Friday, July 23, 2010


    [end of Arrow third installment]

Proof of Lemma 3:        Assume xDy for i [i,e,, i is decisive for x against y]

    The proof now proceeds in two stages. First, for an environment [x,y,z], constructed by adding some randomly chosen third element z to x and y, we show that i is a dictator over [x,y, z].

    Then we show how to extend this result step by step to the conclusion that i is a dictator over the entire environment S of admissible alternatives.

First Stage: Proof that i is a dictator over the environment [x,y,z]

(step i)    Construct a set of individual orderings over [x,y,z] as follows.

    Ri: x > y > z [i.e., individual i's ordering of the three]

    All the other Rj: yPjx yPjz Rj[x,z] unspecified

    In other words, we will prove something that is true regardless of how everyone other than i ranks x against z.

(step ii)    xPiy by construction. But, by hypothesis xDy for i. Therefore xPy

    In words, i is assumed to strongly prefer x to y, and since by hypothesis i is decisive for x against y, the society also strongly prefers x to y.

(step iii)     For all i, yPiz, by construction Therefore, yPz, by Condition P, and xPz by Lemma 1(c). In words, since everyone strongly prefers y to z, so does the society. And since the society strongly prefers x to y and y to z, it strongly prefers x to z [since Axiom II, which is used to prove Lemma 1(c), stipulates that the SWF is transitive.]

(step iv) So xPz when xPiz, regardless of how anyone else ranks x and z. [check the construction of the individual orderings in step (i) ]

(step v)    Hence xḎz for i, which is to say that i dictates over the ordered pair (x,z)

(step vi)    Now consider (y,z) and assume the following set of individual orderings:

    Ri:    y > x > z

    All the other Rj: yPjx zPjx and Rj[y,z] unspecified.

(step vii)    yPix for all i. Therefore yPx by Condition P

(step viii)    xḎz for i, by (v). Hence xPz.

(step ix)    So yPz by Lemma 1(c). Thus yḎz for i.

    In words, we have now shown that i dictates over the ordered pair (y,z). Let us take a minute to review what is going on here. We are trying to prove that if i is decisive for a single ordered pair, (x,y), then i is a dictator over an environment consisting of x, y, and some randomly chosen z. If we can show that i is a dictator for every ordered pair in the environment [x,y,z] then we shall have shown that i is a dictator over that environment. There are six ordered pairs that can be selected from the environment, namely (x,y), (x,z), (y,x), (y,z), (z,x), and (z,y). So we must establish that i dictates over every single one of these ordered pairs. We have already established that i dictates over (x,z) in step (v) and over (y,z) in step (ix).

(step x) We can now extend this argument to the other four ordered pairs that can be selected from the environment [x,y,z]. In particular, let us do this for the ordered pair (y,x). Construct the following set of orderings:

    Ri:    y > z > x

    All the other Rj: zPjy zPjx Rj[x,y] unspecified.

(step xi)      zPix for all i. Hence zPx by Condition P

(step xii)    yḎz for i by (step ix). Hence yPz

(step xiii)    So yPx by Lemma 1(c). Thus yḎx for i.

    So we have proved [or can do so, by just iterating these steps a few more times] that i dictates over every ordered pair in [x,y,z], and therefore i is a dictator over the environment [x,y,z]. So much for Stage One of the proof of Lemma 3. Now, take a deep breath, review what has just happened to make sure you understand it, and we will continue to:

Stage Two: The extension of our result to the entire environment, S, of available alternatives. Keep in mind that S, however large it may be, is finite.

    Assume xDy for i [our initial assumption -- just repeating for clarity] and also assume the result of Stage One. Now consider any ordered pair of alternatives (z,w) selected from the environment S. There are just seven possibilities.

    1. x = z     w is a third alternative

    2. x = w     z is a third alternative

    3. y = z     w is a third alternative

    4. y = w     z is a third alternative

    5. x = z     y = w

    6. y = z    x = w

    7. Neither z nor w is either x or y

Case 1:        We have an environment consisting of three alternatives: [x=z, y, w]. Stage One shows that if xDy for i, then x=zḎw for i.

Case 2, 3, 4:    Similarly

Case 5:        Trivial

Case 6:        Add any other element v to form the environment [x=w, y=z, v]. From x=wDy=z for i, it follows that y=zḎx=w for i. [In words, just in case you are getting lost: In the case in which y is element z and x is element w, from the fact that i is decisive for x against y, which is to say for w against z, , it follows that i dictates over y and x, which is to say over z and w. This is just a recap of Stage One.

case 7: This is the only potentially problematic one case, and it needs a little explaining. We are starting from the assumption that i is decisive for x against y, and we want to show that i is a dictator over some totally different of alternatives z and w, so we are going to creep up on that conclusion, as it were. First we will add one of those two other alternatives, z, to the two alternatives x and y to form the environment [x,y,z]. From Stage One, if xDy for i then xḎz for i. But trivially, since xḎz for i, it follows that xDz for i. [The point is that if i dictates over x and z, then of course i is decisive for x against z].

    Now add w to x and z to form the environment [x,z,w]. Since xDz for i, it follows that zḎw for i, by Stage One. In words, if i is decisive for x against z, then in the environment [x,z,w], i dictates over z and w. This follows from Stage One. What this shows is just how powerful Lemma 3 really is.

    Thus we have demonstrated that xDy for i implies zḎw for i, for all z and w in S. In other words, if i is decisive for some ordered pair (x,y), then i is a dictator over S. But Condition 5 stipulates that no individual may be a dictator. Therefore:

    An acceptable Social Welfare Function does not permit any individual to be decisive for even a single ordered pair of alternatives in the environment S of available alternatives.

    Can we all say Ta-Da? This is the heavy lifting in Arrow's theorem. Using this Lemma, we can now fairly quickly prove that there is no SWF satisfying Axioms I and II and all four Conditions, 1', 3, P, and D.


  1. I have problems understanding, why this proof works: steps i-v seem to prove the following: if xDy for i and individual preference orderings are thus-and-thus, then xḎz. So is step viii ok? In step vi you are assuming a different set of individual orderings and the following doesn't seem to be a correct inference

    1.if xDy for i and individual preference orderings are thus-and-thus, then xḎz. (step v)
    2. xDy for i (assumed t the beginning) and now some different individual preference orderings (step vi)
    3. therefore, xḎz (step viii)

    Besides, I dont see why this proof would work for all sets of individual prefence orderings: in all your constructions of different preference individual orderings, you either assume that aPib for all i, or that aPjb for all j who aren't i, and bPiA, so that you can always infer aPb, either with Condition P or the assumption that aDb for i. Yet since condition P and the assumtption aDb are both if-then statements, they are true when their antecedents are false. So the assumption aDb for i is true, even when (some of) the others agree with i. But in that case, you cant infer via modus ponens that aPb (since the antecedent is false). But how do you prove in these cases, that aḎb for i for all a and b?

  2. Simo, This is the problem with trying to do this online. For me, at any rate, it would be much easier if we could talk about it face to face. About the first point: The trick in the proof is that if, assuming some particular set of preference rankings, Arrow can prove that i dictates over some ordered pair, then i is a dictator over that ordered pair regardless of how everybody else's rankings change. i is not just a dictator over the ordered pair for some special set of rankings. This is another way of saying how powerful Lemma 3 is, and therefore how powerful an assumption it is to say that i is decisive over a single ordered pair.

    As for the last part of the comment,I think [I emphasize think] the point is that in P, for example, the "if then" is an entailment, not a truth functional connective. Does that help at all?

  3. I was stuck at a similar point. Can I check I understand?

    At first glance, it looks like the proof is invalid. The condition for i to dictate over (x,z) is that in any setting where i strongly prefers x to z, society strongly prefers x to z. But, looking at steps (i)-(v), it seems like all we have is that when i strongly prefers x to z *and a bunch of further assumptions are in place* then society prefers x to z.

    But I'm thinking the trick concerns Condition 3 of the assumptions. By that assumption, how society ranks x against z is determined by individuals' ranking of x against z. So let's look just at {x,z}, assume that i strongly prefers x over z, and assume nothing about what anyone else thinks. Now, because of the independence assumption (condition 3) we're allowed to fill in the details any way we choose as to preferences over options not in {x,z} (where individuals rank y, for example)---and this can't change anything about the societal ranking of x and z.

    What the proof does is fix on a particular way of adding in details about where y fits in to each person's preferences, and showing that in that one case, the assumptions force society to rank x over z. But, as we said, the one case is representative for all cases, since the societal ordering can't depend on these sort of extraneous details. So society prefers x to z in all rankings.

    Is that how the argument goes? Sorry if this is nonsense...