[ker2014-2.git] / report / ass2-1.tex

\chapter{Probabilistic representation and reasoning (and burglars)}
\section{Formal description}
In our representation of the model we introduced a \textit{Noisy OR} to
represent the causal independence of \textit{Burglar} and \textit{Earthquake}
on \textit{Alarm}. The representation of the network is displayed in
Figure~\ref{bnetwork21}

\begin{figure}[H]
        \caption{Bayesian network alarmsystem}
        \label{bnetwork21}
        \centering
        \includegraphics[scale=0.5]{d1.eps}
\end{figure}

Days were chosen as unit to model the story. Calculation of the probability of
a\textit{Burglar} event happening at some day is then (assuming a gregorian
calendar and leap days):
$$\frac{1}{365 + 0.25 - 0.01 - 0.0025}=\frac{1}{365.2425}$$

The resultant probability distributions can be found in Table~\ref{probdist},
in order to avoid a unclear graph.

\begin{table}[H]
        \label{probdist}
        \begin{tabular}{|l|l|}
                \hline
                & Earthquake\\
                \hline
                T & $0.0027$\\
                F & $0.9973$\\
                \hline
        \end{tabular}
        %
        \begin{tabular}{|l|l|}
                \hline
                & Burglar\\
                \hline
                T & $0.0027$\\
                F & $0.9973$\\
                \hline
        \end{tabular}
        
        \begin{tabular}{|l|ll|}
                \hline
                & \multicolumn{2}{c|}{$I_1$}\\
                Earthquake & T & F\\
                \hline
                T & $0.2$ & $0.8$\\
                F & $0$ & $1$\\
                \hline
        \end{tabular}
        \begin{tabular}{|l|ll|}
                \hline
                & \multicolumn{2}{c|}{$I_2$}\\
                Burglar & T & F\\
                \hline
                T & $0.95$ & $0.05$\\
                F & $0$ & $1$\\
                \hline
        \end{tabular}
        \begin{tabular}{|ll|ll|}
                \hline
                && \multicolumn{2}{c|}{Alarm}\\
                $I_1$ & $I_2$ & T & F\\
                \hline
                T & T & $1$ & $0$\\
                T & F & $1$ & $0$\\
                F & T & $1$ & $0$\\
                F & F & $0$ & $1$\\
                \hline
        \end{tabular}
        
        \begin{tabular}{|l|ll|}
                \hline
                & \multicolumn{2}{c|}{Watson}\\
                Alarm & T & F\\
                \hline
                T & $0.8$ & $0.2$\\
                F & $0.4$ & $0.6$\\
                \hline
        \end{tabular} 
        \begin{tabular}{|l|ll|}
                \hline
                & \multicolumn{2}{c|}{Gibbons}\\
                Alarm & T & F\\
                \hline
                T & $0.99$ & $0.01$\\
                F & $0.04$ & $0.96$\\
                \hline
        \end{tabular}
        \begin{tabular}{|l|ll|}
                \hline
                & \multicolumn{2}{c|}{Radio}\\
                Earthquake & T & F\\
                \hline
                T & $0.9998$ & $0.0002$\\
                F & $0.0002$ & $0.9998$\\
                \hline
        \end{tabular}
\end{table}

\textit{If there is a burglar present (which could happen once every ten
years), the alarm is known to go off 95\% of the time.} We modelled this by
setting the value for Burglar True and $I_2$ True on 0,95.\\
\textit{There’s a 40\% chance that Watson is joking and the alarm is in fact
off.} This is modelled by putting the value for Watson True and Alarm F on 0,4.
As Holmes expects Watson to call in 80\% of the time, the value for alarm True
and Watson True is set 0,2. Because the rows have to sum to 1, the other values
are easily calculated.\\
\textit{She may not have heard the alarm in 1\% of the cases and is thought to
erroneously report an alarm when it is in fact off in 4\% of the cases.} We
modelled this by assuming that when Mrs. Gibbons hears the alarm, she calls
Holmes. Meaning that the value for Gibbons False and Alarm true is 0,01. As
she reports when the alarm is in fact off in 4\% of the cases, the value for
Gibbons True and alarm False is 0,04.\\
 

\section{Implementation}
We implemented the distributions in \textit{AILog}, see Listing~\ref{alarm.ail}

\begin{listing}[H]
        \label{alarm.ail}
        \caption{Alarm.ail}
        \inputminted[linenos,fontsize=\footnotesize]{prolog}{./src/alarm.ail}
\end{listing}

\section{Queries}
Now that we have modelled the story with the corresponding probabilities, we
can have ailog calculate some other probabilities given a some observations.
Down below we wrote down some probabilties and the associated ailog output.\\
The chance that a burglary happens given that Watson calls is greater than the
chance that a burglary happens without this observations, as is observerd by
the difference between a and b. This makes sense as Watson calls rightly in
80\% of the time. So when Holmes receives a call by Watson, the chance that the
alarm goes of increases.\\
When we compare b to c, the same mechanisme holds. There are more observations
that give evidence for a burglary as both Watson and Gibbons have called in the
case of c.\\
When you take a look at the last case, d, you see that the probability has
decreased compared to c. This can be explained by an observation that is added
on top of the observations of b; the radio. The variable Radio means that the
newcast tells that there was an earhquake. As that is also a reason why the
alarm could go of, but has nothing to do with a burglary, it decreases the
probability of a burglary.
%We kunnen misschien de kans uitrekenen dat Watson en Gibbons allebei foutief bellen? Daar mis je denk ik info over of niet?
\begin{enumerate}[a)]
        \item $P(\text{Burglary})=
                0.002737757092501968$
        \item $P(\text{Burglary}|\text{Watson called})=
                0.005321803679438259$
        \item $P(\text{Burglary}|\text{Watson called},\text{Gibbons called})=
                0.11180941544755249$
        \item $P(\text{Burglary}|\text{Watson called},\text{Gibbons called}
                , \text{Radio})=0.01179672476662423$
\end{enumerate}

\begin{listing}[H]
        \begin{minted}[fontsize=\footnotesize]{prolog}
ailog: predict burglar.
Answer: P(burglar|Obs)=0.002737757092501968.
  [ok,more,explanations,worlds,help]: ok.

ailog: observe watson.
Answer: P(watson|Obs)=0.4012587986186947.
  [ok,more,explanations,worlds,help]: ok.

ailog: predict burglar.
Answer: P(burglar|Obs)=[0.005321803679438259,0.005321953115441623].
  [ok,more,explanations,worlds,help]: ok.

ailog: observe gibbons.
Answer: P(gibbons|Obs)=[0.04596053565368094,0.045962328885721306].
  [ok,more,explanations,worlds,help]: ok.

ailog: predict burglar.
Answer: P(burglar|Obs)=[0.11180941544755249,0.1118516494624678].
  [ok,more,explanations,worlds,help]: ok.

ailog: observe radio.
Answer: P(radio|Obs)=[0.02582105837443645,0.025915745316785182].
  [ok,more,explanations,worlds,help]: ok.

ailog: predict burglar.
Answer: P(burglar|Obs)=[0.01179672476662423,0.015584580594335082].
  [ok,more,explanations,worlds,help]: ok.
        \end{minted}
\end{listing}

ToDO write down the most probable explanation for the observed evidence

\section{Comparison with manual calculation}
ToDO: english.
When we let ailog calculate the probability of alarm. %Wat is Obs hier??
Querying the \textit{Alarm} variable gives the following answer:
\begin{minted}{prolog}
        ailog: predict alarm.
        Answer: P(alarm|Obs)=0.0031469965467367292.
                            
          [ok,more,explanations,worlds,help]: ok.
\end{minted}

Using the formula for causal independence with a logical OR:\\
$P(Alarm|C_1, C_2) = P(i_1|C_1)+P(i_2|C_2)(1-P(i_1|C_1))$ we can calculate the
probability of the \textit{Alarm} variable using variable elimination. This
results in the following answer:\\
$P(Alarm|burglar, earthquake) =
P(i_1|burglar)+P(i_2|earthquake)(1-P(i_1|burglar)) =
0.2*0.0027+0.95*0.0027*(1-0.2*0.0027)=0.00314699654673673$ \\
TODOOOOOOOOOOO %Ik weet niet of we i_1 en i_2 nog door iets anders vervangen
% moeten worden.
When you compare the output of ailog and of the variable elimination, you see
that they are similar.

\newpage
\section{Burglary problem with extended information}
$P(burglary)\cdot\left(
        P(\text{first house is holmes'})+
        P(\text{second house is holmes'})+
        P(\text{third house is holmes'})\right)=\\
0.5102041\cdot\left(
        \frac{1}{10000}+
        \frac{9999}{10000}\cdot\frac{1}{9999}+
        \frac{9999}{10000}\cdot\frac{9998}{9999}\cdot\frac{1}{9998}\right)=
\frac{3}{19600}\approx0.000153$

\section{Bayesian networks}
A bayesian network representation of the burglary problem with a multitude of
houses and burglars is possible but would be very big and tedious because all
the constraints about the burglars must be incorporated in the network.
The network would look something like in figure~\ref{bnnetworkhouses}


\begin{tabular}{|l|l|}
        \hline
        Joe &\\
        \hline
        T & $\nicefrac{5}{7}$\\
        F & $\nicefrac{2}{7}$\\
        \hline
\end{tabular}
\begin{tabular}{|l|l|}
        \hline
        William &\\
        \hline
        T & $\nicefrac{5}{7}$\\
        F & $\nicefrac{2}{7}$\\
        \hline
\end{tabular}
\begin{tabular}{|l|l|}
        \hline
        Jack & \\
        \hline
        T & $\nicefrac{5}{7}$\\
        F & $\nicefrac{2}{7}$\\
        \hline
\end{tabular}
\begin{tabular}{|l|l|}
        \hline
        Averall & \\
        \hline
        T & $\nicefrac{5}{7}$\\
        F & $\nicefrac{2}{7}$\\
        \hline
\end{tabular}

\begin{tabular}{|llll|ll|}
        \hline
        & & & & Burglary &\\
        Joe & William & Jack & Averall & T & F\\
        \hline
        F& F& F& F & $0$ & $1$\\
        F& F& F& T & $0$ & $1$\\
        F& F& T& F & $0$ & $1$\\
        F& F& T& T & $0$ & $1$\\
        F& T& F& F & $0$ & $1$\\
        F& T& F& T & $0$ & $1$\\
        F& T& T& F & $0$ & $1$\\
        F& T& T& T & $0$ & $1$\\
        T& F& F& F & $0$ & $1$\\
        T& F& F& T & $0$ & $1$\\
        T& F& T& F & $1$ & $0$\\
        T& F& T& T & $0$ & $1$\\
        T& T& F& F & $1$ & $0$\\
        T& T& F& T & $0$ & $1$\\
        T& T& T& F & $1$ & $0$\\
        T& T& T& T & $1$ & $0$\\
        \hline
\end{tabular}
\begin{tabular}{|lll|}
        \hline
        & Holmes &\\
        Burglary & T & F\\
        \hline
        T & $0.000153$ & $0.999847$\\
        F & $0$ & $1$\\
        \hline
\end{tabular}

\begin{figure}[H]
        \caption{Bayesian network of burglars and houses}
        \label{bnnetworkhouses}
        \centering
        \includegraphics[scale=0.5]{d2.eps}
\end{figure}