**Mood:**incredulous

**Topic:**Social Choice

Consider Arrow’s example on p. 27 of “Social Choice and Individual Values” where he shows a deficiency of the Borda count: *“Let *R_{1}, ... *R _{n} and R_{1}^{'}, *...

*R*i

_{n}^{'}be two sets of individual orderings and let C(S) and C'(S) be the corresponding social choice functions. If, for all individuals*and all x and y in a given environment S, x R*(independence of irrelevant alternatives). The reasonableness of this condition can be seen by consideration of the possible results in a method of choice which does not satisfy Condition 3, the rank-order method of voting frequently used in clubs. With a finite number of candidates, let each individual rank all the candidates, i.e., designate his first-choice candidate, second-choice candidate, etc. Let preassigned weights be given to the first, second, etc., choices, the higher weight to the higher choice, and then let the candidate with the highest weighted sum of votes be elected. In particular, suppose that there are three voters and four candidates,

_{i}Y if and only if x R/ y, then C(S) and C'(S) are the same*x, y ,z,*and

*w.*Let the weights for the first, second, third, and fourth choices be 4, 3, 2, and

**1,**respectively. Suppose that individuals 1 and 2 rank the candidates in the order

*x, y, z*and

*w,*while individual 3 ranks them in the order

*z, w, x,*and

*y.*Under the given electoral system,

*x*is chosen. Then, certainly, if

*y*is deleted from the ranks of the candidates, the system applied to the remaining candidates should yield the same result, especially since, in this case,

*y*is inferior to

*x*according to the tastes of every individual; but, if

*y*is in fact deleted, the indicated. electoral system would yield a tie between

*x*and

*z.”*

However, if you use a corresponding economic example, there is no such problem.

Let x be 1 unit of work eg 1 hour at task X.

Let y be 1 unit of work eg 1 hour at task Y.

Let z be 1 unit of work eg 1 hour at task Z.

Let w be 1 unit of work eg 1 hour at task W.

Let’s stipulate that each individual must work the same number of hours.

The general idea is to assign the tasks in such a way as to maximize social utility where utility is defined as sum over individual utilities and individual utility is defined as 4 x (time spent on most preferred task) + 3 x (time spent on second most preferred task) etc. ie a general Borda ranking. Let individuals 1 and 2 have the preference ranking xyzw and individual 3 have the ranking zwxy. Since 1 and 2 have the same preferences, we let them spend half their time on their first preference x and half their time on their second preference y. Let 3 do z since it’s his first choice. That leaves w. Let each individual do one third of w. Then the individual utility for 1 and 2 is (½)x4 + (½)x3 + (1/3)x(1) = 3 and (5/6). The individual utility for 3 is 4x1 + 3(1/3) = 5. Social utility is 12 and (2/3). The maximum possible utility is 16.

Now, if task y is deleted as in Arrow’s example, we have the preferences for 1 and 2 as xzw and 3 as zwx. Now let 1 and 2 each do (½)x and (½)w and 3 do all of z. Each person does 1 hour of work. Total possible utility = 12. For 1 and 2, utility = (1/2)x3 + (1/2)x1 = 2.Utility for 3 = 3. Total utility = 7. Alternatively, let 1 and 2 do (½)x and (1/4)w and (1/4)z. Utility = 3x(1/2) + 2x(1/4) + 1x(1/4) = 9/4 = 2 and (1/4). Let 3 do (1/2)z and (1/2)w. Utility = 3x(1/2) + 2x(1/2) = 5/2 = 2 and ½. Total utility = 7.

The point is that eliminating y does not upset the work schedules in the same way that it upsets political rankings because the work can be divided among the workers in many possible ways. Hence Arrow’s Impossibility Theorem doesn’t apply.