[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 (*)
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
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