Nash Equilibrium and Manipulation in a Mean
Rule Experiment
Veronica Block∗, Klaus Nehring†, Clemens Puppe‡
March 12, 2012
Abstract
In this study we test voting behavior when applying different voting rules. In particular we use the mean and the median rule when
preferences are single-peaked. We give an algorithm to calculate the
Nash Equilibrium and test experimentally whether individual behavior
accords to the Nash prediction.
∗
Dept.
Germany.
†
Dept.
‡
Dept.
Germany.
of Economics and Business Engineering, Karlsruhe Institute of Technology,
[email protected]
of Economics, University of California at Davis, U.S.A. [email protected]
of Economics and Business Engineering, Karlsruhe Institute of Technology,
[email protected]
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1
Notations
A = [0, M ]
I = {1, . . . , n}
ui : A → R
x∗i ∈ A
(x∗1 , . . . , x∗n ) =: x∗ ∈ An
x̄
xi ∈ A
x−i = (x1 , . . . , xi−1 , xi+1 , . . . , xn ) ∈ An−1
interval of feasible alternatives (M < ∞)
set of individuals
utility function of i ∈ I
peak of individual i ∈ I
ordered vector of peaks,i.e. x∗i ≤ x∗i+1
P
mean of x, i.e. x̄ = n1 ni=1 xi
choice of i ∈ I
choice of j ∈ I\{i}
ui is a single-peaked function w.r.t. x∗i , i.e.
ui (x∗i ) ≥ ui (x) ∀x ∈ A
and for all x̄, ȳ ∈ A with x∗i ≥ x̄ > ȳ or ȳ > x̄ ≥ x∗i hold
ui (x∗i ) ≥ ui (x̄) > ui (ȳ).
N
xN = (xN
1 , . . . , xn ) is a Nash Equilibrium (NE) if for all i ∈ I the choice
is optimal given the choice of all other individuals:
ui (x̄N ) ≥ ui (ȳ) with y−i = xN
−i
x∗ is strictly ordered if: x∗i < x∗i+1 for i ∈ {1, . . . , n − 1}.
The following intervals A1 , . . . , An play a crucial rule in the algorithm :
M
M
(n − i), (n − i + 1)
Ai :=
n
n
Obviously,
A=
n
[
Ai
i=1
and
n
\
i=1
Ai =
n [
M
i=1
n
(n − i + 1)
1
6= ∅.
2
Theoretical Analysis of Nash-Equilibria under the
mean rule
Theorem 1. Let (u1 , . . . , un ) be single-peaked functions w.r.t. (x∗1 , . . . , x∗n )
and let x∗ be strictly ordered. Then, there exists a unique Nash Equilibrium xN .
xN is of the form ( 0, . . . , 0, xN
, M, . . . , M ), where M is the upper end| {z } i0 | {z }
(i0 −1)−times
(n−i0 )−times
point of the feasible interval.1
N
Proof. First, we show by contradiction that a Nash-Equilibrium (xN
1 , . . . , xn )
N
is ordered by value, i.e. xN
i ≤ xi+1 for all i ∈ {1, . . . , n − 1}.
Assume there exists a Nash-Equilibrium y = (y1 , . . . , yn ) with yj > yj+1 for
some j.2
Case 1: ȳ > x∗j , i.e. the mean of the NE is larger than j’s peak.
As yj > yj+1 ≥ 0, there exists ε > 0, such that yj − ε ∈ [0, M ], (choose
ε small enough, i.e. ε < n(ȳ − x∗j ), see below). So, by choosing yj − ε
instead of yj and given all other individuals’ choices y−j remain the
same, the utility of j changes.
X
ε
1
yi )) = uj ȳ −
uj ( (yj − ε +
n
n
i6=j
As x∗j < ȳ −
ε
n
< ȳ, it follows by single-peakedness of uj that
ε
uj ȳ −
> uj (ȳ)
n
We see that j’s utility increases which means that she can manipulate.
Case 2: ȳ ≤ x∗j , i.e. the mean of the NE is smaller than j’s peak.
The proof is analog to case 1. Here, individual j + 1 can increase
her utility by choosing a value yj+1 + ε which is greater than yj+1 .
yj+1 + ε ∈ [0, M ] exists, as M ≥ yj > yj+1 .
Hence, y cannot be a Nash-Equilibrium and we conclude that every NashEquilibrium must be of the form (x1 , . . . , xn ) with xi < xi+1 .
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For one specific i0 ∈ I and her choice xN
i0 ∈ [0, M ].
This condition seems to be weak, at first. But if there exist k, l with k < l and yk > yl ,
then there also exist some consecutive individuals j, j + 1 with yj > yj+1 .
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Second, we show that in a Nash-Equilibrium there can be at most one
value xi0 distinct from zero or M, i.e. xN
i ∈ {0, M } for all i\{i0 }.
For contradiction let us assume that in a Nash-Equilibrium y there exist k, l
with yk , yl ∈
/ {0, M }. Without loss of generality, k < l. Hence, x∗k < x∗l .
Obviously, the mean value cannot equal both peaks, i.e. x∗k 6= ȳ or x∗l 6= ȳ.
Wlog x∗k 6= ȳ. We show that yk ∈
/ {0, M } cannot be optimal, as k’s optimal
response x̃k is calculated as follows:
1
(n · ȳ − yk + x̃k )
n
! x̃k − yk
x∗k − ȳ =
n
!
x∗k =
(1)
(2)
By assumption, the left hand side is nonzero, so x̃k 6= yk , i.e. yk in nonoptimal which contradicts y being a NE. In most cases, x̃k is not in the
interval [0, 100]. Because of single-peaked preferences, k is better of by choosing the endpoint of the interval, which is closer to x̃k than any other point
in (0, 100).
In other words, in an unrestricted domain, for every individual the choice is
non-optimal unless the mean equals her peak. In a restricted domain, in this
case, the choice has to be one of the endpoints of the interval.
Now, we know, that the Nash-Equilibrium must be of the form
xN = ( 0, . . . , 0, xN
, M, . . . , M )
| {z } i0 | {z }
(i0 −1)−times
(n−i0 )−times
It remains to figure out, which is the individual i0 , that achieves her peak
when applying the mean rule. The following theorem gives us a concrete
algorithm.
Theorem 2. Determination of the Nash Equilibrium
N
If every individual i chooses her xN
i according to the following algorithm, x
is a Nash Equilibrium.
• Case 1 ("x∗i < Ai "): x∗i <
M
n (n
− i). Then xN
i = 0.
• Case 2 ("x∗i > Ai "): x∗i >
M
n (n
− i + 1). Then xN
i = M.
∗
• Case 3 ("x∗i ∈ Ai "): M
n (n − i) ≤ xi ≤
∗
Then xN
i = nxi + M (i − n).
3
M
n (n
− i + 1).
Proof. According to Theorem 1, there exists at most one individual with a
choice unequal the endpoints of the interval. So let us assume, we found i0
and calculate her optimal choice. We know by Theorem 1, that all individuals
with a smaller index choose 0, and all individuals with a higher index choose
M . In the following equation i0 ’s peak equals the mean given these choices.
x∗i0 =
1
[(i0 − 1) · 0 + xN
i0 + (n − i0 ) · M ]
n
(3)
This leads us to the optimal choice of i0 :
∗
xN
i0 = nxi0 + M (i0 − n)
This term is smaller than 0 iff x∗i0 <
iff x∗i0 > M
n (n − i0 + 1) (case 2).
M
n (n
(4)
− i0 ) (case 1) and larger than M
Remark: If x∗ is not strictly ordered, i.e. there exist k, l with x∗k = x∗l ,
the Nash Equilibrium may not be unique. More precisely, if the peaks belong
to the intervals in case 1 or 2 of Theorem 2, the Nash Equilibrium is still
unique.
But there exist multiple NE, if for one of those individuals case 3 holds. Let
us call this individual k. Then, the mean of the Nash Equilibrium equals
both peaks, i.e.
x̄N = x∗k = x∗l
Hence, a vector y is a NE, if the sum of the choices of k and l remains the
N
same, i.e. yk + yl = xN
k + xl and all other individuals’ choices remain the
same, i.e. yi = xN
i for all i ∈ I\{k, l}.
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Experimental Results of the mean rule vs. the
median rule
To be written.
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