mirror of
https://github.com/NotXia/unibo-ai-notes.git
synced 2025-12-16 19:32:21 +01:00
Moved FAIKR3 in year1
This commit is contained in:
19
src/year1/fundamentals-of-ai-and-kr/metadata.json
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19
src/year1/fundamentals-of-ai-and-kr/metadata.json
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{
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"name": "Fundamentals of Artificial Intelligence and Knowledge Representation",
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"year": 1,
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"semester": 1,
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"pdfs": [
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{
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"name": "FAIKR module 1",
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"path": "module1/faikr1.pdf"
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},
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{
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"name": "FAIKR module 2",
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"path": "module2/faikr2.pdf"
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},
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{
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"name": "FAIKR module 3",
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"path": "module3/faikr3.pdf"
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}
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]
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}
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1
src/year1/fundamentals-of-ai-and-kr/module3/ainotes.cls
Symbolic link
1
src/year1/fundamentals-of-ai-and-kr/module3/ainotes.cls
Symbolic link
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../../../ainotes.cls
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18
src/year1/fundamentals-of-ai-and-kr/module3/faikr3.tex
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18
src/year1/fundamentals-of-ai-and-kr/module3/faikr3.tex
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\documentclass[11pt]{ainotes}
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\title{Fundamentals of Artificial Intelligence and Knowledge Representation\\(Module 3)}
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\date{2023 -- 2024}
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\def\lastupdate{{PLACEHOLDER-LAST-UPDATE}}
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\begin{document}
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\makenotesfront
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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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\input{sections/_exact_inference.tex}
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\input{sections/_approx_inference.tex}
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\eoc
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\end{document}
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|
||||
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|
||||
\chapter{Approximate inference}
|
||||
|
||||
\begin{description}
|
||||
\item[Stochastic simulation] \marginnote{Stochastic simulation}
|
||||
Class of methods that draw $N$ samples from the distribution
|
||||
and estimate an approximate posterior $\hat{\mathcal{P}}$.
|
||||
|
||||
\begin{description}
|
||||
\item[$\delta$-stochastic absolute approximation]
|
||||
Given $\delta \in ]0, 0.5[$ and $\varepsilon \in ]0, 0.5[$, a $\delta$-stochastic absolute approximation has error:
|
||||
\[ \left\vert \prob{X | \matr{E}} - \hat{\mathcal{P}}(X | \matr{E}) \right\vert \leq \varepsilon \]
|
||||
Moreover, the method might fail (with greater error) with probability $\delta$.
|
||||
|
||||
\item[$\delta$-stochastic relative approximation]
|
||||
Given $\delta \in ]0, 0.5[$ and $\varepsilon \in ]0, 0.5[$, a $\delta$-stochastic relative approximation has error:
|
||||
\[ \frac{\left\vert \prob{X | \matr{E}} - \hat{\mathcal{P}}(X | \matr{E}) \right\vert}{\prob{X | \matr{E}}} \leq \varepsilon \]
|
||||
Moreover, the method might fail (with greater error) with probability $\delta$.
|
||||
\end{description}
|
||||
|
||||
\begin{theorem}
|
||||
Approximate inference is NP-hard for any $\delta, \epsilon < 0.5$.
|
||||
\end{theorem}
|
||||
|
||||
\item[Consistency] \marginnote{Consistency}
|
||||
A sampling method is consistent if:
|
||||
\[ \lim_{N \rightarrow \infty} \hat{\mathcal{P}}(x) = \prob{x} \]
|
||||
\end{description}
|
||||
|
||||
|
||||
|
||||
\section{Sampling from an empty network}
|
||||
\marginnote{Sampling from an empty network}
|
||||
|
||||
Sample each variable in topological order (i.e. from parents to children).
|
||||
|
||||
The probability $\mathcal{S}$ of sampling a specific event $x_1, \dots, x_n$ is given by the
|
||||
probability of the single events knowing their parents:
|
||||
\[ \mathcal{S}(x_1, \dots, x_n) = \prod_{i=1}^n \prob{x_i | \texttt{parents}(X_i)} = \prob{x_1, \dots, x_n} \]
|
||||
|
||||
\begin{theorem}
|
||||
Sampling from an empty network is consistent.
|
||||
|
||||
\begin{proof}
|
||||
Let $N$ be the number of samples and
|
||||
$\mathcal{N}(x_1, \dots, x_n)$ the number of times the event $x_1, \dots, x_n$ has been sampled.
|
||||
\[
|
||||
\begin{split}
|
||||
\lim_{N \rightarrow \infty} \hat{\mathcal{P}}(x_1, \dots, x_n) &=
|
||||
\lim_{N \rightarrow \infty} \frac{\mathcal{N}(x_1, \dots, x_n)}{N} \\
|
||||
&= \mathcal{S}(x_1, \dots, x_n) =
|
||||
\prob{x_1, \dots, x_n}
|
||||
\end{split}
|
||||
\]
|
||||
\end{proof}
|
||||
\end{theorem}
|
||||
|
||||
\begin{example}
|
||||
Given the following Bayesian network:
|
||||
\begin{center}
|
||||
\includegraphics[width=0.5\textwidth]{img/_approx_infer_example.pdf}
|
||||
\end{center}
|
||||
|
||||
A possible sampling order is \texttt{Cloudy}, \texttt{Sprinkler}, \texttt{Rain}, \texttt{WetGrass}.
|
||||
|
||||
Assuming that a random generator gives the sequence of probabilities $(0.4, 0.8, 0.1, 0.5)$,
|
||||
the sample will be:
|
||||
\[ \langle \prob{C}, \prob{S | C}, \prob{R | C}, \prob{W | S, R} \rangle \]
|
||||
\[ \langle C=\texttt{false}, \prob{S | C=\texttt{false}}, \prob{R | C=\texttt{false}}, \prob{W | S, R} \rangle \]
|
||||
\[ \langle C=\texttt{false}, S=\texttt{false}, R=\texttt{true}, \prob{W | S=\texttt{false}, R=\texttt{true}} \rangle \]
|
||||
\[ \langle C=\texttt{false}, S=\texttt{false}, R=\texttt{true}, W=\texttt{true} \rangle \]
|
||||
|
||||
Note that the adopted convention is the following:
|
||||
if $r$ it the probability given by a random generator and $\prob{X} = p$, $X = \texttt{true}$ if $r \leq p$.
|
||||
\end{example}
|
||||
|
||||
|
||||
|
||||
\section{Rejection sampling}
|
||||
\marginnote{Rejection sampling}
|
||||
|
||||
Given a known evidence $\matr{E}$, rejection sampling works as sampling from an empty network
|
||||
but removes any sample that does no agree with the evidence.
|
||||
|
||||
Obviously if $\prob{\matr{E}}$ is low, the majority of the samples will be discarded and
|
||||
more iterations are required to reach the desired number of samples.
|
||||
|
||||
\begin{theorem}
|
||||
Rejection sampling is consistent.
|
||||
|
||||
\begin{proof}
|
||||
Let $\mathcal{N}(\matr{X})$ be the number of times the event $\matr{X}$ has been sampled.
|
||||
\[
|
||||
\begin{split}
|
||||
\hat{\mathcal{P}}(\matr{X} | \matr{E}) &=
|
||||
\frac{\mathcal{N}(\matr{X}, \matr{E})}{\mathcal{N}(\matr{E})} \\
|
||||
&\approx \frac{\prob{\matr{X}, \matr{E}}}{\prob{\matr{E}}} =
|
||||
\prob{\matr{X} | \matr{E}}
|
||||
\end{split}
|
||||
\]
|
||||
The approximation derives from the consistency of sampling from an empty network.
|
||||
\end{proof}
|
||||
\end{theorem}
|
||||
|
||||
|
||||
|
||||
\section{Likelihood weighting}
|
||||
\marginnote{Likelihood weighting}
|
||||
|
||||
Given a known evidence $\matr{E}$, likelihood weighting samples non-evidence variables and
|
||||
weights each sample by the likelihood of the evidence.
|
||||
|
||||
The probability $\mathcal{S}$ of sampling a specific event $\matr{Z}$ and evidence $\matr{E}$ is given by the
|
||||
probability of the single events in $\matr{Z}$ knowing their parents:
|
||||
\[ \mathcal{S}(\matr{Z}, \matr{E}) = \prod_{z_i \in \matr{Z}} \prob{z_i | \texttt{parents}(z_i)} \]
|
||||
|
||||
The weight of a sample $(\matr{Z}, \matr{E})$ is given by the
|
||||
probability of the single events in $\matr{E}$ knowing their parents:
|
||||
\[ \text{w}(\matr{Z}, \matr{E}) = \prod_{e_i \in \matr{E}} \prob{e_i | \texttt{parents}(e_i)} \]
|
||||
|
||||
\begin{theorem}
|
||||
Likelihood weighting is consistent.
|
||||
|
||||
\begin{proof}
|
||||
The weighted sampling probability is given by:
|
||||
\[
|
||||
\begin{split}
|
||||
\mathcal{S}(\matr{Z}, \matr{E}) \cdot \text{w}(\matr{Z}, \matr{E})
|
||||
&= \prod_{z_i \in \matr{Z}} \prob{z_i | \texttt{parents}(z_i)} \cdot \prod_{e_i \in E} \prob{e_i | \texttt{parents}(e_i)} \\
|
||||
&= \prob{\matr{Z}, \matr{E}}
|
||||
\end{split}
|
||||
\]
|
||||
This is a consequence of the global semantics of Bayesian networks.
|
||||
\end{proof}
|
||||
\end{theorem}
|
||||
|
||||
\begin{example}
|
||||
Given the following Bayesian network:
|
||||
\begin{center}
|
||||
\includegraphics[width=0.5\textwidth]{img/_approx_infer_example.pdf}
|
||||
\end{center}
|
||||
|
||||
Knowing that $S=\texttt{true}$ and $W=\texttt{false}$,
|
||||
we sample in the order: \texttt{Cloudy}, \texttt{Rain}.
|
||||
|
||||
Assuming that a random generator gives the sequence of probabilities $(0.4, 0.1)$,
|
||||
the sample will be:
|
||||
\[ \langle \prob{C}, S=\texttt{true}, \prob{R | C}, W=\texttt{false} \rangle \]
|
||||
\[ \langle C=\texttt{true}, S=\texttt{true}, \prob{R | C=\texttt{true}}, W=\texttt{false} \rangle \]
|
||||
\[ \langle C=\texttt{true}, S=\texttt{true}, R=\texttt{true}, W=\texttt{false} \rangle \]
|
||||
|
||||
The weight associated to the sample is given by the probability of the evidence:
|
||||
\[
|
||||
\begin{split}
|
||||
\text{w} &= \prob{S=\texttt{true} | C=\texttt{true}} \cdot \prob{W=\texttt{false} | S=\texttt{true}, R=\texttt{true}} \\
|
||||
&= 0.1 \cdot (1 - 0.99) = 0.001
|
||||
\end{split}
|
||||
\]
|
||||
\end{example}
|
||||
|
||||
|
||||
|
||||
\section{Markov chain Monte Carlo}
|
||||
\marginnote{Markov chain Monte Carlo}
|
||||
|
||||
Sampling on a Markov chain where states contain an assignment to all variables.
|
||||
|
||||
Adjacent states of the Markov chain differ by only one variable.
|
||||
Therefore, the probability of an edge connecting two states is given by the probability of the updated variable known its Markov blanket:
|
||||
\[
|
||||
\prob{x_i | \texttt{markov\_blanket}(X_i)} =
|
||||
\prob{x_i | \texttt{parents}(X_i)} \cdot \prod_{Z_j \in \texttt{children}(x_i)} \prob{z_j | \texttt{parents}(Z_j)}
|
||||
\]
|
||||
|
||||
\begin{theorem}
|
||||
Markov chain Monte Carlo is consistent.
|
||||
|
||||
Note: nevertheless, it is difficult to tell if convergence has been achieved.
|
||||
|
||||
\begin{proof}
|
||||
Consequence of the fact that a long-run on a Markov chain converges to the posterior probability of the states.
|
||||
\end{proof}
|
||||
\end{theorem}
|
||||
|
||||
\begin{description}
|
||||
\item[Compiled network]
|
||||
A naive implementation of Markov chain Monte Carlo requires to repeatedly compute the probabilities with the Markov blanket.
|
||||
A solution is to compile the network into a model-specific inference code.
|
||||
\end{description}
|
||||
|
||||
\begin{example}
|
||||
Given the evidence $S=\texttt{true}$ and $W=\texttt{true}$,
|
||||
the structure of the Markov chain can be defined as follows:
|
||||
\begin{center}
|
||||
\includegraphics[width=0.5\textwidth]{img/_markov_chain_sampling.pdf}
|
||||
\end{center}
|
||||
\end{example}
|
||||
@ -0,0 +1,557 @@
|
||||
\chapter{Bayesian networks}
|
||||
|
||||
|
||||
\section{Bayes' rule}
|
||||
|
||||
\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}
|
||||
\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 have been 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 are no active trails 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 the 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}
|
||||
|
||||
\subsection{Algorithm}
|
||||
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=0.15\linewidth]{img/_monty_hall1.pdf}
|
||||
\caption{First interaction}
|
||||
\end{subfigure}
|
||||
\begin{subfigure}{.3\textwidth}
|
||||
\centering
|
||||
\includegraphics[width=0.45\linewidth]{img/_monty_hall2.pdf}
|
||||
\caption{Second interaction (note that $P \perp G$)}
|
||||
\end{subfigure}
|
||||
\begin{subfigure}{.3\textwidth}
|
||||
\centering
|
||||
\includegraphics[width=0.45\linewidth]{img/_monty_hall3.pdf}
|
||||
\caption{Third interaction}
|
||||
\end{subfigure}
|
||||
\end{figure}
|
||||
\end{example}
|
||||
|
||||
The nodes of the resulting network can be classified as:
|
||||
\begin{descriptionlist}
|
||||
\item[Initial evidence] The initial observation.
|
||||
\item[Testable variables] Variables that can be verified.
|
||||
\item[Operable variables] Variables that can be changed by intervening on them.
|
||||
\item[Hidden variables] Variables that "compress" more variables to reduce the parameters.
|
||||
\end{descriptionlist}
|
||||
|
||||
\begin{example} \phantom{}\\
|
||||
\begin{minipage}{.4\linewidth}
|
||||
\begin{description}
|
||||
\item[Initial evidence] Red.
|
||||
\item[Testable variables] Green.
|
||||
\item[Operable variables] Orange.
|
||||
\item[Hidden variables] Gray.
|
||||
\end{description}
|
||||
\end{minipage}
|
||||
\begin{minipage}{.5\linewidth}
|
||||
\begin{center}
|
||||
\includegraphics[width=\linewidth]{img/_car_example.pdf}
|
||||
\end{center}
|
||||
\end{minipage}
|
||||
\end{example}
|
||||
|
||||
|
||||
\subsection{Structure learning}
|
||||
\marginnote{Structure learning}
|
||||
Learn the network from the available data.
|
||||
\begin{description}
|
||||
\item[Constraint-based]
|
||||
Independence tests to identify the constraints of the edges.
|
||||
\item[Score-based]
|
||||
Define a score to evaluate the network.
|
||||
\end{description}
|
||||
|
||||
|
||||
|
||||
\section{Causal networks}
|
||||
When building a Bayesian network, a correct ordering of the nodes
|
||||
that respects the causality allows to obtain more compact networks.
|
||||
|
||||
\begin{description}
|
||||
\item[Structural equation] \marginnote{Structural equation}
|
||||
Given a variable $X_i$ with values $x_i$, its structural equation is a function $f_i$
|
||||
such that it represents all its possible values:
|
||||
\[ x_i = f_i(\text{other variables}, U_i) \]
|
||||
$U_i$ represents unmodeled variables or error terms.
|
||||
|
||||
\item[Causal network] \marginnote{Causal network}
|
||||
Restricted class of Bayesian networks that only allows causally compatible ordering.
|
||||
|
||||
An edge exists between $X_j \rightarrow X_i$ iff $X_j$ is an argument of
|
||||
the structural equation $f_i$ of $X_i$.
|
||||
|
||||
\begin{example} \phantom{}\\[0.5em]
|
||||
\begin{minipage}{.3\linewidth}
|
||||
\centering
|
||||
\includegraphics[width=\linewidth]{img/_causal_network_example1.pdf}
|
||||
\end{minipage}
|
||||
\begin{minipage}{.6\linewidth}
|
||||
The structural equations are:
|
||||
\[
|
||||
\begin{split}
|
||||
\texttt{cloudy} &= f_C(U_C) \\
|
||||
\texttt{sprinkler} &= f_S(\texttt{Cloudy}, U_S) \\
|
||||
\texttt{rain} &= f_R(\texttt{Cloudy}, U_R) \\
|
||||
\texttt{wet\_grass} &= f_W(\texttt{Sprinkler}, \texttt{Rain}, U_W) \\
|
||||
\texttt{greener\_grass} &= f_G(\texttt{WetGrass}, U_G)
|
||||
\end{split}
|
||||
\]
|
||||
\end{minipage}\\[0.5em]
|
||||
|
||||
If the sprinkler is disabled, the network becomes:\\[0.5em]
|
||||
\begin{minipage}{.3\linewidth}
|
||||
\centering
|
||||
\includegraphics[width=\linewidth]{img/_causal_network_example2.pdf}
|
||||
\end{minipage}
|
||||
\begin{minipage}{.6\linewidth}
|
||||
The structural equations become:
|
||||
\[
|
||||
\begin{split}
|
||||
\texttt{cloudy} &= f_C(U_C) \\
|
||||
\texttt{sprinkler} &= f_S(U_S) \\
|
||||
\texttt{rain} &= f_R(\texttt{Cloudy}, U_R) \\
|
||||
\texttt{wet\_grass} &= f_W(\texttt{Rain}, U_W) \\
|
||||
\texttt{greener\_grass} &= f_G(\texttt{WetGrass}, U_G)
|
||||
\end{split}
|
||||
\]
|
||||
\end{minipage}
|
||||
\end{example}
|
||||
|
||||
\item[do-operator] \marginnote{do-operator}
|
||||
The do-operator allows to represent manual interventions on the network.
|
||||
The operation $\texttt{do}(X_i = x_i)$ makes the structural equation of $X_i$
|
||||
constant (i.e. $f_i = x_i$, without arguments, so there won't be inward edges to $X_i$).
|
||||
|
||||
\begin{example} \phantom{}\\[0.5em]
|
||||
\begin{minipage}{.3\linewidth}
|
||||
\centering
|
||||
\includegraphics[width=\linewidth]{img/_do_operator_example1.pdf}
|
||||
\end{minipage}
|
||||
\begin{minipage}{.65\linewidth}
|
||||
By applying $\texttt{do}(\texttt{Sprinkler} = \texttt{true})$, the structural equations become:
|
||||
\[
|
||||
\begin{split}
|
||||
\texttt{cloudy} &= f_C(U_C) \\
|
||||
\texttt{sprinkler} &= \texttt{true} \\
|
||||
\texttt{rain} &= f_R(\texttt{Cloudy}, U_R) \\
|
||||
\texttt{wet\_grass} &= f_W(\texttt{Sprinkler}, \texttt{Rain}, U_W) \\
|
||||
\texttt{greener\_grass} &= f_G(\texttt{WetGrass}, U_G)
|
||||
\end{split}
|
||||
\]
|
||||
\end{minipage}\\[0.5em]
|
||||
|
||||
\begin{minipage}{.3\linewidth}
|
||||
\centering
|
||||
\includegraphics[width=\linewidth]{img/_do_operator_example2.pdf}
|
||||
\end{minipage}
|
||||
\begin{minipage}{.65\linewidth}
|
||||
Note that Bayesian networks are not capable of modelling manual interventions.
|
||||
In fact, intervening and observing a variable are different concepts:
|
||||
\[ \prob{\texttt{WetGrass} \mid \texttt{do}(\texttt{Sprinkler} = \texttt{true})} \]
|
||||
\[ \neq \]
|
||||
\[ \prob{\texttt{WetGrass} \mid \texttt{Sprinkler} = \texttt{true}} \]
|
||||
\end{minipage}
|
||||
\end{example}
|
||||
\end{description}
|
||||
|
||||
|
||||
|
||||
\section{Compact conditional distributions}
|
||||
|
||||
Use canonical distributions (standard patterns) to reduce
|
||||
the number of variables in a conditional probability table.
|
||||
|
||||
|
||||
\subsection{Noisy-OR}
|
||||
\marginnote{Noisy-OR}
|
||||
Noisy-OR distributions model a network of non-interacting causes with a common effect.
|
||||
A node $X$ has $k$ parents $U_1, \dots, U_k$ and possibly a leak node $U_L$ to capture unmodeled concepts.
|
||||
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
\includegraphics[width=0.3\textwidth]{img/_noisy_or_example.pdf}
|
||||
\caption{Example of noisy-OR network}
|
||||
\end{figure}
|
||||
|
||||
Each node $U_i$ has a failure (inhibition) probability $q_i$:
|
||||
\[ q_i = \prob{\lnot x \mid u_i, \lnot u_j \text{ for } j \neq i} \]
|
||||
The CPT can be built by computing the probabilities as:
|
||||
\[ \prob{\lnot x \mid \texttt{Parents($X$)}} = \prod_{j:\, U_j = \texttt{true}} q_j \]
|
||||
In other words:
|
||||
\[ \prob{\lnot x \mid u_1, \dots, u_n} =
|
||||
\prob{\lnot x \mid u_1} \cdot \prob{\lnot x \mid u_2} \cdot \text{\dots} \cdot \prob{\lnot x \mid u_n} \]
|
||||
|
||||
Because only the failure probabilities are required, the number of parameters is linear in the number of parents.
|
||||
|
||||
\begin{example}
|
||||
We have as causes \texttt{Cold}, \texttt{Flu} and \texttt{Malaria} and as effect \texttt{Fever}.
|
||||
For simplicity there are no leak nodes.
|
||||
The failure probabilities are:
|
||||
\[
|
||||
\begin{split}
|
||||
q_\texttt{cold} &= \prob{\lnot \texttt{fever} \mid \texttt{cold}, \lnot\texttt{flu}, \lnot\texttt{malaria}} = 0.6 \\
|
||||
q_\texttt{flu} &= \prob{\lnot \texttt{fever} \mid \lnot\texttt{cold}, \texttt{flu}, \lnot\texttt{malaria}} = 0.2 \\
|
||||
q_\texttt{malaria} &= \prob{\lnot \texttt{fever} \mid \lnot\texttt{cold}, \lnot\texttt{flu}, \texttt{malaria}} = 0.1
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
Known the failure probabilities, the entire CPT can be computed:
|
||||
\begin{center}
|
||||
\begin{tabular}{c|c|c|rc|c}
|
||||
\hline
|
||||
\texttt{Cold} & \texttt{Flu} & \texttt{Malaria} & \multicolumn{2}{c|}{$\prob{\lnot\texttt{fever}}$} & $1-\prob{\lnot\texttt{fever}}$ \\
|
||||
\hline
|
||||
F & F & F & & 0.0 & 1.0 \\
|
||||
F & F & T & $q_\texttt{malaria} =$ & 0.1 & 0.9 \\
|
||||
F & T & F & $q_\texttt{flu} =$ & 0.2 & 0.8 \\
|
||||
F & T & T & $q_\texttt{flu} \cdot q_\texttt{malaria} =$ & 0.02 & 0.98 \\
|
||||
T & F & F & $q_\texttt{cold} =$ & 0.6 & 0.4 \\
|
||||
T & F & T & $q_\texttt{cold} \cdot q_\texttt{malaria} =$ & 0.06 & 0.94 \\
|
||||
T & T & F & $q_\texttt{cold} \cdot q_\texttt{flu} =$ & 0.12 & 0.88 \\
|
||||
T & T & T & $q_\texttt{cold} \cdot q_\texttt{flu} \cdot q_\texttt{malaria} =$ & 0.012 & 0.988 \\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\end{example}
|
||||
|
||||
|
||||
\subsection{Hybrid Bayesian networks}
|
||||
\marginnote{Hybrid Bayesian networks}
|
||||
|
||||
Network with discrete and continuous random variables.
|
||||
Continuous variables must be converted into a finite representation.
|
||||
Possible approaches are:
|
||||
\begin{description}
|
||||
\item[Discretization] \marginnote{Discretization}
|
||||
Values are divided into a fixed set of intervals.
|
||||
This approach may introduce large errors and large CPTs.
|
||||
|
||||
\item[Finitely parametrized canonical families] \marginnote{Finitely parametrized canonical families}
|
||||
There are two cases to handle using this approach:
|
||||
\begin{descriptionlist}
|
||||
\item[Continuous child]
|
||||
Given the continuous variables $X$ and $C$ and a discrete (boolean, for simplicity) variable $D$,
|
||||
we want to compute the distribution $\textbf{P}(X \mid C, D)$.
|
||||
|
||||
The discrete parent is handled by enumeration, by computing the probability over the domain of $D$.
|
||||
|
||||
For the continuous parent, an arbitrarily chosen distribution over the values of $X$ is used.
|
||||
A common choice is the \textbf{linear Gaussian} \marginnote{Linear Gaussian}
|
||||
whose mean is a linear combination of the values of the parents and the variance is fixed.
|
||||
|
||||
A network with all continuous linear Gaussian distributions has the property
|
||||
of having a multivariate Gaussian distribution as joint distribution.
|
||||
Moreover, if a continuous variable has some discrete parents, it defines a conditional Gaussian distribution
|
||||
where, fixed the values of the discrete variables, the distribution over the continuous variable is a multivariate Gaussian.
|
||||
|
||||
\begin{example}
|
||||
Let \texttt{Subsidy} and \texttt{Buys} be discrete variables and
|
||||
\texttt{Harvest} and \texttt{Cost} be continuous variables.
|
||||
\begin{center}
|
||||
\includegraphics[width=0.3\textwidth]{img/_linear_gaussian_example.pdf}
|
||||
\end{center}
|
||||
|
||||
To compute $\textbf{P}(\texttt{Cost} \mid \texttt{Harvest}, \texttt{Subsidy})$,
|
||||
we split the probabilities over the values of the discrete variable \texttt{Subsidy}
|
||||
and use a linear Gaussian for \texttt{Harvest}.
|
||||
We therefore have that:
|
||||
\[
|
||||
\begin{split}
|
||||
\prob{\texttt{C} = \texttt{c} \mid \texttt{Harvest} = \texttt{h}, \texttt{Subsidy} = \texttt{true}}
|
||||
&= \mathcal{N}(a_t h + b_t, \sigma_t)(c) \\
|
||||
\prob{\texttt{C} = \texttt{c} \mid \texttt{Harvest} = \texttt{h}, \texttt{Subsidy} = \texttt{false}}
|
||||
&= \mathcal{N}(a_f h + b_f, \sigma_f)(c)
|
||||
\end{split}
|
||||
\]
|
||||
where $a_t$, $b_t$, $\sigma_t$, $a_f$, $b_f$ and $\sigma_t$ are parameters.
|
||||
\end{example}
|
||||
|
||||
\item[Discrete child with continuous parents]
|
||||
Given the continuous variable $C$ and a discrete variable $X$,
|
||||
the probability of $X$ given $C$ in obtained by using a threshold function.
|
||||
For instance, probit or sigmoid distributions can be used.
|
||||
\end{descriptionlist}
|
||||
\end{description}
|
||||
|
||||
|
||||
\subsection{Other methods}
|
||||
|
||||
\begin{description}
|
||||
\item[Dynamic Bayesian network] \marginnote{Dynamic Bayesian network}
|
||||
Useful to model the evolution through time.
|
||||
A template variable $X_i$ is instantiated as $X_i^{(t)}$ at each time step.
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
\includegraphics[width=0.3\textwidth]{img/_dynamic_bn_example.pdf}
|
||||
\caption{Example of dynamic Bayesian network}
|
||||
\end{figure}
|
||||
|
||||
\item[Density estimation] \marginnote{Density estimation}
|
||||
Parameters of the conditional distribution can be learned using:
|
||||
\begin{description}
|
||||
\item[Bayesian learning] calculate the probability of each hypothesis.
|
||||
\item[Approximations] using the maximum-a-posteriori and maximum-likelihood hypothesis.
|
||||
\item[Expectation-maximization algorithm{\normalfont.}]
|
||||
\end{description}
|
||||
|
||||
\item[Undirected graphical models] \marginnote{Undirected graphical models}
|
||||
Markov networks are an alternative to probabilistic graphical models (as Bayesian networks).
|
||||
Markov networks are undirected graphs with factors (instead of probabilities) and
|
||||
are able to naturally capture independence relations.
|
||||
\end{description}
|
||||
@ -0,0 +1,118 @@
|
||||
\chapter{Exact inference}
|
||||
|
||||
|
||||
\section{Inference by enumeration}
|
||||
\marginnote{Inference by enumeration}
|
||||
|
||||
Method to sum out a joint probability without explicitly representing it
|
||||
by using CPT entries.
|
||||
|
||||
Enumeration follows a depth-first exploration and has a space complexity of $O(n)$
|
||||
and time complexity of $O(d^n)$.
|
||||
It must be noted that some probabilities appear multiple times but
|
||||
require to be recomputed because of the definition of the algorithm.
|
||||
|
||||
\begin{example}[Burglary]
|
||||
Given the Bayesian network:
|
||||
\begin{center}
|
||||
\includegraphics[width=0.15\textwidth]{img/_burglary_net.pdf}
|
||||
\end{center}
|
||||
We want to compute $\textbf{P}(B \mid j, m)$:
|
||||
\[
|
||||
\begin{split}
|
||||
\textbf{P}(B \mid j, m) &= \frac{\textbf{P}(B, j, m)}{\prob{j, m}} \\
|
||||
&= \alpha \textbf{P}(B, j, m) \\
|
||||
&= \alpha \sum_{e} \sum_{a} \textbf{P}(B, j, m, e, a) \\
|
||||
&= \alpha \sum_{e} \sum_{a} \textbf{P}(B) \prob{e} \textbf{P}(a \mid B, e) \prob{j \mid a} \prob{m \mid a} \\
|
||||
&= \alpha \textbf{P}(B) \sum_{e} \prob{e} \sum_{a} \textbf{P}(a \mid B, e) \prob{j \mid a} \prob{m \mid a} \\
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
This can be represented using a tree:
|
||||
\begin{center}
|
||||
\includegraphics[width=0.75\textwidth]{img/_burglary_enumeration.pdf}
|
||||
\end{center}
|
||||
\end{example}
|
||||
|
||||
|
||||
\section{Inference by variable elimination}
|
||||
\marginnote{Inference by variable elimination}
|
||||
Method that carries out summations right-to-left and stores intermediate results (called factors).
|
||||
|
||||
\begin{description}
|
||||
\item[Pointwise product of factors] $f(X, Y) \times g(Y, Z) = p(X, Y, Z)$
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
\includegraphics[width=0.5\textwidth]{img/_pointwise_factors.pdf}
|
||||
\caption{Example of pointwise product}
|
||||
\end{figure}
|
||||
|
||||
\item[Summing out]
|
||||
To sum out a variable $X$ from a product of factors:
|
||||
\begin{enumerate}
|
||||
\item Move constant factors outside (i.e. factors that do not depend on $X$).
|
||||
\item Compute the pointwise product of the remaining terms.
|
||||
\end{enumerate}
|
||||
|
||||
\begin{example}
|
||||
\[
|
||||
\begin{split}
|
||||
\sum_X f_1 \times \dots \times f_k &= f_1 \times \dots \times f_i \sum_X f_{i+1} \times \dots \times f_k \\
|
||||
&= f_1 \times \dots \times f_i \times f_X
|
||||
\end{split}
|
||||
\]
|
||||
\end{example}
|
||||
\end{description}
|
||||
|
||||
\begin{example}[Burglary]
|
||||
Given the Bayesian network:
|
||||
\begin{center}
|
||||
\includegraphics[width=0.15\textwidth]{img/_burglary_net.pdf}
|
||||
\end{center}
|
||||
We want to compute
|
||||
$\textbf{P}(B \mid j, m) = \alpha \textbf{P}(B) \sum_{e} \prob{e} \sum_{a} \textbf{P}(a \mid B, e) \prob{j \mid a} \prob{m \mid a}$.
|
||||
|
||||
We first work on the summation on $A$.
|
||||
We introduce as factors the entries of the CPT:
|
||||
\[ \textbf{P}(B \mid j, m) = \alpha \textbf{P}(B) \sum_{e} \prob{e} \sum_{a} f_A(a, b, e) f_J(a) f_M(a) \]
|
||||
Note that $j$ and $m$ are not parameters of the factors $f_J$ and $f_M$ because they are already given.
|
||||
We then sum out on $A$:
|
||||
\[ \textbf{P}(B \mid j, m) = \alpha \textbf{P}(B) \sum_{e} \prob{e} f_{AJM}(b, e) \]
|
||||
|
||||
Now, we repeat the same process and sum out $E$:
|
||||
\[ \textbf{P}(B \mid j, m) = \alpha \textbf{P}(B) f_{EAJM}(b) \]
|
||||
|
||||
At last, we factor $\textbf{P}(B)$:
|
||||
\[ \textbf{P}(B \mid j, m) = \alpha f_B(b) f_{EAJM}(b) \]
|
||||
\end{example}
|
||||
|
||||
|
||||
\subsection{Irrelevant variables}
|
||||
\marginnote{Irrelevant variables}
|
||||
A variable $X$ is irrelevant if summing over it results in a probability of $1$.
|
||||
|
||||
\begin{theorem}
|
||||
Given a query $X$, the evidence $\matr{E}$ and a variable $Y$:
|
||||
\[ Y \notin (\texttt{Ancestors($\{ X \}$)} \cup \texttt{Ancestors($\matr{E}$)}) \rightarrow Y \text{ is irrelevant} \]
|
||||
\end{theorem}
|
||||
|
||||
\begin{theorem}
|
||||
Given a query $X$, the evidence $\matr{E}$ and a variable $Y$:
|
||||
\[ Y \text{ d-separated from } X \text{ by } \matr{E} \rightarrow Y \text{ is irrelevant} \]
|
||||
\end{theorem}
|
||||
|
||||
|
||||
\subsection{Complexity}
|
||||
\begin{description}
|
||||
\item[Singly connected networks]
|
||||
Network where any two nodes are connected with at most one undirected path.
|
||||
Time and space complexity is $O(d^k n)$.
|
||||
\item[Multiply connected networks] The problem is NP-hard.
|
||||
\end{description}
|
||||
|
||||
|
||||
\section{Clustering algorithm}
|
||||
\marginnote{Clustering algorithm}
|
||||
|
||||
Method that joins individual nodes to form clusters.
|
||||
Allows to estimate the posterior probabilities for $n$ variables with complexity $O(n)$.
|
||||
@ -0,0 +1,48 @@
|
||||
\chapter{Introduction}
|
||||
|
||||
|
||||
\section{Uncertainty}
|
||||
\begin{description}
|
||||
\item[Uncertainty] \marginnote{Uncertainty}
|
||||
A task is uncertain if it has:
|
||||
\begin{itemize}
|
||||
\item Partial observations
|
||||
\item Noisy or wrong information
|
||||
\item Uncertain outcomes of the actions
|
||||
\item Complex models
|
||||
\end{itemize}
|
||||
|
||||
A purely logic approach leads to:
|
||||
\begin{itemize}
|
||||
\item Risks falsehood: unreasonable conclusion when applied in practice.
|
||||
\item Weak decisions: too many conditions required to make a conclusion.
|
||||
\end{itemize}
|
||||
\end{description}
|
||||
|
||||
|
||||
\subsection{Handling uncertainty}
|
||||
\begin{descriptionlist}
|
||||
\item[Default/non-monotonic logic] \marginnote{Default/non-monotonic logic}
|
||||
Works on assumptions.
|
||||
An assumption can be contradicted by the evidence.
|
||||
|
||||
\item[Rule-based systems with fudge factors] \marginnote{Rule-based systems with fudge factors}
|
||||
Formulated as premise $\rightarrow_\text{prob.}$ effect.
|
||||
Have the following issues:
|
||||
\begin{itemize}
|
||||
\item Locality: how can the probability account all the evidence.
|
||||
\item Combination: chaining of unrelated concepts.
|
||||
\end{itemize}
|
||||
|
||||
\item[Probability] \marginnote{Probability}
|
||||
Assign a probability given the available known evidence.
|
||||
|
||||
Note: fuzzy logic handles the degree of truth and not the uncertainty.
|
||||
\end{descriptionlist}
|
||||
|
||||
\begin{description}
|
||||
\item[Decision theory] \marginnote{Decision theory}
|
||||
Defined as:
|
||||
\[ \text{Decision theory} = \text{Utility theory} + \text{Probability theory} \]
|
||||
where the utility theory depends on one's preferences.
|
||||
\end{description}
|
||||
@ -0,0 +1,236 @@
|
||||
\chapter{Probability}
|
||||
|
||||
\begin{description}
|
||||
\item[Sample space] \marginnote{Sample space}
|
||||
Set $\Omega$ of all possible worlds.
|
||||
\begin{descriptionlist}
|
||||
\item[Event] \marginnote{Event}
|
||||
Subset $A \subseteq \Omega$.
|
||||
\item[Sample point/Possible world/Atomic event] \marginnote{Sample point}
|
||||
Element $\omega \in \Omega$.
|
||||
\end{descriptionlist}
|
||||
|
||||
\item[Probability space] \marginnote{Probability space}
|
||||
A probability space/model is a function $\prob{\cdot}: \Omega \rightarrow [0, 1]$ assigned to a sample space such that:
|
||||
\begin{itemize}
|
||||
\item $0 \leq \prob{\omega} \leq 1$
|
||||
\item $\sum_{\omega \in \Omega} \prob{\omega} = 1$
|
||||
\item $\prob{A} = \sum_{\omega \in A} \prob{\omega}$
|
||||
\end{itemize}
|
||||
|
||||
\item[Random variable] \marginnote{Random variable}
|
||||
A function from an event to some range (e.g. reals, booleans, \dots).
|
||||
|
||||
\item[Probability distribution] \marginnote{Probability distribution}
|
||||
For any random variable $X$:
|
||||
\[ \prob{X = x_i} = \sum_{\omega \text{ s.t. } X(\omega)=x_i} \prob{\omega} \]
|
||||
|
||||
\item[Proposition] \marginnote{Proposition}
|
||||
Event where a random variable has a certain value.
|
||||
\[ a = \{ \omega \,\vert\, A(\omega) = \texttt{true} \} \]
|
||||
\[ \lnot a = \{ \omega \,\vert\, A(\omega) = \texttt{false} \} \]
|
||||
\[ (\texttt{Weather} = \texttt{rain}) = \{ \omega \,\vert\, B(\omega) = \texttt{rain} \} \]
|
||||
|
||||
\item[Prior probability] \marginnote{Prior probability}
|
||||
Prior/unconditional probability of a proposition based on known evidence.
|
||||
|
||||
\item[Probability distribution (all)] \marginnote{Probability distribution (all)}
|
||||
Gives all the probabilities of a random variable.
|
||||
\[ \textbf{P}(A) = \langle \prob{A=a_1}, \dots, \prob{A=a_n} \rangle \]
|
||||
|
||||
\item[Joint probability distribution] \marginnote{Joint probability distribution}
|
||||
The joint probability distribution of a set of random variables gives
|
||||
the probability of all the different combinations of their atomic events.
|
||||
|
||||
Note: Every question on a domain can, in theory, be answered using the joint distribution.
|
||||
In practice, it is hard to apply.
|
||||
|
||||
\begin{example}
|
||||
$\textbf{P}(\texttt{Weather}, \texttt{Cavity}) = $
|
||||
\begin{center}
|
||||
\small
|
||||
\begin{tabular}{|c | c|c|c|c|}
|
||||
\cline{2-5}
|
||||
\multicolumn{1}{c|}{} & \texttt{Weather=sunny} & \texttt{Weather=rain} & \texttt{Weather=cloudy} & \texttt{Weather=snow} \\
|
||||
\hline
|
||||
\texttt{Cavity=true} & 0.144 & 0.02 & 0.016 & 0.02 \\
|
||||
\hline
|
||||
\texttt{Cavity=false} & 0.576 & 0.08 & 0.064 & 0.08 \\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\end{example}
|
||||
|
||||
\item[Probability density function] \marginnote{Probability density function}
|
||||
The probability density function (PDF) of a random variable $X$ is a function $p: \mathbb{R} \rightarrow \mathbb{R}$
|
||||
such that:
|
||||
\[ \int_{\mathcal{T}_X} p(x) \,dx = 1 \]
|
||||
\begin{descriptionlist}
|
||||
\item[Uniform distribution] \marginnote{Uniform distribution}
|
||||
\[
|
||||
p(x) = \text{Unif}[a, b](x) =
|
||||
\begin{cases}
|
||||
\frac{1}{b-a} & a \leq x \leq b \\
|
||||
0 & \text{otherwise}
|
||||
\end{cases}
|
||||
\]
|
||||
\item[Gaussian (normal) distribution] \marginnote{Gaussian (normal) distribution}
|
||||
\[ \mathcal{N}(\mu, \sigma^2) = \frac{1}{\sigma\sqrt{2\pi}}e^{\frac{-(x-\mu)^2}{2\sigma^2}} \]
|
||||
|
||||
$\mathcal{N}(0, 1)$ is the standard Gaussian.
|
||||
\end{descriptionlist}
|
||||
|
||||
\item[Conditional probability] \marginnote{Conditional probability}
|
||||
Probability of a prior knowledge with new evidence:
|
||||
\[ \prob{a \vert b} = \frac{\prob{a \land b}}{\prob{b}} \]
|
||||
The product rule gives an alternative formulation:
|
||||
\[ \prob{a \land b} = \prob{a \vert b}{\prob{b}} = \prob{b \vert a}{\prob{a}} \]
|
||||
|
||||
\begin{description}
|
||||
\item[Chain rule] \marginnote{Chain rule}
|
||||
Successive application of the product rule:
|
||||
\[
|
||||
\begin{split}
|
||||
\textbf{P}(X_1, \dots, X_n) &= \textbf{P}(X_1, \dots, X_{n-1}) \textbf{P}(X_n \vert X_1, \dots, X_{n-1}) \\
|
||||
&= \textbf{P}(X_1, \dots, X_{n-2}) \textbf{P}(X_{n-1} \vert X_1, \dots, X_{n-2}) \textbf{P}(X_n \vert X_1, \dots, X_{n-1}) \\
|
||||
&= \prod_{i=1}^{n} \textbf{P}(X_i \vert X_1, \dots, X_{i-1})
|
||||
\end{split}
|
||||
\]
|
||||
\end{description}
|
||||
|
||||
\item[Independence] \marginnote{Independence}
|
||||
Two random variables $A$ and $B$ are independent ($A \perp B$) iff:
|
||||
\[
|
||||
\textbf{P}(A \vert B) = \textbf{P}(A) \,\text{ or }\,
|
||||
\textbf{P}(B \vert A) = \textbf{P}(B) \,\text{ or }\,
|
||||
\textbf{P}(A, B) = \textbf{P}(A)\textbf{P}(B)
|
||||
\]
|
||||
|
||||
\item[Conditional independence] \marginnote{Conditional independence}
|
||||
Two random variables $A$ and $B$ are conditionally independent iff:
|
||||
\[ \textbf{P}(A \,\vert\, C, B) = \textbf{P}(A \,\vert\, C) \]
|
||||
\end{description}
|
||||
|
||||
|
||||
|
||||
\section{Inference with full joint distributions}
|
||||
Given a joint distribution, the probability of any proposition $\phi$
|
||||
can be computed as the sum of the atomic events where $\phi$ is true:
|
||||
\[ \prob{\phi} = \sum_{\omega:\, \omega \models \phi} \prob{\omega} \]
|
||||
|
||||
\begin{example}
|
||||
Given the following joint distribution:
|
||||
\begin{center}
|
||||
\begin{tabular}{|c|c|c|c|c|}
|
||||
\cline{2-5}
|
||||
\multicolumn{1}{c|}{} & \multicolumn{2}{c|}{\texttt{toothache}} & \multicolumn{2}{c|}{$\lnot$\texttt{toothache}} \\
|
||||
\cline{2-5}
|
||||
\multicolumn{1}{c|}{} & \texttt{catch} & $\lnot$\texttt{catch} & \texttt{catch} & $\lnot$\texttt{catch} \\
|
||||
\hline
|
||||
\texttt{cavity} & 0.108 & 0.012 & 0.072 & 0.008 \\
|
||||
\hline
|
||||
$\lnot$\texttt{cavity} & 0.016 & 0.064 & 0.144 & 0.576 \\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
|
||||
We have that:
|
||||
\begin{itemize}
|
||||
\item $\prob{\texttt{toothache}} = 0.108 + 0.012 + 0.016 + 0.064 = 0.2$
|
||||
\item $\prob{\texttt{cavity} \vee \texttt{toothache}} = 0.108 + 0.012 + 0.072 + 0.008 + 0.016 + 0.064 = 0.28$
|
||||
\item $\prob{\lnot\texttt{cavity} \,\vert\, \texttt{toothache}} = \frac{\prob{\lnot\texttt{cavity} \land \texttt{toothache}}}{\prob{\texttt{toothache}}} =
|
||||
\frac{0.016 + 0.064}{0.2} = 0.4$
|
||||
\end{itemize}
|
||||
\end{example}
|
||||
|
||||
\begin{description}
|
||||
\item[Marginalization] \marginnote{Marginalization}
|
||||
The probability that a random variable assumes a specific value is given by
|
||||
the sum off all the joint probabilities where that random variable assumes the given value.
|
||||
\begin{example}
|
||||
Given the joint distribution:
|
||||
\begin{center}
|
||||
\small
|
||||
\begin{tabular}{|c | c|c|c|c|}
|
||||
\cline{2-5}
|
||||
\multicolumn{1}{c|}{} & \texttt{Weather=sunny} & \texttt{Weather=rain} & \texttt{Weather=cloudy} & \texttt{Weather=snow} \\
|
||||
\hline
|
||||
\texttt{Cavity=true} & 0.144 & 0.02 & 0.016 & 0.02 \\
|
||||
\hline
|
||||
\texttt{Cavity=false} & 0.576 & 0.08 & 0.064 & 0.08 \\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
We have that $\prob{\texttt{Weather}=\texttt{sunny}} = 0.144 + 0.576$
|
||||
\end{example}
|
||||
\item[Conditioning] \marginnote{Conditioning}
|
||||
Adding a condition to a probability (reduction and renormalization).
|
||||
|
||||
\item[Normalization] \marginnote{Normalization}
|
||||
Given a conditional probability distribution $\textbf{P}(A \vert B)$,
|
||||
it can be formulated as:
|
||||
\[ \textbf{P}(A \vert B) = \alpha\textbf{P}(A, B) \]
|
||||
where $\alpha$ is a normalization constant.
|
||||
In fact, fixed the evidence $B$, the denominator to compute the conditional probability is the same for each probability.
|
||||
|
||||
\begin{example}
|
||||
Given the joint distribution:
|
||||
\begin{center}
|
||||
\begin{tabular}{|c|c|c|c|c|}
|
||||
\cline{2-5}
|
||||
\multicolumn{1}{c|}{} & \multicolumn{2}{c|}{\texttt{toothache}} & \multicolumn{2}{c|}{$\lnot$\texttt{toothache}} \\
|
||||
\cline{2-5}
|
||||
\multicolumn{1}{c|}{} & \texttt{catch} & $\lnot$\texttt{catch} & \texttt{catch} & $\lnot$\texttt{catch} \\
|
||||
\hline
|
||||
\texttt{cavity} & 0.108 & 0.012 & 0.072 & 0.008 \\
|
||||
\hline
|
||||
$\lnot$\texttt{cavity} & 0.016 & 0.064 & 0.144 & 0.576 \\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
|
||||
We have that:
|
||||
\[
|
||||
\textbf{P}(\texttt{Cavity} \vert \texttt{toothache}) =
|
||||
\langle
|
||||
\frac{\prob{\texttt{cavity}, \texttt{toothache}, \texttt{catch}}}{\prob{\texttt{toothache}}},
|
||||
\frac{\prob{\lnot\texttt{cavity}, \texttt{toothache}, \lnot\texttt{catch}}}{\prob{\texttt{toothache}}}
|
||||
\rangle
|
||||
\]
|
||||
\end{example}
|
||||
|
||||
\item[Probability query] \marginnote{Probability query}
|
||||
Given a set of query variables $\bm{Y}$, the evidence variables $\vec{e}$ and the other hidden variables $\bm{H}$,
|
||||
the probability of the query can be computed as:
|
||||
\[
|
||||
\textbf{P}(\bm{Y} \vert \bm{E}=\vec{e}) = \alpha \textbf{P}(\bm{Y}, \bm{E}=\vec{e})
|
||||
= \alpha \sum_{\vec{h}} \textbf{P}(\bm{Y}, \bm{E}=\vec{e}, \bm{H}=\vec{h})
|
||||
\]
|
||||
The problem of this approach is that it has exponential time and space complexity
|
||||
that makes it not applicable in practice.
|
||||
|
||||
To reduce the size of the variables, conditional independence can be exploited.
|
||||
\begin{example}
|
||||
Knowing that $\textbf{P} \models (\texttt{Catch} \perp \texttt{Toothache} \vert \texttt{Cavity})$,
|
||||
we can compute the distribution $\textbf{P}(\texttt{Toothache}, \texttt{Catch}, \texttt{Cavity})$ as follows:
|
||||
\[
|
||||
\begin{split}
|
||||
\textbf{P}&(\texttt{Toothache}, \texttt{Catch}, \texttt{Cavity}) = \\
|
||||
&= \textbf{P}(\texttt{Toothache} \,\vert\, \texttt{Catch}, \texttt{Cavity})
|
||||
\textbf{P}(\texttt{Catch} \,\vert\, \texttt{Cavity}) \textbf{P}(\texttt{Cavity}) \\
|
||||
&= \textbf{P}(\texttt{Toothache} \,\vert\, \texttt{Cavity})
|
||||
\textbf{P}(\texttt{Catch} \,\vert\, \texttt{Cavity}) \textbf{P}(\texttt{Cavity})
|
||||
\end{split}
|
||||
\]
|
||||
$\textbf{P}(\texttt{Toothache}, \texttt{Catch}, \texttt{Cavity})$ has 7 independent values that grows exponentially
|
||||
($2 \cdot 2 \cdot 2 = 8$ values, but one of them can be omitted as a probability always sums up to 1).
|
||||
|
||||
$\textbf{P}(\texttt{Toothache} \,\vert\, \texttt{Cavity}) \textbf{P}(\texttt{Catch} \,\vert\, \texttt{Cavity}) \textbf{P}(\texttt{Cavity})$
|
||||
has 5 independent values that grows linearly ($4 + 4 + 2 = 10$, but a value of $\textbf{P}(\texttt{Cavity})$ can be omitted.
|
||||
The conditional probabilities require two tables (one for each prior) each with 2 values,
|
||||
but for each table a value can be omitted, therefore requiring $2$ independent values per conditional probability instead of $4$).
|
||||
\end{example}
|
||||
\end{description}
|
||||
|
||||
|
||||
|
||||
Reference in New Issue
Block a user