[EM] Relative margin definition and example

Elisabeth Varin/Stephane Rouillon stephane.rouillon at sympatico.ca
Wed May 22 06:17:36 PDT 2002


               Sorry, too many things to do...

               Ok this is going to be long, not that much complex,
               and would be easier with some context.
               Relative margin is defined at 4) and is equal to:
               margin / (pairwise comparisons different than "?")

               1) I am working actually on a multiple winner partisan
               election method. But I still use single member districts
               so
               people do not have to listen to more candidates than
               there are parties (and independent candidates).
               But I need weigths as an output of each single district
               to obtain a full proportionnality. I was using IRV,
               keeping the support in favour of a last preference
               when it is eliminated as their output weight. I have just

               finished an identical inner motor using ranked pair.
               I will introduce it later in this mail.

               2) Neither margins nor winning votes. After
               reading all what was to my disposal I favour the
               use of relative margins. Why, because it seems the only
               fair way to respond to a universal ballot. So what is
this
               universal ballot?

               3) This came when I could not admit Mr. Ossipoff
               reasoning about winning votes being superior to margins.
               The case is:
               510: A > B
               490: B > A
               compared to
               200: B > C
               0: C > B
               800: B = C
               Winning votes would lock A>B before B>C.
               Margins would do the contrary.
               When I look at this my human analysis says that
               margins is the good way to order the locking procedure.

               But what if
               510: A > B
               490: B > A
               compared to
               19: B > C
               0: C > B
               981: B = C
               Would B>C be a more probabilistic ranking than A>B ?
               Margins locks A>B first. The question is what means the
gesture
               of the 981 people would did not rank B and C. Did they
simply
               truncate
               B and C because they were not interested by them and had
not the time
               to look at this? Or did they have an opinon, which would
be B and C
               are
               clones, put anyone it is the same, I have no preference?
               So I think we need to differentiate these 2 cases:
               A ? B when I have no opinion and you can rely on other
voters choice.
               A = B when my opinion is that both are equally good or
bad.
               So if 981: B = C for real, then A > B should get locked
first.
               But if 981: B ? C, then B > C is more probable than A >
B.

               You can check, when there is no "?" nor "=", winning
votes,
               margins and relative margins produce the same result.
               Adding the equal (=) and undefinite (?) option for
voters,
               relative margins leads to the most probabilistic locking
               procedure.

               Finally, another element comes from approval and is
               present with Demorep1's representation. Voters want
               to manifest who they approve and who they disapprove.
               This cannot be represented using only ">", "=" and "?" .
               Disapproving some candidates is equivalent to
               admit a standard replacement (because we need to have one

               representative) instead. Each voter can assess his own
               criteria for an average replacement politician.
               We will call him Z. We can now
               represent any ballot. Note we have gained some
flexibility
               in ordering approved and disliked candidates.

               A FPTP ballot in favour of A:
               A > Z > B ? C ? D ? E

               An approval ballot for ABC:
               A = B = C > Z > D ? E

               A prefential ballot A1 B2 C3 D4 E5:
               A > B > C > D > E > Z

               A truncated preferential ballot , being unable to make
its mind about
               its first choice:
               A ? B > C > D > Z > E

               We even have some freedom degrees with the use of "=" and
"?" around
               Z. We will not use them for the moment.

               4) So now we have all this possibilities to gather the
right
               information

               with this universal ballot, how to use it? Relative
margins seems the
               right
               probabilistic measure to lock pairwise results. This is
an example:

               A>B B>A A=B A?B
               32 24 12 32

               Winning votes: A>B (32)
               Margins: A>B (8)
               Relative margins: A>B (8/68 = 11,9%)

               Comparing several cases, relative margins seems
intuitively
               to show the support in favour of an option. It works well

               even to compare pairwise with different electorate size,
               which makes sense.

               5) Now trying to rank all candidates to produce a path
with
               the least probability of inversion error is the goal of
Mr.
               Tideman procedure. Whatever parameter used (winning
votes,
               margins, relative margins), there are some cases were the
order
               produced does not minimize the summation of the
inversions.
               As an example you can check:
               A>Z (22), A>C (4), A>B (13), D>A (16),
               B>Z (30), B>D (15), C>B (10), Z>C (3),
               D>C (7) and Z>D (1).
               Tideman produces C>B>D>A>Z with an error of 28
(13+7+4+3+1).
               Best order with minimized error is:
               A>B>D>Z>C with an error of 27 (16+10+1).
               Tideman just ensures to get the smallest bigger inversion
error (13).
               The ultimate solution needs a branch and bound tree, it
is
               non-linear,
               but it can be done.

               Instead, some people would argue that minimizing the sum
of
               the error is not the best objective, minimizing the
bigger error is
               their goal.
               It is logic if we think that one voter could be wrong on
several
               pairwise
               and we do not want his or her errors to count several
times, so
               Mr. Tideman goal minimizes the biggest error made by
different
               persons.
               A well suited goal that I gladly rally to.

               6) So I need weights and I do not want those weights to
be affected
               by
               candidates cloning. Relative margins give me information
about
               relative supports. How to translate my well ordered
ranked pair
               path into a 100% summation weight to compare its result
with
               the other districts ? I have tried before to generalize
the
               Truncation Resistance Criteria (TRC) or Secret Preference

               Criteria (SPC) to a Weigthed Truncation Resistance
Criteria:
               the goal was that further preferences should not modify
               the weight accorded to your favorite, but it could make
               your favourite loose by enhancing another candidate
weight.
               This criteria does not fit with the support relation: if
xA and xB
               represent the weights for A and B and rAB the relative
margin of A>B
               then xB = xA (1-rAB)/(1+rAB). The same for each pairwise
victory.
               With the additional: Sigma x + "none" votes = 100%,
               we obtain a first linear system to produce weights.
               But WTRC is not verified.

               7) The problem is, I am not even sure there is a
weighting procedure
               that would
               verify WTRC. Strategy will always be a possibility it
seems for the
               moment.
               So at least I turned on to put my method out of reach of
               cloning effects. The best element to counter cloning and
crowding is
               rallying.
               So let us try a IRV kind of model: we simply change the
way to select
               the eliminated candidate, it is the last of the ranked
pair path.
               But using ranked pair in this manner brings back Mr.
Ossipoff ghost
               of
               TRC.
               The fact is Mr. Ossipoff is right: truncating preferences
can harm my
               favourite
               using margins or relative margins. But what he does not
say is: NOT
               truncating
               can harm my favourite using winning votes. So strategy is
still an
               issue, but
               one we cannot avoid for the moment...
               Using ranked pair with relative margins we eliminate the
last
               candidate
               of the path,
               keep is weight and evaluate next round. The weight is the
number of
               votes
               where the next prefered candidate is Z or none. It means
the voters
               give
               their approbation
               (support) to the last approvable available candidate. But
their
               ballot
               is
               still active. Their last choices, preferences after Z can
still
               influence the
               future elimination order. The path does not change after
each round.
               So we finally found a way to influence the choice
               between disapproved candidates, while still not approving
any of
               them.
               Demorep1 was right and I was wrong. Ordering disapproved
               candidates is a useful contribution. Note the latest
               eliminated candidate has not for sure the highest weight.

               This seems to be the nearest I can come to satisfy SPC. I
do not
               think
               it is full proof mathematically, but the philosophy goes
in the right
               direction
               at least. Of course Z receive support when removed only
for ballots
               having no letter before Z. For example a universal Blank
ballot:
               Z > A ? B ? C ? D ? E would count as 1 vote for Z (a
blank vote).
               This is useful for my multiple-winner election method.

               8) Ties can be solved by solving each scenario and
               averaging the final results.

               9)I ask for help in trying some examples.
               Demorep1 last example seems a good start:
               26: A > E > Z >...
               25: B > E > Z >...
               24: C > E > Z >...
               23: D > E > Z >...
               1: E > Z >...

               Steph, engineer.

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