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Add FAIKR3 Bayesian net reasoning
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@@ -9,5 +9,6 @@
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\input{sections/_intro.tex}
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\input{sections/_probability.tex}
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\input{sections/_bayesian_net.tex}
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\end{document}
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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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\[
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\textbf{P}(\texttt{Cause}, \texttt{Effect}_1, \dots, \texttt{Effect}_n) =
|
||||
\left(\prod_i \textbf{P}(\texttt{Effect}_i \,\vert\, \texttt{Cause})\right) \textbf{P}(\texttt{Cause})
|
||||
\]
|
||||
The number of parameters is linear.
|
||||
|
||||
\begin{example}
|
||||
Knowing that $\textbf{P} \models (\texttt{Catch} \perp \texttt{Toothache} \vert \texttt{Cavity})$:
|
||||
\[
|
||||
\begin{split}
|
||||
\textbf{P}&(\texttt{Cavity} \,\vert\, \texttt{toothache} \land \texttt{catch}) \\
|
||||
&= \alpha\textbf{P}(\texttt{toothache} \land \texttt{catch} \,\vert\, \texttt{Cavity})\textbf{P}(\texttt{Cavity}) \\
|
||||
&= \alpha\textbf{P}(\texttt{toothache} \,\vert\, \texttt{Cavity})
|
||||
\textbf{P}(\texttt{catch} \,\vert\, \texttt{Cavity})\textbf{P}(\texttt{Cavity}) \\
|
||||
\end{split}
|
||||
\]
|
||||
\end{example}
|
||||
\end{description}
|
||||
|
||||
|
||||
\section{Bayesian network reasoning}
|
||||
|
||||
\begin{description}
|
||||
\item[Bayesian network] \marginnote{Bayesian network}
|
||||
Graph for conditional independence assertions and a compact specification of full joint distributions.
|
||||
\begin{itemize}
|
||||
\item Directed acyclic graph.
|
||||
\item Nodes represent variables.
|
||||
\item The conditional distribution of a node is given by its parents
|
||||
\[ \textbf{P}(X_i \,\vert\, \texttt{parents}(X_i)) \]
|
||||
In other words, if there is an edge from $A$ to $B$, then $A$ (cause) influences $B$ (effect).
|
||||
\end{itemize}
|
||||
|
||||
\begin{description}
|
||||
\item[Conditional probability table (CPT)] \marginnote{Conditional probability table (CPT)}
|
||||
In the case of boolean variables, the conditional distribution of a node can be represented using
|
||||
a table by considering all the combinations of the parents.
|
||||
|
||||
\begin{example}
|
||||
Given the boolean variables $A$, $B$ and $C$, with $C$ depending on $A$ and $B$, we have that:\\
|
||||
\begin{minipage}{.48\linewidth}
|
||||
\centering
|
||||
\includegraphics[width=0.35\linewidth]{img/_cpt_graph.pdf}
|
||||
\end{minipage}
|
||||
\begin{minipage}{.48\linewidth}
|
||||
\centering
|
||||
\begin{tabular}{c|c|c|c}
|
||||
A & B & $\prob{c \vert A, B}$ & $\prob{\lnot c \vert A, B}$ \\
|
||||
\hline
|
||||
a & b & $\alpha$ & $1-\alpha$ \\
|
||||
$\lnot$a & b & $\beta$ & $1-\beta$ \\
|
||||
a & $\lnot$b & $\gamma$ & $1-\gamma$ \\
|
||||
$\lnot$a & $\lnot$b & $\delta$ & $1-\delta$ \\
|
||||
\end{tabular}
|
||||
\end{minipage}
|
||||
\end{example}
|
||||
\end{description}
|
||||
|
||||
\item[Reasoning patterns] \marginnote{Reasoning patterns}
|
||||
Given a Bayesian network, the following reasoning patterns can be used:
|
||||
\begin{descriptionlist}
|
||||
\item[Causal] \marginnote{Causal reasoning}
|
||||
To make a prediction. From the cause, derive the effect.
|
||||
\begin{example}
|
||||
Knowing $\texttt{Intelligence}$, it is possible to make a prediction of $\texttt{Letter}$.
|
||||
\begin{center}
|
||||
\includegraphics[width=0.5\linewidth]{img/_causal_example.pdf}
|
||||
\end{center}
|
||||
\end{example}
|
||||
|
||||
\item[Evidential] \marginnote{Evidential reasoning}
|
||||
To find an explanation. From the effect, derive the cause.
|
||||
\begin{example}
|
||||
Knowing $\texttt{Grade}$, it is possible to explain it by estimating\\$\texttt{Intelligence}$.
|
||||
\begin{center}
|
||||
\includegraphics[width=0.7\linewidth]{img/_evidential_example.pdf}
|
||||
\end{center}
|
||||
\end{example}
|
||||
|
||||
\item[Explain away] \marginnote{Explain away reasoning}
|
||||
Observation obtained "passing through" other observations.
|
||||
\begin{example}
|
||||
Knowing $\texttt{Difficulty}$ and $\texttt{Grade}$,
|
||||
it is possible to estimate \\$\texttt{Intelligence}$.
|
||||
|
||||
Note that if $\texttt{Grade}$ was not known,
|
||||
$\texttt{Difficulty}$ and $\texttt{Intelligence}$ would be independent.
|
||||
\begin{center}
|
||||
\includegraphics[width=0.75\linewidth]{img/_explainaway_example.pdf}
|
||||
\end{center}
|
||||
\end{example}
|
||||
\end{descriptionlist}
|
||||
|
||||
\item[Independence] \marginnote{Bayesian network independence}
|
||||
Intuitively, an effect is independent from a cause,
|
||||
if there is another cause in the middle whose value is already known.
|
||||
\begin{example}
|
||||
\phantom{}
|
||||
|
||||
\begin{minipage}{.3\linewidth}
|
||||
\centering
|
||||
\includegraphics[width=0.85\linewidth]{img/_independence_example.pdf}
|
||||
\end{minipage}
|
||||
\begin{minipage}{.6\linewidth}
|
||||
\[ \textbf{P} \models (\texttt{L} \perp \texttt{D}, \texttt{I}, \texttt{S} \,\vert\, \texttt{G}) \]
|
||||
\[ \textbf{P} \models (\texttt{S} \perp \texttt{L} \,\vert\, \texttt{G}) \]
|
||||
\[ \textbf{P} \models (\texttt{S} \perp \texttt{D}) \text{ but }
|
||||
\textbf{P} \models (\texttt{S} \,\cancel{\perp}\, \texttt{D} \,\vert\, \texttt{G}) \text{ (explain away)} \]
|
||||
\end{minipage}
|
||||
\end{example}
|
||||
|
||||
|
||||
\item[V-structure] \marginnote{V-structure}
|
||||
Effect with two causes.
|
||||
If the effect is not in the evidence, the causes are independent.
|
||||
|
||||
\begin{figure}[H]
|
||||
\centering
|
||||
\includegraphics[width=0.2\textwidth]{img/_v_structure.pdf}
|
||||
\caption{V-structure}
|
||||
\end{figure}
|
||||
|
||||
\item[Active two-edge trail] \marginnote{Active two-edge trail}
|
||||
The trail $X \leftrightharpoons Z \leftrightharpoons Y$ is active either if:
|
||||
\begin{itemize}
|
||||
\item $X$, $Z$, $Y$ is a v-structure $X \rightarrow Z \leftarrow Y$
|
||||
and $Z$ or one of its children is in the evidence.
|
||||
\item $Z$ is not in the evidence.
|
||||
\end{itemize}
|
||||
In other words, influence can flow from $X$ to $Y$ passing by $Z$.
|
||||
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
\includegraphics[width=0.65\textwidth]{img/_active_trail.pdf}
|
||||
\caption{Example of active and non-active two-edge trails}
|
||||
\end{figure}
|
||||
|
||||
\item[Active trail] \marginnote{Active trail}
|
||||
A trail $X_1 \leftrightharpoons \dots \leftrightharpoons X_n$ is active iff
|
||||
each two-edge trail $X_{i-1} \leftrightharpoons X_i \leftrightharpoons X_{i+1}$ along the trail is active.
|
||||
|
||||
\item[D-separation] \marginnote{D-separation}
|
||||
Two sets of nodes $\vec{X}$ and $\vec{Y}$ are d-separated given the evidence $\vec{Z}$ if
|
||||
there is no active trail between any $X \in \vec{X}$ and $Y \in \vec{Y}$.
|
||||
|
||||
\begin{theorem}
|
||||
Two d-separated nodes are independent.
|
||||
In other words, two nodes are independent if there is no active trail between them.
|
||||
\end{theorem}
|
||||
|
||||
\item[Independence algorithm] \phantom{}
|
||||
\begin{description}
|
||||
\item[Blocked node]
|
||||
A node is blocked if it blocks the flow.
|
||||
This happens if one and only one of the following conditions are met:
|
||||
\begin{itemize}
|
||||
\item The node is in the middle of an unmarked v-structure.
|
||||
\item The node is in the evidence.
|
||||
\end{itemize}
|
||||
\end{description}
|
||||
To determine if $X \perp Y$ given the evidence $\vec{Z}$:
|
||||
\begin{enumerate}
|
||||
\item Traverse the graph bottom-up marking all nodes in $\vec{Z}$ or
|
||||
having a child in $\vec{Z}$.
|
||||
\item Find a path from $X$ to $Y$ that does not pass through a blocked node.
|
||||
\item If $Y$ is not reachable from $X$, then $X$ and $Y$ are independent.
|
||||
Otherwise $X$ and $Y$ are dependent.
|
||||
\end{enumerate}
|
||||
|
||||
\begin{example}
|
||||
To determine if $J \perp D$:
|
||||
\begin{center}
|
||||
\includegraphics[width=0.5\textwidth]{img/_d_sep_example.pdf}
|
||||
\end{center}
|
||||
As a path has been found, $J \,\cancel{\perp}\, D$.
|
||||
\end{example}
|
||||
|
||||
|
||||
\item[Global semantics] \marginnote{Global semantics}
|
||||
Given a Bayesian network, the full joint distribution can be defined as
|
||||
the product of the local conditional distributions:
|
||||
\[ \prob{x_1, \dots, x_n} = \prod_{i=1}^{n} \prob{x_i \,\vert\, \texttt{parents}(X_i)} \]
|
||||
|
||||
\begin{example}
|
||||
Given the following Bayesian network:
|
||||
|
||||
\begin{minipage}{.3\linewidth}
|
||||
\centering
|
||||
\includegraphics[width=0.7\linewidth]{img/_global_semantics_example.pdf}
|
||||
\end{minipage}
|
||||
\begin{minipage}{.6\linewidth}
|
||||
\[
|
||||
\begin{split}
|
||||
&\prob{j \land m \land a \land \lnot b \land \lnot e} \\
|
||||
&= \prob{\lnot b} \prob{\lnot e} \prob{a \,\vert\, \lnot b, \lnot e}
|
||||
\prob{j \,\vert\, a} \prob{m \,\vert\, a}
|
||||
\end{split}
|
||||
\]
|
||||
\end{minipage}
|
||||
\end{example}
|
||||
|
||||
\item[Local semantics]
|
||||
Each node is conditionally independent of its non-descendants given its parents.
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
\includegraphics[width=0.35\textwidth]{img/_local_independence.pdf}
|
||||
\caption{Local independence}
|
||||
\end{figure}
|
||||
|
||||
\begin{theorem}
|
||||
Local semantics $\iff$ Global semantics
|
||||
\end{theorem}
|
||||
|
||||
|
||||
\item[Markov blanket]
|
||||
Each node is conditionally independent of all other nodes
|
||||
if its Markov blanket (parents, children, children's parents) is in the evidence.
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
\includegraphics[width=0.35\textwidth]{img/_markov_blanket.pdf}
|
||||
\caption{Markov blanket}
|
||||
\end{figure}
|
||||
\end{description}
|
||||
|
||||
|
||||
|
||||
\section{Building Bayesian networks}
|
||||
|
||||
The following algorithm can be used to construct a Bayesian network of $n$ random variables:
|
||||
\begin{enumerate}
|
||||
\item Choose an ordering of the variables $X_1, \dots, X_n$.
|
||||
\item For $i=1, \dots, n$:
|
||||
\begin{itemize}
|
||||
\item Add $X_i$ to the network.
|
||||
\item Select the parents of $X_i$ from $X_1, \dots, X_{i-1}$ such that:
|
||||
\[ \textbf{P}(X_i \,\vert\, \texttt{parents}(X_i)) =
|
||||
\textbf{P}(X_i \,\vert\, X_1, \dots, X_{i-1}) \]
|
||||
\end{itemize}
|
||||
\end{enumerate}
|
||||
By construction, this algorithm guarantees the global semantics.
|
||||
|
||||
\begin{example}[Monty Hall]
|
||||
The variables are:
|
||||
\begin{itemize}
|
||||
\item $G$: the choice of the guest.
|
||||
\item $H$: the choice of the host.
|
||||
\item $P$: the position of the prize.
|
||||
\end{itemize}
|
||||
Note that $P \perp G$.
|
||||
Let the order be fixed as follows: $P$, $G$, $H$.
|
||||
|
||||
\begin{figure}[h]
|
||||
\begin{subfigure}{.3\textwidth}
|
||||
\centering
|
||||
\includegraphics[width=0.2\linewidth]{img/_monty_hall1.pdf}
|
||||
\caption{First interaction}
|
||||
\end{subfigure}
|
||||
\begin{subfigure}{.3\textwidth}
|
||||
\centering
|
||||
\includegraphics[width=0.6\linewidth]{img/_monty_hall2.pdf}
|
||||
\caption{Second interaction (note that $P \perp G$)}
|
||||
\end{subfigure}
|
||||
\begin{subfigure}{.3\textwidth}
|
||||
\centering
|
||||
\includegraphics[width=0.6\linewidth]{img/_monty_hall3.pdf}
|
||||
\caption{Third interaction}
|
||||
\end{subfigure}
|
||||
\end{figure}
|
||||
\end{example}
|
||||
@@ -234,142 +234,3 @@ can be computed as the sum of the atomic events where $\phi$ is true:
|
||||
|
||||
|
||||
|
||||
\section{Bayesian networks}
|
||||
|
||||
\begin{description}
|
||||
\item[Bayes' rule] \marginnote{Bayes' rule}
|
||||
\[ \prob{a \,\vert\, b} = \frac{\prob{b \,\vert\, a} \prob{a}}{\prob{b}} \]
|
||||
|
||||
\item[Bayes' rule and conditional independence]
|
||||
Given the random variables $\texttt{Cause}$ and\\
|
||||
$\texttt{Effect}_1, \dots, \texttt{Effect}_n$, with $\texttt{Effect}_i$ independent from each other,
|
||||
we can compute $\textbf{P}(\texttt{Cause}, \texttt{Effect}_1, \dots, \texttt{Effect}_n)$ as follows:
|
||||
\[
|
||||
\textbf{P}(\texttt{Cause}, \texttt{Effect}_1, \dots, \texttt{Effect}_n) =
|
||||
\left(\prod_i \textbf{P}(\texttt{Effect}_i \,\vert\, \texttt{Cause})\right) \textbf{P}(\texttt{Cause})
|
||||
\]
|
||||
The number of parameters is linear.
|
||||
|
||||
\begin{example}
|
||||
Knowing that $\textbf{P} \models (\texttt{Catch} \perp \texttt{Toothache} \vert \texttt{Cavity})$:
|
||||
\[
|
||||
\begin{split}
|
||||
\textbf{P}&(\texttt{Cavity} \,\vert\, \texttt{toothache} \land \texttt{catch}) \\
|
||||
&= \alpha\textbf{P}(\texttt{toothache} \land \texttt{catch} \,\vert\, \texttt{Cavity})\textbf{P}(\texttt{Cavity}) \\
|
||||
&= \alpha\textbf{P}(\texttt{toothache} \,\vert\, \texttt{Cavity})
|
||||
\textbf{P}(\texttt{catch} \,\vert\, \texttt{Cavity})\textbf{P}(\texttt{Cavity}) \\
|
||||
\end{split}
|
||||
\]
|
||||
\end{example}
|
||||
|
||||
\item[Bayesian network] \marginnote{Bayesian network}
|
||||
Graph for conditional independence assertions and a compact specification of full joint distributions.
|
||||
\begin{itemize}
|
||||
\item Directed acyclic graph.
|
||||
\item Nodes represent variables.
|
||||
\item The conditional distribution of a node is given by its parents
|
||||
\[ \textbf{P}(X_i \,\vert\, \texttt{parents}(X_i)) \]
|
||||
In other words, if there is an edge from $A$ to $B$, then $A$ (cause) influences $B$ (effect).
|
||||
\end{itemize}
|
||||
|
||||
\begin{description}
|
||||
\item[Conditional probability table (CPT)] \marginnote{Conditional probability table (CPT)}
|
||||
In the case of boolean variables, the conditional distribution of a node can be represented using
|
||||
a table by considering all the combinations of the parents.
|
||||
|
||||
\begin{example}
|
||||
Given the boolean variables $A$, $B$ and $C$, with $C$ depending on $A$ and $B$, we have that:\\
|
||||
\begin{minipage}{.48\linewidth}
|
||||
\centering
|
||||
\includegraphics[width=0.35\linewidth]{img/_cpt_graph.pdf}
|
||||
\end{minipage}
|
||||
\begin{minipage}{.48\linewidth}
|
||||
\centering
|
||||
\begin{tabular}{c|c|c|c}
|
||||
A & B & $\prob{c \vert A, B}$ & $\prob{\lnot c \vert A, B}$ \\
|
||||
\hline
|
||||
a & b & $\alpha$ & $1-\alpha$ \\
|
||||
$\lnot$a & b & $\beta$ & $1-\beta$ \\
|
||||
a & $\lnot$b & $\gamma$ & $1-\gamma$ \\
|
||||
$\lnot$a & $\lnot$b & $\delta$ & $1-\delta$ \\
|
||||
\end{tabular}
|
||||
\end{minipage}
|
||||
\end{example}
|
||||
\end{description}
|
||||
|
||||
\item[Reasoning patterns] \marginnote{Reasoning patterns}
|
||||
Given a Bayesian network, the following reasoning patterns can be used:
|
||||
\begin{descriptionlist}
|
||||
\item[Causal] \marginnote{Causal reasoning}
|
||||
To make a prediction. From the cause, derive the effect.
|
||||
\begin{example}
|
||||
Knowing $\texttt{Intelligence}$, it is possible to make a prediction of $\texttt{Letter}$.
|
||||
\begin{center}
|
||||
\includegraphics[width=0.5\linewidth]{img/_causal_example.pdf}
|
||||
\end{center}
|
||||
\end{example}
|
||||
|
||||
\item[Evidential] \marginnote{Evidential reasoning}
|
||||
To find an explanation. From the effect, derive the cause.
|
||||
\begin{example}
|
||||
Knowing $\texttt{Grade}$, it is possible to explain it by estimating\\$\texttt{Intelligence}$.
|
||||
\begin{center}
|
||||
\includegraphics[width=0.65\linewidth]{img/_evidential_example.pdf}
|
||||
\end{center}
|
||||
\end{example}
|
||||
|
||||
\item[Explain away] \marginnote{Explain away reasoning}
|
||||
Observation obtained "passing through" other observations.
|
||||
\begin{example}
|
||||
Knowing $\texttt{Difficulty}$ and $\texttt{Grade}$,
|
||||
it is possible to estimate \\$\texttt{Intelligence}$.
|
||||
|
||||
Note that if $\texttt{Grade}$ was not known,
|
||||
$\texttt{Difficulty}$ and $\texttt{Intelligence}$ would be independent.
|
||||
\begin{center}
|
||||
\includegraphics[width=0.70\linewidth]{img/_explainaway_example.pdf}
|
||||
\end{center}
|
||||
\end{example}
|
||||
\end{descriptionlist}
|
||||
|
||||
\item[Global semantics] \marginnote{Global semantics}
|
||||
Given a Bayesian network, the full joint distribution can be defined as
|
||||
the product of the local conditional distributions:
|
||||
\[ \prob{x_1, \dots, x_n} = \prod_{i=1}^{n} \prob{x_i \,\vert\, \texttt{parents}(X_i)} \]
|
||||
|
||||
\begin{example}
|
||||
Given the following Bayesian network:
|
||||
|
||||
\begin{minipage}{.3\linewidth}
|
||||
\centering
|
||||
\includegraphics[width=0.7\linewidth]{img/_global_semantics_example.pdf}
|
||||
\end{minipage}
|
||||
\begin{minipage}{.6\linewidth}
|
||||
\[
|
||||
\begin{split}
|
||||
&\prob{j \land m \land a \land \lnot b \land \lnot e} \\
|
||||
&= \prob{\lnot b} \prob{\lnot e} \prob{a \,\vert\, \lnot b, \lnot e}
|
||||
\prob{j \,\vert\, a} \prob{m \,\vert\, a}
|
||||
\end{split}
|
||||
\]
|
||||
\end{minipage}
|
||||
\end{example}
|
||||
|
||||
\item[Independence] \marginnote{Bayesian network independence}
|
||||
Intuitively, an effect is independent from a cause,
|
||||
if there is another cause in the middle whose value is already known.
|
||||
\begin{example}
|
||||
\phantom{}
|
||||
|
||||
\begin{minipage}{.3\linewidth}
|
||||
\centering
|
||||
\includegraphics[width=0.75\linewidth]{img/_independence_example.pdf}
|
||||
\end{minipage}
|
||||
\begin{minipage}{.6\linewidth}
|
||||
\[ \textbf{P} \models (\texttt{L} \perp \texttt{D}, \texttt{I}, \texttt{S} \,\vert\, \texttt{G}) \]
|
||||
\[ \textbf{P} \models (\texttt{S} \perp \texttt{L} \,\vert\, \texttt{G}) \]
|
||||
\[ \textbf{P} \models (\texttt{S} \perp \texttt{D}) \text{ but }
|
||||
\textbf{P} \models (\texttt{S} \,\cancel{\perp}\, \texttt{D} \,\vert\, \texttt{G}) \text{ (explain away)} \]
|
||||
\end{minipage}
|
||||
\end{example}
|
||||
\end{description}
|
||||
Reference in New Issue
Block a user