[EM] XA (Andy Jennings)

[Jameson Quinn]

One issue I have with XA is that it makes numerical votes inherently
meaningful; it is entirely possible to change the election outcome by
adding or subtracting a constant from all ballots. I'm wondering if this is
fixable.

What if you transformed all ratings into percentiles first? Let's call this
system empirical chiastic approval, EXA. So if you had something like
(candidates W-Z and grades A-F)

3: WA XB YC ZF
2: WF XC YD ZA

... that would be an empirical grade distribution of 5As 3Bs 5Cs 2Ds 5Fs,
or in percentiles A75 B60 C35 D25 F0. So the EXA score is W60, X60, Y35,
Z40. (This example gave a tie because I deliberately made it so there would
be round numbers. In general, a tie would be highly unlikely. Clearly, in
order to minimize strategy, this tie should be broken in favor of W, the
candidate who had the most excess goodwill on their weakest
positive-influence ballot.)

I think there are probably strategies involving manipulating the empirical
distribution (non-semi-honestly), but I doubt that anyone would have the
fine-grained info necessary to pull such a strategy off.

If you condition on a given percentile distribution, it satisfies the same
kind of criteria that XA does, including "individual non-strategy".

Interestingly, though not very usefully, this is a voting system which
would work just fine with allowing a negative infinity to positive infinity
ballot scale. (it would work even better than Bucklin in that sense.)

2016-11-01 16:54 GMT-04:00 Forest Simmons <fsimmons at pcc.edu>:

> It is more likely that two candidates will have the same median score (an
> MJ tie situation) than having the same XA score.
>
> Part of the reason is that the XA scores depend continuously on the
> distribution of ratings, while the median can be a discontinuous function
> of the distribution.
>
> From another point of view, the graph of y = x is more likely to be
> perpendicular to the graph of the distribution function F(x) = Probability
> that on a random ballot candidate X will have a rating of at least x.  An
> orthogonal intersection minimizes error due to random perturbations.
>
> The graph of F stair steps down from some point on the y axis between (0,
> 0) and (0, 1) to some point on the vertical segment connecting (1, 0) to
> (1, 1).  If the distribution is uniform, then the graph of F is the
> diagonal line segment connecting (0, 1) to (1, 0), perpendicular to the
> line y = x.
>
> The median point (used in MJ and other Bucklin variants) is the
> intersection of the graph of F with the vertical line given by x = 1/2,
> cutting the square with diagonal corners at (0, 0) and (1, 1) in half.
>
> The midrange Approval value is the intersection of the graph of F with the
> horizontal line y = 1/2.
>
> The XA value is the intersection of the graph of F and line y = x, which
> bisects the right angle formed by  x = 1/2 and y = 1/2 at the intersection
> (1/2, 1/2).
>
> So XA can be thought of as a method half way between midrange Approval and
> score based Bucklin.
>
> More later ...
>
> Forest
>
>
>
>
>> From: Andy Jennings <elections at jenningsstory.com>
>> To: Michael Ossipoff <email9648742 at gmail.com>
>> Cc: "election-methods at electorama.com"
>>         <election-methods at electorama.com>
>> Subject: Re: [EM] XA
>>
>> On Mon, Oct 31, 2016 at 7:13 PM, Michael Ossipoff <email9648742 at gmail.com
>> >
>> wrote:
>>
>> > What makes XA do that more effectively than MJ? What's the main
>> advantage
>> > that distinguishes how XA does that from how MJ does it, or the results,
>> > from the voters' strategic standpoint?
>>
>>
>> Michael,
>>
>> As Rob said, the median is not terribly robust if the distribution of
>> votes
>> is two-peaked:
>> http://www.rangevoting.org/MedianVrange.html#twopeak
>> And I'm afraid many of our contentious political elections are two-peaked,
>> at least in the current environment.
>>
>> With MJ, I like the fact that if the medians for all candidates will fall
>> between B and D, then I can use the range outside that for honest
>> expression.  Yet in the back of my head, I know that if everyone tries to
>> "use the range outside that for honest expression", then the medians won't
>> be in that range anymore and it seems like a slippery slope to everyone
>> using only the two extreme grades.
>>
>> XA solves this problem by making the more extreme grades more difficult to
>> achieve.  As Rob said, in the case where everyone grades at the extremes,
>> the XA will match the mean.
>>
>> On the other hand, I admit that:
>> 1) with the median, 50% would have to give the top grade for a candidate
>> to
>> receive that grade.  And 50% would have to give the bottom grade for a
>> candidate to receive that grade.  I consider both of these very unlikely.
>> 2) MJ is not just "the median", it has a tie-breaking scheme which
>> mitigates this somewhat.
>>
>> ~ Andy
>>
>>
>
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>
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