[EM] clarification of proof in my previous post

[Simmons, Forest]

In the proof paragraph after step 6 below, the constancy of the sign of the difference of the probability above and below X before and after the averaging step is only true for X=A_k , precisely where it is needed.
 
I had written ...

By the way, your method of adjusting only the relative likelihood of the current winner to the other candidates is growing on me.  Here's a specific version adapted to ordinal ballots:

1.  Calculate the random ballot lottery L_0 .

2.  Use Joe Weinstein's weighted median (approve X iff L_0 gives more probability to lower ranked than higher ranked candidates) strategy on each ballot to determine the approval winner A_0 .

3.  Let L'_0 be the lottery that gives A_0 one hundred percent of the probability, and let L_1 be the average of L_0 and L'_0.

4.  Use weighted median based on L_1 to determine A_1.

5. etc.

6.  In finitely many steps the sequence  A_0, A_1, ... will converge to (i.e. get stuck on) the winning candidate A.

Proof of the assertion in step 6 is the same as Jobst's proof in the case of Cardinal ballots, because in Joe Weinstein's strategy the difference in total probability above X and below X has the same sign before and after the averaging step, so A_k's approval doesn't change from stage k to stage (k+1).

[Therefore the current approval champ beats the previous one only by setting a new approval record.  This cannot happen more than a finite number of times when there are a finite number of voters.]

Forest

________________________________

From: Jobst Heitzig [mailto:heitzig-j at web.de]
Sent: Wed 8/3/2005 11:03 PM
To: Simmons, Forest
Cc: election-methods-electorama.com at electorama.com
Subject: Re: [EM] 0-info approval voting, repeated polling, and adjusting priors



Dear Forest!

You wrote:
> At each successive stage we would base the new lottery calculation on a
> weighted average of all the old cutoffs.  In other words, the cutoff on
> each ballot is adjusted slightly towards the most recent lottery
> expected value before calculating the new lottery probabilities.

One must know the individual cardinal utility functions for this, it
seems. An alternative would be to switch from expected utility 0-info
strategy to median utility 0-info strategy (aka Weinstein's strategy) so
that only rankings would have to be known.

Jobst



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