[EM] XA (Andy Jennings)

[Jameson Quinn]

I thought about the fact that percentiles are asymmetrical. Obviously the
choices are: top value, bottom value, median value, or any of the above
with renormalization. So in my exaple, the A's/F's respectively could be
100/25, 75/0, 87.5/12.5, or any of the foregoing but renormalized to 100/0.
(Note that this renormalization would actually affect the middle grades
slightly differently depending on which you started with.) Since I have no
intuition for why any of these would be better than any other, I just went
with the standard convention for percentiles, which is bottom without
renormalization.

2016-11-02 11:57 GMT-04:00 Toby Pereira <tdp201b at yahoo.co.uk>:

> Your percentiles seem to be inherently asymmetrical with A at 75 and F at
> 0. Obviously there are several different methods that people use - your one
> seems to start at 0 and go up to 100 - (1/N) for N values, and then if
> there are ties, always award the lowest percentile. So in your case with 20
> values, they go from 0-95 but because there are five As, they would occupy
> 95, 90, 85, 80 and 75, so you award them 75.
>
> Is that something you consciously decided?
>
>
> ------------------------------
> *From:* Jameson Quinn <jameson.quinn at gmail.com>
> *To:* Forest Simmons <fsimmons at pcc.edu>
> *Cc:* EM <election-methods at lists.electorama.com>
> *Sent:* Tuesday, 1 November 2016, 21:28
> *Subject:* Re: [EM] XA (Andy Jennings)
>
> 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.c om <election-methods at electorama.com>"
>         <election-methods at electorama.c om
> <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/Med ianVrange.html#twopeak
> <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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