[EM] Re: Weighted Median Approval

[Chris Benham]

  Mike, Dave, Forest, Kevin, anyone interested,

(Thanks Dave for  pointing out my simple-arithmetic stumble. Yes, 4 - 2 
= 2, not 1. Fortunately  it didn't cause me to
announce a wrong winner.)

I previously wrote (Sun.Apr.11):

> Plain  WMA, as  I  have defined it,  is descended from an earlier 
> version (from Joe Weinstein, Forest tells us) in which each ballot
> approves  as many of  the highest-ranked candidates as possible 
> without their combined weight exceeding half the total weight,
> and then only approves the next ranked candidate if the weight of 
> candidates ranked below this (pivot) candidate is greater  than
> the weight of candidates ranked above it. If the two weights are 
> equal, then the ballot half-approves that candidate.
> The problem with this is that  it fails 3-candidate Condorcet. To 
> distinguish it, this earlier version could perhaps be called
> "Above Median Weighted Approval" (AMWA). 

I had in mind this relatively simple type of example:

40: A>B>C
25: B>A>C
35: C>B>A
100 ballots. B is the CW.

Weights   A:40     B:25     C:35
AMWA  approvals
40: A      (not B as in WMA, because A  has a greater weight than C.)
25: BA    (because C  has a greater weight than B)
35: CB
AMWA final scores       A: 65     B: 60    C: 35,     A  wins.

WMA easily elects B (with 100% "approval").  Bucklin  and  "Highest 
Median Rank" (HMR) also elect B.
Quota-Limited Trickle-Down (QLTD)  elects A.

However I was wrong to imply that WMA meets 3-candidate Condorcet.
An example adapted from one of  Woodall's:
300: A
200: A>C>B
300: B>C>A
200: B>A>C
300: C>A>B
199: C>B>A
1499 ballots. C is  CW.

WMA final scores     A: 1000      B: 699     C: 999,   A wins. (AMWA is 
the same, except that C scores 799).
WMA-STV  eliminates B, and then elects C.
Bucklin and its close relatives elect A.
I  adapted this from Woodall's proof that Condorcet is incompatible with 
Later-no-help (and also mono-raise-random,
mono-raise-delete, mono-sub-top, mono-sub-plump)  in his paper 
"Monotonicity and Single-Seat Election Rules",
page 13, "Election 6".
http://groups.yahoo.com/group/election-methods-list/files/wood1996.pdf
So  WMA, in common with Bucklin, QLTD, and HMR fail  3-candidate Condorcet.
I can't see how WMA-STV can fail to meet it.

This is "Election 2" from the same paper (page 11):
12: A>B>C>D>E>F
11: C>A>B>D>E>F
10: B>C>A>D>E>F
27: D>E>F
60 ballots. Smith set comprises ABC.

Bucklin, QLTD, HMR agree with WMA  (and RP and BP) in electing A. 
WMA-STV  elects C.

Weights   A:12    B:10   C:11   D:27   E:0   F:0
WMA approvals
12: ABC
11: CAB
10: BCA
27: DEF

ABC are all tied on 33, but we break the tie in favour of the candidate 
with the greatest "weight", A.
Also using weights to break ties to fix the WMA-STV elimination 
schedule, we eliminate first F and E and then D  and then B,
and then C wins  (C>A, 21-12).

Now we add  6 A>D ballots:
12: A>B>C>D>E>F
06:A>D
11: C>A>B>D>E>F
10: B>C>A>D>E>F
27: D>E>F
66 ballots. Smith set is ABCD, Schwartz set is ABC.

Now Bucklin, HMR, and QLTD all elect D. Adding ballots all with A ranked 
first, causing A to lose, demonstrates that those
methods fail  Mono-add-top.

Weights    A:18    B:10    C:11    D:27    E:0    F:0
WMA approvals
12: ABC
06: AD
11: CAB
10: BCA
27: DEF
WMA final scores     A :39     B:33    C:33    D:33    E:27    F:27,   
 A  wins.
WMA-STV eliminates F and E,  and then B (the "lightest" 33), and then C, 
and  then A wins (A>D, 39-27).

It seems to me that WMA and  WMA-STV meet  Mono-add-top.

Chris Benham