Re: [EM] XA (Andy Jennings)
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@jenningsstory.com
To: Michael Ossipoff email9648742@gmail.com
Cc: "election-methods@electorama.com"
election-methods@electorama.com
Subject: Re: [EM] XA
On Mon, Oct 31, 2016 at 7:13 PM, Michael Ossipoff email9648742@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:
- 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. - MJ is not just "the median", it has a tie-breaking scheme which
mitigates this somewhat.
~ Andy
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@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@jenningsstory.com
To: Michael Ossipoff email9648742@gmail.com
Cc: "election-methods@electorama.com"
election-methods@electorama.com
Subject: Re: [EM] XA
On Mon, Oct 31, 2016 at 7:13 PM, Michael Ossipoff <email9648742@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:
- 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. - MJ is not just "the median", it has a tie-breaking scheme which
mitigates this somewhat.
~ Andy
Election-Methods mailing list - see http://electorama.com/em for list info
I guess I was assuming normalized ballots all along, in other words you
would have at least one candidate that you approved unconditionally and at
least one that you would not approve no matter what the other voters did,
at least in an election setting.
On the other hand if you are just contributing to a five star rating guide
...
On Tue, Nov 1, 2016 at 2:28 PM, Jameson Quinn jameson.quinn@gmail.com
wrote:
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@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@jenningsstory.com
To: Michael Ossipoff email9648742@gmail.com
Cc: "election-methods@electorama.com"
election-methods@electorama.com
Subject: Re: [EM] XA
On Mon, Oct 31, 2016 at 7:13 PM, Michael Ossipoff <
email9648742@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:
- 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. - MJ is not just "the median", it has a tie-breaking scheme which
mitigates this somewhat.
~ Andy
Election-Methods mailing list - see http://electorama.com/em for list
info
I don't think you understand my point. Say candidate A has 60 votes at 0, 5
votes at 35, and 35 votes at 100; while candidate B has 63 at 0, 5 at 40,
and 32 at 100. B wins with 37. Now add 5 to the middle scores of both;
nothing fundamental changes about how any voter views either candidate, but
now A wins with 40.
My system of EXA (or percentile chiastic approval, PXA) would build ratings
scales automatically, so that an electorate that liked numbers around 80
could get the same results as one that liked numbers around 20.
2016-11-01 20:32 GMT-04:00 Forest Simmons fsimmons@pcc.edu:
I guess I was assuming normalized ballots all along, in other words you
would have at least one candidate that you approved unconditionally and at
least one that you would not approve no matter what the other voters did,
at least in an election setting.
On the other hand if you are just contributing to a five star rating guide
...
On Tue, Nov 1, 2016 at 2:28 PM, Jameson Quinn jameson.quinn@gmail.com
wrote:
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@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@jenningsstory.com
To: Michael Ossipoff email9648742@gmail.com
Cc: "election-methods@electorama.com"
election-methods@electorama.com
Subject: Re: [EM] XA
On Mon, Oct 31, 2016 at 7:13 PM, Michael Ossipoff <
email9648742@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:
- 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. - MJ is not just "the median", it has a tie-breaking scheme which
mitigates this somewhat.
~ Andy
Election-Methods mailing list - see http://electorama.com/em for list
info
I see. Very good!
On Tue, Nov 1, 2016 at 5:54 PM, Jameson Quinn jameson.quinn@gmail.com
wrote:
I don't think you understand my point. Say candidate A has 60 votes at 0,
5 votes at 35, and 35 votes at 100; while candidate B has 63 at 0, 5 at 40,
and 32 at 100. B wins with 37. Now add 5 to the middle scores of both;
nothing fundamental changes about how any voter views either candidate, but
now A wins with 40.
My system of EXA (or percentile chiastic approval, PXA) would build
ratings scales automatically, so that an electorate that liked numbers
around 80 could get the same results as one that liked numbers around 20.
2016-11-01 20:32 GMT-04:00 Forest Simmons fsimmons@pcc.edu:
I guess I was assuming normalized ballots all along, in other words you
would have at least one candidate that you approved unconditionally and at
least one that you would not approve no matter what the other voters did,
at least in an election setting.
On the other hand if you are just contributing to a five star rating
guide ...
On Tue, Nov 1, 2016 at 2:28 PM, Jameson Quinn jameson.quinn@gmail.com
wrote:
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@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@jenningsstory.com
To: Michael Ossipoff email9648742@gmail.com
Cc: "election-methods@electorama.com"
election-methods@electorama.com
Subject: Re: [EM] XA
On Mon, Oct 31, 2016 at 7:13 PM, Michael Ossipoff <
email9648742@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:
- 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. - MJ is not just "the median", it has a tie-breaking scheme which
mitigates this somewhat.
~ Andy
Election-Methods mailing list - see http://electorama.com/em for list
info
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@gmail.com>
To: Forest Simmons fsimmons@pcc.edu
Cc: EM election-methods@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@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@jenningsstory.com
To: Michael Ossipoff email9648742@gmail.com
Cc: "election-methods@electorama.c om"
<election-methods@electorama.c om>
Subject: Re: [EM] XA
On Mon, Oct 31, 2016 at 7:13 PM, Michael Ossipoff email9648742@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:
- 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. - MJ is not just "the median", it has a tie-breaking scheme which
mitigates this somewhat.
~ Andy
Election-Methods mailing list - see http://electorama.com/em for list info
Election-Methods mailing list - see http://electorama.com/em for list info
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@gmail.com>
To: Forest Simmons fsimmons@pcc.edu
Cc: EM election-methods@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@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@jenningsstory.com
To: Michael Ossipoff email9648742@gmail.com
Cc: "election-methods@electorama.c om"
<election-methods@electorama.c om>
Subject: Re: [EM] XA
On Mon, Oct 31, 2016 at 7:13 PM, Michael Ossipoff email9648742@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:
- 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. - MJ is not just "the median", it has a tie-breaking scheme which
mitigates this somewhat.
~ Andy
Election-Methods mailing list - see http://electorama.com/em for list info
Election-Methods mailing list - see http://electorama.com/em for list info
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@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@gmail.com
To: Forest Simmons fsimmons@pcc.edu
Cc: EM election-methods@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@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@jenningsstory.com
To: Michael Ossipoff email9648742@gmail.com
Cc: "election-methods@electorama.c om election-methods@electorama.com"
<election-methods@electorama.c om
election-methods@electorama.com>
Subject: Re: [EM] XA
On Mon, Oct 31, 2016 at 7:13 PM, Michael Ossipoff email9648742@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:
- 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. - MJ is not just "the median", it has a tie-breaking scheme which
mitigates this somewhat.
~ Andy
Election-Methods mailing list - see http://electorama.com/em for list info
Election-Methods mailing list - see http://electorama.com/em for list info
On Tue, Nov 1, 2016 at 1:54 PM, Forest Simmons fsimmons@pcc.edu wrote:
It is more likely that two candidates will have the same median score (an
MJ tie situation) than having the same XA score.
I know you reformulated your continuity argument, which I agree with.
But as for this point, I'm not sure that XA is less likely to have ties.
Chiastic medians will be more densely clustered around 50 than will normal
medians. (For any given distribution function, compare the intersection
with the diagonal to the intersection with the vertical midpoint.
Horizontally, the former will be between 50 and the latter.)
For a discrete grading scale, say integers from 0 to 100, this would
increase the probability for collisions. On the other hand, the diagonal
might intersect with a horizontal segment of the distribution function and
could give a chiastic median which is not in the discrete grading scale,
but corresponds to a percentage of the electorate. (If exactly 75.43% of
the electorate gave a grade of 76 or above, then the chiastic median is
75.43.) I guess this may make collisions less common. I'm not sure
exactly how these two forces would balance out. It seems like you're
definitely most likely to see collisions on the integers.
For a continuous grading scale, say all real numbers from 0 to 100, any two
grades will collide with probability 0, so the medians will almost never
collide. Whereas the chiastic medians have some probability of colliding
if they are returning discrete percentages of the electorate. (The
horizontal segments of the distribution function are at discrete heights,
so if the diagonal hits one of them, then there could be collisions.)
So perhaps it depends on the grading scale and the presumed distribution of
votes...
~ Andy
Jameson, could you please explain your EXA (or PXA) method in a little more
detail? I had some trouble following your example.
I am also uncomfortable with the emphasis on making the ratings numerical.
Frango ut patefaciam -- I break so that I may reveal
On Tue, Nov 1, 2016 at 2:28 PM, Jameson Quinn jameson.quinn@gmail.com
wrote:
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@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@jenningsstory.com
To: Michael Ossipoff email9648742@gmail.com
Cc: "election-methods@electorama.com"
election-methods@electorama.com
Subject: Re: [EM] XA
On Mon, Oct 31, 2016 at 7:13 PM, Michael Ossipoff <
email9648742@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:
- 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. - MJ is not just "the median", it has a tie-breaking scheme which
mitigates this somewhat.
~ Andy
Election-Methods mailing list - see http://electorama.com/em for list
info
Election-Methods mailing list - see http://electorama.com/em for list info