05 May 2018

## Another Arrow in Your Quiver

Here are my class notes on Arrow’s Impossibility Theorem (thanks to Transformer for hosting the PDF). You have to be able to read formal set theory notation, but other than that there’s no prior knowledge required.

#### 7 Responses to “Another Arrow in Your Quiver”

1. Keshav Srinivasan says:

Bob, one criticism I have of a lot of presentations of Arrow’s theorem including yours, Steve’s and Amartya Sen’s, is that they state the IIA condition rather imprecisely, using vague English words like “depend” and “affect”. I wish more presentations would use the formal statement of the condition: “For two preference profiles (R_1, …, R_N) and (S_1, …, S_N) such that for all individuals i, alternatives a and b have the same order in R_i as in S_i, alternatives a and b have the same order in F(R_1, R_2, …, R_N) as in F(S_1, S_2, …, S_N).”

I think if more people understood that formal statement, there would be less confusion about how IIA is applied in the proof.

• Rick Hull says:

Hi Keshav,

I am sure that with sufficient effort I can understand your statement, and in a perfect world, I would already have your (widely understood) program loaded in my head. But Bob, Steve, and Amartya are doing people like myself a favor by presenting the theorem in more fuzzy yet comprehensible terms.

I also wish that more people (like myself) understood your formal statement, but I also wish I had a Ferrari. If your goal is to minimize annoyance at slowpokes like me, then please continue feeding highly formalized statements to the logical computer. As a programmer, I share your lament 😉

2. DZ says:

I followed conceptually until the last paragraph where you walk from the decisive subset G down to a decisive individual. Is the idea that G is a decisive subset over N, and then some subset of G is decisive over G…etc.? In other words, it’s not that the dictator is directly decisive over N, but that they’re decisive over some larger decisive subset that’s decisive over some still larger decisive subset…on and on until you reach G and then N?

• Bob Murphy says:

DZ I’m not exactly understanding what you’re asking, but: Did you agree with the Contraction Lemma? I.e. do you agree that if some group G is decisive (and it has at least two members), then we can find a proper, nonempty subset of G that is also decisive?

• DZ says:

Yes I do agree, but I think I’m verifying that the proper, non-empty subset of G is decisive over G and not N. In other words, if N is very large and there is some G that is a decisive subset over N, you can break G into subset A, with A decisive over G. Then keep going and break A into a subset B, with B decisive over A…continually repeating this until you get smaller and smaller subsets ending at the dictator. So the dictator kind of has control over N through a long chain of decisive subsets leading eventually back to the subset G which is decisive over N.

• Bob Murphy says:

DZ, I think you’re getting a little mixed up. A group of individuals is decisive over elements from the set of possible outcomes, not over other individuals. It’s not “grammatical” to say a group G is decisive over N.

So if some group G is decisive over a,b, it means that if everybody in group G is agreed on how to rank a,b, then that is what the social ranking will be too, regardless of how the people outside of group G rank a,b.

• DZ says:

Ok thanks. This made me go back and read more carefully and I see where I was messing up. I missed the ‘at least three elements’ assumption, which had me trying to interpret the conclusion in a two-element scenario. I was trying to think of ways some group G could be decisive over a particular social ranking of a,b, and then how a subset of G could also be decisive over a particular social ranking of a,b. Led me astray. Thanks again for helping me through.