# [EM] D2MAC analysis for two factions

Jobst Heitzig heitzig-j at web.de
Thu Mar 8 06:15:20 PST 2007

```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  (*)

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

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