[EM] XA (Andy Jennings)

[Forest Simmons]

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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