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Inefficiency of Nash Equilibria

TLDR
It is shown that Nash Equilibria of smooth games generally tend to be inefficient in the Pareto sense.
Abstract
It is shown that Nash Equilibria of smooth games generally tend to be inefficient in the Pareto sense.

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NOT FOR QUOTATION
WITHOUT PERMISSION
OF THE AUTHOR
INEFFICIENCY OF NASH EQUILIBRIA
Pradeep Dubey
July 1983
WP-83-74
Working
Papers
are interim reports on work of the
International Institute for Applied Systems Analysis
and have received only limited review.
Views
or
opinions expressed herein do not necessarily repre-
sent those of the Institute or of
its
National Member
Organizations.
INTERNATIONAL INSTITUTE FOR APPLIED SYSTEMS ANALYSIS
A-2361 Laxenburg, Austria

PREFACE
The main theme of this paper
is
that Nash Equilibria
of games with smooth payoff functions generally tend to
be inefficient (in the Pareto sense).
Andrze
j
P
.
Wierzbicki
Chairman
System
&
Decision Sciences

INEFFICIENCY OF NASH EQUILIBRIA
BY
PRADEEP DUBEY
1.
Introduction. The main theme of this paper
is
that Nash Equilibria
(N.E.) of games with smooth payoff functions are generally Pareto-inefficient.
Suppose that a strategic game with n players
is
given by n maps
i
i
i
=
1
,...,
n
,
where S
is
the strategy-set and u the payoff function
i
of player
i
.
Our result states that if the functions u are
c2
,
then
generically (for an open dense set of payoffs): (a) the
set
of N.E.
is
finite, (b) if an
N.E.
is
efficient, then at least one player
is
on a vertex
of his strategy-set,
(c) if an N.E.
is
strong, then at most one player
is
off a vertex of his strategy-set. Note that (b) implies generic inefficiency
if the strategy-sets are vertex-free
(e.g., manifolds) or if vertices can
a priori be ruled out of N.E. in the given case. The result applies to the
multi-matrix games of Nash (section
4).
Here a vertex corresponds to a
pure strategy and, given the special structure of payoff functions,
(c)
can be strenthened to: if an N.E.
is
strong every player
is
using a pure
strategy.
That the outcome of noncooperation (N.E.)
is
generally incompatible
with cooperation (efficiency) has been part of the "folklore" of Game
Theory as exemplified in the
paradigm
of
the "Prisoner's ~ilemma".
The pur-
pose of the paper
is
to put this on a rigorous footing.
This paper
is
a rewrite of an old version
[2]
.
It
is
a pleasure to
thank
J.3.
Rogawski for several comments, and in particular for the example
in section
3,
which
is
due in
its
entirety to him.

2.
The Main Theorem
i
k(i)
Let
N
=
{l
,
.. .
,
n)
,
n
2
2
,
be the set of players, and S
C
R
9
i
k(i)
21
,
the strategy-set of player
i
.
Here S
is
the unit simplex,
i
k(i)
k(i)
:
z
x;
i.e., S
=
{x
e
R,
1
.
(For more general strategy-sets,
1~
j
=1
J
1
n
see Remark
1.
)
Fix neighborhoods
vi
of
si
and put
V
=
V
x.
. .
x
V
.
Let
U
be the linear space of all
c2
functions from
V
to the reals
2
1
>
endowed with the
C
-norm, i.e., for all u in
U
,
2
llull
=
supill u(s)ll
,
II~u(s)ll
,
IID
u(s)ll
:
s
V)
.
Our space of noncoopera-
1
i
tive games
will
be (u)~
;
for any
u
=
(u
,
.
. .
,
un)
a
(u)~
,
u
is
the
payoff function of player
i
.
i
1
n
For any
s
=
{s
:
i
~N)QS
=
S
x...x
S
,
TCN
,
and
i
e
=
{e
:
i
c
T)
E
X
si
,
let
(sic)
denote the element of S obtained
ia T
i i
from
s
by replacing
s
by e for each
i
T
.
1
Assume u
=
(u
,
. . .
,
,un)
6
(u)"
is
fixed.
A
point
s
6
S
is
called
i
(1)
T-efficient if there does not exist any point e
E
X
S such that
i6T
ui(s/e) 2ui(s) for all
i
eT
uj (S (e)
>
uj (s)
for some
j
g
T
(2) a Nash Equilibrium (R.E.) if
it
is
T-efficient for all subsets T
consisting of one element
(3)
efficient if
it
is
T-efficient for T
=
N
(4)
a strons Nash Equilibrium if
it
is
T-efficient for all subsets
TCN.
1
>
i.e.,
it
is
required that the norm llull of u be finite in order that
u€
U
.
(The case when u
is
required to be defined only on S
is
more natural, but
will
follow from the current case--see Remark 2.)

Let N(u)
,
E(u)
,
G(u) denote the sets of Nash, efficient, strong Nash
points of the game u
.
Theorem There
is
an open dense
set
Uo
of (u)" such that, for u
U
:
0
(a) N(u)
is
a finite
set
1
(b) if
s
=
(s
,
.
.
.
,
sn) N(u)
fl
E(u)
,
then at least one sj
is
a
vertex
1
(c) if
s
=
(S
,
. .
.
,
sn)
E
G
(u)
,
then at most one
si
is
not a vertex.
i
Proof First
we
focus on the case when
s
is
in the interior of
i
S
(1
,,
n)
.
Let r(j)
=
C
k(i)
.
Then
V
may be viewed as a
i<j
pr(n) of dimension
r
(n)
.
For
subset of the Euclidean space
L.
1
s
=
(S
,
.
.
.
,
sn)
E
V
,
s1
gives the first k(1) coordinates of
s
,
s
2
the next
k(2) coordinates,
i.e.,
the coordinates from r(1)
+
1
till
r(2)
,
etc. Consider the derivative map
given by
:
1
n
where u
=
(u
,,
u
)
.
For fixed u
,
is
the map obtained by restricting
D
:

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