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Say I have a starting situation with Pareto efficient allocations between 2 individuals, now a third one comes in and changes the allocations. Now a fourth one, etc. Up untill a nth individual comes in. Do we now have to resolve (n-1) equations or n^(n-1) equilibria? How does the "difficulty" scale for n individuals? I don't know if that makes sense? Thanks.

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  • $\begingroup$ It does not, please clarify. $\endgroup$ – Giskard Jun 13 '17 at 5:26
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If you are doing pairwise comparisons, then with $n$ individuals there will be $$\dfrac{n(n-1)}{2} $$ such comparisons. You can write this as $\binom{n}{2}$ or as $\frac12n^2-\frac12n$ if you prefer.

$1$ for two people, $3$ for three, $6$ for four and so on, known as triangular numbers

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  • $\begingroup$ It makes sense thank you, I somehow thought every new allocation would change the older allocations so that there would a massive number of equilibria before you get to the final one. $\endgroup$ – NaturalFrequency Jun 13 '17 at 12:58
  • $\begingroup$ @Henry It seems I am somewhat slow today. Having read the answer I still do not understand what the question was. Could you please explain? $\endgroup$ – Giskard Jun 13 '17 at 17:45
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    $\begingroup$ @denesp Presumably to check Pareto optimality and equilibrium, you need to ensure there are no possible exchanges of allocations between pairs of individuals which make both happier. I took the question to be how many such pairs there were to be checked, given there are $n$ individuals $\endgroup$ – Henry Jun 13 '17 at 22:31
  • $\begingroup$ @Henry I see. Thank you. On a side note: It is not enough to check pairs. Consider the case of three goods and three consumers with $U(x,y,z) = \min(x,y,z)$ and the initial distribution $$ (3,0,0) \hskip 10pt (0,3,0) \hskip 10pt (0,0,3). $$ Pairwise no improvements are possible but clearly they could all get $(1,1,1)$ which is better. $\endgroup$ – Giskard Jun 14 '17 at 6:45
  • $\begingroup$ @denesp If you can compare multiples, then the answer is $1$ (just compare everybody together). $\endgroup$ – Henry Jun 14 '17 at 7:11

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