[EM] D2MAC analysis for two factions

[Jobst Heitzig]

Hello folks,

here's a little mathematics to show that:

Under D2MAC, in a situation with individual utilities

  a*n voters: A 1, C alpha, B 0
  b*n voters: B 1, C beta,  A 0,

where  0 < b < a < 1  and  1/2 < alpha,beta < 1,
the voting strategy

  a*n voters: A favourite, C also approved
  b*n voters: B favourite, C also approved 

leads to the sure election of the compromise C
and it is a Nash equilibrium if and only if

    alpha > (1+a)/2
and  beta > (1+b)/2.


Analysis:

As D2MAC is monotonic, it will never be strategically advantageous for 
the A-voters not to specify A as "favourite" or to specify B as "also 
approved", and likewise for the B-voters. 

That is, in a game-theoretic analysis is seems suitable to consider 
these two factions the two "players" of a game in which the every pure 
strategy for "player A" (=the a*n A-voters) consists in choosing some  
x  between 0 and  a  and then voting

  x*n voters:     A favourite, C also approved
  (a-x)*n voters: A favourite only

Likewise, the pure strategies for "player B" (=the b*n B-voters) are to 
vote

  y*n voters:     B favourite, C also approved
  (b-y)*n voters: B favourite only

for some  y  between 0 and  b.

(Alternatively, we could also assume that each A-voter approves of C 
with an independent probability of  x/a,  and every B-voter with 
probability  y/b.  In a large electorate, this will make no essential 
difference.)

In order to determine which  x  gives the best expected utility for 
player A when some  y  is assumed (and vice versa), we first compute 
the...

Winning probabilities given  a,b,alpha,beta,x,y:
------------------------------------------------

(i) if x+y <= b (so that C is the least approved):

  PA(x,y) = x(1-y)   + (a-x) = a - xy
  PB(x,y) = y(1-x)   + (b-y) = b - xy
  PC(x,y) = xy     + yx      = 2xy

(ii) if b <= x+y <= a (so that approval says A>C>B):

  PA(x,y) = x(1-y)   + (a-x) = a - xy
  PB(x,y) = y(1-x-y) + (b-y) = b - xy - y²
  PC(x,y) = xy     + y(x+y)  = 2xy + y²

(iii) if a <= x+y (so that C is the most approved):

  PA(x,y) = x(1-x-y) + (a-x) = a - xy - x² 
  PB(x,y) = y(1-x-y) + (b-y) = b - xy - y²
  PC(x,y) = x(x+y) + y(x+y)  = 2xy + x² + y²

(I will be sloppy with the limiting cases in which x+y=a or x+y=b here 
since we did not specify a tie-breaking rule yet.)

This determines the...

Expected utilities for players A and B:
---------------------------------------

(i) if x+y <= b:

  EuA(x,y) = PA(x,y) + alpha*PC(x,y) = a + (2alpha-1)xy 
    (monotonic in x and y)
  EuB(x,y) = PB(x,y) +  beta*PC(x,y) = b + (2beta-1)xy
    (monotonic in x and y)
    
(ii) if b <= x+y <= a:

  EuA(x,y) = a + (2alpha-1)xy + alpha*y²
    (monotonic in x and y)
  EuB(x,y) = b + (2beta-1)xy - (1-beta)y²
    (monotonic in x but not always in y)

(iii) if a <= x+y:

  EuA(x,y) = a + (2alpha-1)xy - (1-alpha)x² + alpha*y²
    (monotonic in y but not always in x)
  EuB(x,y) = b + (2beta-1)xy + beta*x² - (1-beta)y²
    (monotonic in x but not always in y)

The condition under which EuA is monotonic in x in case (iii) can be 
determined by differentiating EuA by x:
     0 < (2alpha-1)y - 2(1-alpha)x
i.e. x < y*(2alpha-1)/(2-2alpha)

Likewise, the condition under which EuB is monotonic in y in cases (ii) 
and (iii) is: 
     0 < (2beta-1)x - 2(1-beta)y
i.e. y < x*(2beta-1)/(2-2beta)


Full cooperation?
-----------------

Let us now ask under what conditions "full cooperation" (i.e. x=a, y=b) 
is a Nash equilibrium. This is equivalent to

    EuA(a,b) > EuA(x,b) for all x<a
and EuB(a,b) > EuB(a,y) for all y<b.

From what we know about the monotonicity, we first need 

    a < b*(2alpha-1)/(2-2alpha)
and b < a*(2beta-1)/(2-2beta)

which is equivalent to

    alpha > (1+a)/2
and  beta > (1+b)/2  (*)

In addition, we need 

    EuA(a,b) > max { EuA(a-b,b) (case ii), EuA(0,b) (case i) } 
and EuB(a,b) > EuB(a,0)

and all these conditions also suffice.

The latter are equivalent to

    (2alpha-1)b² - (1-alpha)a² > 0
and (2alpha-1)ab - (1-alpha)a² + alpha*b² > 0
and (2beta-1)ab  -  (1-beta)b² +  beta*a² > 0.

Since all these are already implied by (*), we conclude:


CONCLUSION: 
-----------
Under D2MAC with individual utilities

  a*n voters: A 1, C alpha, B 0
  b*n voters: B 1, C beta,  A 0,

the voting strategy

  a*n voters: A favourite, C also approved
  b*n voters: B favourite, C also approved 

leads to the sure election of the compromise C
and it is a Nash equilibrium if and only if

    alpha > (1+a)/2
and  beta > (1+b)/2.


Yours, Jobst