From: jmorriso@ee.ubc.ca (John Paul Morrison)
Subject: Re: Voting Systems (was Re: Proportional representation)
Organization: University of BC, Electrical Engineering
Date: Mon, 9 Nov 1992 02:28:17 GMT

In article <1dk6adINN5jb@iskut.ucs.ubc.ca> ramsay@unixg.ubc.ca (Keith Ramsay) writes:
>In article <1992Nov8.074732.7435@ee.ubc.ca> jmorriso@ee.ubc.ca 
>(John Paul Morrison) writes:
>[Regarding the Arrow impossibility theorem:]
>>I would like to see how this following system is affected. I believe
>>that the theorem you mention can't apply in this case.
>[...]
>>This is similar to the way shareholders elect a board (but not really,
>>so don't bother picking on this point).
>>
>>1. A voter can only vote for one person, no prioritizing
>>2. The candidate is elected if he gets at least one vote.
>>3. The candidate is actually a delegate, so in parliament, he or she
>>   represents as many votes as they received.
>
>There's no way to express the preference, for instance, of those who
>might mainly want some kind of coherence and consensus of purpose in
>the parliament, for efficiency, and aren't as concerned about getting
>their own first choice of representative into it.

You don't read very closely do you? Any candidate who gets a vote, takes
that vote with him to the parliament. The voter is saying: here I trust
you, I transfer my vote to you. I am delegating my responsibility to you.
The more votes (ie the more people who trust that delegate) the delegate
gets, the more influence he has. (Since a delegate who represents 2,000
people has more voting power than a delegate who has onlt 500 people
supporting him).

What you don't understand, is that in this hypothetical system, the votes
of each delegate are not equal
>
>It also fails another condition of the theorem, in that it is possible
>for a person to cause an option A to be selected instead of an option
>B, by deciding to try to get C instead of D. Suppose there are a
>number of us who'd like to vote for Smith, but for whatever reason
>we'd rather she not have too much power. If, however, she already has
>sufficient votes, then we'd rather vote for Jones.

This isn't the case. Your logic system requires that out of the set 
{A,B,C,D}, exactly one item is selected. In the system I am suggesting
(to demonstrate that this line of reasoning does not apply) you actually
have a "fuzzy" set where ALL of the set {A,B,C,D} is chosen. But the
weights are different:  they might be 700A, 200B, 50C, 50D (sums to
1000. or 0.7, 0.2, 0.05, .05 of the local vote). 
Thus each delegate represents exactly the number of voters who
voted for him. Put A,B,C,D in a larger parliament with AA, BB, CC, DD etc
and A still represents 700 votes. IF AA..DD get 500, 300, 100, and 100,
(sums to 1000) the total parliament would represent 2000 votes. But
A still represents 35% of the popular vote.
>
>Let A, B, C, and D be possible compositions of the legislature, where
>A differs from B only by Smith getting some of the votes Jones got in
>case B, and likewise for C and D, where A and B are alternatives in
>which Smith gets enormous power, and C and D are alternatives in which
>Smith gets either a modest amount or none at all. The condition
>described by the theorem, requiring that changing preferences between
>A and B will not affect whether C or D is chosen, is violated by this
>voting system. By trying to ensure Smith gets a modicum of influence,
>we could end up giving her more than we had wanted, or conversely.

It is completely independent! You haven't proven anything.

If there are 100 voters:

suppose delegate A gets 50 votes, B gets 30, c gets 15 and d gets 5
but then some voters take away their vote from And give it to B,
giving    A = 40, B = 40, C = 15 and D = 5
C and D are NOT affected. In both cases, all 4 delegats go to parliament,
however A's vote will only be worth 40 instead of 50. D still has 5 votes
to cast in the parliament.


>
>The conditions about "independence of preferences" are in some sense
>the stiffest constraints placed by the theorem, and are often
>violated. It is often possible for it to be the case that if you
>prefer A to B to C, you are well advised to vote as if you prefer B
>most of all, so as to make sure that B rather than C occurs.  This
>"sophistical voting" is what the Arrow conditions are formulated to
>preclude, and the upshot is that there always is a price paid for
>avoiding it.


the Arrow theorem assumes that only one candidate is elected. The example
posted does not follow Arrows assumptions, so it doesn't apply.

IF you prefer A to B to C, voting for A will always help A and never C,
because in this system, a delegate's power depends only on the votes
he receives. IF A has 100 votes, B has 100 and C has 100 votes, and
you vote for your first choice A, A will  get 101 votes, when ALL of
A, B and C go to parliament, A will still have 101 votes, while C
will always have 100 votes.


This does not fall into range where Arrow's theorem applies, because it
Assumes only one of ABC is selected, but this system selects all of ABC,
and A,B, and C have different weights. Arrows's reasoning requires
that if A is selected, A has weight 1.0 and B and C get 0.0 (even
if B and C get voted for), but in this example, A would get a weight of
a, B gets b, and C gets c, where a+b+c = 1.0.



hope this cleats it up. it seems that Arrow's theorem doesn't apply if
you have a "fuzzy" (ie statistically weighted) system

>
>Keith Ramsay
>ramsay@unixg.ubc.ca


-- 
__________________________________________________________________________
 John Paul Morrison                     |  
 University of British Columbia, Canada |  
 Electrical Engineering                 |   .sig file without a cause
 jmorriso@ee.ubc.ca              VE7JPM |  
________________________________________|_________________________________