the boolean variables $A$, $B$ and $C$, with $C$ depending on $A$ and $B$, we have that:\\
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\begin{minipage}{.48\linewidth}
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\centering
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\includegraphics[width=0.35\linewidth]{img/_cpt_graph.pdf}
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\end{minipage}
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\begin{minipage}{.48\linewidth}
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\centering
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\begin{tabular}{c|c|c|c}
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A & B & $\prob{c \vert A, B}$ & $\prob{\lnot c \vert A, B}$ \\
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\hline
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a & b & $\alpha$ & $1-\alpha$ \\
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$\lnot$a & b & $\beta$ & $1-\beta$ \\
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a & $\lnot$b & $\gamma$ & $1-\gamma$ \\
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$\lnot$a & $\lnot$b & $\delta$ & $1-\delta$ \\
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\end{tabular}
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\end{minipage}
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\end{example}
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\end{description}
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\item[Reasoning patterns] \marginnote{Reasoning patterns}
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Given a Bayesian network, the following reasoning patterns can be used:
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\begin{descriptionlist}
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\item[Causal] \marginnote{Causal reasoning}
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To make a prediction. From the cause, derive the effect.
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\begin{example}
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Knowing $\texttt{Intelligence}$, it is possible to make a prediction of $\texttt{Letter}$.
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\begin{center}
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\includegraphics[width=0.5\linewidth]{img/_causal_example.pdf}
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\end{center}
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\end{example}
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\item[Evidential] \marginnote{Evidential reasoning}
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To find an explanation. From the effect, derive the cause.
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\begin{example}
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Knowing $\texttt{Grade}$, it is possible to explain it by estimating\\$\texttt{Intelligence}$.
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\begin{center}
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\includegraphics[width=0.7\linewidth]{img/_evidential_example.pdf}
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\end{center}
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\end{example}
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\item[Explain away] \marginnote{Explain away reasoning}
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Observation obtained "passing through" other observations.
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\begin{example}
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Knowing $\texttt{Difficulty}$ and $\texttt{Grade}$,
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it is possible to estimate \\$\texttt{Intelligence}$.
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Note that if $\texttt{Grade}$ was not known,
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$\texttt{Difficulty}$ and $\texttt{Intelligence}$ would be independent.
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\begin{center}
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\includegraphics[width=0.75\linewidth]{img/_explainaway_example.pdf}
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\end{center}
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\end{example}
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\end{descriptionlist}
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\item[Independence] \marginnote{Bayesian network independence}
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Intuitively, an effect is independent from a cause,
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if there is another cause in the middle whose value is already known.
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\begin{example}
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\phantom{}
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\begin{minipage}{.3\linewidth}
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\centering
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\includegraphics[width=0.85\linewidth]{img/_independence_example.pdf}
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\end{minipage}
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\begin{minipage}{.6\linewidth}
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\[ \textbf{P} \models (\texttt{L} \perp \texttt{D}, \texttt{I}, \texttt{S} \,\vert\, \texttt{G}) \]
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\[ \textbf{P} \models (\texttt{S} \perp \texttt{L} \,\vert\, \texttt{G}) \]
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\[ \textbf{P} \models (\texttt{S} \perp \texttt{D}) \text{ but }
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\textbf{P} \models (\texttt{S} \,\cancel{\perp}\, \texttt{D} \,\vert\, \texttt{G}) \text{ (explain away)} \]
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\end{minipage}
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\end{example}
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\item[V-structure] \marginnote{V-structure}
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Effect with two causes.
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If the effect is not in the evidence, the causes are independent.
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\begin{figure}[H]
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\centering
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\includegraphics[width=0.2\textwidth]{img/_v_structure.pdf}
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\caption{V-structure}
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\end{figure}
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\item[Active two-edge trail] \marginnote{Active two-edge trail}
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The trail $X \leftrightharpoons Z \leftrightharpoons Y$ is active either if:
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\begin{itemize}
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\item $X$, $Z$, $Y$ is a v-structure $X \rightarrow Z \leftarrow Y$
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and $Z$ or one of its children is in the evidence.
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\item $Z$ is not in the evidence.
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\end{itemize}
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In other words, influence can flow from $X$ to $Y$ passing by $Z$.
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.65\textwidth]{img/_active_trail.pdf}
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\caption{Example of active and non-active two-edge trails}
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\end{figure}
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\item[Active trail] \marginnote{Active trail}
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A trail $X_1 \leftrightharpoons \dots \leftrightharpoons X_n$ is active iff
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each two-edge trail $X_{i-1} \leftrightharpoons X_i \leftrightharpoons X_{i+1}$ along the trail is active.
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\item[D-separation] \marginnote{D-separation}
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Two sets of nodes $\vec{X}$ and $\vec{Y}$ are d-separated given the evidence $\vec{Z}$ if
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there is no active trail between any $X \in \vec{X}$ and $Y \in \vec{Y}$.
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\begin{theorem}
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Two d-separated nodes are independent.
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In other words, two nodes are independent if there is no active trail between them.
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\end{theorem}
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\item[Independence algorithm] \phantom{}
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\begin{description}
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\item[Blocked node]
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A node is blocked if it blocks the flow.
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This happens if one and only one of the following conditions are met:
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\begin{itemize}
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\item The node is in the middle of an unmarked v-structure.
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\item The node is in the evidence.
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\end{itemize}
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\end{description}
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To determine if $X \perp Y$ given the evidence $\vec{Z}$:
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\begin{enumerate}
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\item Traverse the graph bottom-up marking all nodes in $\vec{Z}$ or
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having a child in $\vec{Z}$.
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\item Find a path from $X$ to $Y$ that does not pass through a blocked node.
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\item If $Y$ is not reachable from $X$, then $X$ and $Y$ are independent.
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Otherwise $X$ and $Y$ are dependent.
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\end{enumerate}
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\begin{example}
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To determine if $J \perp D$:
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\begin{center}
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\includegraphics[width=0.5\textwidth]{img/_d_sep_example.pdf}
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\end{center}
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As a path has been found, $J \,\cancel{\perp}\, D$.
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\end{example}
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\item[Global semantics] \marginnote{Global semantics}
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Given a Bayesian network, the full joint distribution can be defined as
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the product of the local conditional distributions:
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\[ \prob{x_1, \dots, x_n} = \prod_{i=1}^{n} \prob{x_i \,\vert\, \texttt{parents}(X_i)} \]
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\begin{example}
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Given the following Bayesian network:
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\begin{minipage}{.3\linewidth}
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\centering
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\includegraphics[width=0.7\linewidth]{img/_global_semantics_example.pdf}
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\end{minipage}
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\begin{minipage}{.6\linewidth}
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\[
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\begin{split}
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&\prob{j \land m \land a \land \lnot b \land \lnot e} \\
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&= \prob{\lnot b} \prob{\lnot e} \prob{a \,\vert\, \lnot b, \lnot e}
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\prob{j \,\vert\, a} \prob{m \,\vert\, a}
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\end{split}
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\]
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\end{minipage}
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\end{example}
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\item[Local semantics]
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Each node is conditionally independent of its non-descendants given its parents.
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.35\textwidth]{img/_local_independence.pdf}
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\caption{Local independence}
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\end{figure}
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\begin{theorem}
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Local semantics $\iff$ Global semantics
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\end{theorem}
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\item[Markov blanket]
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Each node is conditionally independent of all other nodes
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if its Markov blanket (parents, children, children's parents) is in the evidence.
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.35\textwidth]{img/_markov_blanket.pdf}
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\caption{Markov blanket}
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\end{figure}
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\end{description}
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\section{Building Bayesian networks}
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The following algorithm can be used to construct a Bayesian network of $n$ random variables:
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\begin{enumerate}
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\item Choose an ordering of the variables $X_1, \dots, X_n$.
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\item For $i=1, \dots, n$:
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\begin{itemize}
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\item Add $X_i$ to the network.
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\item Select the parents of $X_i$ from $X_1, \dots, X_{i-1}$ such that:
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\[ \textbf{P}(X_i \,\vert\, \texttt{parents}(X_i)) =
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\textbf{P}(X_i \,\vert\, X_1, \dots, X_{i-1}) \]
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\end{itemize}
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\end{enumerate}
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By construction, this algorithm guarantees the global semantics.
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\begin{example}[Monty Hall]
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The variables are:
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\begin{itemize}
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\item $G$: the choice of the guest.
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\item $H$: the choice of the host.
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\item $P$: the position of the prize.
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\end{itemize}
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Note that $P \perp G$.
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Let the order be fixed as follows: $P$, $G$, $H$.
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\begin{figure}[h]
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\begin{subfigure}{.3\textwidth}
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\centering
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\includegraphics[width=0.2\linewidth]{img/_monty_hall1.pdf}
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\caption{First interaction}
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\end{subfigure}
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\begin{subfigure}{.3\textwidth}
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\centering
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\includegraphics[width=0.6\linewidth]{img/_monty_hall2.pdf}
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\caption{Second interaction (note that $P \perp G$)}
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\end{subfigure}
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\begin{subfigure}{.3\textwidth}
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\centering
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\includegraphics[width=0.6\linewidth]{img/_monty_hall3.pdf}
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\caption{Third interaction}
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\end{subfigure}
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\end{figure}
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\end{example}
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