1 \chapter{Probabilistic representation and reasoning (and burglars)
}
2 \section{Formal description
}
3 In our representation of the model we chose to introduce a
\textit{Noisy OR
} to
4 represent the causal independence of
\textit{Burglar
} and
\textit{Earthquake
}
5 on
\textit{Alarm
}. The visual representation of the network is visible in
6 Figure~
\ref{bnetwork21
}
9 \caption{Bayesian network, visual representation
}
12 \includegraphics[scale=
0.5]{d1.eps
}
15 As for the probabilities for
\textit{Burglar
} and
\textit{Earthquake
} we chose
16 to calculate them using days the unit. Calculation for the probability of a
17 \textit{Burglar
} event happening at some day is then this(assuming a gregorian
18 calendar and leap days).
19 $$
\frac{1}{365 +
0.25 -
0.01 -
0.0025}=
\frac{1}{365.2425}$$
21 This gives the following probability distributions visible in
26 \begin{tabular
}{|l|l|
}
35 \begin{tabular
}{|l|l|
}
44 \begin{tabular
}{|l|ll|
}
46 &
\multicolumn{2}{c|
}{$I_1$
}\\
53 \begin{tabular
}{|l|ll|
}
55 &
\multicolumn{2}{c|
}{$I_2$
}\\
62 \begin{tabular
}{|ll|ll|
}
64 &&
\multicolumn{2}{c|
}{Alarm
}\\
65 $I_1$ & $I_2$ & T & F\\
74 \begin{tabular
}{|l|ll|
}
76 &
\multicolumn{2}{c|
}{Watson
}\\
83 \begin{tabular
}{|l|ll|
}
85 &
\multicolumn{2}{c|
}{Gibbons
}\\
92 \begin{tabular
}{|l|ll|
}
94 &
\multicolumn{2}{c|
}{Radio
}\\
97 T & $
0.9998$ & $
0.0002$\\
98 F & $
0.0002$ & $
0.9998$\\
103 \section{Implementation
}
104 This distribution results in the
\textit{AILog
} code in Listing~
\ref{alarm.ail
}
109 \inputminted[linenos,fontsize=
\footnotesize]{prolog
}{./src/alarm.ail
}
113 Using the following queries the probabilities or as follows:\\
114 \begin{enumerate
}[a)
]
115 \item $P(
\text{Burglary
})=
116 0.002737757092501968$
117 \item $P(
\text{Burglary
}|
\text{Watson called
})=
118 0.005321803679438259$
119 \item $P(
\text{Burglary
}|
\text{Watson called
}\wedge\text{Gibbons called
})=
121 \item $P(
\text{Burglary
}|
\text{Watson called
}\wedge\text{Gibbons called
}
122 \wedge\text{Radio
})=
0.01179672476662423$
126 \begin{minted
}[fontsize=
\footnotesize]{prolog
}
127 ailog: predict burglar.
128 Answer: P(burglar|Obs)=
0.002737757092501968.
129 [ok,more,explanations,worlds,help
]: ok.
131 ailog: observe watson.
132 Answer: P(watson|Obs)=
0.4012587986186947.
133 [ok,more,explanations,worlds,help
]: ok.
135 ailog: predict burglar.
136 Answer: P(burglar|Obs)=
[0.005321803679438259,
0.005321953115441623].
137 [ok,more,explanations,worlds,help
]: ok.
139 ailog: observe gibbons.
140 Answer: P(gibbons|Obs)=
[0.04596053565368094,
0.045962328885721306].
141 [ok,more,explanations,worlds,help
]: ok.
143 ailog: predict burglar.
144 Answer: P(burglar|Obs)=
[0.11180941544755249,
0.1118516494624678].
145 [ok,more,explanations,worlds,help
]: ok.
147 ailog: observe radio.
148 Answer: P(radio|Obs)=
[0.02582105837443645,
0.025915745316785182].
149 [ok,more,explanations,worlds,help
]: ok.
151 ailog: predict burglar.
152 Answer: P(burglar|Obs)=
[0.01179672476662423,
0.015584580594335082].
153 [ok,more,explanations,worlds,help
]: ok.
157 \section{Comparison with manual calculation
}
158 Querying the
\textit{Alarm
} variable gives the following answer
159 \begin{minted
}{prolog
}
160 ailog: predict alarm.
161 Answer: P(alarm|Obs)=
0.0031469965467367292.
163 [ok,more,explanations,worlds,help
]: ok.
166 Using formula: $P(i_1|C_1)+P(i_2|C_2)(
1-P(i_1|C_1))$ we can calculate the
167 probability of the
\textit{Alarm
} variable using variable elimination. This
168 results in the following answer:
169 $$
0.2*
0.0027+
0.95*
0.0027*(
1-
0.2*
0.0027)=
0.00314699654673673$$
173 \section{Burglary problem with extended information
}
174 $P(burglary)
\cdot\left(
175 P(
\text{first house is holmes'
})+
176 P(
\text{second house is holmes'
})+
177 P(
\text{third house is holmes'
})
\right)=\\
180 \frac{9999}{10000}\cdot\frac{1}{9999}+
181 \frac{9999}{10000}\cdot\frac{9998}{9999}\cdot\frac{1}{9998}\right)=
182 \frac{3}{19600}\approx0.000153$
184 \section{Bayesian networks
}
185 A bayesian network representation of the burglary problem with a multitude of
186 houses and burglars is possible but would be very big and tedious because all
187 the constraints about the burglars must be incorporated in the network.
188 The network would look something like in figere~
\ref{bnnetworkhouses
}
189 accompanied with the probability distributions below.
191 \begin{tabular
}{|l|l|
}
195 T & $
\nicefrac{5}{7}$\\
196 F & $
\nicefrac{2}{7}$\\
199 \begin{tabular
}{|l|l|
}
203 T & $
\nicefrac{5}{7}$\\
204 F & $
\nicefrac{2}{7}$\\
207 \begin{tabular
}{|l|l|
}
211 T & $
\nicefrac{5}{7}$\\
212 F & $
\nicefrac{2}{7}$\\
215 \begin{tabular
}{|l|l|
}
219 T & $
\nicefrac{5}{7}$\\
220 F & $
\nicefrac{2}{7}$\\
224 \begin{tabular
}{|llll|ll|
}
227 Joe & William & Jack & Averall & T & F\\
229 F& F& F& F & $
0$ & $
1$\\
230 F& F& F& T & $
0$ & $
1$\\
231 F& F& T& F & $
0$ & $
1$\\
232 F& F& T& T & $
0$ & $
1$\\
233 F& T& F& F & $
0$ & $
1$\\
234 F& T& F& T & $
0$ & $
1$\\
235 F& T& T& F & $
0$ & $
1$\\
236 F& T& T& T & $
0$ & $
1$\\
237 T& F& F& F & $
0$ & $
1$\\
238 T& F& F& T & $
0$ & $
1$\\
239 T& F& T& F & $
1$ & $
0$\\
240 T& F& T& T & $
0$ & $
1$\\
241 T& T& F& F & $
1$ & $
0$\\
242 T& T& F& T & $
0$ & $
1$\\
243 T& T& T& F & $
1$ & $
0$\\
244 T& T& T& T & $
1$ & $
0$\\
247 \begin{tabular
}{|lll|
}
252 T & $
0.000153$ & $
0.999847$\\
258 \caption{Bayesian network of burglars and houses
}
259 \label{bnnetworkhouses
}
261 \includegraphics[scale=
0.5]{d2.eps
}