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%%%%%%%%%%%%%% NOTES %%%%%%%%%%%%%%%%%%%  ---------------------------------------------  \appendix  \section{Irrational individuals?}  Theoretically, proxy voting comes with an interesting feature: individual judgment sets (when including the proxy's judgment on each delegated issue) may turn out to be inconsistent.   As social choice theory (and judgment aggregation) standardly assume that individual preferences (and individual judgment sets), all the main results in those fields (i.e. Arrow's theorem) rely on this individual rationality assumption. Allowing for inconsistent individual judgment sets hence opens a new level of generality in judgment aggregation and social choice theory.   As an illustration, recall the doctrinal paradox, which famously shows that consistent individual judgment sets sometimes leads to an inconsistent collective outcome:  \begin{table}  \begin{center}  \begin{tabular}{ l | c c c c }  & p & q & p \rightarrow q & \\   \hline  a & 1 & 0 & 0 & \\   b & 0 & 0 & 1 & \\   c & 1 & 1 & 1 & \\   \hline  majority & 1 & 0 & 1 &   \end{tabular}   \end{center}  \end{table}  Consider now the following modification of the above case: assume that agent $c$ votes himself on issue $p$ but delegates to agent $b$ his vote on $q$ and on $p\rightarrow q$. We obtain the following situation, where agent's $c$ judgments are inconsistent:   \begin{table}   \begin{center}  \begin{tabular}{ l | c c c c }  & p & q & p \rightarrow q & \\   \hline  a & 1 & 0 & 0 & \\   b & 0 & 0 & 1 & \\   c & 1 & \bf{0} & 1 & \\   \hline  majority & 1 & 0 & 1 &   \end{tabular}   \end{center}  \end{table}  This raises new research questions: What are the conditions for the collective outcome of a vote to be rational, when individuals can be irrational? In particular, under what conditions does liquid democracy preserve majority consistency?  Moreover, what are the criteria for \emph{rational delegation}?\footnote{It seems intuitive to impose the following constraint: if agent $i$ delegates issue $r$ to an agent $j$, he should also delegate to $j$ all the remaining issues which depend solely on the truth of $r$, given $i$'s judgment on non-delegated issues. In our example, this means that it would not be rational for agent $c$ to delegate to agent $b$ on issue $q$ but not on issue $p\rightarrow q$, given that $c$ judges $p$ true.} And assuming some criteria of rational delegation, is it possible to go from the above cases of discursive dilemma (with consistent individuals) to a liquid case where inconsistent individuals outputs a consistent collective outcome, via only rational delegation?  \section{DeGroot influence}  By multiplying the two matrices, a new opinion matrix is generated, representing the opinion state of agents after one influence step (for instance after they exchange information with each other). DeGroot, using the associativity of matrix multiplication, shows that whether the opinions of a group of agents converge in the limit depends on the underlying influence graph. If multiplying the influence matrix by itself converge in the limit, then any distribution of opinions of the agents will converge too.  %as Jackson puts it: no matter what beliefs p(0) the agents start with, they end up with limiting beliefs corresponding to the entries of $p(\infinite)=lim_t T_tp(0)$.   More precisely, convergence depends on (each connected and closed component) of the influence graph being aperiodic (that is, the greatest common divisor of all directed cycles lenghts is $1$.): ``convergence in overall society hold if an only if each closed and strongly connected set of nodes converges, which happens if and only if each such set is aperiodic." \cite[p.233]{jackson08social}  %Moreover, any strongly connected and closed group reaches a consensus for any initial opinion vector if and only if it is aperiodic (jackson, p.234). (corollary 8.1) An overall consensus is reached iff there is exactly one strongly connected and closed group of agents and it is aperiodic. (Corollary 8.2, Berger) A consensus is reached iff there exists n such that some column of the influence matrix multiplied $n$ times by itself has all positive entries. Once a column is all positive, it stays positive forever! (a column with all positive entries represent the fact that at some point one agent influences all other agents, showing that all agents must have a path to some agent and so there is exactly one strongly connected closed group.   \section{A limit case of DeGroot processes}  Consider the limit cases of DeGroot influence, where both the opinion of each agent and the influence weight belong to $\{0,1\}$ instead of $[0,1]$. We will call such cases Boolean deGroot process (BdG).  In that case the trust network is a graph which is functional (or serial and deterministic): every agent has exactly one ``guru" (possibly itself).   %We have said that Liquid democracy can be regarded as this limit case, but NOT REALLY: THERE IS A DIFFERENCE: IN DEGROOT, THE EFFECT OF INFLUENCE ALWAYS APPEARS "ONE TIME-STEP LATER". IN LIQUID DEMOCRACY, EVERYTHING IS SIMULTANEOUS.   The influence matrix $P$ will read $1$ for entry $P(i,j)$ of the matrix when agent $i$ follows (copies) $j$'s opinion, and $o$ for all other entries of the same row.   The opinion matrix $Q$ will read $1$ for entry $p(i,m)$ which the agent judges true, and $o$ for other entries in the same row.   For simplicity, we will start with considering the case where there is only one issue $p$.   % In Jackson (8.3.1) opinion leaders are agents who are influenced only by their own opinion, and hence have fixed opinion!   %We mentioned at some point that the network of who delegates to whom is a forest. But this is not the case... unless we impose additional constraints on delegation.   On the deterministic frames, $\Box p\leftrightarrow \diamond p$ is a validity. The rule ``copy your guru" is just the same as the rule of ``influence by unanimity" (which states: switch opinion on $p$ if and only if all of your influencers have the opposite opinion than you).  %or if we wanna use Johan's different definition: have opinion that p iff all of your neighbors do. Note that this rule is "asymmetric" in the sense that it does not behave the same with respect to $p$ than with respect to $\lnot p$.  A model is stable if and only if the formula $p \leftrightarrow\Box p$ is a validity. That is to say $p$ is a fixpoint of $\Box p$.   - Since $\Box p$ is a positive formula (and hence $\Box$ denotes a monotonic function from sets of states to sets of states) the Knaster-Tarski fixpoint theorem applies: so we know that a smallest and largest fixpoints exist (and possibly more in-between). The smallest fixpoint is the empty-set (nothing is $p$) and the largest is the whole domain (everything is $p$).   \section{Convergence}  NOTE FOR LATER: WE CAN GIVE THE PARALLEL RESULTS FOR THE CASE OF THE UNANIMITY RULE ON (NON-NECESSARILY-FUNCTIONAL) SYMMETRIC INFLUENCE GRAPHS. WE CAN GIVE THE CLASS OF INFLUENCE PROFILES WHICH GUARANTEE CONVERGENCE OF ALL OPINIONS PROFILES, AND WE CAN GIVE THE CLASS OF OPINION PROFILES WHICH CONVERGE ON THE OTHER INFLUENCE PROFILES:   \begin{proposition}  Let $\G=(G_{p_1},\dots,G_{p_m})$ be a SYMMETRIC AND NON-NECESSARILY FUNCTIONAL influence profile and $\O=(O_1,\dots,O_n)$ be an opinion profile.  The opinion diffusion with unanimity aggregation rule converges for $\O$ on $\G$ iff:  \begin{itemize}  \item[] FOR ALL $j\in\{1,\ldots,m\}$, $G_pj$ IS NOT TWO-COLORABLE, OR  \item[] THERE IS A $j\in\{1,\ldots,m\}$, SUCH THAT $\O$ PROPERLY COLORS $G_pj$, THAT IS SUCH THAT FOR ALL $i,j\in\N$ WITH $iRj$, $O{p_i}=1$ if and only if $O{p_j}=1$.  \end{itemize}  \end{proposition}