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\title{Belief Function Theory Applied to US Stock Market}
\author{Nima Salehy, Major Professor: Dr. Giray \"{O}kten}
\institute{Financial Mathematics, Department of Mathematics, Florida State University}
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\begin{document}
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\begin{block}{Belief Function Theory}
Consider the following results of an election survey:
\vspace{-1.5cm}
\begin{center}
   	\begin{tabular}{|c|c|c|c|c|}
   		\hline
   		Social Democrats ($S$) & Labor ($L$) & Conservatives ($R$) & $S$ or $L$ & Undecided \\ \hline
   		$40\%$ & $10\%$ & $30\%$ & $5\%$ & $15\%$  \\ \hline
   	\end{tabular}
\end{center}
\begin{enumerate}
\vspace{.3cm}
\item \textbf{Frame of discernment} is a nonempty finite set $\Theta$ of mutually exclusive and exhaustive outcomes. In this case, $\Theta=\{S, L, R\}$.
\vspace{.1cm}
\item \textbf{Mass function} over $\Theta$ is a function $m:2^\Theta\longrightarrow [0,1]$ that assigns a mass value to each $A\subseteq\Theta$ such that: $$(1) \ m(\emptyset)=0, \hspace{0.7cm} (2) \ \sum_{A\subseteq\Theta}m(A)=1.$$
\vspace{-.7cm}
Mass values for the election example:
\vspace{.7cm}
\begin{center}
	\begin{tabular}{|c||c|c|c|c|c|c|}
		\hline
		$A$ & $\{S\}$ & $\{L\}$ & $\{R\}$ & $\{S, L\}$ & $\{S, L, R\}$ & otherwise \\ \hline
		$m(A)$ & $0.4$ & $0.1$ & $0.3$ & $0.05$ & $0.15$ & $0$  \\ \hline
	\end{tabular}
\end{center}
\vspace{.6cm}
\begin{enumerate}
	\setbeamercolor{item projected}{fg=white,bg=Lavender}
	\item Those subsets with positive mass values are called \textbf{focal sets}. The collection of the focal sets is denoted by $\mathcal{F}_{m}$. Here, $$\mathcal{F}_{m}=\{\{S\}, \{L\}, \{R\}, \{S, L\}, \{S, L, R\}\}.$$
	\item $(\mathcal{F}_{m},m)$ is called a \textbf{body of evidence}.
\end{enumerate}
\vspace{.1cm}
\item \textbf{Belief function} is a function $bel:2^\Theta\longrightarrow [0,1]$, corresponding to a specific mass function $m$, that assigns to every proposition $A\subseteq\Theta$ the total amount of belief committed to that proposition, including all the subsets of $A$: $$bel(A)=\sum_{B\subseteq A}m(B).$$
Therefore, the corresponding belief function for the election example can easily be obtained. For instance: $$bel(\{S, L\})= m(\{S\}) + m(\{L\}) + m(\{S, L\}) =0.4 + 0.1 +0.05 = 0.55.$$
\vspace{-1.7cm}
\item \textbf{Plausibility function} is denoted by $pl$. $pl(A)$ gives the maximum amount of belief that could be assigned to the proposition $A$, if all evidence collected in the future supports the proposition.

\begin{enumerate}
	\setbeamercolor{item projected}{fg=white,bg=Lavender}
	\item In our election example the current belief for $\{S, R\}$ is $0.7$. In the future, if the undecideds with mass value $0.15$ support $\{S\}$, $\{R\}$ or $\{S, R\}$, then the belief for $\{S, R\}$ would increase to $0.7 + 0.15 = 0.85$. If moreover, those who currently support $\{S, L\}$, with mass value $0.05$, move to support $\{S\}$ only, then the total belief for $\{S, R\}$ would increase to $0.7 + 0.15 + 0.05 = 0.9$.
	
	\item In other words, the plausibility of the proposition $\{S, R\}$ is the maximum amount of belief that could be assigned to a proposition, which is $1$, minus the amount of belief for its complement, $\{L\}$, i.e., $$pl(\{S, R\})=1-bel(\overline{\{S, R\}})=1-bel(\{L\})=1-0.1=0.9.$$
	\item Here is the formal definition:
	\begin{center}
		\begin{tabular}{rl}
			$pl:2^\Theta$ & $\longrightarrow \ [0,1]$\\
			$pl(A) =$ & $1-bel(\bar{A}).$
		\end{tabular}
	\end{center}
\end{enumerate}
\vspace{.1cm}
\item \textbf{What will happen to these functions when there is no uncertainty?}
\begin{enumerate}
	\setbeamercolor{item projected}{fg=white,bg=Lavender}
	\item The focal sets are the singletons.
	\item Belief and plausibility functions become identical, and turn into a probability measure.
\end{enumerate}
\end{enumerate}
\end{block}
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\begin{block}{Dempster's Rule of Combination}
Assume, another election survey is done and we have:
\begin{center}
	\begin{tabular}{|c|c|c|c|}
		\hline
		Social Democrats ($S$) & Labor ($L$) & Conservatives ($R$) & Undecided \\ \hline
		$30\%$ & $10\%$ & $25\%$ & $35\%$  \\ \hline
	\end{tabular}
\end{center}

Thus, the mass values are assigned as follows:
\begin{center}
	\begin{tabular}{|c||c|c|c|c|c|}
		\hline
		$A$ & $\{S\}$ & $\{L\}$ & $\{R\}$ & $\{S, L, R\}$ & otherwise \\ \hline
		$m(A)$ & $0.3$ & $0.1$ & $0.25$ & $0.35$ & $0$  \\ \hline
	\end{tabular}
\end{center}

\begin{enumerate}
\item \textbf{Can we take these two surveys, and merge them, in a sense, to obtain their sum?} \textit{Dempster} first introduced a method to combine evidence coming from different sources. Let us illustrate the idea for the case of two bodies of evidence:
\begin{enumerate}
\setbeamercolor{item projected}{fg=white,bg=Lavender}
\item Suppose $(\mathcal{F}_{m_{1}},m_{1})$ and $(\mathcal{F}_{m_{2}},m_{2})$ are two distinct bodies of evidence which are defined over the same frame of discernment $\Theta$.
\item Let $\mathcal{F}_{m_{1}}=\{B_{1},\ldots, B_{t}\}$. Since the mass values add up to $1$, $m_{1}(B_{i})$, $1\leq i\leq t$, can be depicted as segments of a line segment of length $1$. Similarly, if $\mathcal{F}_{m_{2}}=\{C_{1},\ldots, C_{l}\}$, then the mass values $m_{2}(C_{j})$, $1\leq j\leq l$, can also be depicted as segments of a line segment of length 1:

\begin{center}
	\begin{tikzpicture}[scale=1.65]
	\draw[thick,->] (0,0) -- (8,0) node[anchor=west] {$m_1$};
	\draw[thick,->] (0,0) -- (0,8) node[anchor=east] {$m_2$};
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	\draw (1,0) -- (1,7);
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	\draw (7,0) -- (7,7);
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	\draw (0,3) -- (7,3);
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	\draw (0,6) -- (7,6);
	\draw (0,7) -- (7,7);
	\filldraw[fill=gray!40!white, draw=black] (3,3) rectangle (4,4);
	\draw[thick,->] (7.9,6) parabola (3.5,3.5);
	\draw (18.3,6.6) -- (18.3,6.6) node[anchor=north east] {mass value of measure $m_1(B_{i})m_2(C_{j})$};
	\draw (15.9,5.7) -- (15.9,5.7) node[anchor=north east] {committed to $B_{i}\cap C_{j}$};
	\draw [decorate,decoration={brace,amplitude=5pt},xshift=-4pt,yshift=0pt] (0,0) -- (0,1) node [black,midway,xshift=-1.5cm]  {\scriptsize $m_{2}(C_{1})$};
	\draw [decorate,decoration={brace,amplitude=5pt},xshift=-4pt,yshift=0pt] (0,3) -- (0,4) node [black,midway,xshift=-1.5cm]  {\scriptsize $m_{2}(C_{j})$};
	\draw [decorate,decoration={brace,amplitude=5pt},xshift=-4pt,yshift=0pt] (0,6) -- (0,7) node [black,midway,xshift=-1.5cm]  {\scriptsize $m_{2}(C_{l})$};
	\draw [decorate,decoration={brace,amplitude=5pt},xshift=-4pt,yshift=0pt] (0,2) -- (0,2) node [black,midway,xshift=-1.5cm]  {$\vdots$};
	\draw [decorate,decoration={brace,amplitude=5pt},xshift=-4pt,yshift=0pt] (0,5) -- (0,5) node [black,midway,xshift=-1.5cm]  {$\vdots$};
	\draw [decorate,decoration={brace,mirror,amplitude=5pt},xshift=0pt,yshift=-4pt] (0,0) -- (1,0) node [black,midway,yshift=-.9cm]  {\scriptsize $m_{1}(B_{1})$};
	\draw [decorate,decoration={brace,mirror,amplitude=5pt},xshift=0pt,yshift=-4pt] (3,0) -- (4,0) node [black,midway,yshift=-.9cm]  {\scriptsize $m_{1}(B_{i})$};
	\draw [decorate,decoration={brace,mirror,amplitude=5pt},xshift=0pt,yshift=-4pt] (6,0) -- (7,0) node [black,midway,yshift=-.9cm]  {\scriptsize $m_{1}(B_{t})$};
	\draw [decorate,decoration={brace,mirror,amplitude=5pt},xshift=0pt,yshift=-4pt] (2,0) -- (2,0) node [black,midway,yshift=-.9cm]  {$\cdots$};
	\draw [decorate,decoration={brace,mirror,amplitude=5pt},xshift=0pt,yshift=-4pt] (5,0) -- (5,0) node [black,midway,yshift=-.9cm]  {$\cdots$};
	%\foreach \x in {7}
	\draw (7 cm,1pt) -- (7 cm,-1pt) node[anchor=south west] {\footnotesize $1$};
	\foreach \y in {7}
	\draw (1pt,7 cm) -- (-1pt,7 cm) node[anchor=south west] {\footnotesize $1$};
	\end{tikzpicture}
\end{center}
\item Therefore, the total mass value committed to $A$ is
$\displaystyle\sum_{B_{i} \cap C_{j}=A}m_{1}(B_{i})m_{2}(C_{j}).$

\item Here, we should discard all the rectangles that are committed to the empty set, since we need the mass value of the empty set to be zero. Moreover, we want a normalization constant so that the new mass values add up to $1$: 
$$k\defeq\Big(1-\displaystyle\sum_{B_{i}\cap C_{j}=\emptyset}m_{1}(B_{i})m_{2}(C_{j})\Big)^{-1}=\Big(\displaystyle\sum_{B_{i}\cap C_{j}\ne\emptyset}m_{1}(B_{i})m_{2}(C_{j})\Big)^{-1}.$$
\end{enumerate}
\item The mass function $m_{1\oplus 2}=m_{1}\oplus m_{2}$, combined according to Dempster's rule $\oplus$, is defined as
$$
m_{1\oplus 2}(A)=\left\{\begin{array}{ll}
k\displaystyle\sum_{B\cap C=A}m_{1}(B)m_{2}(C) & \ if \ A \ne\emptyset, \\[0.6cm]
0 & \ if \ A=\emptyset.
\end{array} \right.
$$
\item Let us get back to our election example. The combined mass function is:
\vspace{.5cm}
\begin{center}
	\begin{tabular}{|c||c|c|c|c|c|c|}
		\hline
		$A$ & $\{S\}$ & $\{L\}$ & $\{R\}$ & $\{S, L\}$ & $\{S, L, R\}$ & otherwise \\ \hline
		$m_{1\oplus 2}(A)$ & $0.4758$ & $0.0967$ & $0.3234$ & $0.026$ & $0.0781$ & $0$  \\ \hline
	\end{tabular}
\end{center}
\vspace{.7cm}
\item Dempster's rule becomes computationally intractable as the number of bodies of evidence grows. \textbf{Monte Carlo methods} can be used in these cases.
\end{enumerate}
\end{block}
\end{column}

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\begin{block}{Application to US Stock Market}
\begin{enumerate}
\item \textbf{What is the most promising industry in the US in the next month?}
\begin{enumerate}
\setbeamercolor{item projected}{fg=white,bg=Lavender}
\item There are 12 industries in the US stock Market which form the frame of discernment $\Theta$.
\item Collect \textit{analyst ratings} from \url{http://www.MarketBeat.com}.
\item Each analyst rating form a mass function. Combine them using Dempster's rule.
\item Pick the industry with the highest mass value, and invest $1$ dollar.
\item Sell the stocks $25$ business days later and calculate the excess return.
\end{enumerate}
\item \textbf{What are the most promising stocks in the next month?}
\vspace{-1.2cm}
\begin{enumerate}
\setbeamercolor{item projected}{fg=white,bg=Lavender}
\item To find an answer (portfolio), repeat the above procedure with the modified \textit{frame of discernment} and \textit{mass functions}.
\item On each business day, choose top $x\%$ of the stocks with the highest mass value to invest.
\end{enumerate}
\item Repeat the process for each business day from 2/5/2015 to 10/17/2016. Finally, take the average to find the average of the excess returns on a day.
\item \textbf{Results:} The average of excess returns are depicted when $n$ analysts ratings are considered ($n=5, 10$).
\end{enumerate}
\begin{figure}[ht]
	\begin{center}
		\includegraphics[width=9.3in]{m5crop.png}
		\vspace{-.4cm}
		\caption{Average of excess returns when $n=5$.}
	\end{center}
\end{figure}
\begin{figure}[ht]
	\begin{center}
		\includegraphics[width=9.3in]{m10crop.png}
		\vspace{-.4cm}
		\caption{Average of excess returns when $n=10$.}
	\end{center}
\end{figure}
\end{block}

\begin{block}{References}	
\small{\begin{thebibliography}{99}
\bibitem{Rei2014} T. Reineking, {\em Belief Functions: Theory and Algorithms}, Doctoral dissertation, Staats-und Universitätsbibliothek Bremen, 2014.

\bibitem{Shafer1976} G. Shafer, {\em A mathematical theory of evidence}, Princeton Univ. Press. Princeton, NJ, 1976.

\bibitem{Xu2014} Y. Xu, L. Wu, X. Wu and Z. Xu, {\em Belief fusion of predictions of industries in China's stock market}, Belief Functions: Theory and Applications, Springer International Publishing, 348-355, 2014.
\end{thebibliography}}
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