[EM] XA (Andy Jennings)

Wed Nov 2 09:15:04 PDT 2016

Of course, one could go in the opposite direction, and make XA more mean-like than median-like. 
The overall societal rating for a candidate under XA (with scores out of 100) is the score that is the same as 100 minus its percentile. So if the 60th percentile score for a candidate is 40, then 40 becomes its overall societal rating.
But where the median is the 50th percentile, we could generalise the mean and say that the mean in the 50th "permeantile". So under this new system, the overall societal rating for a candidate (with scores out of 100) is the score that is the same as 100 minus its "permeantile". So if the 60th permeantile score for a candidate is 40, then 40 becomes its overall societal rating.
But how do you calculate a permeantile? Well, a formula that seems to work is that the pth permeantile is the point that becomes the mean if you multiply the weight of everything below it by a factor of (100-p)^2 and everything above it by p^2. So if you have a uniform data set from 0 to 1 and you want to find the 25th permeantile, you'd find that this formula made 0.25 the 25th permeantile. Which is what it should be. You'd think that this should be a standard statistical tool, but I've never found a mention of it - maybe partly because it's difficult to know what to search for.

      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 ZF2: 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.c om>
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
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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