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\usepackage{booktabs} % Allows the use of \toprule, \midrule and \bottomrule in tables

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%	TITLE PAGE
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\title[Presentation]{IST Modifications} % The short title appears at the bottom of every slide, the full title is only on the title page

\author{Songyu He} % Your name
\institute[UCSD] % Your institution as it will appear on the bottom of every slide, may be shorthand to save space
{
	\\ % Your institution for the title page
	\medskip
	\textit{} % Your email address
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\begin{document}

	\begin{frame}
		\titlepage % Print the title page as the first slide
	\end{frame}
	
		\begin{frame}
		\tableofcontents
	\end{frame}
	
	\section{Background} % Sections can be created in order to organize your presentation into discrete blocks, all sections and subsections are automatically printed in the table of contents as an overview of the talk
	%------------------------------------------------

%--------------------------------------page 2-----------------------------------
\begin{frame}{Compatibility Issue}
	\begin{itemize}
		\item should we put plain or discounted monetary value?\pause
		\begin{itemize}
			\item[-] discounting utility(DU) and salience discounting share analogous purpose:
			\begin{itemize}
				\item[>] both make future periods less important\pause
				\item[>] evidence from riskless CTB suggests existence of DU w.o. salience\pause
 				\item[>] under current framework of rank-based CTB, it's hard to differentiate these two: \\
 				\pause DU is insignificant, we don't know whether discounting affect salience function\pause

			\end{itemize}\pause
		\item But we can use the mechanism of salience to create CTB environments in which discounting would affect states' ranks!
		\end{itemize}
	\end{itemize}\end{frame}
%--------------------------------------page 3-----------------------------------
\begin{frame}{Reverting CTB}
			$$\max\limits_{c_1,c_2}\:\:   p_1v(c_1) + \delta^{k} p_2 v(c_2)$$
			$$s.t.\; (1+r)c_1 + c_2 = 100$$
			\begin{itemize}
				\item[-] as $r\rightarrow \infty$, $c_1$ vanishes\pause
				\item[-] we can find $r^*>0$ where riskless and risky data converges\pause
				\begin{itemize}
					\item[] $r^*$ represents the threshold at which states have same salience level\pause
					\item[] 	let $p_1=1, \; p_2<1$ to smooth the converge \pause
				\end{itemize}
			\item if DU and salience are not entangled, then $r^*$ should be fixed\pause
			\item o.w. $r^*$ should be decreasing in time
			\end{itemize}
\end{frame}

\begin{frame}{Math}
		$$\max\limits_{c_1,c_2}\:\:   (\pi_a +\pi_f)c_1^\alpha +  (\pi_a +\pi_s)\delta^{k}  c_2^\alpha$$
		
		$$\ln\dfrac{c_1}{c_2} = \dfrac{k}{\alpha-1}\delta +\dfrac{1}{\alpha-1}\ln(1+r) + \dfrac{1}{\alpha - 1}\ln(\dfrac{\pi_a+\pi_s}{\pi_a+\pi_f})$$\pause
		
		\begin{itemize}
			\item if $p_1 = 1$, as $r\geq 1$, $\dfrac{\pi_a+\pi_s}{\pi_a+\pi_f}\rightarrow \ln(\dfrac{p_2}{p_1}) + \ln(\dfrac{\theta}{1+\theta})$\pause
			
		\item if $p_1 < 1$, as $r$ increasing, $\dfrac{\pi_a+\pi_s}{\pi_a+\pi_f}\rightarrow \ln(\dfrac{p_2}{p_1}) + \ln(\dfrac{\theta+\theta^2}{1+\theta})$\pause
		\end{itemize}
	\bigskip
	\begin{itemize}
		\item find $(1, p_2)$, $(q_1, q_2)$ s.t. $q_2/q_1< p_2$ and two data converge at large $r$\pause
		\item variating $k$ and check whether the $r^*$ changes
	\end{itemize}
	
\end{frame}

\begin{frame}{attributes vs. uncertainty -- a rough version}
	\begin{itemize}
		\item attributes are consumptions in different time perioods. 
		\begin{itemize}
			\item[-] a cover $\mathcal{T}$ of $\{1,2,...,n\}$\pause
		\end{itemize}
	\item Under each realized state, evaluate attributes by 
	
	$$\sum\limits_{\tau\in\mathcal{T}}\gamma_\tau\sum\limits_{t\in\tau}\delta^t u(c_t)$$
	\item $\gamma_\tau$ is the decision weight on attribute $\tau$
	\end{itemize}
\end{frame}


\begin{frame}{Estimation}
	\begin{itemize}
		\item Recall: Under certainty
              $$\ln\dfrac{c_1}{c_2} = \dfrac{k}{\alpha-1}\delta +\dfrac{1}{\alpha-1}\ln(1+r) $$\pause
              \item if $c_1>$ cutoff
                            $$\ln\dfrac{c_1}{c_2} = \dfrac{k}{\alpha-1}\delta +\dfrac{1}{\alpha-1}\ln(1+r) + \dfrac{1}{\alpha-1}\ln(\dfrac{1+\theta^2}{\theta+\theta^2})$$\pause
              \item if $c_1<$ cutoff 
                            $$\ln\dfrac{c_1}{c_2} = \dfrac{k}{\alpha-1}\delta +\dfrac{1}{\alpha-1}\ln(1+r) + \dfrac{1}{\alpha-1}\ln(\dfrac{\theta+\theta^2}{1+\theta^2})$$
	\end{itemize}
\end{frame}
\begin{frame}{Estimation}
	\begin{itemize}
		\item Recall: Under certainty
		$$\ln\dfrac{c_1}{c_2} = \dfrac{k}{\alpha-1}\delta +\dfrac{1}{\alpha-1}\ln(1+r) $$
		\item if $c_1>$ cutoff
		$$\ln\dfrac{c_1}{c_2} = \dfrac{k}{\alpha-1}\delta +\dfrac{1}{\alpha-1}\ln(1+r) + \dfrac{1}{\alpha-1}\ln(\dfrac{1+\theta^2}{\theta+\theta^2})$$
		\item if $c_1<$ cutoff 
		$$\ln\dfrac{c_1}{c_2} = \dfrac{k}{\alpha-1}\delta +\dfrac{1}{\alpha-1}\ln(1+r) + \dfrac{1}{\alpha-1}\ln(\dfrac{\theta+\theta^2}{1+\theta^2})$$
		\item if $c_1 =$ cutoff
		$$\ln\dfrac{c_1}{c_2} = \dfrac{k}{\alpha-1}\delta +\dfrac{1}{\alpha-1}\ln(1+r) $$
	\end{itemize}
\end{frame}
\begin{frame}{estimation}
	\begin{itemize}
		\item if we don't adjust utility curvature: cutoff $ = \dfrac{m}{2(1+r)}$\pause
		\item if adjust: cutoff $< \dfrac{m}{2(1+r)}$\pause
		\item For those who choose middle point, the rank will be fixed as $$\ddd_l>\ddd_s>\ddd_a$$\pause
		\item counter-intuitive because:
		\begin{itemize}
			\item if $r>1$, when no risk, I put everything into my later account\pause
			\item if there is uncertainty, I split because I fear if the later account is not made
		\end{itemize}
	\end{itemize}
\end{frame}

\begin{frame}{estimation}
	\begin{itemize}
		\item In Cheung, $\approx 30\%$ choose exactly middle... leading $\theta> 1$. \pause 
		\item should we use instrument variable? + an extra reduced form analysis?
		\item use choice made under certainty as an indicator of the rank
	\end{itemize}
\end{frame}

\begin{frame}{estimation}
	\begin{itemize}
		\item let $Y_1^* = \ln(\dfrac{c_1+w}{c_2 + w})$ made under RISKY condition
		\item let $Y_2^* = \ln(\dfrac{c_1+w}{c_2 + w})$ made under RISKLESS condition\pause
		
		\begin{align*}
		Y_2^* &= \dfrac{1}{1-\alpha}(\ln r + k\ln \delta) + u\pause\\
		Y_1^* &= Y_2^* +(-) \dfrac{1}{1-\alpha}\ln(\dfrac{1+\theta^2}{\theta +\theta^2}) + v\pause\\
		&v\perp u \: XXX
		\end{align*}
	\end{itemize}
\end{frame}

\begin{frame}{estimation}
			\begin{align*}
	Y_2^* &= \dfrac{1}{1-\alpha}(\ln r + k\ln \delta) + u\\
	Y_1^* &= Y_2^* +(-) \dfrac{1}{1-\alpha}\ln(\dfrac{1+\theta^2}{\theta +\theta^2}) + v\\
	\end{align*}
	\begin{itemize}
	\item ignore sensoring issue, we want to estimate $f(Y_1^*, Y_2^*|r, k)$\pause
	\item That is
	\begin{align*}
&f(Y_1^*|Y_2^*, r, k)f(Y_2^*|r,k) = \dfrac{1}{\sigma_u\sigma_v}\\
&\phi(Y_2^* - \dfrac{1}{1-\alpha}(\ln r + k\ln \delta)) 
\cdot\phi(Y_1^* -Y_2^* - \dfrac{1}{1-\alpha}\ln(\dfrac{1+\theta^2}{\theta +\theta^2}) )
	\end{align*}

\end{itemize}
\end{frame}

\begin{frame}
	\begin{table}
	

\begin{small}


\begin{tabular}{l*{3}{c}}
	\hline\hline
	&\multicolumn{1}{c}{(1)}&\multicolumn{1}{c}{(2)}&\multicolumn{1}{c}{(3)}\\
	\hline
	alpha       &            &            &            \\
	\_cons      &       0.937&       0.933&       0.945\\
	&     (0.004)&     (0.004)&     (0.004)\\
	\hline
	delta       &            &            &            \\
	\_cons      &       1.000&       1.000&       0.999\\
	&     (0.000)&     (0.000)&     (0.000)\\
	\hline
	theta       &            &            &            \\
	\_cons      &       0.900&       0.976&       0.896\\
	&     (0.014)&     (0.017)&     (0.015)\\
	\hline
	lnsigma     &            &            &            \\
	\_cons      &       1.275&       1.322&       1.277\\
	&     (0.032)&     (0.030)&     (0.037)\\
	\hline
	lnsigma2    &            &            &            \\
	\_cons      &       1.293&       1.480&       1.327\\
	&     (0.028)&     (0.036)&     (0.030)\\
	\hline
	
	N           &    2240.000&    3108.000&    1764.000\\
	ll          &   -6054.067&   -8764.415&   -4799.602\\
	\hline\hline
\end{tabular}
\end{small}
	\end{table}

\end{frame}

\begin{frame}
	\begin{table}
		
		
		\begin{small}
			
			
\begin{tabular}{l*{3}{c}}
	\hline\hline
	&\multicolumn{1}{c}{(1)}&\multicolumn{1}{c}{(2)}&\multicolumn{1}{c}{(3)}\\
	\hline
	alpha       &            &            &            \\
	\_cons      &       0.937&       0.933&       0.945\\
	&     (0.004)&     (0.004)&     (0.004)\\
	\hline
	delta       &            &            &            \\
	\_cons      &       1.000&       1.000&       0.999\\
	&     (0.000)&     (0.000)&     (0.000)\\
	\hline
	theta       &            &            &            \\
	\_cons      &       0.889&       1.101&       0.938\\
	&     (0.013)&     (0.021)&     (0.015)\\
	\hline
	lnsigma     &            &            &            \\
	\_cons      &       1.275&       1.322&       1.277\\
	&     (0.032)&     (0.030)&     (0.037)\\
	\hline
	lnsigma2    &            &            &            \\
	\_cons      &       1.281&       1.469&       1.350\\
	&     (0.027)&     (0.036)&     (0.032)\\
	\hline
	\hline
	N           &    2240.000&    3108.000&    1764.000\\
	ll          &   -6041.165&   -8747.630&   -4820.106\\
	\hline\hline
\end{tabular}
		\end{small}
	\end{table}
	
\end{frame}

\begin{frame}
	\begin{table}
		
		
		\begin{small}
			
			
	\begin{tabular}{l*{3}{c}}
		\hline\hline
		&\multicolumn{1}{c}{(1)}&\multicolumn{1}{c}{(2)}&\multicolumn{1}{c}{(3)}\\
		
		\hline
		alpha       &            &            &            \\
		\_cons      &       0.937&       0.933&       0.945\\
		&     (0.004)&     (0.004)&     (0.004)\\
		\hline
		delta       &            &            &            \\
		\_cons      &       1.000&       1.000&       0.999\\
		&     (0.000)&     (0.000)&     (0.000)\\
		\hline
		theta       &            &            &            \\
		\_cons      &       0.891&       0.918&       0.892\\
		&     (0.013)&     (0.015)&     (0.015)\\
		\hline
		lnsigma     &            &            &            \\
		\_cons      &       1.275&       1.322&       1.277\\
		&     (0.032)&     (0.030)&     (0.037)\\
		\hline
		lnsigma2    &            &            &            \\
		\_cons      &       1.283&       1.469&       1.320\\
		&     (0.027)&     (0.035)&     (0.029)\\
		\hline\hline
		N           &    2240.000&    3108.000&    1764.000\\
		ll          &   -6042.742&   -8747.160&   -4792.979\\
		\hline\hline
	\end{tabular}
		\end{small}
	\end{table}
	
\end{frame}

\begin{frame}{censoring}
	\begin{itemize}
			\item only consider censoring under riskless condition
			\begin{align*}
Y_2^* &= Y_1^* -\dfrac{1}{1-\alpha}\ln(\dfrac{1+\theta^2}{\theta +\theta^2})  -v\\
Y_1^* &= \dfrac{1}{1-\alpha}(\ln r + k\ln \delta) +(-) \dfrac{1}{1-\alpha}\ln(\dfrac{1+\theta^2}{\theta +\theta^2}) + v 
+ u\\
Y_2& \in [\underline{y}, \bar{y}]
\end{align*}\pause
		\item $f(Y_2|Y_1^*, r, k)f(Y_1^*|r, k )$
		\item $f(Y_2|Y_1^*, r, k) = \Phi(Y_1^* -\dfrac{\sigma_v^2}{\sigma_u^2+\sigma_v^2} \hat{e} - \bar{y} )*\chi_{Y_2^*>= \bar y}\phi(...)$ $(1 - \Phi(Y_1^* +\dfrac{\sigma_u^2}{\sigma_u^2+\sigma_v^2} \hat{e} -  \underline{y}) )*\chi_{Y_2^*<= \underline{y}}$
	\end{itemize}
	
\end{frame}

\begin{frame}
	\begin{table}
		
		
		\begin{small}
			
			
\begin{tabular}{l*{3}{c}}
	\hline\hline
	&\multicolumn{1}{c}{(1)}&\multicolumn{1}{c}{(2)}&\multicolumn{1}{c}{(3)}\\
	\hline
	alpha       &            &            &            \\
	\_cons      &       0.957&       0.961&       0.966\\
	&     (0.002)&     (0.002)&     (0.001)\\
	\hline
	delta       &            &            &            \\
	\_cons      &       1.000&       0.999&       1.001\\
	&     (0.000)&     (0.000)&     (0.000)\\
	\hline
	theta       &            &            &            \\
	\_cons      &       0.849&       0.873&       0.867\\
	&     (0.007)&     (0.006)&     (0.006)\\
	\hline
	lnsigma     &            &            &            \\
	\_cons      &       1.582&       1.501&       1.743\\
	&     (0.018)&     (0.019)&     (0.017)\\
	\hline
	lnsigma2    &            &            &            \\
	\_cons      &       0.970&       0.901&       1.600\\
	&     (0.030)&     (0.032)&     (0.022)\\
	\hline
	\hline
	N           &    2240.000&    1764.000&    3108.000\\
	ll          &   -1.04e+04&   -8611.544&   -1.52e+04\\
	\hline\hline
\end{tabular}
		\end{small}
	\end{table}
	
\end{frame}
\begin{frame}{Treatment Effect Interpretation}
	\begin{itemize}
		\item According to our method:
		$$\dfrac{u'(c_t)}{u'(c_{t+k})} = \delta^k * r * \dfrac{p_2}{p_1}$$\pause
		$$\dfrac{u'(c_t^{50})}{u'(c_{t+k}^{50})} - \dfrac{u'(c_t^{100})}{u'(c_{t+k}^{100})}  = \delta^k*r*\Big(\dfrac{p_2}{p_1} - \dfrac{\pi_2}{\pi_1}\Big)$$\pause
		\item Under CRRA what we expect:
		$$\ln\dfrac{c_t^{50}}{c_{t+k}^{50}} -\ln\dfrac{c_t^{100}}{c_{t+k}^{100}} = \dfrac{1}{\alpha - 1}\ln\dfrac{1+\theta}{\theta +\theta^2} < 0 $$
	\end{itemize}
\end{frame}

\begin{frame}{Treatment Effect Interpretation}
$$\ln\dfrac{c_t^{50}}{c_{t+k}^{50}} -\ln\dfrac{c_t^{100}}{c_{t+k}^{100}} = \dfrac{1}{\alpha - 1}\ln\dfrac{1+\theta}{\theta +\theta^2} $$
\begin{itemize}
	\item If we use OLS on $\ln\dfrac{c_t}{c_{t+k}}$ : coefficients are $\hat{\alpha} \approx 0.8, \ln\dfrac{1+\theta}{\theta+\theta^2} \approx -0.513$\pause
	\item If we use OLS on $\ln\dfrac{c_t^{50}}{c_{t+k}^{50}}$: $\hat{\alpha} \approx 0.75, \ln\dfrac{1+\theta}{\theta+\theta^2} \approx -0.445$. \\ \pause If we ignore salience: $\hat{\alpha} \approx 0.81$
\end{itemize}
\end{frame}

\begin{frame}{Treatment Effect Interpretation}
	\begin{itemize}
		\item $r$ $k$ $\pi$ are exogeneous... but salience rank is not!
		\item As $r$ increase, average salience rank shifts
		\item Therefore, what we saw is a variantion distributing to two sources\pause
		\bigskip
		\item This problem is severe here because the structural model is not the true model
		\item The reason the difference works is because the difference eliminate the correlation on the second stage
	\end{itemize}
\end{frame}
\section{Experiment}
\begin{frame}{Overlook}
	\begin{itemize}
		\item What we have so far... A: (18,2) vs. B: (10,10). Total: 173
		\item EDU: (A,A,A) 12; (B,B,B) 63 -- Total 75
		\item IST: (A,B,A) 12; (B,B,A) 18 -- Total 30
		\item others: (A,B,B) 40; (B,A,A) 8; (B,A,B) 11; (A,A,B) 9 -- Total 68
	\end{itemize}\pause
\bigskip
	\begin{itemize}
		\item If we have CERT as 1st task, results are more consistent.
		\item CERT as first provokes a less ``risk-averse'' preference:
		\begin{itemize}
			\item[]	$c_3$:  34/91 vs. 16/82 -- diff: 18\%
			\item[] $c_2$: 20/91 vs. 20/82 -- diff: -2.4\%
			\item[] $c_1$: 39/91 vs. 34/82 -- diff: 1.4\%
		\end{itemize}
	\end{itemize}
\end{frame}

\begin{frame}{Preference for Consistency?}
	\begin{itemize}
		\item How to deal with (A,B,B)?
		\begin{itemize}
			\item[-] generally 40/173 $\approx$ 0.23
			\item[-] CERT first: 22/91 $\approx$ 0.24
		\end{itemize}
	\end{itemize}
\end{frame}

\begin{frame}{New Data}
	\begin{itemize}
		\item 23 recruited, 18 past attention check.  \pause
		\item Q1(CERT): (18, 2) vs. (9,10) -- 14 chose (18,2) \pause
		\item Q2(IND): (18,2) vs. (8,9) -- 15 chose (18,2)\pause
		\item Q3(IND): (18,2) vs. (7,10) -- 14 chose (18,2)\pause
		\item Q4(IND):(18,2) vs. (8,10) -- 15 chose (18,2)\pause
		\bigskip
		\item 50\% subjects made at least one switch between IND and reordered. 

	\end{itemize}
\end{frame}

\begin{frame}{IND vs. Reordered}
	\begin{itemize}
		\item Q2 and Q5: (18,2) vs. (8,9) -- $\begin{bmatrix}
			 13&2\\1&2
		\end{bmatrix}$\pause\bigskip
		\item Q3 and Q6: (18,2) vs. (7,10) --  $\begin{bmatrix}
			11&3\\2&2
		\end{bmatrix}$\pause\bigskip
		\item Q4 and Q7: (18,2) vs. (8,10) --  $\begin{bmatrix}
	10&5\\1&2
\end{bmatrix}$
		
	\end{itemize}
\end{frame}


\begin{frame}{The General Model?}
	
	(100, 0) vs. (50, 50), assuming indifferent under CERT, choose (50,50) under IND if: $$\pi_l (\delta v(50))^\alpha - \pi_s (v(100) - v(50))^\alpha\geq 0$$
	\begin{itemize}
		\item[1.] Different from others -- (My mistake)
		\item[-] Destroy separability in this way can't explain switches between CERT \& IND. 
		If $(x_1, 100 - x_1)\succ (x_2, 100 - x_2)$, $x_1 > x_2$
		Then $$v(x_1) - v(x_2)\geq \delta v(x_2) - \delta v(x_1)$$
		
	\end{itemize}

\end{frame}
\begin{frame}{The General Model?}
	
	(100, 0) vs. (50, 50), assuming indifferent under CERT, choose (50,50) under IND if: $$\pi_l (\delta v(50))^\alpha - \pi_s (v(100) - v(50))^\alpha\geq 0$$
	\begin{itemize}
		\item[1.] Other Evidence
		\item[-] $(15, 5,5,5)$ vs. $(0,10,10,10)$ 58\% under normal form. 39\% under reordered\pause
		
		$$\begin{bmatrix}
			15 & 5 & 5&5\\0&10&10&10
		\end{bmatrix}$$
	\end{itemize}
\end{frame}

\begin{frame}{The General Model?}
	
	(100, 0) vs. (50, 50), assuming indifferent under CERT, choose (50,50) under IND if: $$\pi_l (\delta v(50))^\alpha - \pi_s (v(100) - v(50))^\alpha\geq 0$$
	\begin{itemize}
		\item[1.] Other Evidence
		\item[-] $(15, 5,5,5)$ vs. $(0,10,10,10)$ 58\% under normal form. 39\% under reordered\pause
		
		$$\begin{bmatrix}
			15 & 5 & 5&5\\0&10&10&10
		\end{bmatrix}$$
	\end{itemize}
\end{frame}

\begin{frame}{General Model}
	\begin{tabular}{lcccc} 
		Option & \multicolumn{4}{c} { State } \\
		& $B$ & $N$ & $F$ & $S$ \\
		\hline$L_{(100,0)}$ & (100,0) & (0,0) & (100,0) & (0,0) \\
		$L_{(100,0)}^{\prime}$ & (100,0) & (100,0) & (0,0) & (0,0) \\
		$L_{(50,50)}$ & (50,50) & (0,0) & (50,0) & (0,50)
	\end{tabular}
\end{frame}


\begin{frame}{Results}
% Please add the following required packages to your document preamble:
% \usepackage[table,xcdraw]{xcolor}
% If you use beamer only pass "xcolor=table" option, i.e. \documentclass[xcolor=table]{beamer}
\begin{table}[]
	\begin{tabular}{|l|c|c|c|c|}
		\hline
		\textbf{CERT vs. IND} & \multicolumn{1}{l|}{\textbf{A, A}} & \multicolumn{1}{l|}{\textbf{A, B}} & \multicolumn{1}{l|}{\textbf{B, A}} & \multicolumn{1}{l|}{\textbf{B, B}} \\ \hline
		\textbf{(18,2) vs. (12,9)} & 1                                  & {\color[HTML]{3166FF} 2}           & 3                                  & 26                                 \\ \hline
		\textbf{(18,2) vs. (11,9)} & 5                                  & {\color[HTML]{3166FF} 8}           & 3                                  & 16                                 \\ \hline
		\textbf{(18,2) vs. (10,9)} & 12                                 & {\color[HTML]{3166FF} 16}          & 1                                  & 3                                  \\ \hline
		\textbf{(18,2) vs. (9,9)}  & 13                                 & {\color[HTML]{3166FF} 15}          & 2                                  & 2                                  \\ \hline
		\textbf{(18,2) vs. (8,9)}  & 21                                 & {\color[HTML]{3166FF} 10}          & 1                                  & 0                                  \\ \hline
		\textbf{(18,2) vs. (7,9)}  & 23                                 & {\color[HTML]{3166FF} 8}           & 0                                  & 1                                  \\ \hline
		\textbf{Total}  & 75                              &{\color[HTML]{3166FF} 59}          & 10      &  48                           \\ \hline
	\end{tabular}
\end{table}

Number of violations:
\begin{table}[]
	\begin{tabular}{|c|c|c|c|c|c|c|c|}
		\hline
		& 0 & 1  & 2 & 3 & 4 & 5 & 6 \\ \hline
		Count & 3 & 11 & 7 & 5 & 3 & 1 & 2 \\ \hline
	\end{tabular}
\end{table}

\end{frame}

\begin{frame}{Results}
	% Please add the following required packages to your document preamble:
	% \usepackage[table,xcdraw]{xcolor}
	% If you use beamer only pass "xcolor=table" option, i.e. \documentclass[xcolor=table]{beamer}
% Please add the following required packages to your document preamble:
% \usepackage[table,xcdraw]{xcolor}
% If you use beamer only pass "xcolor=table" option, i.e. \documentclass[xcolor=table]{beamer}
\begin{table}[]
	\begin{tabular}{|l|c|c|c|c|}
		\hline
		IND vs. REOR               & \multicolumn{1}{l|}{\textbf{A, A}} & \multicolumn{1}{l|}{\textbf{A, B}} & \multicolumn{1}{l|}{\textbf{B, A}} & \multicolumn{1}{l|}{\textbf{B, B}} \\ \hline
		\textbf{(18,2) vs. (12,9)} & 2                                  & {\color[HTML]{000000} 1}           & {\color[HTML]{FE0000} 3}           & 26                                 \\ \hline
		\textbf{(18,2) vs. (11,9)} & 6                                  & {\color[HTML]{000000} 2}           & {\color[HTML]{FE0000} 5}           & 19                                 \\ \hline
		\textbf{(18,2) vs. (10,9)} & 9                                  & {\color[HTML]{000000} 4}           & {\color[HTML]{FE0000} 1}           & 18                                 \\ \hline
		\textbf{(18,2) vs. (9,9)}  & 14                                 & {\color[HTML]{000000} 1}           & {\color[HTML]{FE0000} 4}           & 13                                 \\ \hline
		\textbf{(18,2) vs. (8,9)}  & 17                                 & {\color[HTML]{000000} 5}           & {\color[HTML]{FE0000} 6}           & 4                                  \\ \hline
		\textbf{(18,2) vs. (7,9)}  & 18                                 & {\color[HTML]{000000} 5}           & {\color[HTML]{FE0000} 5}           & 4                                  \\ \hline
		\textbf{Total}  & 66                                & {\color[HTML]{000000} 18}           & {\color[HTML]{FE0000} 24}           &  84                                \\ \hline
	\end{tabular}
\end{table}
Number of violations:
\begin{table}[]
	\begin{tabular}{|c|c|c|c|c|c|c|c|}
		\hline
		& 0 & 1  & 2 & 3 & 4 & 5 & 6 \\ \hline
		Count & 8 & 14 & 4 & 4 & 2 & 0 & 0 \\ \hline
	\end{tabular}
\end{table}
\end{frame}

\begin{frame}{Next Steps...}
	\begin{itemize}
		\item CERT vs. IND: IND gives too much edge to the hedging choice..\pause
		\item IND vs. REORDER: $\alpha$ and salience works in contradicting ways.\pause
		\item If we fix marginal distributions and state probabilities, these two must work in opposite ways. 
	\end{itemize}
\end{frame}

\begin{frame}{Next Steps...}
	\begin{itemize}
		\item Do not use IND but use a mirror version of REORDER...\pause
		
	\end{itemize}
\begin{tabular}{lccc} 
	& $N$ & $F$ & $S$ \\
	\hline
	$L_{(100,0)}$  & $(0,0)$ & $(0,20)$ & $(80,0)$ \\
	$L_{(50,50)}$ & $(50,0)$ & $(0,50)$ & $(0, 0)$
\end{tabular}\pause
\bigskip

\begin{tabular}{lccc} 
	& $N$ & $F$ & $S$ \\
	\hline
	$\bar L_{(100,0)}$  & $(80,0)$ & $(0,20)$ & $(0,0)$ \\
	$\bar L_{(50,50)}$ & $(50,0)$ & $(0,0)$ & $(0,50)$
\end{tabular}\pause
\bigskip
\begin{itemize}
	\item Directional: If $\bar L_{(100,50)}>\bar L_{50,50}$ then $L_{(100,50)}'> L_{50,50}'$\pause
	\item First Pair: $\pi_1 [v(80)]^\alpha -\pi_2 [v(50)]^\alpha - \pi_3[v(50) - v(20)]^\alpha$\pause
	\item Second Pair: $-\pi_1 [v(50)]^\alpha +\pi_2 [v(20)]^\alpha + \pi_3[v(80) - v(50)]^\alpha$\pause
	\item subtract: 
	\begin{align*}
		&\pi_1 [v(80)]^\alpha -\pi_2 [v(50)]^\alpha - \pi_3[v(80) - v(50)]^\alpha\\
	+ & \pi_1 [v(50)]^\alpha -\pi_2 [v(20)]^\alpha - \pi_3[v(50) - v(20)]^\alpha 
\end{align*}
\end{itemize}
\end{frame}

\begin{frame}{Detailed Examples}
	\begin{tabular}{lccc} 
		& $N$ & $F$ & $S$ \\
		\hline
		$(18,2)$  & $(0,0)$ & $(0,2)$ & $(18,0)$ \\
		$(12,9)$ & $(0,9)$ & $(12,0)$ & $(0, 0)$
	\end{tabular}\pause
	\bigskip
	
	\begin{tabular}{lccc} 
		& $N$ & $F$ & $S$ \\
		\hline
		$(18,2)$  & $(18,0)$ & $(0,2)$ & $(0,0)$ \\
		$(12,9)$ & $(0,9)$ & $(0,0)$ & $(12,0)$
	\end{tabular}\pause
	\begin{align*}
		&\pi_1 [v(18)]^\alpha -\pi_2 [v(9)]^\alpha - \pi_2[v(18) - v(9)]^\alpha\\
		+ & \pi_1 [v(12)]^\alpha -\pi_3 [v(2)]^\alpha - \pi_3[v(12) - v(2)]^\alpha 
	\end{align*}
	\begin{itemize}
		\item we want A B\pause
	\end{itemize}
\end{frame}
\begin{frame}{Detailed Examples}
	\begin{tabular}{lccc} 
		& $N$ & $F$ & $S$ \\
		\hline
		$(18,2)$  & $(0,0)$ & $(0,2)$ & $(18,0)$ \\
		$(10,9)$ & $(0,9)$ & $(10,0)$ & $(0, 0)$
	\end{tabular}\pause
	\bigskip
	
	\begin{tabular}{lccc} 
		& $N$ & $F$ & $S$ \\
		\hline
		$(18,2)$  & $(18,0)$ & $(0,2)$ & $(0,0)$ \\
		$(10,9)$ & $(0,9)$ & $(0,0)$ & $(10,0)$
	\end{tabular}\pause
	\begin{align*}
	&\pi_1 [v(18)]^\alpha -\pi_2 [v(9)]^\alpha - \pi_2[v(18) - v(9)]^\alpha\\
	+ & \pi_1 [v(10)]^\alpha -\pi_3 [v(2)]^\alpha - \pi_3[v(10) - v(2)]^\alpha 
\end{align*}
\begin{itemize}
	\item we want A B\pause
\end{itemize}
\end{frame}
\end{document}