[sean.whetten]

Hi.

I just finished digesting the (Advanced) lecture about this theorem. After some thought, I found an intuitive way to think about the proof (at least, it is intuitive to me), which is to see that the Independence of Irrelevant Alternatives and the (unstated) stipulation that the output from the SWF must be transitive are incompatible with each other.

This incompatibility is the primary driver that was used to derive the Contagion lemma. To do so, we used a bunch of information that we knew about the “irrelevant” alternatives and the transitivity of the output to decide that x >s y, but then said “the actual mapping function is not allowed to use the information that we just used to derive this ordering, because it’s “irrelevant”.”

Here’s my thought. The irrelevant alternatives condition implies that we could deconstruct F into subfunctions that each take the rankings from everyone for only 2 of the choices and outputs the social ranking for those choices, then reconstruct all the results of the subfunctions back into the results of F. e.g. say that there are 3 choices (a,b,c). If irrelevant alternatives holds, then we can deconstruct F into Fab, Fac, Fbc, and the outputs from each of the deconstructed functions should be compatible (re, transitivity) for any case.

It was easy for me to convince myself that there’s no way for the deconstructed functions to ensure transitivity unless they pick a dictator by considering this example:

Assume that the following is true:
For some N, the SWF is a >s b >s c (Or, more formally, F(N) = a >s b >s c and Fab(N) = a >s b and Fac(N) = a >s c and Fbc(N) = b >s c)
There exists an element of N (i) whose preferences are a >i c >i b

Let’s replace i with i’ whose preferences are c >i a >i b. According to the Independence stipulation, we are allowed to change the relative ranking of a compared to c, but not the relative rankings of a compared to b nor c compared to b, because their relative rankings vis-a-vis i and i’ didn’t change. Or, more formally, Fab(N) = Fab(N’) = a >s b and Fbc(N) = Fbc(N’) = b >s c, but Fac(N) is unknown compared to Fac(N’)

If new information DOES change the relative ranking output derived from Fac, then the new SWF can have one of (c >s a, c >~s a, a ~s c, or a ~< c), and also must have a >s b and b >s c. However, any of these changes to Fac would make the SWF intransitive. A contradiction. So, we can conclude that the new information CANNOT change the relative ranking for the SWF in this case and that it doesn’t matter what i’s preference between a and c is. (Or, more formally, if Fab(N) = a >s b and Fbc(N) = b >s c then Fac(N) must have no effect on F). However, Fac(N) has no way of “knowing” this based on its inputs!

More generally, any particular deconstructed function has no way of knowing (from it’s inputs alone) whether it has to be fixed (because output from the other ones has already implied that it is, as in the example), or whether it is allowed to change. The only solution that will ensure that the result of F is transitive is for all of the deconstructed functions to agree on a dictator (whose preferences are known to be transitive) and always output his preferences.

So, to ensure transitivity you have to either 1. take the “irrelevant” alternatives into account (e.g. breaking the Irrelevant Alternatives condition), or, 2. Pick some guy and just use his preferences (e.g. break the No Dictator condition).

[jmherbener]

Arrow’s Impossibility Theorem has generated much commentary. Here’s an overview: https://plato.stanford.edu/entries/arrows-theorem/

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