<META HTTP-EQUIV="Content-Type" CONTENT="text/html; charset=iso-8859-1">
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2//EN">
<HTML>
<HEAD>
<META NAME="Generator" CONTENT="MS Exchange Server version 6.5.7226.0">
<TITLE>Re: [EM] 0-info approval voting, repeated polling, and adjusting priors</TITLE>
</HEAD>
<BODY>
<DIV id=idOWAReplyText86448 dir=ltr>
<DIV dir=ltr><FONT face=Arial color=#000000 size=2>You're right Jobst. So
in general, instead of directly adjusting the cutoff value as a weighted average
of previous cutoff values, one should take a weighted average of the lotteries,
and then (on the basis of this averaged lottery) calculate the cutoff
value using the expected value in the case of cardinal
ballots and Joe Weinstein's weighted median in the case of ordinal
ballots.</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>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:</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>1. Calculate the random ballot
lottery L_0 .</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>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 .</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>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.</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>4. Use weighted median based on L_1
to determine A_1.</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>5. etc.</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>6. In finitely many steps the
sequence A_0, A_1, ... will converge to (i.e. get stuck on) the winning
candidate A.</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>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).</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>[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.]</FONT></DIV>
<DIV dir=ltr><FONT face=Arial size=2></FONT> </DIV>
<DIV dir=ltr><FONT face=Arial size=2>Forest</FONT></DIV></DIV>
<DIV dir=ltr><BR>
<HR tabIndex=-1>
<FONT face=Tahoma size=2><B>From:</B> Jobst Heitzig
[mailto:heitzig-j@web.de]<BR><B>Sent:</B> Wed 8/3/2005 11:03 PM<BR><B>To:</B>
Simmons, Forest <BR><B>Cc:</B>
election-methods-electorama.com@electorama.com<BR><B>Subject:</B> Re: [EM]
0-info approval voting, repeated polling, and adjusting
priors<BR></FONT><BR></DIV>
<DIV>
<P><FONT size=2>Dear Forest!<BR><BR>You wrote:<BR>> At each successive stage
we would base the new lottery calculation on a<BR>> weighted average of all
the old cutoffs. In other words, the cutoff on<BR>> each ballot is
adjusted slightly towards the most recent lottery<BR>> expected value before
calculating the new lottery probabilities.<BR><BR>One must know the individual
cardinal utility functions for this, it<BR>seems. An alternative would be to
switch from expected utility 0-info<BR>strategy to median utility 0-info
strategy (aka Weinstein's strategy) so<BR>that only rankings would have to be
known.<BR><BR>Jobst<BR><BR></FONT></P></DIV>
</BODY>
</HTML>