A Short Proof of Arrow’s Impossibility Theorem
In 1951, Kenneth J. Arrow produced a long, elaborate, hyper-mathematical proof that social choice was impossible based on five criteria that he postulated. He could have proven the same result with a lot less effort and a lot less mathematics, but this would not have impressed the world of academia nearly as much. A much shorter proof follows.
We assume that there are m alternatives and n voters. Each voter ranks the m alternatives according to his/her preferences. A social welfare function (SWF) then converts this data into a ranking that holds for the entire society. The social choice must meet Arrow’s 5 criteria or the SWF is invalid. Of course, someone else might have chosen a different set of criteria, but let’s assume Arrow’s criteria for now. Arrow goes on to prove that there is no social welfare function that will produce results that meet his 5 criteria in every case.
If only one case can be found for which a SWF does not produce a result which meets all of Arrow’s criteria, that SWF is invalid. Therefore, the social welfare function must work for the case, m = n = 3. Let’s say Voter 1’s choice is ABC, voter 2’s choice is BCA and voter 3’s choice is CAB. This represents the voting paradox which has been known for hundreds of years. The possible social choices are the following: ABC, ACB, BAC, BCA, CAB and CBA. In order for there to be a valid social choice, one of these outcomes must satisfy all of Arrow’s 5 criteria.
Let us consider Arrow’s Condition 3, the Independence of Irrelevant Alternatives. This Condition states that if the social welfare function produces the outcome XYZ and then one of the alternatives is eliminated the same SWF applied to the 2 remaining alternatives must produce the same outcome as would be produced by blotting out the eliminated alternative in the 3 alternative case. For example, if the SWF produces an outcome XYZ in the case in which there are 3 alternatives, it must produce the outcome XY if Z is eliminated, XZ if Y is eliminated and YZ if X is eliminated.
In the above case, assume that the SWF produces the outcome ABC and when C is eliminated it produces the result AB. Therefore, in the case n = 2, to be consistent, it must always produce the result XY if the individual votes are XY, XY, YX. Now let us assume B is eliminated. The SWF must produce the outcome AC, but that would be inconsistent since the individual votes are AC, CA, CA. Therefore, no SWF could produce the outcome ABC and be consistent with itself.
In the same way consider the other possible outcomes: ACB, BAC, BCA, CAB, CBA. In every case we find that the SWF cannot produce that outcome and be consistent with itself. Therefore, there is no SWF, consistent with Arrow’s Condition 3, that will provide an outcome for the case m = n = 3 and individual votes ABC, BCA, CAB.
Therefore, social choice is impossible since there is no SWF that will provide an outcome that meets one of Arrow’s Criteria in one case.