# [EM] What kind of monotonicity whould we exspect from a PR method?

Juho Laatu juho4880 at yahoo.co.uk
Tue Aug 9 05:40:58 PDT 2011

```Just another example vote set FYI.

2 AB
2 AC
1 B
1 C
4 D

Natural winners are maybe A and D.

100 AB
100 AC
1 B
1 C
4 D

Natural winners are maybe B and C. Is it a problem that additional support to A (and B and C) meant that A was not elected?

(A was top ranked by all the new voters. B and C were top ranked by 50% of the new voters.)

Juho

On 8.8.2011, at 22.40, fsimmons at pcc.edu wrote:

> It seems that if a PR method chose slate {X, Y} for a two winner election, and only X or Y received
> increased support in the rankings or ratings, then {X, Y} should still be chosen by the method.
>
> But consider the following approval profile (for a two winner election):
>
> 3 X
> 1 XY
> 2 Y
> 2 Z
>
> It seems pretty clear that the slate {X, Y} should be elected, and that is the PAV decision.
>
> Now suppose that X gets additional approval on some ballots but the Y and Z approvals stay the same:
>
> 2 X
> 3 XY
> 2 Z
>
> Now PAV elects {X, Z}, and this seems like the right choice, because this slate completely covers the
> electorate, unlike any other pair.  Candidate Y has more approvals than Z, but everybody that approves
> Y also approves X, so given that X is part of the slate, Y would only contribute half a satisfaction point
> per ballot, while Z adds a full point per ballot.  Since 2>1.5, Z wins over Y for the remaining position on
> the slate.
>
> This violates the strong monotonicity ideal of the first paragraph, but does not violate a weaker version
> that says if only one candidate X of the winning slate gets additional support on some ballots (and all
> other candidates have the same or less support as before on all ballots) then that one candidate X
> should be a part of the new slate.
>
> Now let's look at this example from the point of view of the Ultimate Lottery:
>
> In the before scenario, the Ultimate Lottery probabilities x, y, and z for the respective candidates X, Y,
> and Z are obtained by maximizing the product
>
>   x^3*(x+y)*y^2*z^2  subject to the constraint  x+y+z=1
>
> The solution is exactly (x,y,z)=(45%, 30%, 25%) .
>
> After the increase in support for x the Ultimate Lottery probabilities are obtained by maximizing the
> product
>
>   x^3*(x+y)^3*z^3  subject to the same constraint  x+y+z=1.
>
> The solution is precisely  (x, y, z) = (75%, 0, 25%) .
>
> Note that (in keeping with the strong ideal expressed at the beginning of this message) the only
> candidate to increase in probability was X, the one that received increased support.  It did so at the
> expense of Y whose probability decreased to zero. So Z passed up Y without any change in its
> probability.  That's basically why Z took Y's place on the slate without any increased support on the
> ballots.
>
> So this helps us understand (in the PR election) why the weaker member of the two winner slate
> changes from Y to Z, and why we cannot expect the strong monotone property for a finite winner PR
> election; the discretization in going from the ideal proportion of the Ultimate Lottery to a finite slate
> allows only a crude approximation to the ideal proportion.
>
> In other words, it is just one of the classical apportionment problems in disguise.
>
> How do other PR methods stack up with regard to monotonicity?
>
> Since IRV is non-montone, automatically STV fails even the weak montonicity sartisfied by PAV.
>
> How about the other common methods?
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