[EM] Criterion: 0-info mean consistency (0IMC)

[Jobst Heitzig]

Dear folks!

Although I don't believe that all voters are utility maximizers, I still
think that voters who *are* utility maximizers should be encouraged by
the election method to vote sincerely. Hence I started studying the
following minimal criterion:


Def. 0-INFO MEAN CONSISTENCY (0IMC).
Under zero information, voting sincerely must maximize expected utility.


More precisely and mathematically, this is meant that, given any utility
function u and no information about the other voters, there must be some
numbers a,b,c such that

   (i) voting X>Y,X=Y,X<Y if u(X)>u(Y),u(X)=u(Y),u(X)<u(Y), respectively
       (if pairwise information is asked for),

  (ii) approving of X if and only if u(X)>a
       (if approval information is asked for), and

 (iii) rating X as r(X):=b+c*u(X)
       (if ratings information is asked for)

results in an expected utility at least as large as any other way of voting.


It seems to be natural to request that at least when there's no
information one should not have incentive to vote insincerely, right?
Probably this has been discussed here already? Anyway, let's study this
a bit:

As we know, in Approval Voting, 0-info strategy consists of approving of
all candidates with above-mean utility, hence Approval Voting meets (ii)
and thus fulfills 0IMC.

But unfortunately, it is easy to fail the criterion. Even our beloved
winning votes Condorcet with only 3 candidates *fails* it. Assume that
A,B,C have utilities 1, 0.9 and 0, respectively. Then it is better to
vote A=B>C than A>B>C since it is more important to prevent that the
cycle A>B>C>A is resolved by breaking B>C than to support A over B. This
can be seen like this:


Mathematical analysis showing that winning votes Condorcet fails 0IMC:
----------------------------------------------------------------------
Assume 3 candidates A,B,C and a voter who assigns to them utilities
u(A)=a, u(B)=b, and u(C)=c.
To assess the difference in expected utility under winning votes
Condorcet caused by different ways of voting, we first focus on one
pair, say A and B. For this analysis, we only need to determine the
conditional expected utility of voting A>B, A=B, or A<B, given that this
voter's individual decision influences the result at all. Assuming the
number of voters is large, our voter's decision about A and B only
matters in one of the following situations (XY denotes the number of
other voters ranking X over Y):

  (1a) BC>CA=AB      > BA,AC,CB
  (1b) BC>CA=AB+1>AB > BA,AC,CB
  (2a) CA>BC=AB      > BA,AC,CB
  (2b) CA>BC=AB+1>AB > BA,AC,CB
  (3a) AC>CB=BA      > AB,BC,CA
  (3b) AC>CB=BA+1>BA > AB,BC,CA
  (4a) CB>AC=BA      > AB,BC,CA
  (4b) CB>AC=BA+1>BA > AB,BC,CA
  (5a) AC>BC > AB=BA      > CB,CA
  (5b) AC>BC > AB=BA+1>BA > CB,CA
  (5c) AC>BC > BA=AB+1>AB > CB,CA
  (6a) BC>AC > AB=BA      > CB,CA
  (6b) BC>AC > AB=BA+1>BA > CB,CA
  (6c) BC>AC > BA=AB+1>AB > CB,CA
  (7) some situation with two or more (near)
      equalities between AB, BA, AC, CA, BC, and CB.

Assuming a large number of voters, situation (7) is of negligible
probability compared to situations (1a) to (6c) since they involve only
one equality. Moreover, these probabilities fulfil

  P(1a)=P(2a)=P(3a)=P(4a) =: alpha,
  P(1b)=P(2b)=P(3b)=P(4b) =: beta,
  P(5a)=P(6a)             =: gamma,
  P(5b)=P(6b)=P(5c)=P(6c) =: delta

with some yet unknown numbers alpha,beta,gamma,delta which depend on the
number of voters.

Under winning votes Condorcet, the results of voting A>B, A=B, or A<B
are (where X/Y means X and Y win with equal probability):

        A>B  A=B  A<B
 ---------------------
  (1a)   A   A/B  A/B
  (1b)  A/B   B    B

  (2a)   C   B/C  B/C
  (2b)  B/C   B    B

  (3a)  A/B  A/B   B
  (3b)   A    A   A/B

  (4a)  A/C  A/C   C
  (4b)   A    A   A/C

  (5a)   A   A/B   B
  (5b)   A    A   A/B
  (5c)  A/B   B    B

  (6a)   A   A/B   B
  (6b)   A    A   A/B
  (6c)  A/B   B    B

Summing up, twice the conditional expected utility of voting A>B, A=B,
or A<B is

  2Eu(A>B) = alpha( 4a + 1b + 3c ) +  beta( 5a + 2b + 1c )
           + gamma( 4a + 0b + 0c ) + delta( 6a + 2b + 0c )

  2Eu(A=B) = alpha( 3a + 3b + 2c ) +  beta( 4a + 4b + 0c )
           + gamma( 2a + 2b + 0c ) + delta( 4a + 4b + 0c )

  2Eu(B>A) = alpha( 1a + 4b + 3c ) +  beta( 2a + 5b + 1c )
           + gamma( 0a + 4b + 0c ) + delta( 2a + 6b + 0c )

As one would expect, A>B gives better expected utility than B>A if and
only if a>b, that is, if A has greater individual utility than B.

However, the expected utility of A=B need not be the average of the
other two but can be larger than both that of A>B and B>A, contrary to
what one would expect! This can happen if C's utility c is small enough
compared to the difference in utility of A and B. More precisely, voting
A=B maximizes expected utility if  2E(A=B)-2E(A>B)  and  2E(A=B)-2E(B>A)
 are both non-negative, that is, if

  2 (alpha + beta + gamma + delta) a
     >= (alpha + beta + 2gamma + 2delta) b + (alpha + beta) c

and

  2 (alpha + beta + gamma + delta) b
     >= (alpha + beta + 2gamma + 2delta) a + (alpha + beta) c .

My estimation is that the quotient  (gamma+delta)/(alpha+beta)  is
approximately sqrt(2) when the number of voters is large. If this is
correct, then voting A=B would for example be better than A>B or B>A
whenever c=0, a=1, and

  b >= (1+2sqrt(2))/(2+2sqrt(2))=(3-sqrt(2))/2 = 0.7928...

Now, returning to the question what to vote on all three candidates,
assume without loss of generality that a>b>c. Then, by the above
inequalities, it is always optimal to vote A>C and B>C on these two pairs.

Hence, in order to maximize expected utility under winning votes
Condorcet with zero information, one should

      vote   sincerely  A>B>C  if  (b-a)/(c-a) <= 0.7928,
  but vote insincerely  A=B>C  if  (b-a)/(c-a) >= 0.7929.


---

Guess what a similar analysis says about margins Condorcet!?

Well, in margins Condorcet, we *do* seem to have  Eu(A=B) =
(Eu(A>B)+Eu(A<B))/2  and margins Condorcet *does* seem to fulfil 0-info
mean consistency...

Yours, Jobst