[EM] IIA Theory

[Craig Carey]

At 01:00 07.10.99 , Markus Schulze wrote:
>I prefer the following wordings:
>
[A]
>Deterministic Independence from Irrelevant Alternatives:
>
>   Suppose, that candidate A would have not been elected
>   if candidate B hadn't run. Then -if candidate B does run-
>   candidate A must not be elected.
>
[B]
>Stochastic Independence from Irrelevant Alternatives:
>
>   Suppose, that candidate A would have been elected with
>   the probability p if candidate B hadn't run. Then
>   -if candidate B does run- the probability that candidate
>   A is elected must not be larger than p.
>
>Markus Schulze

I show that my IFPP passes wording [A], at least for 1 and
 2 winner 3 candidate elections. Wording [B] probably ought
 be rejected.

===============================================================

Regarding statement [A].

>Deterministic Independence from Irrelevant Alternatives:
>
>   Suppose, that candidate A would have not been elected
>   if candidate B hadn't run. Then -if candidate B does run-
>   candidate A must not be elected.

Reword without changing it, and swap A with B:

  ====================================================
   Suppose, that candidate B loses when candidate A
   does not run. Then when candidate A does run,
   candidate B continues to lose.
  ====================================================

(I have doubts about the rule surviving, as preferences
 for A may go before preferences for B, or before
 preferences that give support to opponents, and the
 matter becomes beyond understanding in general if/since
 votes lose their power when preferences for losers
 are inserted before other preferences.)

V =
   A.  Zaa |
   AB  Zab | a
   AC  Zac |
   B.  |
   BA  | b
   BC  |
   C.  |
   CA  | c
   CB  |

U =
   B b+Zb
   C c+Zc

bU = (B wins) = (c<b)
cU = (C wins) = (b<c)

X = (For All a,b,c, a+b+c=1).[(b loses U) => (b loses V)]

Note that X is correct for both the 1 winner and 2 winner cases.

X = (For All a,b,c, a+b+c=1)[-bU => -bV]
X = (For All a,b,c, a+b+c=1)[bU or -bV]
-X = (There Exists a,b,c, a+b+c=1).-[bU or -bV]
-X = (There Exists a,b,c, a+b+c=1) . -bU . bV
-X = (There Exists a,b,c, a+b+c=1) . Q


CASE: IFPP, 3 candidates, 1 winner:

aV = (b+c<2a)[(b+Zcb<a+Zca) or (2b<a+c)][(c+Zbc<a+Zba) or (2c<a+b)]
bV = (c+a<2b)[(c+Zac<b+Zab) or (2c<b+a)][(a+Zca<b+Zcb) or (2a<b+c)]
cV = (a+b<2c)[(a+Zba<c+Zbc) or (2a<c+b)][(b+Zab<c+Zac) or (2b<c+a)]

Write  Tcb = (c+Zac<b+Zab), Tab = (a+Zca<b+Zcb).
bV = (c+a<2b).[Tcb or (2c<b+a)].[Tab or (2a<b+c)]

Q = -bU . bV
Q = (b<c) . (c+a<2b).[Tcb or (2c<b+a)].[Tab or (2a<b+c)]

Note: (3b<3c)(c+a<2b) => (b+a<2c),  (x or y)=(x.-y or y)
Note: (3b<3c)(2c+2a<4b) => (2a<b+c)

Q = (b<c).(c+a<2b).((a+b<2c)(2a<b+c)) .
                     [Tcb.(b+a<2c) or (2c<b+a)].[Tab or (2a<b+c)],
Q = (b<c).(c+a<2b).(2c<b+a)

Note: (c+a<2b).(2c<b+a) => (3c<3b)

Q = False
X = True
Therefore 1 winner 3 canddiate IFPP satisifes the rule.
The rule apparently can be imposed sucessfully to
 one winner 3 candidate methods.

--------------------------------------------------------------
CASE: IFPP, 3 candidates, 2 winners:

Q = -bU . bV, (same as before)
Use IFPP duality principle/property:

-bV = (c+a>2b)[(c+Zac>b+Zab) or (2c>b+a)][(a+Zca>b+Zcb) or (2a>b+c)]
bU = True

Q = -bU . bV = False . bV = False.

So for both 1 winnners and 2, IFPP satisfies the rule.

My guess is that even IFPP will start to be failed by
 the Schulze-IIA rule.


===============================================================

Regarding [B], the use of the word "chance" is a 'probable'
 mistake isn't it?. Here is the wording here:

Q: What method has semi-random results and requires a strict rule?.

Q: Mr Schulze!: is that rule "[B]" intended to be applied to
 elections that do not use preferential voting (partly random
 FPTP, or games or athletes contests?).

[B]
>Stochastic Independence from Irrelevant Alternatives:
>
>   Suppose, that candidate A would have been elected with
>   the probability p if candidate B hadn't run. Then
>   -if candidate B does run- the probability that candidate
>   A is elected must not be larger than p.
>
>Markus Schulze

This is a rule that is imprecise due to the use of the word
 probaility.

When different types of preferential voting papers are made
 to correspond to points of a simplex, and a count of papers
 is a point inside (after normalising to get the total count
 down to 1.0), then the word "chance" does not correspond to
 the word "hypervolume", but instead to "the integral over
 the volume of the probability of the election's voting paper
 ratios being that which is represented by the point in the
 election simplex.

Here is some suggestions on fixing the problem created by the
 referring to an unknown probability distribution function:

(a) One or more statisticians uses the most modern data fitting
 techniques (or any technique which respectability), and
 supplies a good probability function that subequently be
 processed with other pure logic theory axioms.

 A problem is that there can be a lot of corners in the simplex
 representing votes for an electorate. If there are 6 candidates,
 there will be 756 corners, so a lot of elections. (The progression
 follows in this way: 0,2,9,40,125,756,... M(n)=n.(M(n-1)+1)).

(b) Eliminate all ideas of probability and regard a candidate
 as either winning or losing, once the papers have been counted.

(c) A possibility is to use no data fitting and use bribes to
 modify the voting behaviour of voters so that they did
 collectively simulated a uniform random distribution (with
 electoral reform groups explaining the rationale).
 Finding out how minimise bribe payments without later having the
 whole batch of elections rejected seems to be a problem that
 might require further analysis. Then a class of formulae could be
 certified on an country by country basis.

Note:
 This IIA rule specifies tangent slopes to the win/lose regions.
 It makes specifications about what must not happen to the outcome
 when  papers are altered.
 So the important function is the derivative of the probaility
 function, which of course would equire more elections in an
 attempt to get the standard errors (standard deviations) down
 low enough to the satisfaction of the mathematicians whose theories
 were on the verge of being rejected.

Hint to theoreticians Schulze and Catchpole:
 Have you been reading too many books or something: or this
 is motivated out of consideration the semi-random STV?
 (the Irish have a slightly-random STV?)

I have to rush this message out, and no time to fix up the
 errors in it.

Craig Carey.