Exact spatial model probabilities?
On Wed, Jan 26, 2022 at 1:30 PM Forest Simmons forest.simmons21@gmail.com
wrote:
I should probably give more of an explanation for the choice of the atomic
electo-potential, i.e. the scalar potential function V(r) for one voter
located at R:
V(r)=||r-R||
The force=-grad V(r) represents the influence of the voter pushing or
pulling in one direction or another. The true winner should be at a point
where the resultant force of the entire electorate is balanced at as close
to zero as possible.
So what is the force of one voter under this choice of potential V(r)?
It is -grad V(r) anchored at the voter position R.
I don't think that it really makes sense for a voter's force to follow a
1/r^2 law. That would mean that two candidates that are both extremely
close to you can have vastly different forces while two highly dissimilar
candidates that are far from you get nearly the same force. That doesn't
seem like a realistic model. If I am a Bernie Sanders voter I would be
mostly indifferent if I got AOC instead, but I would have a strong view
between a center-right and a far-right party.
It might be more accurate to see the force from a voter as linear with
distance --- every voter holds an elastic band that follows Hooke's law.
This can still be expressed in the language of potentials if that's useful.
Cheers,
Dr. Daniel Carrera
Postdoctoral Research Associate
Iowa State University
On Wed, Jan 26, 2022 at 10:19 PM Forest Simmons forest.simmons21@gmail.com
wrote:
Practical suggestions ... convert the second order vector differential
equations into a system of two first order vector ODE's ...
dr/dt=v,
dv/dt=-grad V(r)
Solve numerically using RK4 or even simple Euler's method ... after all
we're just using the path to sample the space. Try several initial
conditions ... nothing too far outside the support of the voter
distribution.
I'm pretty sure that that would just create an orbit. Since V(r) is not a
point mass we might expect a complicated non-periodic or quasi-periodic
orbit, but an orbit nonetheless. If we knew that V(r) had a simple bowl
shape we could just do a hill-climbing algorithm, but in general V(r) could
have all sorts of hills, valleys, and saddle points.
One simple example: If you have a cluster of left-wing voters and a cluster
of right-wing voters, and they are well segregated, you could end up just
orbiting one of the two clusters. That could say something interesting
about why it might be hard for a politician to be a centrist (politician
can't climb out of her respective local potential to move to the center)
but that's not what you are looking for.
Cheers,
Dr. Daniel Carrera
Postdoctoral Research Associate
Iowa State University
On 27.01.2022 08:12, Daniel Carrera wrote:
On Wed, Jan 26, 2022 at 6:53 AM Kristofer Munsterhjelm
<km_elmet@t-online.de mailto:km_elmet@t-online.de> wrote:
So when figuring out the proportion who'd vote A>B>C in the continuous
(V->infty) case, you could first create a Voronoi map for all the
candidates. The volume of this region gives you how many voters would
vote A first. Then you'd take the region for A and calculate the Voronoi
map within it for every candidate but A excluded. The subregion that's
now closest to B gives you the number of voters who vote A>B>...;
and so on.
You've convinced me that there's a lot of value in doing the exact
geometric solution. I would be interested to see simulations with very
large numbers of voters. Real elections can have 10^7 voters. I haven't
figured out how to compute the volume of a Voronoi cell, but I did have
a thought: What if we simplify the problem by reducing the number of
candidates and issues?
For real elections, we don't really care about an arbitrary number of
characters or an arbitrary number of issues. Political views are often
described in just a two-dimensional space, with one "economic" axis
(left <-> right) and one "social" axis (libertarian <-> authoritarian).
That alone is enough to describe most of politics. The number of parties
can be larger, but honestly, the countries that have >3 prominent
parties are few and they probably use PR anyway. If all we do is find
exact solutions for 3 parties (e.g. two major parties and a challenger)
or 4 parties (e.g. two major parties + Green + Libertarian) and a
2-dimensional issue space, that would be a very relevant case that would
tell us a lot about the real world elections we care about.
Finding the area of a Voronoi cell in a 2D space with just 3-4 cells
doesn't sound difficult. I haven't done the math, but it feels like high
school math problem.
It isn't very difficult. Fortune's algorithm can get the cells in O(c
log c) time, and then you can find the area of each (given its vertices)
by sorting the vertices in a particular order, in O(v log v) time. IIRC,
the number of vertices of a 2d Voronoi cell is bounded, so that is in
essence constant time.
A side note: I seem to recall that Patterns of Democracy (by Lijphart)
says that as a reasonable approximation, the number of dimensions in a
country's political realm is the number of effective parties, plus one.
(And anecdotally there definitely seems to be more than one dimension to
politics here.)
So if a good single-winner method can support multiparty democracy by
itself, then higher dimensions may be relevant eventually. Still, any
two-party system that's evolving to a higher party system will pass
through lower dimensionality spaces before getting to higher ones as
parties find out if it's safe to split and specialize.
(There's some evidence that single-member districts can support more
than two parties, e.g. the use of top two runoff in France.)
However, that doesn't solve the integral. My idea was to get the (as
close to exact as possible) probability that some random chosen voter
will vote A>B>C. Say that this is p(A>B>C).
Since the voters are independent, this then gives a probability that
some random election of V voters will be of a particular type; e.g. you
can figure out the probability that a three-candidate election will be
e_s = (A>B>C, A>B>C, B>C>A). Let the probability that we will encounter
election e be p(e). Then the probability of the example election is just
p(e_s) = p(A>B>C) * p(A>B>C) * p(B>C>A).
I'm a bit confused by the notation. If p(A>B>C) is the probability that
a random voter picks A>B>C then what is p(A>B>C) * p(A>B>C) * p(B>C>A)?
I also don't get what you mean by "e_s = (A>B>C, A>B>C, B>C>A)" being an
election.
The former is just a statement of statistical independence. Suppose that
we roll three ordinary dice, call them x, y, and z. Then the probability
of the result being all ones is p(x=1) * p(y=1) * p(z=1) due to
independence - the different outcomes don't constrain each other.
So in a similar way, if we let p(A>B>C) be the probability that a random
variable representing a given voter turns out to be A>B>C, then due to
statistical independence, the probability that the first voter votes
A>B>C, the second votes A>B>C, and the third votes B>C>A, is simply
p(A>B>C) * p(A>B>C) * p(B>C>A). And if the election is defined by this
particular voting pattern, then you have the probability that a randomly
chosen election will be this one.
e_s = (A>B>C, A>B>C, B>C>A) just means that the election e_s has these
ballots, in this order:
1: A>B>C
1: A>B>C
1: B>C>A.
(Again on a side note: since it doesn't matter in what order the voters
vote, you could save some computations by using a multinomial - e.g. if
you find out that method M is susceptible to manipulation with election
(x, y, z), you automatically know it's susceptible to manipulation with
election (x, z, y) too. But it would make the code more complex.)
-km
On Thu, Jan 27, 2022 at 4:20 AM Kristofer Munsterhjelm km_elmet@t-online.de
wrote:
I'm a bit confused by the notation. If p(A>B>C) is the probability that
a random voter picks A>B>C then what is p(A>B>C) * p(A>B>C) * p(B>C>A)?
I also don't get what you mean by "e_s = (A>B>C, A>B>C, B>C>A)" being an
election.
The former is just a statement of statistical independence. Suppose that
we roll three ordinary dice, call them x, y, and z. Then the probability
of the result being all ones is p(x=1) * p(y=1) * p(z=1) due to
independence - the different outcomes don't constrain each other.
Ok. The part that wasn't obvious to me was that "e_s = (A>B>C, A>B>C,
B>C>A)" was the election with exactly those ballots in that order. You
might want to instead define an election so that it's only about the
combination of ballots and not the permutation; so if you shuffle the
ballots it's the same election.
e_s = {2: A>B>C, 1:B>C>A}
=> p(e_s) = Choose(3,2)*p(A>B>C)*P(B>C>A)
(Again on a side note: since it doesn't matter in what order the voters
vote, you could save some computations by using a multinomial - e.g. if
you find out that method M is susceptible to manipulation with election
(x, y, z), you automatically know it's susceptible to manipulation with
election (x, z, y) too. But it would make the code more complex.)
One way to avoid the combinatorial part is to define the election so that
the ballots are sorted in some canonical order. For example, sort the
possible ballots alphabetically:
b1 = ABC
b2 = ACB
b3 = BAC
b4 = BCA
b5 = CAB
b6 = CBA
Let nk be the number of voters that cast the ballot bk. Then:
p(e) = Prod_k p(bk)^nk
Cheers,
Dr. Daniel Carrera
Postdoctoral Research Associate
Iowa State University
On Thu, Jan 27, 2022 at 4:20 AM Kristofer Munsterhjelm km_elmet@t-online.de
wrote:
Finding the area of a Voronoi cell in a 2D space with just 3-4 cells
doesn't sound difficult. I haven't done the math, but it feels like high
school math problem.
It isn't very difficult. Fortune's algorithm can get the cells in O(c
log c) time, and then you can find the area of each (given its vertices)
by sorting the vertices in a particular order, in O(v log v) time. IIRC,
the number of vertices of a 2d Voronoi cell is bounded, so that is in
essence constant time.
A side note: I seem to recall that Patterns of Democracy (by Lijphart)
says that as a reasonable approximation, the number of dimensions in a
country's political realm is the number of effective parties, plus one.
(And anecdotally there definitely seems to be more than one dimension to
politics here.)
You know, if we embrace the idea that the number of dimensions scales with
the number of parties and make it part of the model, that simplifies the
possible shapes of the Voronoi cells:
a) 3 candidates in a 2D space, as long as they don't fall on a line, must
have Voronoi cells that all meet at a point.
b) 3 candidates in a 3D space, as long as they don't fall on a line, must
have Voronoi cells that all meet at a line.
c) 3 candidates in a 4D space, as long as they don't fall on a line, must
have Voronoi cells that all meet on a plane.
More generally, my intuition is that N candidates in a space with at least
N-1 dimensions cannot "surround" each other and that limits the possible
set of shapes their Voronoi cells can have. Furthermore, I suspect that if
the space has exactly N-1 dimensions, the Voronoi cells probably all meet
at a point and the cells look similar to triangles/pyramids/etc. They're
not exactly triangles/pyramids because the one side of the cell ends at the
boundary of the space and we haven't defined what shape that's going to
have. Nonetheless, my point is that we don't need to compute the volume of
a fully generic Voronoi cell in higher dimension; just a relatively simple
subtype.
Cheers,
Dr. Daniel Carrera
Postdoctoral Research Associate
Iowa State University
El jue., 27 de ene. de 2022 1:44 a. m., Daniel Carrera dcarrera@gmail.com
escribió:
On Wed, Jan 26, 2022 at 1:30 PM Forest Simmons forest.simmons21@gmail.com
wrote:
I should probably give more of an explanation for the choice of the
atomic electo-potential, i.e. the scalar potential function V(r) for one
voter located at R:
V(r)=||r-R||
You must have read this as V(r)=1/||r-R||, which would yield the inverse
square law of force,
But the V(r) actually written gives a force of -grad||r-R||=(R-r)/||R-r|| a
unit force pointed towards the voter position. Each voter exerts the same
amount influence, but different direction depending on the position of the
voter.
The force=-grad V(r) represents the influence of the voter pushing or
pulling in one direction or another. The true winner should be at a point
where the resultant force of the entire electorate is balanced at as close
to zero as possible.
So what is the force of one voter under this choice of potential V(r)?
It is -grad V(r) anchored at the voter position R.
I don't think that it really makes sense for a voter's force to follow a
1/r^2 law. That would mean that two candidates that are both extremely
close to you can have vastly different forces while two highly dissimilar
candidates that are far from you get nearly the same force. That doesn't
seem like a realistic model. If I am a Bernie Sanders voter I would be
mostly indifferent if I got AOC instead, but I would have a strong view
between a center-right and a far-right party.
It might be more accurate to see the force from a voter as linear with
distance --- every voter holds an elastic band that follows Hooke's law.
This can still be expressed in the language of potentials if that's useful.
Hook's law gives more influence to the distant voter.
Think about voters distributed along a straight line ... Hook's law would
choose the mean voter candidate instead of the median. The extreme voters
would have undue influence.
Cheers,
Dr. Daniel Carrera
Postdoctoral Research Associate
Iowa State University
On Thu, Jan 27, 2022 at 3:55 PM Forest Simmons forest.simmons21@gmail.com
wrote:
El jue., 27 de ene. de 2022 1:44 a. m., Daniel Carrera dcarrera@gmail.com
escribió:
On Wed, Jan 26, 2022 at 1:30 PM Forest Simmons <
forest.simmons21@gmail.com> wrote:
I should probably give more of an explanation for the choice of the
atomic electo-potential, i.e. the scalar potential function V(r) for one
voter located at R:
V(r)=||r-R||
You must have read this as V(r)=1/||r-R||, which would yield the inverse
square law of force,
Ugh... indeed I did. My bad.
But the V(r) actually written gives a force of -grad||r-R||=(R-r)/||R-r||
a unit force pointed towards the voter position. Each voter exerts the same
amount influence, but different direction depending on the position of the
voter.
...
Hook's law gives more influence to the distant voter.
Think about voters distributed along a straight line ... Hook's law would
choose the mean voter candidate instead of the median. The extreme voters
would have undue influence.
Yeah, I'm convinced. Your constant force idea is much better.
Cheers,
Dr. Daniel Carrera
Postdoctoral Research Associate
Iowa State University