Rent-seeking group contests with one-sided private information

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Rent-seeking group contests with one-sided private

information

R.J. Everhardt (S1407627) 20 October 2010

Abstract

We consider a rent-seeking group contest in which one group has private inform-ation about its valuinform-ation of the prize. This valuinform-ation might be either high or low. All other groups have a publicly known and identical valuation of the prize. We derive the unique equilibrium of the efforts played and show the conditions under which groups play a positive or zero effort.

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1 Introduction

Tullock (1980) introduced the fundamental game-theoretic model for what later would be known as contest theory. Players compete for a prize that just one player will win. Each player can choose to make an effort, that will be non-refundable. The probability that a player wins the prize depends on his effort and that of the other players. Since Tullock (1980), a vast amount of papers appeared to extend the research on contest theory. Some papers on contest theory speak in terms of rent-seeking. The players are termed rent-seekers and the prize is the rent. Rent dissipation is the aggregated effort exerted by the rent-seekers. Most papers focus on the characteristics of the effort that players make. Nitzan (1994) and Konrad (2009) give an overview of popular extensions of the basic model.

For example, players could have asymmetric valuations for the prize. This is, among others, treated by Hillman and Riley (1989) and Stein (2002). The latter concludes that some players could be inactive, i.e. make a zero effort, in the (pure strategy Nash) equilibrium. Stein (2002) derives the conditions under which a player is active or inactive. Nitzan (1991) introduced a model for group contests, where players form groups and jointly compete for the prize. One group will win the prize, which is then divided among the players in that group. When the winning group divides the prize equally between group members, Nitzan (1991) concludes that both larger group size and a smaller number of groups result in a smaller aggregated effort.

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and Shogren (1998a,b). The other players are always active in the model of Schoonbeek and Winkel (2006).

This paper considers a group contest model with one-sided private information. It generalises Schoonbeek and Winkel (2006) in the sense that players compete in groups. Players in a group have the same valuation for the prize. One group has private inform-ation about its valuinform-ation, which can be low or high. The other groups share an identical and publicly know valuation, that can be different from the high and low valuation of the group with private information. The winning group divides the prize equally between its members. The model could be applied to situations where groups of incumbents face a group of entrants. Examples are sports teams that face a new competitor in their league, university departments that have to compete for a grant with a newly formed department and countries that lobby at the UN security council where one of them is an unpredictable country like North Korea.

We will show that a unique symmetric (pure strategy Nash) equilibrium exists for our model. We derive the exact effort each group will make and show the necessary and sufficient conditions under which each group will or will not play a positive effort. For the player with private information in the two-group model we also show how this condition can be rewritten in terms of relative resolve. We show that groups that do not possess private information increase their effort if the group with private information is less likely to play and that always at least one group is active. We also show that larger groups make smaller efforts and that for groups of equal size all will play or neither. Furthermore, we derive similarities and differences between our model with private information and a benchmark model with complete information.

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2 Relevant models from the literature on contest theory

2.1 The standard Tullock contest

Tullock (1980) considers a contest where n risk neutral players compete for a prize, which only one player will win. All players choose their non-refundable non-negative effort level. For player i (i = 1, 2, . . . , n) this effort is xi. The probability pi that player

i wins the contest is

pi = ( xs i Pn j=1xsj if Pn j=1xj > 0, 1 n otherwise, (1)

where s > 0. For our models we use s = 1, a setting often used in the literature. A Tullock contest with this setting is also referred to as the ‘lottery contest’.

2.2 Asymmetric valuation

Analogous to Stein (2002), we now assume for the standard Tullock contest as defined in section 2.1 that s = 1 and that player i (i = 1, 2, . . . , n) values the prize at Vi > 0.

Without loss of generality, we can assume that V1 ≥ V2 ≥ . . . ≥ Vn > 0. The expected

payoff πi of player i reads

πi = piVi− xi. (2)

It can be verified that (2) is concave for each xi. As player i wants to maximise his

expected payoff, we derive the first-order condition to find this maximum:

Vi    Pn j=1xj − xi Pn j=1xj 2   − 1 ≤ 0, (3)

which holds with equality if xi > 0. Let r be the number of active players. So for

i = 1, 2, . . . , r, we can rewrite (3) into

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from which it follows that r X j=1 xj = r n X j=1 xj−   n X j=1 xj   2 r X j=1 1 Vj (5) ⇔ n X j=1 xj = r − 1 Pr j=1V1j . (6)

We plug (6) into (4) to obtain the unique equilibrium effort ˆxi of player i = 1, 2, . . . , r:

ˆ xi= r − 1 Pr j=1V1j 1 − r − 1 ViPrj=1V1j ! . (7)

Note that for i = r + 1, r + 2, . . . , n we have that player i is not active, i.e. ˆxi= 0. From

(4) and (6) we can also derive the probability that in equilibrium, for i = 1, 2, . . . , r, player i will win the contest:

pi= ˆ xi Pn j=1xˆj (8) = Pn j=1xˆj −( Pn j=1xˆj) 2 Vi Pn j=1xˆj (9) = 1 − Pn j=1xˆj Vi (10) = 1 − r − 1 ViPrj=1V1j (11)

Note that, as V1 ≥ V2 ≥ . . . ≥ Vn, we have that p1 ≥ p2 ≥ . . . ≥ pn. This implies

that players with higher valuations will have make a larger effort and will have a larger probability of winning the prize than players with a lower valuation. Stein (2002) shows that r = max i ≥ 2 : Vi > i − 2 Pi−1 j=1 V1j (12)

and that FOC (3) also holds for i = r + 1, r + 2, . . . , n.

2.3 Group contests

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one group. The non-refundable non-negative effort of player j (j = 1, 2, . . . , ni) of group

i is denoted xij. The aggregated effort of group i is xi = Pnj=1i xij. We again use the

standard Tullock success function of (1), with s = 1. The expected payoff πij of player

j from group i now is

πij = pi V ni − x_ij (13) = xi niPmj=1xj − xij. (14)

It can be verified that (13) is concave for each xij. Nitzan (1991) shows that every group

will make a positive effort: xi > 0 for i = 1, 2, . . . , m. This implies, as players in a

group are identical in this model, that every player makes a positive effort: xij > 0 for

i = 1, 2, . . . , m and j = 1, 2, . . . , ni. As player j from group i wants to maximise his

expected payoff, we derive the first-order condition to find this maximum:

V Pm j=1xj− xi ni Pm j=1xj 2 − 1 = 0. (15)

We can rewrite (15) to obtain

xi = m X j=1 xj  1 −ni V m X j=1 xj  . (16)

By summing (16) over i = 1, 2, . . . , m, we get

m X j=1 xj = m X j=1 xj  m − m X j=1 nj Pm j=1xj V   (17) ⇔ m X j=1 xj = V (m − 1) Pm j=1nj . (18)

We plug (18) into (16) to obtain the unique equilibrium effort ˆxi of group i = 1, 2, . . . , m:

xi= V (m − 1) Pm j=1nj 1 −nPi(m − 1)m j=1nj ! . (19)

From (19) we can derive that

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From this, Nitzan (1991) shows that if n1 ≤ n2 ≤ . . . ≤ nm, we have that p1 ≥ p2 ≥

. . . ≥ pn> 0 and that x1 ≥ x2≥ . . . ≥ xm> 0. So groups with few members will make

a larger effort and have a larger probability of winning the prize than groups with more members.

2.4 Private information

Hurley and Shogren (1998a,b) and Schoonbeek and Winkel (2006) consider one-sided private information models for individual players. Let there be m players. The value of the prize for player 1 can be VH (high) and VL (low), where VH > VL > 0. Before

the start of the contest the valuation of the prize by player 1 is determined, where VH

has probability p and VL probability (1 − p). We assume that 0 < p < 1. The other

players know the distribution of the valuation of player 1, not the realisation. These other players value the prize at V , which is also known by player 1.

Schoonbeek and Winkel (2006) show that a unique symmetric (pure strategy Nash) equilibrium exists for this model. They show that players that do not possess private information will always play a positive effort. They derive necessary and sufficient con-ditions for player 1 playing a positive of zero effort. We recall here their conclusion:

1. Player 1 will play a positive effort both if he has the high valuation and has the low valuation ⇔ VL V > p q VL VH − VL (m−1)VH + (1−p)(m−2)_(m−1) .

2. Player 1 with the high valuation will play a positive effort, while player 1 with the low valuation will play a zero effort ⇔ (m−2)VL

(m−1)VH < VL V ≤ p q VL VH − VL (m−1)VH + (1−p)(m−2) (m−1) .

3. Player 1 will play a zero effort both if he has the high valuation and has the low valuation ⇔ VL

V ≤

(m−2)VL

(m−1)VH.

3 The two-group model

3.1 The model

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player takes part in one group and all players in a group have the same valuation for the prize. The value of the prize for players in group 1 can be VH (high) and VL (low),

where VH > VL> 0. Before the start of the contest the valuation of the prize by group

1 is determined, where VH has probability p and VL probability (1 − p). We assume that

0 < p < 1. Players from group 2 know the distribution of the valuation of group 1, not the realisation. Group 2 values the prize at V , which is also known by players in group 1.

Let t (t = H, L) be the realised type of group 1, let x1jt be the non-negative and

non-refundable effort of player j (j = 1, . . . , n1) from group 1 of type t and let xij

be the non-negative and non-refundable effort of player j (j = 1, . . . , n2) from group

2. Furthermore, let the aggregate effort for group 1 be x1t = Pnj=11 x1jt and let the

aggregate effort for group 2 be x2 =Pn_j=12 x2j. To describe the probability that group i

wins the contest we use the Tullock success function

qi = ( x1t x1t+x2 for i = 1, x2 x1t+x2 for i = 2. (21)

If x1t+ x2= 0, the probability that group i (i = 1, 2) wins is 1/2.

Players of both groups choose their efforts to maximise their expected payoff. If a group wins, all its players will receive an equal share of the prize. The expected payoff π1jt for player j (j = 1, . . . , n1) of group 1 is

π1jt=

x1tVt

(x1t+ x2)n1

− x1jt, (22)

while for player j (j = 1, . . . , n2) of group 2 the expected payoff π2j is

π2j = p x2V (x1H+ x2)n2 + (1 − p) x2V (x1L+ x2)n2 − x_2j. (23) 3.2 The solution

We now investigate symmetric (pure strategy Nash) equilibria of this contest, which we will denote (ˆx1H, ˆx1L, ˆx2). That we restrict our search to symmetric equilibria implies

that all players that are in the same group will choose the same effort level. Note that x1jt and x2j are non-negative and that payoff functions (22) and (23) are concave in

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first-order conditions (FOCs): ˆ x2VH (ˆx1H+ ˆx2)2n1 ≤ 1 (24) ˆ x2VL (ˆx1L+ ˆx2)2n1 ≤ 1 (25) p xˆ1HV (ˆx1H+ ˆx2)2n2 + (1 − p) xˆ1LV (ˆx1L+ ˆx2)2n2 ≤ 1. (26)

Note that (24), (25) and (26) hold with equality if ˆx1H > 0, ˆx1L > 0 and ˆx2 > 0,

respectively.

We split the analysis of the solution into eight distinct parts based on the three decision variables taking a zero or positive value. See table 1 for the analysis, in which we directly show that six of them do not yield an equilibrium.

Case xˆ1H xˆ1L xˆ2 Can this yield an equilibrium?

1 0 0 0 No, for a player in group 2 playing an infinitesimal yields expected payoff of _nV

2 − which is larger than

V

2n2, the payoff of playing 0. Similar argument applies

for players of group 1. 2 0 0 > 0 No, (26) fails.

3 0 > 0 0 No, (25) fails. 4 > 0 0 0 No, (24) fails.

5 0 > 0 > 0 No, it implies that (25) holds with equality. While VH > VL and x1H = 0, (24) fails.

6 > 0 0 > 0 Yes.

7 > 0 > 0 0 No, (24) and (25) fail. 8 > 0 > 0 > 0 Yes.

Table 1: Analysis of potential equilibria in the two-group model

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Case 6. In this case ˆx1H > 0, ˆx1L = 0 and ˆx2 > 0. Let us define qH = ˆx1H/ˆx2. The

first-order conditions (24) to (26) now become

VH (qH+ 1)2xˆ2n1 = 1 (27) VL ˆ x2n1 ≤ 1 (28) p qHV (qH+ 1)2xˆ2n2 = 1. (29)

From (27) and (29) we can conclude that the unique solution is characterised by

qH =

VHn2

pV n1

. (30)

With this, ˆx1H and ˆx2 are also uniquely defined, see (27).

To find the condition under which the players will end up in this case, note that from (27) and (28) it follows that qH+ 1 ≤pVH/VL. Plugging this in (29) yields

µ ≤ p( √ λ − λ)n1 n2 , (31) where µ = VL/V and λ = VL/VH.

Case 8. In this case ˆx1H > 0, ˆx1L > 0 and ˆx2 > 0. Next to the already defined qH

(qH = ˆx1H/ˆx2), we also define qL= ˆx1L/ˆx2. The first-order conditions (24) to (26) now

become VH (qH + 1)2xˆ2n1 = 1 (32) VL (qL+ 1)2xˆ2n1 = 1 (33) p qHV (qH + 1)2xˆ2n2 + (1 − p) qLV (qL+ 1)2xˆ2n2 = 1. (34)

Combining (32) and (33) yields

VH

(qH + 1)2

= VL

(qL+ 1)2

, (35)

which can be rewritten as

qH = (qL+ 1)

p

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Plugging (32), (33) and (36) in (34) yields qL= n2 n1V − p √ VHVL + p VH p √ VHVL + (1−p) VL . (37)

See appendix A for the algebraic details. This qL is unique and therefore qH, ˆx1H, ˆx1L

and ˆx2 are also uniquely defined by (36) and (33).

To find the condition under which the players will end up in this case, note that we must have that qL > 0. As the denominator of (37) is always positive, we must have

that n2 n1V −√ p VHVL + p VH > 0. (38) This is equivalent to µ > p( √ λ − λ)n1 n2 , (39)

where µ = VL/V and λ = VL/VH again.

3.3 Interpretation of the solution

As shown above, case 6 will yield the unique equilibrium if (31) holds and the equilibrium will fall into case 8 if (39) holds. We can rewrite (31) as

n2≤ V pn1( 1 √ VHVL − 1 VH ). (40)

Let us assume group 1 turns out to have the low valuation. We can deduct from (40) the factors that will influence them not to make a positive effort. Each factor follows with our interpretation:

• n₁ is large. The reward for a player of group 1 is small as they have to share the prize with a lot of players. Those players will therefore not make a (big) effort. The effort per player decreases faster than the number of players in group 1 increases. Therefore, the total effort of group 1 is lower for a large n1 as well.

• n2is small. Players from group 2 will have a large reward. They are therefore

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• V is large. Players from group 2 will have a large reward. They are therefore willing to increase their effort. This increase in effort is unattractive for players from group 1.

• VL is small. As their valuation of the prize is low, it is unattractive for players

from group 1 to make a (big) effort.

• p is large. Players from group 2 must always play a positive effort, so they will make a more aggressive effort as group 2 has to deal more often with a high-valued group 1. This is unattractive for players of group 1 if they have the low valuation.

• The impact of a change in VH depends on its value. Taking the derivative of the

right hand side of (40) with respect to VH we obtain V pn1 (VH) 3 2 1 √ VH − 1 2√VL . From this we can conclude that if VH < 4VL, an increase in VH will make group 1 less

likely to make a positive effort, while for VH > 4VL, an increase in VH will make

group 1 more likely to play a positive effort.

To interpret this, assume that the equilibrium falls in case 6. We can then derive from (24) and (26) that if qH > 1, an increase in VH results in an increase in ˆx1H

and a decrease in ˆx1L. If qH < 1, an increase in VH results in an increase in both

ˆ

x1H and ˆx1L. Remember that group 2 will play a positive effort in all situations,

but notice that group 2 is only willing to react aggressively (i.e. increase its effort) if it is the dominant party. We now plug qH = 1 into the condition under which

players will end up in case 6, which is qH+1 ≤ V_VH_L. From this we obtain the turning

point VH = 4VL for the boundary situation of the discriminating condition.

This shows us that if VH increases when VH < 4VL, group 2 increases its effort,

which is unattractive for group 1 with a low valuation and which makes it less inclined to play a positive effort. On the other hand, if VH increases when VH >

4VL, the effort of group 2 will decrease, which is attractive for group 1 and makes

it more willing to play a positive effort if it has the low valuation.

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Similar to Hurley and Shogren (1998a), note that the relative resolve ρ2t of group 2

playing against type t group 1 is

ρ2t=

r V Vt

. (41)

The expected relative resolve of group 2, E(ρ2) is

E(ρ2) = p r V VH + (1 − p)r V VL (42)

and the variance of the expected relative resolve of group 2, Var(ρ2), is

Var(ρ2) = E(ρ22) − E(ρ2)2= p(1 − p)V (

1 VH + 1 VL ) − 2p(1 − p)√ V VLVH . (43)

We derive the following equivalent expression for qL> 0:

qL> 0 (44) ⇔ n2 n1V − p √ VHVL + p VH p √ VHVL+ (1−p) VL > 0 (45) ⇔ V√µ n2 n1V − p √ VHVL+ p VH p q V VH + (1 − p) q V VL > 0 (46) ⇔ √ µn2 n1 − p V √ VHVL + p V VH E(ρ2) > 0 (47) ⇔ √ µn2 n1 + Var(ρ2) + E(ρ2) 2_{− (1 − p)}V VL − p V √ VHVL E(ρ2) > 0 (48) ⇔ √ µn2 n1 + Var(ρ2) + E(ρ2) 2_{− (1 − p)}qV VL − p q V VH E(ρ2) > 0 (49) ⇔ √ µn2 n1 + Var(ρ2) + E(ρ2) 2 E(ρ2) > 1 (50) ⇔ µ > _n E(ρ2) 2 n1 + Var(ρ2) + E(ρ2) 2 !2 . (51)

From (51) we can conclude that if the second moment of ρ2 increases, qL is more easily

positive. Thus, an increase in uncertainty in group 2 about the value group 1 attaches to the prize will make group 1 more inclined to play a positive effort.

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4 The m-group model

4.1 The model

Let us generalise the two-group model to a situation in which there are m groups. Group i (i = 1, 2, . . . , m) has ni (ni ≥ 1) players and every player is in exactly one group. All

players in a group have the same valuation for the prize. For group 1 this valuation can be VH (high) with probability p and VL (low) with probability 1 − p, where 0 < p < 1.

It is given that VH > VL> 0. Before the start of the contest the players of group 1 will

know the realisation of the valuation, while the players of the other groups only know the distribution. Players from group 2 to m value the prize at V , which is also known by players in group 1.

Let t (t = H, L) be the realised type of group 1, let x1jtbe the negative and

non-refundable effort of player j (j = 1, . . . , n1) from group 1 of type t and let xij be the

non-negative and non-refundable effort of player j (j = 1, . . . , ni) from group i = 2, 3, . . . , m.

Furthermore, let the aggregate effort for group 1 of type t be x1t=Pn_j=11 x1jtand let the

aggregate effort for group i = 2, 3, . . . , m be xi =Pnj=1i xij. To describe the probability

that group i wins the contest we use the Tullock success function

qi= ( _x_1t x1t+Pmk=2xk for i = 1, xi x1t+Pmk=2xk for i = 2, 3, . . . , m. (52)

If x1t+Pm_k=2xk= 0, the probability that group i (i = 1, 2, . . . , m) wins is 1/m.

Players of all groups choose their efforts to maximise their expected payoff. If a group wins, all its players will receive an equal share of the prize. The expected payoff of player j (j = 1, . . . , n1) from group 1, π1jt is

π1jt=

x1tVt

(x1t+Pmk=2xk)n1

− x_1jt, (53)

while the expected payoff πij for player j (j = 1, . . . , ni) from group i = 2, 3, . . . , m is

πij = p xiV (x1H+Pm_k=2xk)ni + (1 − p) xiV (x1L+Pm_k=2xk)ni − xij. (54) 4.2 The solution

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that x1jt and xij are non-negative and that payoff functions (53) and (54) are concave

in x1jt and xij, respectively. We derive the following first-order conditions (FOCs):

ˆ xVH (ˆx1H+ ˆx)2n1 ≤ 1 (55) ˆ xVL (ˆx1L+ ˆx)2n1 ≤ 1 (56) p(ˆx1H+ ˆx − ˆxi)V (ˆx1H+ ˆx)2ni + (1 − p)(ˆx1L+ ˆx − ˆxi)V (ˆx1L+ ˆx)2ni ≤ 1 if i > 1 (57)

Note that (55), (56) and (57) hold with equality if ˆx1H > 0, ˆx1L > 0 and ˆxi > 0,

respectively, where i > 1.

We again split the analysis of the solution into eight distinct cases based on the decision variables being zero or taking a positive value. See table 2 for the analysis, in which we directly show that five of the cases do not yield an equilibrium.

Case xˆ1H xˆ1L xˆ Can this yield an equilibrium?

1 0 0 0 No, for a player in group i > 1 playing an infin-itesimal yields expected payoff of _nV

i− which is

larger than _mnV

i, the payoff of playing 0. Similar

argument applies for players of group 1.

2 0 0 > 0 Yes.

3 0 > 0 0 No, it implies that (56) holds with equality. While VH > VL and ˆx1H = 0, (55) fails.

4 > 0 0 0 No, (55) fails.

5 0 > 0 > 0 No, it implies that (56) holds with equality. While VH > VL and ˆx1H = 0, (55) fails.

6 > 0 0 > 0 Yes.

7 > 0 > 0 0 No, (55) and (56) fail. 8 > 0 > 0 > 0 Yes.

Table 2: Analysis of potential equilibria in the m-group model

We will consider the three cases that are left one by one. We will show that there is a unique equilibrium (ˆx1H, ˆx1L, ˆx2, . . . , ˆxm) in all three cases. Afterwards we will show

under which conditions the players will end up in each case.

We assume without loss of generality that groups 2 to m are ordered by group size, where a group with a larger group size will have a higher group number, i.e. n2 ≤ n3≤

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for a, b ∈ {2, 3, . . . , m} we have that

a ≥ b ⇔ na≥ nb ⇔ ˆxa ≤ ˆxb (58)

and

a > b ⇐ na> nb ⇔ ˆxa< ˆxb. (59)

Furthermore, (57) implies that groups with equal size will all make the same effort. From the viewpoint of symmetry this makes sense.

Let r ∈ {1, 2, . . . , m − 1} be such that (i) ˆxr+1 > 0 and (ii) ˆxr+2 = 0 or r + 1 = m.

Note that if r = m − 1, the ˆxr+2 = 0 condition does not exist. The variable r represents

the number of groups that play a positive effort, excluding group 1, the group that possesses private information.

For example, suppose that there are four groups (m = 4) of which group 1, 2 and 3 play a positive effort (ˆx1H > 0, ˆx1L > 0, ˆx2 > 0 and ˆx3 > 0), while group 4 does

not (ˆx4 = 0). In this case, r = 2. By construction of our model, group 1 is the group

that possesses private information, group 2 has fewer or as many members as group 3 (n2 ≤ n3) and group 3 has fewer or as many members as group 4 (n3 ≤ n4). Furthermore,

the total effort of group 2 is larger than or equal to that of group 3 (ˆx2 ≥ ˆx3), while the

total effort of group 3 must be larger than that of group 4 (ˆx3 > ˆx4).

We will now treat the three cases with a possible equilibrium one by one. Let the subscript [c] denote case c, where c ∈ {2, 6, 8}.

Case 2 We know that ˆx[2]1H = 0, ˆx[2]1L = 0 and ˆx[2] > 0. Let i ∈ {2, 3, . . . , m}. The

first-order conditions now become

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From (62) we can deduct that for i ≤ r[2]+ 1 ˆ x[2]i= ˆx[2]− ˆ x2_[2]ni V . (64)

Now we can obtain an expression for ˆx_[2]:

ˆ x[2] = r[2]+1 X j=2 ˆ x[2]j = r_[2]− 1 Pr[2]+1 j=2 nj V. (65)

As defined, r[2] is such that (i) ˆx[2]r[2]+1 > 0 and (ii) ˆx[2]r[2]+2 = 0 or r[2] + 1 = m. By

(64) and (65) this is equivalent to

r[2]= max i : 1 −

i − 1 Pi+1

j=2nj

ni+1> 0. (66)

Note that if (60) to (63) hold, there is a unique equilibrum (ˆx1H, ˆx1L, ˆx2, . . . , ˆxm),

where ˆx1H = ˆx1L = 0 and ˆx2 to ˆxm can be calculated by (64).

Case 6 We know that ˆx[6]1H > 0, ˆx[6]1L = 0 and ˆx[6] > 0. Again, let i ∈ {2, 3, . . . , m}.

The first-order conditions now become ˆ x_[6]VH (ˆx_[6]1H + ˆx_[6])2_n 1 = 1 (67) VL ˆ x_[6]n1 ≤ 1 (68) p(ˆx[6]1H + ˆx[6]− ˆx[6]i)V (ˆx[6]1H+ ˆx[6])2ni + (1 − p)(ˆx[6]− ˆx[6]i)V (ˆx[6])2ni = 1, if i ≤ r_[6]+ 1 (69) p V (ˆx_[6]1H + ˆx_[6])ni + (1 − p) V ˆ x_[6]ni ≤ 1, if i > r_[6]+ 1. (70)

From (67) and (69) we can deduct that for i ≤ r[6]+ 1

ˆ x_[6]i= pp ˆx[6]VHn1+ (1 − p)VH− VH V xˆ[6]ni pn1+ (1 − p)_xV_ˆH [6] . (71)

See appendix A for the algebraic details. As ˆx_[6] = Pr[6]+1

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Solving (72) for ˆx[6] is equivalent to solving the following quadratic equation:  pn1+ VH V r[6]+1 X j=2 nj  y2− r[6]p p VHn1 y − (1 − p)VH(r[6]− 1) = 0, (73)

where y =p ˆx[6] > 0. By applying the quadratic formula we get

ˆ x[6] =   r[6]p √ VHn1+ q r_[6]2 p2_V Hn1+ 4(pn1+V_VH Pr[6]+1 j=2 nj)(1 − p)VH(r[6]− 1) 2(pn1+ V_VH P r[6]+1 j=2 nj)   2 . (74) As defined, r_[6] is such that (i) ˆx_[6]r_[6]₊₁ > 0 and (ii) ˆx_[6]r_[6]₊₂ = 0 or r_[6]+ 1 = m. Thus by (71) and (74) r_[6] = max i : ppVHn1   ip√VHn1+ q i2_p2_V Hn1+ 4(pn1+V_VH Pi+1_j=2nj)(1 − p)VH(i − 1) 2(pn1+V_VH Pi+1_j=2nj)   + (1 − p)VH −VH V   ip√VHn1+ q i2_p2_V Hn1+ 4(pn1+V_VH Pi+1 j=2nj)(1 − p)VH(i − 1) 2(pn1+V_VH Pi+1_j=2nj)   2 ni+1> 0. (75) See appendix A for the algebraic details.

where ˆx1H is uniquely defined by solving (67), ˆx1L = 0 and ˆx2 to ˆxm can be calculated

by (71).

Case 8 We know that ˆx_[8]1H > 0, ˆx_[8]1L > 0 and ˆx_[8] > 0. Again, let i ∈ {2, 3, . . . , m}. The first-order conditions now become

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From (76), (77) and (78) we can deduct that for i ≤ r[8]+ 1 ˆ x_[8]i= p q _x_ˆ [8] VHn1 + (1 − p) q _x_ˆ [8] VLn1 − niˆx[8] n1V p VH + (1−p) VL . (80)

See appendix A for the algebraic details. As ˆx_[8] = Pr[8]+1

j=2 xˆ[8]j, we can use (80) to obtain ˆ x[8] = r[8]p q ˆ x[8] VHn1 + r[8](1 − p) q ˆ x[8] VLn1 − Pr[8]+1 j=2 njˆx[8] n1V p VH + (1−p) VL . (81)

Solving (81) for ˆx_[8] is equivalent to solving the following quadratic equation:

  p VH +1 − p VL + Pr[8]+1 j=2 nj n1V  y2− _r [8]p √ VHn1 +r[8]√(1 − p) VLn1 y = 0, (82)

where y =p ˆx_[8] > 0. It follows easily that

ˆ x[8] = _r [8]p √ VHn1 + r_√[8](1−p) VLn1 2 p VH + 1−p VL + Pr[8]+1 j=2 nj n1V !2. (83)

As defined, r[8] is such that (i) ˆx[8]r[8]+1 > 0 and (ii) ˆx[8]r[8]+2 = 0 or r[8]+ 1 = m.

Thus by (80) and (83) r_[8] = max i : Pi+1 j=2nj i + V n1 i p VH + 1 − p VL − ni+1> 0. (84)

See appendix A for the algebraic details.

where ˆx1H and ˆx1L are uniquely defined by solving (76) and (77), respectively. The

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4.2.1 Discriminating between the three possible cases

For this section and corresponding proofs, we introduce the following notation for i ∈ {1, . . . , m − 1}: ˆ x[2](i) = i − 1 Pi+1 j=2nj V, (85) ˆ x_[6](i) =   ip√VHn1+ q i2_p2_V Hn1+ 4(pn1+V_VH Pi+1 j=2nj)(1 − p)VH(i − 1) 2(pn1+V_VH Pi+1_j=2nj)   2 , (86) ˆ x_[8](i) = ip √ VHn1 + i(1−p) √ VLn1 2 p VH + 1−p VL + Pi+1 j=2nj n1V 2. (87)

Note that ˆx[2] = ˆx[2](r[2]), ˆx[6] = ˆx[6](r[6]) and ˆx[8]= ˆx[8](r[8]).

Note that for all three cases we have that, when their associated FOCs hold, a single equilibrium (ˆx1H, ˆx1L, ˆx2, . . . , ˆxm) is defined. If for a case not all four FOCs hold, the

case cannot yield an equilibrium. The third and fourth FOC, (62) and (63) for case 2, (69) and (70) for case 6, and (78) and (79) for case 8, always hold, as they are implied by the selection of r_[2] in (66), r_[6] in (75) and r_[8] in (84), respectively. See lemma 9 for a proof of this for the fourth FOC in all three cases.

Lemma 1. Considering the two FOCs of each case that do not hold for all para-meter values, (60), (61), (67), (68), (76) and (77), we know that a game, defined by (m, VH, VL, V, p, ni), i ∈ {1, 2, . . . , m}, will have an equilibrium in

• case 2 ⇔ ˆx_[2](r_[2]) ≥ VH/n1,

• case 6 ⇔ V_H/n1> ˆx[6](r[6]) ≥ VL/n1,

• case 8 ⇔ V_L/n1 > ˆx[8](r[8]).

Before we prove that for every game exactly one equilibrium exists, we provide twelve auxiliary lemmas. Proof of all lemmas can be found in appendix B.

Lemma 2. The function ˆx[2](r[2]) represents the aggregated effort of groups 2 to m for

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ˆ

x[2](r[2]), but for ˆx[2](i):

ˆ x[2](i) ≥ VH n1 ⇔ i − 1 VH ≥ Pi+1 j=2nj V n1 . (88)

case 6. Let i ∈ {1, . . . , m − 1}. We can prove the following equivalency, not only for ˆ

x[6](r[6]), but for ˆx[6](i):

ˆ x[6](i) < VH n1 ⇔ i − 1 VH < Pi+1 j=2nj V n1 . (89)

x[6](r[6]), but for ˆx[6](i):

ˆ x[6](i) ≥ VL n1 ⇔ (i − 1) p √ VHVL +1 − p VL ≥ p VH + Pi+1 j=2nj V n1 −√ p VHVL . (90)

x[8](r[8]), but for ˆx[6](i):

ˆ x[8](i) < VL n1 ⇔ (i − 1) p √ VHVL +1 − p VL < p VH + Pi+1 j=2nj n1V −√ p VHVL . (91)

Lemma 6. The number of active groups (of groups 2 to m) in case 2, represented by r_[2], is selected in such a way that it maximises ˆx_[2](i):

ˆ

x_[2](i) ≤ ˆx_[2](i + 1), for i ∈ {1, . . . , r_[2]− 1} (92)

ˆ

x_[2](i) ≥ ˆx_[2](i + 1), for i ∈ {r_[2], . . . , m − 2}. (93)

Lemma 7. The number of active groups (of groups 2 to m) in case 6, represented by r_[6], is selected in such a way that it maximises ˆx_[6](i):

ˆ

x_[6](i) ≤ ˆx_[6](i + 1), for i ∈ {1, . . . , r_[6]− 1} (94)

ˆ

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Lemma 8. The number of active groups (of groups 2 to m) in case 8, represented by r[8], is selected in such a way that it maximises ˆx[8](i):

ˆ

x_[8](i) ≤ ˆx_[8](i + 1), for i ∈ {1, . . . , r_[8]− 1} (96)

ˆ

x_[8](i) ≥ ˆx_[8](i + 1), for i ∈ {r_[8], . . . , m − 2}. (97)

Lemma 9. The selection of r[2], r[6] and r[8] in (66), (75) and (84), respectively, implies

that the fourth FOC for each case, (63), (70) and (79), respectively, holds.

Lemma 10. There cannot be an equilibrium in both case 2 and case 6. In mathematics,

ˆ

x[2](r[2]) ≥ VH/n1 ⇒ ˆx[6](r[6]) ≥ VH/n1. (98)

ˆ

x_[2](r_[2]) ≥ VH/n1⇒ ˆx[8](r[8]) ≥ VL/n1. (99)

ˆ

x[8](r[8]) < VL/n1 ⇒ ˆx[6](r[6]) < VL/n1. (100)

Lemma 13. There must be an equilibrium in at least one case. In mathematics,

ˆ

x[6](r[6]) < VL/n1 ⇒ ˆx[8](r[8]) < VL/n1, (101)

ˆ

x[6](r[6]) ≥ VH/n1 ⇒ ˆx[2](r[2]) ≥ VH/n1. (102)

Our central proposition directly follows from lemmas 10 to 13:

Proposition 1. Every game, defined by (m, VH, VL, V, p, ni), i ∈ {1, 2, . . . , m}, has

ex-actly one equilibrium that can be expressed as (ˆx1H, ˆx1L, ˆx2, . . . , ˆxm).

4.3 Interpretation of the solution

Factors influencing group 1 to play Let us assume that group 1 turns out to have the high valuation. Group 1 will play a zero effort if the unique equilibrium falls in case 2, so by lemma 1 when ˆx[2](r[2]) ≥ V_nH₁. The group will play a positive effort if

the equilibrium falls in cases 6 or 8, i.e. ˆx_[2](r_[2]) < VH

n1. We can therefore deduct from

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• VH is small. As their valuation of the prize is low, it is unattractive for players

• V is large. Players from other groups will have a large reward. They are therefore willing to increase their effort. This increase in effort is unattractive for players from group 1.

• n1 is large. The reward for a player of group 1 is small as he has to share the prize

with a lot of players. Those players will therefore not make a (big) effort. The effort per player decreases faster than the number of players in group 1 increases. Therefore, the total effort of group 1 is lower for a large n1 as well.

Note that the decision of other groups whether to play a positive effort is independent of the parameters n1, V and VH, as can be seen in (66). That decision is dependent on the

the group sizes n2, n3, . . . , nm, which are discrete parameters. This makes it impossible

to determine, by means of differential calculus, the impact of the group sizes on the decision of group 1 with a high valuation to be either active or inactive.

If group 1 turns out to have the low valuation, group 1 will play a zero effort if the unique equilibrium falls in cases 2 or 6. By lemma 1 and proposition 1, we have that case 2 or 6 is active when ˆx[8](r[8]) ≥ VnL1. The group will play a positive effort if the

equilibrium falls in case 8, i.e. ˆx_[8](r_[8]) < VL

n1. We can therefore deduct the factors that

will influence members of group 1 not to make a positive effort from the negation of lemma 5. Note that it implies

VL/n1≤ ˆx[8](r[8]) ⇔ r[8] p √ VHVL +1 − p VL ≥ p VH + Pr[8]+1 j=2 nj V n1 +1 − p VL . (103)

The decision of other groups whether to play a positive effort depends on the discrete group size parameters n1, n2, n3, . . . , nmand on the continuous parameters VH, VL, V and

p, as can be seen in (84). It is therefore impossible to determine, by means of differential calculus, the impact of the group sizes on the decision of group 1 with a low valuation to be either active or inactive. As (84) is a strict inequality and while the function that is maximised is continuous in the four continuous parameters, we have that a sufficiently small change in those parameters does not change r[8]. We can therefore deduct from

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influence the members of group 1 not to make a positive effort. Each factor follows with our interpretation:

• V_H decreases, for r[8] ≥ 2. This can be seen by bringing all terms in (103) to the

left hand side and taking the derivative with respect to VH. This derivative is p (VH) 3 2 1 √ VH − r[8] 2√VL

, which is negative for r[8] ≥ 2. Players from groups 2 to m

will make a more aggressive effort as group 1 on average has a lower valuation. This is unattractive for players from group 1. See section 3.3 for the analysis with r_[8] = 1. This is a bit different, as in that situation all groups without private information (i.e. group 2) must play a positive effort at all times.

• VL decreases. As their valuation of the prize is lower, it is unattractive for players

• V increases. Players from other groups will have a larger reward. They are there-fore willing to increase their effort. This increase in effort is unattractive for players from group 1.

• p decreases, for r_[8] ≥ 2. This can be seen by bringing all terms in (103) to the left hand side and taking the derivative with respect to p. This derivative is

r_[8] √ VHVL − 1 VH −r[8]− 1 VL = (r_[8]− 1) 1 √ VHVL − 1 VL +√ 1 VHVL − 1 VH (104) ≤ 2 − q VH VL − q VL VH √ VHVL (105) = √ VH− √ VL √ VH − √ VH− √ VL √ VL √ VHVL (106) < 0, (107)

where the first inequality only holds for r[8] ≥ 2. Players from groups 2 to m will

make a more aggressive effort as group 1 on average has a lower valuation. This is unattractive for players from group 1. See section 3.3 for the analysis with r[8] = 1.

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Effort of other groups increases when group 1 is less likely to play As shown in lemma 1, when the equilibrium falls in case 2 and thus group 1 does not play a positive effort, the aggregated equilibrium effort of the other groups is larger than in case 6, when there is a chance that group 1 plays, namely if it has the high valuation. The aggregated equilibrium effort of groups 2 to m in case 6 is in turn larger than the aggregated effort of those groups in case 8, when it is certain that group 1 plays, both with a high and a low valuation. This makes sense, as groups 2 to m have a larger chance of obtaining the prize when group 1 does not compete.

Larger groups make smaller efforts By (59) we know that for groups 2 to m, a larger group makes a smaller effort. This implies that for larger groups the decrease in expected return per person outweighs the decrease in the effort per person. Of all groups excluding group 1, group 2 has the largest probability of winning the prize.

If groups 2 to m have equal group size, all will play In all three possible cases of the m-group model, the aggregated effort of groups 2 to m is positive, so by (58) at least group 2 makes a positive effort. As shown in section 4.2, groups with the same size will all make the same effort. Combined, this implies that when groups 2 to m have equal group size, all groups will make a positive effort.

Connecting the m-group model to the two-group model The findings for the m-group contest with m = 2 should also hold for the two-group model. In all three relevant cases of the m-group model, the aggregated effort of groups 2 to m is positive. This implies, by (58), that always at least group 2 makes a positive effort. We now assume that m = 2. As we have just shown that at least group 2 makes a positive effort and as there are just two groups, we have that r_[2] = r_[6]= r_[8] = 1.

From lemma 2 we can deduct that case 2 will not yield an equilibrium, as i = 1 in (88), which leads to a contradiction if case 2 would yield an equilibrium.

We have an equilibrium in case 6 when ˆx[6](r[6]) ≥ VnL1. By (74) we know that this is

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⇔ ppVHVL≥ VLp + VLVHn2 V n1 (109) ⇔ p( √ VHVL− VL)n1 VHn2 ≥ VL V , (110)

which is equal to condition (31). Furthermore, note that

ˆ x[6]1H ˆ x[6](r[6]) = q ˆ x[6](r[6])VH n1 − ˆx[6](r[6]) ˆ x[6](r[6]) (111) = s VH n1xˆ[6](r[6]) − 1 (112) =r VH n1 (pn1+V_VHn2) p√VHn1 − 1 (113) = 1 +VHn2 pV n1 − 1 (114) = VHn2 pV n1 , (115)

where the first equality is due to (67) and the third due to (74). Note that (115) is equal to (30).

We have an equilibrium in case 8 when ˆx[8](r[8]) < V_nL₁. By lemma 5 this is for m = 2

equivalent to p VH + n2 n1V −√ p VHVL > 0 (116) ⇔ n2 n1V > √ p VHVL − p VH + (117) ⇔ VL V > p r VL VH − VL VH ! n1 n2 , (118)

which is equal to (39). Furthermore, note that

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= p VH + 1−p VL + n2 n1V − p √ VHVL −1−p_V L p √ VHVL+ 1−p VL (122) = p VH + n2 n1V − p √ VHVL p √ VHVL + 1−p VL , (123)

where the first equality is due to (77) and the second due to (83). Note that (123) is equal to (37). Finally, we see that

ˆ x_[8]1H ˆ x[8](r[8]) = q ˆ x[8](r[8])VH n1 − ˆx[8](r[8]) ˆ x[8](r[8]) (124) =r VH n1 p VH + 1−p VL + n2 n1V p √ VHn1 + 1−p √ VLn1 ! − 1 (125) =r VH VL _x_ˆ [8]1L ˆ x[8](r[8]) + 1 − 1, (126)

where the first equality is due to (76), the second due to (83) and the third due to (120). Note that (126) is equal to (36).

With this we have shown that, as should be expected, the equilibrium two-group model coincides with the equilibrium of the m-group model with m = 2.

Connecting the m-group model to the model of Schoonbeek and Winkel (2006) If we add the restriction to our model that every group has exactly one member (i.e. ni = 1, where i = 1, 2, . . . , m), we should have the same results as Schoonbeek and

Winkel (2006) found. We have shown earlier in this section that if groups 2 to m have equal group size, all will play. Note that we thus have in this situation that n1 = 1 and

r[2] = r[6] = r[8] = Pr[2]+1 j=2 nj = Pr[6]+1 j=2 nj = Pr[8]+1 j=2 nj = m − 1. If we now compare

lemmas 2, 3, 4 and 5 with the results from section 2.4, we see that the conditions under which group 1 plays a zero or positive effort are the same for both models. It could also be shown that the exact equilibrium of our model with one member per group coincides with the model of Schoonbeek and Winkel (2006).

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model with complete information, that we can compare to our m-group model with private information. Note that the assumptions for our m-group model with private information do not allow VH = VL. The reason for this is that case 6 cannot yield

an equilibrium in that setting. Let us assume for the moment that VH = VL. Then,

referring to table 2, note that for case 6 we have that (55) holds with equality. While VH = VL, ˆx1H > 0 and ˆx1L = 0, (56) fails. Note that the results for case 2 and 8 are

valid for the parameter setting VH = VL.

In the benchmark model, let ˜VL= ˜VH be the valuation of group 1 and let ˜r be the

number of groups besides group 1 that play a positive effort. If group 1 has the high valuation in the private information model, we have that ˜VL = ˜VH = VH. Analogously,

if group 1 has the low valuation in the private information model, we have that ˜VL =

˜

VH = VL. The other parameters in the benchmark model (V , p and n1, n2. . . , nm) are

equal to those in the private information model. Note that p is a superfluous parameter for the benchmark model as the equilibrium is independent of it. As shown above, the benchmark model can only yield an equilibrium in case 2 and case 8.

The following proposition is proved in appendix B:

Proposition 2. We compare the equilibrium effort of group 1 in the private information model with the benchmark model:

1. If group 1 has the high valuation, it plays a positive effort in the benchmark model if and only if it plays a positive effort in the private information model.

2. If group 1 has the low valuation and ˜r[8] = 1, group 1 will always be active in the

benchmark model, while it can be inactive in the model with private information.

3. If group 1 has the low valuation while ˜r_[8]≥ 2 and r_[8]= 1, group 1 can be active and inactive for both the private information model as the benchmark model, without one model implying activity or inactivity for the other.

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Proposition 2 implies that the decision of group 1 to be either active or inactive is independent of having private information if its valuation of the prize is high. When its valuation is low, having private information does change its decision. With one other active group (both in the model with and without private information), having private information gives group 1 the possibility to be inactive, something that is not possible without private information. If more than one other group is active (both in the model with and without private information), missing private information makes groups 2 to m less aggressive (i.e. they exert a smaller aggregated effort), which makes group 1 more often inclined to play a positive effort.

Note that proposition 2 does not state anything about the actual value of the equi-librium effort of group 1, only whether it is zero or positive.

4.4 A numerical example

Consider our m-group model with ten groups (m = 10) and the following group sizes:

n = 3 1 2 4 4 5 6 7 8 90. (127)

We let groups 2 to 10 value the prize at V = 1, while let group 1 have the following distribution of its valuation of the prize: with probability p = 0.3 it values the prize at VH = 6 and with probability 1 − p = 0.7 it values the prize at VL= 4.

With help of (66), (75) and (84) we can calculate the number of groups excluding group 1 that play a positive effort: r[2] = 2, r[6] = 2 and r[8] = 1. Furthermore, by

(65), (74) and (83) we can calculate the aggregated effort of those groups for each case: ˆ

x_[2] = 1₃, ˆx_[6] ≈ 0.295 and ˆx_[8] ≈ 0.239. Note that, as the aggregated efforts ˆx_[2], ˆx_[6] and ˆx[8] are all smaller than V_nL₁ = 4₃, we have that only case 8 can yield an equilibrium.

Thus by (76), ˆx1H ≈ 0.452 and by (77), ˆx1L ≈ 0.325. The equilibrium effort of group

2 can be derived from (80). This results in ˆx2 ≈ 0.239. As r[6] = 1, we have that

ˆ

x3 = ˆx4= . . . = ˆx10= 0.

If we now set V = 5, the equilibrium falls in case 6 with r[6] = 2. As the groups

that do not possess private information now have a higher valuation of the prize, their aggregated effort has increased: ˆx[6] ≈ 1.663. Finally, if we set V = 6, case 2 yields the

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5 Conclusion

We investigated a model for a Tullock group contest in which players compete in groups for a prize. One group possesses private information about its valuation of the prize while all other groups share a common publicly known valuation for it. We started out with two groups and later generalised the model to m groups. For both models we have shown that a unique symmetric (pure strategy Nash) equilibrium exists for all possible parameter configurations. We have shown under which conditions each group will or won’t play and what equilibrium effort each group (and each group member) will make. Our models yielded some interesting results. When players compete in groups, groups can choose to play a zero effort, although the smallest group that does not possess private information will play a positive effort in every situation. In Schoonbeek and Winkel (2006) it was shown that for players competing individually only the player with private information could play a zero effort. We derived in the two-group model, for the group with private information that has the low valuation, the influence of the parameters on its decision to be either active or inactive. Striking is the effect of the high valuation on this decision: ceteris paribus, the group with private information is the least inclined to play a positive effort if the high valuation is four times the low valuation. Of the groups that do not possess the private information, larger groups play a smaller effort, not only per person but also as a group. This finding is in line with Nitzan (1991). We have related our results for the two-group model to the notion of relative resolve, as introduced by Hurley and Shogren (1998a). With this, we have shown than an increase in uncertainty about the valuation of the group with private information will make that group more inclined to play a positive effort. This is in line with the findings of Schoonbeek and Winkel (2006). Furthermore, we compared the results of our m-group model with results on a benchmark model with complete information. Results of that comparison are quite similar to those in Schoonbeek and Winkel (2006).

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A

Algebraic details of some equations

Equation (37) We plug (32) and (33) in (34), giving

pqHV n1 VHn2

+ (1 − p)qLV n1 VLn2

= 1. (A.1)

Plugging (36) in (A.1), we get

1 = p q VH VL(qL+ 1) − 1 V n1 VHn2 + (1 − p)qLV n1 VLn2 = p q VH VLV n1 VHn2 qL+ p q VH VLV n1 VHn2 − p V n1 VHn2 + (1 − p)V n1 VLn2 qL. (A.2)

Rewriting this with qL as the left hand side results in

qL= 1 − p r VH VLV n1 VHn2 + p V n1 VHn2 p r VH VLV n1 VHn2 + (1 − p) V n1 VLn2 . = n2 n1V − p √ VHVL + p VH p √ VHVL+ (1−p) VL , (A.3) which is (37).

Equation (71) Plugging (67) in (69) yields

p( q ˆ x[6]VH n1 − ˆx[6]i)V ˆ x[6]VH n1 ni + (1 − p)(ˆx[6]− ˆx[6]i)V (ˆx[6])2ni = 1 (A.4)

Rewriting this with ˆx[6]i as the left hand side results in

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Equation (75) Note that ˆx[6]i> 0 if and only if the numerator of (71) is positive, as

the denominator is always positive. Thus, to find the largest i for which ˆx[6]i+1> 0, we

have to find

r_[6] = max i : pqxˆ_[6]VHn1+ (1 − p)VH −

VH

V xˆ[6]ni+1> 0. (A.6) Plugging (74) in (A.6) results in

r[6] = max i : ppVHn1   ip√VHn1+ q i2_p2_V Hn1+ 4(pn1+V_VH Pi+1j=2nj)(1 − p)VH(i − 1) 2(pn1+V_VH Pi+1j=2nj)   + (1 − p)VH −VH V   ip√VHn1+ q i2_p2_V Hn1+ 4(pn1+V_VH Pi+1j=2nj)(1 − p)VH(i − 1) 2(pn1+V_VH Pi+1j=2nj)   2 ni+1> 0, (A.7) which is (75).

Equation (80) Plugging (76) and (77) in (78) gives us

p( q ˆ x[8]VH n1 − ˆx[8]i)V ˆ x[8]VH n1 ni + (1 − p)( q ˆ x[8]VL n1 − ˆx[8]i)V ˆ x[8]VL n1 ni = 1. (A.8)

Rewriting this with ˆx[8]i as the left hand side results in

ˆ x[8]i= p r ˆ x[8]VH n1 V ˆ x[8]VH n1 ni + (1 − p) r ˆ x[8]VL n1 V ˆ x[8]VL n1 ni − 1 p_x[8]VHˆ V n1 ni + (1 − p)ˆ_x[8]VLV n1 ni = p q _x_ˆ [8] VHn1 + (1 − p) q _x_ˆ [8] VLn1 − niˆx[8] n1V p VH + (1−p) VL , (A.9) which is (80).

Equation (84) Note that ˆx[8]i> 0 if and only if the numerator of (80) is positive, as

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have to find r[8] = max i : p s ˆ x_[8] VHn1 + (1 − p) s ˆ x_[8] VLn1 −ni+1xˆ[8] n1V > 0 ⇔ r_[8] = max i : p√1 VH + (1 − p)√1 VL −√ni+1 n1V √ ˆ x > 0. (A.10)

Plugging (83) in (A.10) results in

r[8] = max i : p 1 √ VH + (1 − p)√1 VL −√ni+1 n1V   ip √ VHn1 + i(1−p) √ VLn1 p VH + 1−p VL + Pi+1 j=2nj n1V  > 0 ⇔ r_[8] = max i : 1 − √ni+1 n1V   i √ n1 p VH + 1−p VL + Pi+1 j=2nj n1V  > 0 ⇔ r_[8] = max i : p VH + 1 − p VL + Pi+1 j=2nj n1V −√ni+1 n1V i √ n1 > 0 ⇔ r_[8] = max i : Pi+1 j=2nj i + V n1 i p VH +1 − p VL − n_i+1> 0, (A.11) which is (84).

B

Proofs of lemmas and theorems

Proof of lemma 2. Equation (85) implies that

x_[2](i) = _Pi − 1_i+1

j=2nj

≥ VH V n1

. (A.12)

Equation (A.12) can be rewritten as

i − 1 VH ≥ Pi+1 j=2nj V n1 . (A.13)

Proof of lemma 3. We know from (86) that

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⇔ 4p(pi − p + i − 1 − pi + p) + 4 VH V n1 i+1 X j=2 nj(i − 1) < 4pVH V n1 i+1 X j=2 nj+ 4   VH V n1 i+1 X j=2 nj   2 (A.23) ⇔ (i − 1)  4p + 4VH V n1 i+1 X j=2 nj   < 4p VH V n1 i+1 X j=2 nj + 4   VH V n1 i+1 X j=2 nj   2 (A.24) ⇔ i − 1 VH < Pi+1 j=2nj V n1 . (A.25)

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⇔ (p + VH V n1 i+1 X j=2 nj)(1 − p) VH VL (i − 1) ≥ p2+ 2p VH V n1 i+1 X j=2 nj+ ( VH V n1 i+1 X j=2 nj)2− (p + VH V n1 i+1 X j=2 nj)ip r VH VL (A.31) ⇔ (p + VH V n1 i+1 X j=2 nj)(1 − p) VH VL (i − 1) ≥  p + VH V n1 i+1 X j=2 nj   2 −(p + VH V n1 i+1 X j=2 nj)(i − 1)p r VH VL − (p + VH V n1 i+1 X j=2 nj)p r VH VL (A.32) ⇔ (1 − p)VH VL (i − 1) ≥ p + VH V n1 i+1 X j=2 nj − (i − 1)p r VH VL − pr VH VL (A.33) ⇔ (i − 1) (1 − p)VH VL + pr VH VL ! ≥ p + VH V n1 i+1 X j=2 nj− p r VH VL (A.34) ⇔ i − 1 ≥ p + VH V n1 Pi+1 j=2nj− p q VH VL (1 − p)VH VL + p q VH VL (A.35) ⇔ i − 1 ≥ p VH + Pi+1 j=2nj V n1 − p √ VHVL 1−p VL + p √ VHVL . (A.36)

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Proof of lemma 6. For i = 2, 3, . . . , r_[2] we have by (66) ni+1< Pi+1 j=2nj i − 1 . (A.42) Thus, for i = 2, 3, . . . , r[2], 1 i − 1 > ni+1 Pi+1 j=2nj (A.43) ⇔ 1 − 1 i − 1 < 1 − ni+1 Pi+1 j=2nj (A.44) ⇔ i − 1 − 1 i − 1 < Pi+1 j=2nj− ni+1 Pi+1 j=2nj (A.45) ⇔ i − 1 − 1 Pi+1 j=2nj− ni+1 < _Pi − 1_i+1 j=2nj , (A.46)

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which is an inductive proof that i − 1 Pi+1 j=2nj ≤ r[2]− 1 Pr[2]+1 j=2 nj i ∈ {r_[2]+ 1, r_[2]+ 2, . . . , m − 1}. (A.51)

It is trivial that (A.51) also holds for i = r_[2].

Proof of lemma 7. We define for i = 1, 2, . . . , m − 1,

a(i) = ippVHn1 (A.52) b(i) = i2p2VHn1+ 4  pn1+ VH V i+1 X j=2 nj  (1 − p)VH(i − 1) (A.53) c(i) = 2  pn1+ VH V i+1 X j=2 nj   (A.54)

f (i) = a(i) +pb(i)

c(i) =

q ˆ

x[6](i) (A.55)

For i = 2, 3, . . . , m − 1 we have that

f (i − 1) = a(i − 1) +pb(i − 1) c(i − 1) (A.56) = a(i − 1) +pa(i − 1) 2_{+ 2c(i − 1)(1 − p)V} H(i − 2) c(i − 1) (A.57) = a(i − 1) c(i − 1)+ s a(i − 1)2 (c(i − 1))2 + 2(1 − p)VH(i − 2) c(i − 1) (A.58) = i−1 i a(i) c(i) − 2VH V ni+1 + v u u u t i−1 i 2 a(i)2 c(i) − 2VH V ni+1 2 + 2(1 − p)VH(i − 2) c(i) − 2VH V ni+1 . (A.59)

By (75) we know that for i = r[6]+ 1, . . . , m − 1,

ni+1≥ pV c(i) a(i) +pb(i) r n₁ VH + (1 − p) (c(i)) 2 a(i) +pb(i)2 V, (A.60)

Note that by (A.60), for i = r[6]+ 1, . . . , m − 1

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= c(i)   1 −

2p√n1VH(a(i) +pb(i)) + 2(1 − p)c(i)VH

a(i) +pb(i) 2    (A.62) = c(i) 1 −2 (a(i))2 i + 2 a(i) i pb(i) + 2(1 − p)c(i)VH

(a(i))2+ 2a(i)pb(i) + b(i)

!

(A.63)

= c(i) 1 −1 i

2 (a(i))2+ 2a(i)pb(i) + 2(1 − p)c(i)VHi

2 (a(i))2+ 2a(i)pb(i) + 2(1 − p)c(i)VH(i − 1)

!!

(A.64)

= c(i) 1 −1 i 1 +

2(1 − p)c(i)VH

2 (a(i))2+ 2a(i)pb(i) + 2(1 − p)c(i)VH(i − 1)

!! (A.65) = c(i)   1 − 1 i − 2(1 − p)c(i)VH ia(i) +pb(i)2   . (A.66)

Plugging (A.66) into (A.59) yields for i = r[6]+ 1, . . . , m − 1,

f (i − 1) ≥ i−1 i a(i) + v u u t i−1 i 2 a(i)2_{+ 2(1 − p)V} H(i − 2)c(i) 1 −1_i − 2(1−p)c(i)VH ia(i)+√b(i)2 ! c(i) 1 −1_i − 2(1−p)c(i)VH ia(i)+√b(i)2 ! (A.67) = i−1 i a(i) + v u u t i−1 i 2 a(i)2_{+ 2(1 − p)V} Hc(i) i − 3 +2_i −2(i−2)(1−p)c(i)VH ia(i)+√b(i)2 ! c(i) 1 −1_i − 2(1−p)c(i)VH ia(i)+√b(i)2 ! . (A.68)

We focus for the moment on 2(1 − p)c(i)VH:

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= b(i) i − (a(i))2 i + 2c(i)(1 − p)VH(i − 1) i2 + 2c(i)(1 − p)VH i2 (A.72) = b(i) i − (a(i))2 i + b(i) i2 − (a(i))2 i2 + 2c(i)(1 − p)VH i2 (A.73) = 2b(i) i − b(i) i − (a(i))2 i + 2 b(i) i2 − b(i) i2 − (a(i))2 i2 + 2c(i)(1 − p)VH i2 + 2a(i)pb(i) i − 2 a(i)pb(i) i + 2 a(i)pb(i) i2 − 2 a(i)pb(i) i2 (A.74) = − (a(i)) 2 i + (a(i))2 i2 + 2 a(i)pb(i) i + 2 a(i)pb(i) i2 + b(i) i + b(i) i2 ! + 2a(i)pb(i) i + 2 a(i)pb(i) i2 + 2 b(i) i + 2 b(i) i2 + 2c(i)(1 − p)VH i2 (A.75) = − i + 1 i2

(a(i))2+ 2a(i)pb(i) + b(i) +2 i i + 1 i a(i)pb(i) +2 i i + 1 i b(i) +2c(i)(1 − p)VH i2 (A.76) = − i + 1 i2 a(i) +pb(i)2+2 i i + 1 i p

b(i)a(i) +pb(i)

+2c(i)(1 − p)VH

i2 .

(A.77)

Plugging (A.77) into the square root in the numerator of (A.68) yields

i − 1 i 2 a(i)2+ 2(1 − p)VHc(i)   i − 3 + 2 i − 2(i − 2)(1 − p)c(i)VH ia(i) +pb(i)2    (A.78) = i − 1 i 2 a(i)2+ 2(1 − p)VHc(i) ×   i − 3 + 2 i + (i + 1)(i − 2) i3 −

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= i − 1 i 2 a(i)2+ 2(1 − p)VHc(i) ×   i − 3 + 2 i + 1 i − 1 i2 − 2 i3 −

2(i + 1)(i − 2)pb(i) i3_{a(i) +}_pb(i) −2c(i)(1 − p)(i − 2)VH i3_{a(i) +}_pb(i)2    (A.80) = i − 1 i 2 a(i)2+ i − 1 i 2 2(1 − p)VHc(i)(i − 1) + 2(1 − p)VHc(i)   − 2(i − 1)pb(i) i2_{a(i) +}_pb(i) + 2c(i)(1 − p)VH i2_{a(i) +}_pb(i)2    + 4 i3(1 − p)VHc(i)    2pb(i) a(i) +pb(i) − 2c(i)(1 − p)(i − 1)VH a(i) +pb(i)2 − 1    (A.81) = i − 1 i 2 b(i) − 4 i − 1 i2 p b(i)c(i)(1 − p)VH a(i) +pb(i) + 4 i2 c(i)(1 − p)VH a(i) +pb(i) !2 + 4 i3(1 − p)VHc(i)    2pb(i) a(i) +pb(i) − 2c(i)(1 − p)(i − 1)VH a(i) +pb(i)2 − 1    (A.82) = i − 1 i 2 b(i) − 4 i − 1 i2 p b(i)c(i)(1 − p)VH a(i) +pb(i) + 4 i2 c(i)(1 − p)VH a(i) +pb(i) !2 + 4(1 − p)VHc(i) i3_{a(i) +}_pb(i)2

×2pb(i)a(i) + 2b(i) − 2c(i)(1 − p)(i − 1)VH−

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We can plug (A.85) and into (A.68) to obtain for i = r[6]+ 1, . . . , m − 1, f (i − 1) ≥ i−1 i a(i) + v u u t i−1 i pb(i) − 2(1−p)VHc(i) ia(i)+√b(i) !2 c(i) 1 −1_i − 2(1−p)c(i)VH i a(i)+√b(i) 2 ! (A.86) = i−1 i a(i) + i−1 i pb(i) − 2(1−p)VHc(i) ia(i)+√b(i) c(i) 1 −1_i − 2(1−p)c(i)VH i a(i)+√b(i) 2 ! (A.87) = i−1 i a(i) + i−1 i pb(i) −

a(i) +pb(i) 2(1−p)VHc(i)

ia(i)+√b(i)2 c(i) 1 −1_i − 2(1−p)c(i)VH i a(i)+√b(i) 2 ! (A.88) =

a(i) +pb(i) 1 −1_i − 2(1−p)c(i)VH

ia(i)+√b(i)2 ! c(i) 1 −1_i − 2(1−p)c(i)VH i a(i)+√b(i) 2 ! (A.89) = a(i) +pb(i) c(i) (A.90) = f (i), (A.91)

which, when used as an inductive argument, proves the lemma for i = r[6]+ 1, . . . , m − 1.

By (75) we know that for i = 2, 3, . . . , r_[6],

ni+1< pV c(i) a(i) +pb(i) r n₁ VH + (1 − p) (c(i)) 2 a(i) +pb(i)2 V. (A.92)

Following the rest of the proof above, the inequality signs in equations (A.61), (A.67) and (A.86) change to >, < and <, respectively. When used as an inductive argument, this proves the lemma for i = 1, . . . , r_[6]− 1. Note that the lemma is trivially true for i = r[6].

Proof of lemma 8. For i = 2, 3, . . . , r_[8] we know from (84) that

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Using (A.93), we know that for i = 2, 3, . . . , r[8], (i−1)p √ VHn1 + (i−1)(1−p)_√ VLn1 p VH + 1−p VL + Pi j=2nj n1V = 1 −1_i ip √ VHn1 + i(1−p) √ VLn1 p VH + 1−p VL + Pi+1 j=2nj n1V − ni+1 n1V (A.94) < 1 −1_i_√ip VHn1 + i(1−p) √ VLn1 p VH + 1−p VL + Pi+1 j=2nj n1V − Pi+1 j=2nj in1V − 1 i p VH + 1−p VL (A.95) = 1 −1_i_√ip VHn1 + i(1−p) √ VLn1 1 −1_i p VH + 1−p VL + Pi+1 j=2nj n1V (A.96) = ip √ VHn1 + i(1−p) √ VLn1 p VH + 1−p VL + Pi+1 j=2nj n1V , (A.97)

which, when applied inductively, proves this lemma for i = 1, 2, . . . , r[8]− 1.

Similarly, for i = r_[8]+ 1, . . . , m − 1 we know from (84) that

ni+1≥ Pi+1 j=2nj i + V n1 i p VH +1 − p VL . (A.98)

Using (A.98), we know that for i = r_[8]+ 1, . . . , m − 1,

(i−1)p √ VHn1 + (i−1)(1−p)_√ VLn1 p VH + 1−p VL + Pi j=2nj n1V = 1 −1_i_√ip VHn1 + i(1−p) √ VLn1 p VH + 1−p VL + Pi+1 j=2nj n1V − ni+1 n1V (A.99) ≥ 1 −1_i_√ip VHn1 + i(1−p) √ VLn1 p VH + 1−p VL + Pi+1 j=2nj n1V − Pi+1 j=2nj in1V − 1 i p VH + 1−p VL (A.100) = 1 −1_i ip √ VHn1 + i(1−p) √ VLn1 1 −1_i p VH + 1−p VL + Pi+1 j=2nj n1V (A.101) = ip √ VHn1 + i(1−p) √ VLn1 p VH + 1−p VL + Pi+1 j=2nj n1V , (A.102)

which, when applied inductively, proves this lemma for i = r_[8]+ 1, . . . , m − 1. Trivially, this lemma holds for i = r[8].

Proof of lemma 9. For case 2, note that (66) implies for i > r_[2] that

ni+1≥

V x[2](i)

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which by lemma 6 implies that

ni+1≥

V x[2](r[2])

. (A.104)

Note that (A.104) is equivalent to (63).

For case 6, note that (75) implies for i > r[6] that

ni+1≥ pV r _n 1 VHxˆ[6](i) + (1 − p) V ˆ x_[6](i), (A.105)

ni+1≥ pV r _n 1 VHxˆ[6](r[6]) + (1 − p) V ˆ x_[6](r_[6]). (A.106)

Note that (A.106) is equivalent to (70) if you plug in (67). For case 8, note that (84) implies for i > r[8] that

ni+1≥ V n1 i ip √ VHn1 + i(1−p) √ VLn1 q ˆ x[8](i) (A.107) = p V √ n1 q VHxˆ[8](i) + (1 − p) V √ n1 q VLxˆ[8](i) , (A.108)

ni+1≥ p V√n1 q VHxˆ[8](r[8]) + (1 − p) V √ n1 q VLxˆ[8](r[8]) . (A.109)

Note that (A.109) is equivalent to (79) if you plug in (76) and (77).

Proof of lemma 10. When we have an equilibrium in case 2, we have by lemma 2 that

r[2]− 1 VH ≥ Pr[2]+1 j=1 nj V n1 , (A.110)

which by lemma 3 implies that ˆx_[6](r_[2]) ≥ VH

n1, which by lemma 7 implies that ˆx[6](r[6]) ≥

VH

n1. As this last implication is in contradiction with the assumption that there is an

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Proof of lemma 11. When we have an equilibrium in case 8, we have by lemma 8 that x[8](r[2]) < V_nL₁. By lemma 5 we then have that

(r_[2]− 1) p √ VHVL + 1 − p VL < p VH + Pr[2]+1 j=2 nj n1V −√ p VHVL (A.111) ⇔ r_[2] p √ VHVL + 1 − p VL < p VH +1 − p VL + Pr[2]+1 j=2 nj n1V (A.112) ⇒ r_[2] p VH +1 − p VL < p VH +1 − p VL + Pr[2]+1 j=2 nj n1V (A.113) ⇔ (r_[2]− 1) p VH +1 − p VL < Pr[2]+1 j=2 nj n1V (A.114) ⇒ r[2]− 1 VH < Pr[2]+1 j=2 nj n1V . (A.115)

Note that (A.115) implies by lemma 2 that ˆx[2](r[2]) < V_nH₁. As this last implication is

in contradiction with the assumption that there is an equilibrium for case 2, we cannot have that both cases 2 and 8 yield an equilibrium at the same time.

Proof of lemma 12. When we have an equilibrium in case 8, we have by lemma 8 that x_[8](r_[6]) < VL

n1. Lemma 5 now yields that

(r[6]− 1) p √ VHVL +1 − p VL < p VH + Pr[6]+1 j=2 nj n1V −√ p VHVL , (A.116)

which by lemma 4 implies that ˆx[6](r[6]) < V_nL₁. As this last implication is in contradiction

with the assumption that there is an equilibrium for case 6, we cannot have that both cases 6 and 8 yield an equilibrium at the same time.

Proof of lemma 13. Assume that case 6 does not yield an equilibrium. Then either ˆ

x_[6](r_[6]) ≥ VH

n1 or ˆx[6](r[6]) <

VL

n1.

For the case that ˆx[6](r[6]) ≥ V_nH₁, note that lemma 3 implies that

r[6]− 1 VH ≥ Pr[6]+1 j=1 nj V n1 . (A.117)

Applying lemma 6 implies that

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Applying lemma 2 to (A.118), we know that case 2 yields an equilibrium.

For the case that ˆx[6](r[6]) < V_nL₁, note that lemma 7 implies that ˆx[6](r[8]) < V_nL₁. By

lemmas 4 and 5 we know that case 8 yields an equilibrium.

Proof of proposition 2. Suppose that group 1 has the high valuation. This implies that ˜

VL = ˜VH = VH. The condition for playing a positive effort (case 6 or case 8) is the

negation of (88) with i = r_[2] for the private information model and i = ˜r_[2] for the benchmark model. As both (66), which is the function defining r[2] and ˜r[2], and (88) are

independent of VL and VH, we can conclude that group 1 will play a positive effort in

the benchmark model if and only if it plays a positive effort in the private information model.

For the rest of this proof, suppose that group 1 has the low valuation. For the benchmark model this implies that ˜VL= ˜VH = VL. We can derive from lemma 5 when

group 1 is active (case 8) in the private information model. In the model with perfect information (the benchmark model), group 1 of the low valuation type will be active when ˜ r_[8]− 1 VL < Pr˜[8]+1 j=2 nj n1V . (A.119)

Note that we have by (84) that ˜r_[8] ≥ r_[8]. We can therefore directly conclude that if ˜

r[8] = 1, we have that r[8] = 1 and that group 1 with a low valuation will always be active

in the benchmark model, while it can be inactive in the model with private information. Note that this result also holds for the two-group model (n = 2 and therefore r[8] = 1

and ˜r[8] = 1).

If ˜r_[8]≥ 2 and r_[8] = 1, we compare lemma 5 to (A.119) and can conclude that group 1 can be active and inactive for both the private information model as the benchmark model, without one model implying activity or inactivity for the other.

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the benchmark model, we have Pr˜[8]+1 j=2 nj ˜ r[8] + V n1 ˜ r[8]VL − nr˜[8]+1> 0 (A.120) ⇔ 1 + V n1 VLP ˜ r[8]+1 j=2 nj − nr˜[8]+1 ˜ r_[8] Pr˜[8]+1 j=2 nj > 0 (A.121) ⇔ 1 − ˜r[8]− 1 P˜r[8]+1 j=2 nj n˜r[8]+1− nr˜[8]+1 Pr˜[8]+1 j=2 nj + V n1 VLP ˜ r[8]+1 j=2 nj > 0 (A.122) ⇒ 1 − r˜[8]− 1 Pr˜[8]+1 j=2 nj nr˜[8]+1− n˜r[8]+1 P˜r[8]+1 j=2 nj + 1 ˜ r_[8]− 1 > 0, (A.123)

where (A.123) is due to (A.119). Let’s say that ˜r_[8]> ˜r_[2]. Then by (66) we have that n˜r[8]+1 Pr˜[8]+1 j=2 nj ≥ 1 ˜ r_[8]− 1. (A.124)

If we plug (A.124) in (A.123) and compare that with (66) we know that ˜r_[2] ≥ ˜r_[8], which is in contradiction with what we assumed. Therefore we must have that ˜r[2]≥ ˜r[8]. Note

furthermore, that by (66) we have that ˜r[2] = r[2]. Summarising,

r[2]= ˜r[2] ≥ ˜r[8] ≥ r[8]. (A.125)

By lemma 6, (A.125) and (A.119) we now know that

VL n1V > r˜[8]− 1 Pr˜[8]+1 j=2 nj ⇒ VL n1V > r[8]− 1 Pr[8]+1 j=2 nj . (A.126)

We can derive from lemma 5 that group 1 with a low valuation is active in the private information model when

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Note furthermore, that as 0 < VL< VH, we have for r[8] ≥ 2, r VL VH − 1 !2 > 0 (A.131) ⇔ VL VH − 2r VL VH + 1 > 0 (A.132) ⇔ 2 − 2r VL VH > 1 − VL VH (A.133) ⇔ 2 > 1 − VL VH 1 − q VL VH (A.134) ⇒ r_[8] > 1 − VL VH 1 − q VL VH (A.135) ⇔ r_[8]− 1 >r VL VH r[8]− VL VH . (A.136)

By (A.136) and (A.126) we know for r[8] ≥ 2 when group 1 is active in the benchmark

model that p( q VL VHr[8]− VL VH) VL +(1 − p)(r[8]− 1) VL < r[8]− 1 VL < Pr[8]+1 j=2 nj n1V . (A.137)

Comparing (A.137) to (A.130), this implies the following for the situation where group 1 has a low valuation while ˜r[8] ≥ 2 and r[8] ≥ 2: if group 1 is active in the benchmark

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Rent-seeking group contests with one-sided private information