Social Design Engineering Series

SDES-2020-6

Global Stability of Voluntary Contribution Mechanismwith Heterogeneous Preferences

Tatsuyoshi Saijo
Research Institute for Humanity and Nature
Research Institute for Future Design, Kochi University of Technology
Tokyo Foundation for Policy Research

17th July, 2020

School of Economics and Management
Research Institute for Future Design
Kochi University of Technology
KUT-SDE working papers are preliminary research documents published by the School of Economics and Management jointly with the ResearchCenter for Social Design Engineering at Kochi University of Technology. To facilitate prompt distribution, they have not been formally reviewed
and edited. They are circulated in order to stimulate discussion and critical comment and may be revised. The views and interpretations expressedin these papers are those of the author(s). It is expected that most working papers will be published in some other form.

Global Stability of Voluntary Contribution Mechanismwith Heterogeneous Preferences†
Tatsuyoshi Saijo*
Research Institute for Humanity and Nature andResearch Institute for Future Design, Kochi University of Technology

Abstract
We find a necessary and sufficient condition for global stability of the voluntary
contribution mechanism with heterogeneous quasilinear preferences under a simultaneous
system of difference equations. The condition is that there must exist an outstanding player
whose willingness to contribute dominates the sum of the others’ willingness.
JEL Classification: C62, C72, C92, H41† The Japan Society for the Promotion of Science has financial support for this research (JSPS

KAKENHI Grant Number 24243028, 16K13354 and 17H00980). Research Institute for Humanity
and Nature also has financial support for this research (RIHN Project Number 40410000).
* The author would like to thank Naoki Yoshihara, Ryo Kambayashi, Nori Tarui, Yutaka
Kobayashi, and Jun Feng for their comments and suggestions. Of course, all remaining errors
are the author’s.

11. Introduction
After the Great East Japan Earthquake in 2011, Hamamatsu, a city in the central region ofJapan that did not suffer from the earthquake, but expects a large one within a few decades,
started building anti-tsunami embankments on more than 17 km based on voluntary public
contributions. One company, Ichijo Housing Co., donated JPY 30 billion, and households andother companies, more than 3,000 in total, donated JPY 1 billion. To date, the total donations
will cover the cost of the project.
Oxfam reported that eight super-rich men hold the same amount of wealth as the poorest
half of the world’s population in January 20171, and five out of eight are members of the Giving
Pledge where they committed to contribute a majority of their wealth to philanthropic causes.2
These are examples of the private provision of a public good, with a sizable literaturebody developed over the past few decades. The basic framework is that each player chooses
between consumption of a private good and contribution to it, which is called the voluntary
contribution mechanism (VCM). In this respect, researchers such as Bergstrom, Blume, and Varian(1986), Bernheim (1986), Cornes and Sandler (1996), and Kotchen (2006) analyze the nature ofNash equilibria, such as the existence, uniqueness, and neutrality. However, the stability of the
VCM has not been investigated extensively.
If utility functions are linear, then each player has a dominant strategy with nocontribution, making the system stable. However, if they are nonlinear and all players have thesame quasilinear utility function and endowment, the system is not asymptotically stable undersimultaneous difference equations, and is structurally unstable under simultaneous differential
equations as Saijo (2014) shows. That is, the system of the VCM is intrinsically unstable and has
a free-riding issue as well.
Since assuming that every player has the same utility function and the endowment is astringent condition, this paper considers them different. As such, we use quasilinear utility
functions that are linear with respect to player i’s private good consumption xi and nonlinearwith respect to a public good y, that is, ui ( xi , y )  xi  t i ( y ). The rationale behind this formulationis that the private good represents money and, hence, its marginal utility is constant, but themarginal utility of the public good decreases so that the ti ( y ) part is nonlinear. Furthermore,
players may have different endowments.
Experimental tests such as Chen and Plott (1996), Chen and Tang (1998), Chen andGazzale (2004) and Healy (2006) found that players basically use myopic learning dynamics
such as best-responses to a recent history of actions. Following them, we will use best response1

See https://www.oxfamamerica.org/explore/research-publications/an-economy-for-the-99percent/2 See https://givingpledge.org/#enter2

dynamics as the simplest dynamics. If a utility function is quasilinear, the best response
function is linear and its slope is -1. The difference in players can be identified by the intercept
when no other players contribute, ai, and let the player who has the maximum value of ai beplayer 1. Subsequently, the unique Nash equilibrium is that player 1 contributes a1 and every
other player contributes nothing. Furthermore, a necessary and sufficient condition for global
stability of the simultaneous system of difference equations is a1   j 2 a j , as long as wi  ai ,nwhere wi is the endowment of player i. That is, the system is globally stable if there is an
outstanding player who can contribute significantly compared with other players3. It appears
that the anti-tsunami embankment case of Hamamatsu city and the pledgers of the Giving
Pledge correspond to this case.
Even if a1   j 2 a j , the system is locally stable around the Nash equilibrium, but if thennumber of players increases, the area of initial points that converge to the Nash equilibrium
decreases rapidly. That is, the system is intrinsically unstable as long as either a1   j 2 a j isnsatisfied, namely players are all alike, or the number of players is large.
The remainder of this paper is organized as follows. Section 2 provides a necessary andsufficient condition for global stability of the VCM, and section 3 analyzes instability and thenumber of players using an example. Section 4 studies the Cobb-Douglas utility functions, andsection 5 describes further research scope.
2. Global stability of the VCM
Let x be a private good and y be a public good. The production function of the public
good is y = f(x) = x. That is, for example, one hour of labor input produces one millimeter ofanti-tsunami embankment. Player i has endowment wi and must decide to divide wi into i’s own
consumption of the private good x i and contribution si to the public good. That is, y   1 s jnwhere n  2. This system is called the VCM. We assume that each player has a quasilinear
utility function ui ( x i , y )  x i  t i ( y ). Consequently, player i faces the following problem.
Max ui ( xi , si  s i ) subject to wi  xi  si ,where s i   j  i s j . Let ui ( wi  si , si  s i )  vi (si , s i )  wi  si  ti (si  s i ). A list of contributionssˆ  ( sˆ1 ,..., sˆn ) is a Nash equilibrium if for all i vi ( sˆi , sˆ i )  vi ( si , sˆ i ) for all si  [0, wi ]. The best
response function is defined as

3Olson (1965) pointed out that "(i)n smaller groups marked by considerable degrees of inequality - that is,in groups of members of unequal "size" or extent of interest in the collective good - there is the greatest
likelihood that a collective good will be provided" (p.34) from the viewpoint of collective good provision
rather than from the stability viewpoint.
3

ri ( s i )  arg maxvi ( si , s i )|si  [0, wi ]. 4.

si

As such, the best response functions are linear. In order to show this property, consider the firstorder condition for the maximization problem, i.e., vi / si  1  ti / y  0. Then, totallydifferentiating both sides of the condition, we have the slope of the best response function since  ti y  si    t i y  s i :  ti y  s idrids i 
 1.
  ti y  si
ds i ds i
That is, the best response function is linear with -1 slope. Since player i cannot contribute anegative value, ri ( s i )  max{ s i  ai ,0} , where ai is the intercept. For simplicity, assume thata1  a j  0 for all j  1, and wi  ai for all i. Then, since r1 (0)  a 1 and rj ( a 1 )  0 for all j  1,
( a1 ,0, ,0) is a Nash equilibrium. According to Bergstrom, Blume and Varian (1986, 1992), whoprove the uniqueness of Nash equilibrium in a general setting, and Bergstrom, Blume andVarian (1986), who show the Nash equilibrium for quasilinear utility functions, we have thefollowing proposition.
Proposition 1. Suppose that a1  a j  0 for all j  1 and wi  ai for all i. Then, the unique Nashequilibrium is (a1 ,0,,0).
The Nash equilibrium is not Pareto efficient. Let ui ( xi , y )  xi  ai ln y. 5 Then, theSamuelson condition is  a j  y. Since the public good level at the Nash equilibrium is a1 , itis apparently lower than the Pareto efficient level.
t

At time t, let player i’s choice of contribution be si . We simply assume that player itchooses ri ( s i ) at time t + 1, where t = 1, 2, … That is, we assume that every player chooses thebest response to the sum of strategies chosen by the other players in the immediately preceding
period. Therefore, the stability of the following system must be analyzed.
t 1(1) si max{st i  ai ,0} i = 1, …, n and t=1, 2, ….

Ignoring the maximization part, we can rewrite the system in the following manner.

st 1  Ast  a ,If ui is strictly quasi-concave, the maximizer is unique and ri is continuous as per Berge’s maximum
theorem.
5 The best response function for this utility function is r ( s )  max{  s  a ,0}.
i i
i
i
4

4

 0  1   1
 s1t  a1 
 1 0
 
 
 
t


, s     and a     .
where A 
  1
 snt an   1  1 0 

The stability property is governed by the eigenvalues of A, that is, (1  n,1, ,1) .6 Furthermore,
as Proposition 1 shows, the Nash equilibrium is not located on the linear part of the system and1hence, it is not a solution of s  [I  A] a , where I is the identity matrix.7 That is, understanding
the stability property of (1) is a new challenge.
In order to understand the stability of the system, let us introduce the maximum response
function mII (s1 ) of player II, treating all players other than player 1 as one player and denoting
the player as player II:(n  1)s   n a for 0  s  a
1
1
n
j1 j

1

n
(n  2)s  j1 a j for an  s1  an11mII ( s1 )  , 2s1  a2  a3 for a4  s1  a3

 s1  a2 for a3  s1  a2
0 for a  s21which is a continuous piecewise linear function. This function is not the best response functionof all players other than player 1, but shows the maximum possible sum of contributions given

s1.
Consider the following example. Let ( a1 , a2 , a3 )  (10,6,2) and wi  12 for all i. Then, mII
is a function summing up max{  s1  6, 0} and max{  s1  2,0} vertically. This is a-b-c-e in Figure
1, where the vertical axis, sII , represents the range of mII ( s1 ) for player II and s1 for player 1.
The slope of a-b is -2, and the slope of b-c is -1. If s1 is greater than 6, neither players 2 nor 3
contribute, and, hence, the c-e part is flat.
Let s 1  ( s11 , s21 , s31 )  (1, 5, 0) , which corresponds to s1 in Figure 1. Then, the best response is(5,5,0) which corresponds to q. Consider another initial point such as s 1  (1, 3, 2) , which also
corresponds to s1 . The best response is (5,3,0), which corresponds to p. Let sˆ 1  (1,1, 4) , whichstill corresponds to s1 . Then, the best response is (5,1,0), which corresponds to k. Sinces22  max{ ( s11  s31 )  6, 0} and s32  max{ ( s11  s21 )  2, 0}, for s1  1 the maximum possible
1

response s22 is 5, and the maximum possible response s32 is 1. The sum of 5 and 1 shows mII (1)or d. That is, the range of reactions of players 2 and 3 for s1  1 is from 0 to mII (1) , which is d-f.
6
7

See Saijo, Feng and Kobayashi (2017).
Note that every element of I - A is 1 and, hence I-A, is not invertible.
5Given s  (s1 , s1 ), the set of best responses {( a1  s11 , b ) : b  [0, mII ( s11 )]} to s1 is denoted by
1

1

1

1

M ( s 1 ) , and called the set of maximum possible best responses to s . Since the reaction of player 1 tos1 is 5, M( s1 ) should be g-h. Three best responses, such as q, p, and k, are in M( s1 ). If s1  d,then M( d ) is d’-f’, and if s  f , then M( f ) is d”-e. That is, if s1 is somewhere on d-f, then the1

set of the maximum possible best responses to d-f is square d’-f’-e-d”.

s IIPlayer 1's best
response curve

10
8 a

d'

d
5

1

d"

q
p

b

0

g

q'

s1

M(s1)

r"

r'

mII

r

k

f

f'

c
h
5 6

r'''
k'

f"
h'

e
10 12

s1

Nash equilibrium
Figure 1. Best response areas.
Consider now what is the next set of maximum possible responses to square d’-f’-e-d” inFigure 1. Point d’ becomes line r-f’ and q’ becomes r’-h. Furthermore, r becomes r”-f”, and any
point between r and f’ becomes a line parallel to r”-f” and located between f”-e. That is, themaximum possible best responses to line d’-f’ is rectangle r-f’-e-r’’’. Now, consider a point in d’r-r’’’-d”, such as q. Point q becomes line k-h, and this line is included in rectangle r-f’-e-r’’’ since rc-e is included in r-f’-e-r’’’. Consider any point such as k in r-f’-e-r’’’. Point k becomes line k’-h’ inr-f’-e-r’’’. That is, the set of maximum possible responses to square d’-f’-e-d” is rectangle r-f’-e-r’’’.
It is now easy to see that the set of maximum possible responses to rectangle r-f’-e-r’’’ becomessquare r”-f”-e-r’’’. That is, if this sequence of sets converges to the Nash equilibrium, the original
sequence must converge to it. The strategy for showing the stability of the system is describedas follows.

6First, consider Figure 2 and take any s1. Then, s2 must be in rectangle a-0-e-b in Figure 2,
since the largest maximum possible best response set to s1 is [0,  j2 a j ] and the largestn1

maximum possible best response set to  j 2 s1j is [0, a1 ]. Consider now the set of maximumnpossible best responses to a-0-e-b. The maximum possible best responses to line a-0 covers
square c-f-e-b, the maximum possible best responses to a-0-f-c are in c-f-e-b, and the maximum
possible best responses to c-f-e-b are in c-f-e-b. Consider the next set of maximum possible best
responses to square c-f-e-b. The maximum possible best responses to line c-f are in rectangle d-fe-p, the maximum possible best responses to c-d-p-b are in rectangle d-f-e-p, and the maximum
possible best responses to d-f-p-e are in d-f-p-e. Similarly, the next set of maximum possible best
responses to rectangle d-f-p-e is square g-h-e-p. Repeating the same procedure, g-h-e-p becomesrectangle k-h-e-q, and k-h-e-q becomes square r-w-e-q. Since the base of square r-w-e-q is on theflat part of mII , the next rectangle is line w-e. Then, any best response point on w-e becomes onepoint square e, which is the Nash equilibrium. Stability here is global stability: any initial point
converges to some equilibrium point.8

s II
a1
n

 aj aPlayer 1's best
response curve

c

b

j =2

d

g

p

mII

0

f

k

r q

h a

w a e

2

1

s1

Nash
equilibrium

Figure 2. From square to rectangle.

Let us introduce two concepts. Rec ( s1 ) is the minimum rectangle containing two points( s1 , mII (s1 )) and ( a1 ,0) , where a rectangle is a set containing the interior and the boundary, and8

See Arrow and Hurwicz (1958) and Arrow, Block, and Hurwicz (1959).
7sq ( s1 ) is the minimum square containing two points (s1 , a1  s1 ) and ( a1 ,0) , where a square is a
set containing the interior and the boundary. As such, we have the following lemma.
Lemma 1. Suppose that wi  ai for all i and a1   j2 a j . Let s1 [0, a1 ). Then,n(i) the set of maximum possible best responses to rec ( s1 ) is sq ( a1  mII ( s1 )) and sq ( a1  mII ( s1 )) is aproper subset of rec ( s1 ) ; and(ii) the set of maximum possible best responses to sq ( s1 ) is rec ( s1 ) and rec ( s1 ) is a proper subset ofsq ( s1 ) .
Proof. (i) Since wi  ai for all i, player i can choose a strategy up to ai . Take any s1  [0, a1 ).
Since a1 

 a and the construction of m , player 1’s best response curve is always above
n

j 2

j

II

mII , except for s1  a1 , i.e., mII ( s1 )  a1  s1 and hence s1  a1  m II ( s1 ). Since s1 is the value of thehorizontal axis of the left bottom vertex of rec ( s1 ) , a1  mII ( s1 ) is the value of the left bottom
vertex of sq ( a1  m II ( s1 )). Therefore, the base of sq ( a1  mII ( s1 )) is a proper subset of the base ofrec ( s1 ). Since the height of sq ( a1  mII ( s1 )) is mII ( s1 ) , because a1  ( a1  mII ( s1 ))  mII ( s1 ) , and theheight of rec ( s1 ) is mII ( s1 ) by definition, both have the same height. That is, sq ( a1  mII ( s1 )) is aproper subset of rec ( s1 ) .
In order to show that the set of maximum possible best responses to rec ( s1 ) issq ( a1  mII ( s1 )), consider three possible areas of rec ( s1 ). Consider first any point ( s1 , b ) whereb  [0, mII ( s1 )]. Then, M ( s1 , b )  {( a1  b , b ') : for some b '  [0, mII ( s1 )]}. Since( a1  b )  ( a1  mII ( s1 ))  mII ( s1 )  b  0, M ( s1 , b ) is in sq ( a1  mII ( s1 )) , and hence, the set of
maximum possible best responses to ( s1 , b ) with b  [0, mII ( s1 )] is exactly the same assq ( a1  m II ( s1 )).
Consider any point ( s1 , b ) where s1  [ s1 , a1  mII ( s1 )] and b  [0, mII ( s1 )]. Since mII is a
non-increasing function, m( s1 )  m( s1 ). Since a1  mII ( s1 )  a1  mII ( s1 ), M ( s1 , b ) must be insq ( a1  m II ( s1 )).
Finally, consider any point ( s1 , b ) where s1  [ a1  mII ( s1 ), a1 ] and b  [0, mII ( s1 )].
Applying the same argument above, M ( s1 , b ) must be in sq ( a1  m II ( s1 )).
(ii) Take any s1 . Since a1 

 a , the construction of m , and player 1’s best response curve
n

j2

j

II

is always above mII except for s1  a1 , rec ( s1 ) is a proper subset of sq ( s1 ) . Since mII is a nonincreasing function, m( s1 )  m( s1 ) for all s1  [ s1 , a1 ], and hence, M ( s1 , b ) with s1  [ s1 , a1 ] andb  [0, a1  s1 ] must be in rec ( s1 ) .█Proposition 2 shows a necessary and sufficient condition for stability.

8

Proposition 2. Suppose that a1  a j  0 for all j  1 and wi  ai for all i. Then, the simultaneous systemof difference equations is globally stable if and only if a1   j2 a j .
nProof. (i) The “if” part. Take any initial point s1 . Then, the best response s 2 to s 1 must be inrec(0). By lemma 1, the next set of maximum possible best responses to rec(0) is sq ( a1   j2 a j ) ,nand hence, the height of the square is  j2 a j . Let c  a1   j2 a j  0. Since the slope of mII isnn

at most -1 as far as s1  [0, a2 ] , the sequence from rec(0) to sq ( a1   j2 a j ) and the square to thennext rectangle ends in a finite step, due to the fact that the height of a square shrinks by at least cto the next square. Once the base of a square reaches an interval [ a2 , a1 ] on the s1 axis that is nota point since a1  a 2 , the next rectangle is an interval contained in [ a2 , a1 ] . As such, the best
response to the interval must be the Nash equilibrium.
(ii) The “only if” part. It suffices to show that if a1   j2 a j , the system is unstable. Letns1  ( a1 , a2 ,..., an ) . Then, since a1   j2 a j , the best response is (0, 0, …, 0). The next best responsenis ( a1 , a2 ,..., an ) and hence, the system is not stable. █3. Instability and the number of players: an example
Notice that stability in Proposition 2 is not local, but global. That is, as far as a1   j2 a j isnsatisfied, the system goes to the Nash equilibrium in a few steps starting from any initial point.
When a1   j 2 a j , the system is not globally stable. However, a sequence starting from anninitial point in a rather large area such as a-c-w1-d in Figure 3-(a) converges to the Nashequilibrium. In this sense, the system is locally stable.
The stability on the boundaries of a-c-w1-d in Figure 3-(a) is rather complex, since wetranslate the n dimensional space into the two dimensional space. Let ( a1 , a2 , a3 )  (10,6, 4) andwi  10 for all i. Then, the Nash equilibrium is (10, 0, 0), as shown in Figure 3-(b), but the system
is not stable since a1  a2  a3 . Assume that each player can announce only integers. Consider
s1  (0, 4, 6) , that is a in the figure. Then, s 2  (0, 0, 0) , s 3  (10, 6, 4) and s 4  (0, 0, 0). That is, thissequence alternates between 0 and d, and hence, it does not converge to the Nash equilibrium.
On the other hand, consider s1  (0, 3, 7) that is also a in the figure. As such, s 2  (0, 0, 1),s 3  (9, 5, 4), s 4  (1, 0, 0), s 5  (10, 5, 3), s 6  (2, 0, 0), s 7  (10, 4, 2), s 8  (4, 0, 0), s 9  (10, 2, 0),s 10  (8, 0, 0), and s 11  (10, 0, 0). That is, this sequence converges to the Nash equilibrium. The

upper right number of a is the number of initial points that converge to the Nash equilibrium.
Consider s 1  (5, 9, 1) , that is b in the figure. Then, s 2  (0, 0, 0), s 3  (10, 6, 4), and s 4  (0, 0, 0).
That is, this sequence does not converge to the Nash equilibrium. On the other hand, consider
s 1  (5, 10, 0) , that is also b in the figure. Consequently, s 2  (0, 1, 0), s 3  (9, 6, 3), and s 4  (1, 0, 0).

For the rest of the sequence, see the sequence starting from s 1  (0, 3, 7) , which converges to the
Nash equilibrium.
9

n

 aj

s II

b

j =2

(a1 , aj )j =2Player 1's best
response curve

a1

a

d

a

1
2

c1

Player 1's best
response curve

4

2

6

4

2

8

6

4

2

1

10 8

6

4

2

1
1

d

1

b

10

mII

mII

0

sII

n

e

a2

c

w1

a1

Nash
equilibrium

s1

0

(a)

e

4 5 6

s1

10
Nash
equilibrium(b)

Figure 3. The stable initial point area when the system is not stable.

Consider points that are not in a -0-e-d , where the bar indicates the closure of a-0-e-d. If
each element of s 1 is rather large, such as s 1  (7, 8, 9), then s 2 is at 0, i.e., s 2  (0, 0, 0).
Therefore, the sequence repeats d and 0. However, there is a chance for some s 2 to be in thestable part of a-0-e-d even though s 1 is not in a -0-e -d. For example, consider s 1  (1, 10, 4) , that is
c in the figure. Then, s 2  (0, 1, 0) and the rest are the same in the above paragraph. That is, thesequence converges to the Nash equilibrium.
Consider the number of initial points satisfying 0  s11  10 and 0  s21  s31  10, that is, thenumber of all possible initial points of a -0-e -d. As such, since the number of initial points
satisfying 0  s21  s31  10 is 66,9 and the number of possible choices of player 1 is 11, the numberof all possible initial points of a -0-e-d is 726  66  11. On the other hand, the number excluding
a-d in a -0-e-d is 605  55  11. Furthermore, the number of initial points that converge to the

Nash equilibrium is 680, and the number of initial points that are not in a -0-e-d and converge to
the Nash equilibrium is 45 (the sum of small numbers where the vertical axis value is at least 11
in Figure 3-(b)). That is, the number of initial points that are in a -0-e-d and converge to the Nashequilibrium is 635. Furthermore, since the number of initial points on a-d that converge to the
Nash equilibrium is 31, 605 + 31 - 1 = 635, that is consistent with the previous number where“1” is for the origin that does not converge to the Nash equilibrium. Generally, initial points
that are in a-0-e-d or that are “close” to a-0-e-d converge to the Nash equilibrium.
9

The computation is done in Mathematica 9.
10Although the system is not stable due to the distribution of ai, if the number of initialpoints in a-c-w1-d in Figure 3-(a) is relatively large compared with the total number of all
possible initial points, instability is not a significant issue. Considering the previous example
and assuming that each player can announce only integers, let the number of initial points in
a -0-e-d be a surrogate number that converges to the Nash equilibrium. Table 1 shows thenumber of players n, the number of initial points An satisfying 0  s11  10 and 0   j1 s1j  10,n

An+1/An, and An/11n. Note that 11n are the all possible initial points when the number of players

is n. Then, An/11n is a rough ratio of initial points that converges to the Nash equilibrium. As
the number of players increases, the ratio rapidly decreases. When n is 5, 6, 7 or 8, the ratio is
7%, 2%, 0.5% or 0.1%, respectively. This is because the number of all possible initial points
increases eleven times as the number of players increases, while the number of initial points in
a -0-e-d increases less as the number of players increases (see { An1 / An } in Table 1). That is,instability of the VCM is a serious problem even the number of players is below 8.

Thenumber ofplayers (n)Table 1. Possible stable ratio when the system is not stable.
The number of initialpoints ( An ) satisfyingAn1 / An
An /11n0  s11  10 and
0   j1 s1j  10
n

3

726

726/113 = 0.5456

4

3,146

3,146/726=13/3=4.33

3,146/114 = 0.2149

5

11,011

11,011/3,146=3.5

3,146/115 = 0.0684

6

33,03333,033/11,011=3.0

33,033/116 = 0.0186

788,08888,088/33,033=8/3=2.67

88,088/117 = 0.0045

8

213,928

213,928/88,088=17/7=2.43

213,928/118 = 0.0010

4. The Cobb-Douglas case: Examples
Several experimentalists, such as Andreoni (1993), Chan, Mestelman, Moir, and Muller(1996), Cason, Saijo, and Yamato (2002), and Sutter and Weck-Hannemann (2004), use CobbDouglas utility functions assuming that every players has the same utility function. As such, it
could be useful to summarize the stability property when players have heterogeneous CobbDouglas utility functions.
1i

Let player i’s utility function be vi (si , s i )  ( w  si ) i (si  s i )1ifirst order condition is vi / si  i ((si  si )/( w  si )), where 0  i  1. Then, the (1  i )((w  si )/(si  si )) i  0. That is,( si  s i )/( w  si )  (1  i )/ i , and, hence, si  i s i  (1  i )wi . Therefore the system isst 1  Ast  a ,

11

 0
 
where A   2   n

 1



0



 n

 1 
(1   1 )w1  s1t (1   )w 22 , st     and a  
.
 



 n1 
t





 sn 0 
(1   n )wn 

The asymptotic stability of the system is governed by the eigenvalues of A, assuming that
the Nash equilibrium is in the interior. For example, let ( 1 , 2 , 3 )  (3/4, 1/2, 2/5) and( w1 , w2 , w3 )  (20, 9, 8). Then, det[I-A]  0, the Nash equilibrium is (7/17, 42/17, 62/17), whichis in the interior, and the eigenvalue vector is (1.074, -0.634, -0.441). Hence, the system is not
asymptotically stable since some of the absolute values of eigenvalues are above 1. Let( 1 , 2 , 3 )  (3/5, 1/2, 2/5) with the same endowment vector. Then, det[I-A]  0 and the Nashequilibrium is (167/25, 3/25, 52/25), which is in the interior, and the eigenvalue vector is (0.991,-0.554, -0.437). Hence the system is asymptotically stable. That is, the stability of the system is an
important issue when utility functions are Cobb-Douglas.
5. Concluding remarks
We find a necessary and sufficient condition for global stability of the VCM withheterogeneous quasilinear preferences under simultaneous difference equations. The
implication is that there must exist an eminent player whose willingness to totally contribute
dominates the sum of the willingness to contribute of the rest. Since the anti-tsunami
embankment case is rare, the VCM is not globally stable. Although the VCM is not globallystable, there are initial points that converge to the Nash equilibrium. However, the size of thesepoints relative to all possible initial points diminishes rapidly as the number of players
increases. If we consider that these two points are generic, the VCM perspective is bleak. Thatis, it is intrinsically unstable, and, hence, its applicability problematic.
The validity of our findings can be tested by experiments. Although there are many
experimental results, almost none use heterogeneous and nonlinear payoff functions.
Additionally, the number of players used in VCM experiments is at most five to eight,
excluding experiments with large number of subjects. Consequently, conducting experimental
research in this area is required.
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