[EM] Hay voting bust, busted

[Jobst Heitzig]

Peter de Blanc schrieb:
> On Wed, 2007-02-07 at 00:47 +0100, Jobst Heitzig wrote:
>> Then you should be able to provide a thorough proof that it is optimal 
>> to express rankings proportional to your true utilities, by showing the 
>> respective derivatives to be zero.
>>
>> Please do so, since I still question that they are!
>
> Jobst, I can do this for you but it would take me a while to do. I'm not
> very good at typesetting HTML and the formulae are very ugly and
> complicated.
>
> Let me ask you: in the original writeup for "the n-Substance problem,"
> do you believe that:
>
> 0. the pricing rule given satisfies the criterion given (ie that it is
> optimal to purchase quantities proportional to utility densities)?
>
> All I did to get the formulae from Hay Voting from there was:
>
> 1. I let the substances be transfers of voting mass between candidates
> (there are n choose 2 such possible transfers)
> 2. I calculated exactly how large each transfer would be by assuming
> that the size of the transfer would be proportional to the difference in
> utility between candidates.
> 3. I then calculated how much voting mass each candidate would be left
> with after all the transfers
>
> I thought that each of the steps was adequately justified. If you have a
> problem with one particular step, then it would be easier for me to try
> to clarify that.
>
> - Peter de Blanc
OK, let me do it for you then. Given expressed ratings r1...rn, your 
formula was

pi = (sqrt(n-1) - fi) / n sqrt(n-1)
where  fi = c sum_j (rj-ri) = c (t - n ri)
where c = 1 / sum_ij (ri-rj)²
and  t = sum_j rj

Given true utilities u1...un, this results in expected utility

E = sum_i (pi ui).

If this would me at a local maximum for ri proportional to ui, each 
derivative

gk := (d/drk)E(r1=a*u1,...rn=a*un)

should be zero. But gk is proportional to

(d/drk) sum_i (fi ui)
= c² [ (t - n uk) sum_ij (ui-uj)² + sum_i (t - n ui) 4 (t - n uk) ]
= c² [ (t - n uk) sum_ij (ui-uj)² ]

which is zero for each k if and only if

uk = t/n,

for each k, that is, if all uk are equal. This is certainly not the case 
in general, hence putting ri = a ui for whatever a is *not* optimal.

Yours, Jobst