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Sequential contests with synergy and budget constraints

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Abstract

We study a sequential Tullock contest with two stages and two identical prizes. The players compete for one prize in each stage and each player may win either one or two prizes. The players have either decreasing or increasing marginal values for the prizes, which are commonly known, and there is a constraint on the total effort that each player can exert in both stages. We analyze the players’ allocations of efforts along both stages when the budget constraints (effort constraints) are either restrictive, nonrestrictive or partially restrictive. In particular, we show that when the players are either symmetric or asymmetric and the budget constraints are restrictive, independent of the players’ values for the prizes, each player allocates his effort equally along both stages of the contest.

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Notes

  1. Che and Gale (1998) and Gavious et al. (2003) deal with all-pay auctions with bid caps. The bid cap is a budget constraint that the contest designer imposes on the contestants.

  2. Several papers in the literature (see, for example, Leininger 1993; Morgan 2003; Konrad 2004; Klumpp and Polborn 2006) compare simultaneous (one-stage) and sequential (multi-stage) contests.

  3. Several papers deal with sequential auctions. These include, Pitchik and Schotter (1988) who analyzed sequential first and second price auctions with a budget constraint and two different prizes; Pitchik (2009) who analyzed a sequential auction with a budget constraint under incomplete information, and Brusco and Lopomo (2008, 2009) who considered sequential auctions with a budget constraint and with and without a synergy between the values of the prizes.

  4. We assume tat the players’ values are in the interval \([0,1].\) This is only a normalization. Considering higher values wold not qualitatively affect the results.

  5. The uniqueness of the subgame perfect equilibrium is obtained by the uniqueness of the equilibrium in the one-stage Tullock contest with two players.

  6. The uniqueness of the subgame perfect equilibrium is obtained by the uniqueness of the equilibrium in the one-stage Tullock contest with two players.

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Appendix

Appendix

1.1 The proof of Proposition 1

If the budget constraint is nonrestrictive both of the restrictions in the maximization problem (3) are nonrestrictive such that

$$\begin{aligned} x_{i}^{a}+{\widetilde{x}}_{i}^{b}&< w \\ x_{i}^{a}+{\widetilde{x}}_{i}^{a}&< w. \end{aligned}$$

Then the first-order conditions of the maximization problems in the second stage (1) and (2) are

$$\begin{aligned} v^{b}\frac{{\widetilde{x}}_{j}^{a}}{\left({\widetilde{x}}_{i}^{b}+{\widetilde{x}} _{j}^{a}\right)^{2}}-1&= 0 \\ v^{a}\frac{{\widetilde{x}}_{j}^{b}}{\left({\widetilde{x}}_{i}^{a}+{\widetilde{x}} _{j}^{b}\right)^{2}}-1&= 0. \end{aligned}$$

Because of the symmetry we denote

$$\begin{aligned} {\widetilde{x}}_{i}^{a}&= {\widetilde{x}}_{j}^{a}={\widetilde{x}}^{a} \\ {\widetilde{x}}_{i}^{b}&= {\widetilde{x}}_{j}^{b}={\widetilde{x}}^{b}\!. \end{aligned}$$

The solution of the above two first-order conditions is:

$$\begin{aligned} {\widetilde{x}}^{a}&= \frac{v^{b}(v^{a})^{2}}{(v^{a}+v^{b})^{2}}\nonumber \\ {\widetilde{x}}^{b}&= \frac{v^{a}(v^{b})^{2}}{(v^{a}+v^{b})^{2}}. \end{aligned}$$
(21)

The first-order condition of the maximization problem in the first stage (Eq. (3)) is

$$\begin{aligned} \left[v^{a}+v^{b}\frac{{\widetilde{x}}_{i}^{b}}{{\widetilde{x}}_{i}^{b}+ {\widetilde{x}}_{j}^{a}}-{\widetilde{x}}_{i}^{b}-v^{a}\frac{{\widetilde{x}}_{i}^{a} }{{\widetilde{x}}_{i}^{a}+{\widetilde{x}}_{j}^{b}}+{\widetilde{x}}_{i}^{a}\right]\frac{ x_{j}^{a}}{\left(x_{i}^{a}+x_{j}^{a}\right)^{2}}=1. \end{aligned}$$

where \({\widetilde{x}}_{i}^{a},{\widetilde{x}}_{j}^{a},{\widetilde{x}}_{i}^{b}, {\widetilde{x}}_{j}^{b}\) are given by (21). Because of the symmetry we denote

$$\begin{aligned} x_{i}^{a}=x_{j}^{a}=x^{a}. \end{aligned}$$

Then the solution of the above first-order condition is

$$\begin{aligned} x^{a}=\frac{(v^{b})^{3}+v^{a}(v^{b})^{2}+2v^{b}(v^{a})^{2}}{4(v^{a}+v^{b})^{2}}. \end{aligned}$$

By normalizing (\(v^{a}=1\)) we obtain

$$\begin{aligned} {\widetilde{x}}^{a}-x^{a}=\frac{-v^{b}[(v^{b})^{2}+v^{b}-2]}{4(v^{b}+1)^{2}}. \end{aligned}$$

Since the expression \((v^{b})^{2}+v^{b}-2\) is negative for all \(0<v^{b}<1,\) the difference \({\widetilde{x}}^{a}-x^{a}\) is always positive. Furthermore,

$$\begin{aligned} x^{a}-{\widetilde{x}}^{b}=\frac{v^{b}[(v^{b})^{2}-3v^{b}+2]}{4(v^{b}+1)^{2}}. \end{aligned}$$

Since the expression \((v^{b})^{2}-3v^{b}+2\) is positive for all \(0<v^{b}<1,\) the difference \(x^{a}-{\widetilde{x}}^{b}\) is always positive.

Now we examine the conditions under which the budget constraint is nonrestrictive. If the restrictions are nonrestrictive we have

$$\begin{aligned} x^{a}+{\widetilde{x}}^{b}&= \frac{(v^{b})^{3}+5(v^{b})^{2}+2v^{b}}{4(v^{b}+1)^{2}}<w \\ x^{a}+{\widetilde{x}}^{a}&= \frac{(v^{b})^{3}+(v^{b})^{2}+6v^{b}}{ 4(v^{b}+1)^{2}}<w. \end{aligned}$$

Since \(x^{a}+{\widetilde{x}}^{a}>x^{a}+{\widetilde{x}}^{b}\) we obtain that the constraints are nonrestrictive iff \(w>x^{a}+{\widetilde{x}}^{a}.\) Thus, the condition that implies nonrestrictive budget constraints is

$$\begin{aligned} w>\frac{(v^{b})^{3}+(v^{b})^{2}+6v^{b}}{4(v^{b}+1)^{2}}. \end{aligned}$$

\(\square \)

1.2 The proof of Proposition 2

We proved in Proposition 1 that if \(v^{a}\) is normalized to be 1, the budget constraint is nonrestrictive if

$$\begin{aligned} w>\frac{(v^{b})^{3}+(v^{b})^{2}+6v^{b}}{4(v^{b}+1)^{2}}. \end{aligned}$$

In this case the total effort in the first stage of the contest is

$$\begin{aligned} TE_{1}=2x^{a}=\frac{(v^{b})^{3}+(v^{b})^{2}+2v^{b}}{2(1+v^{b})^{2}} \end{aligned}$$

and the total effort in the second stage of the contest is

$$\begin{aligned} TE_{2}={\widetilde{x}}^{a}+{\widetilde{x}}^{b}=\frac{v^{b}}{1+v^{b}}. \end{aligned}$$

The difference between the total efforts in both stages when the budget constraint is nonrestrictive is

$$\begin{aligned} TE_{1}-TE_{2}=\frac{(v^{b})^{2}(v^{b}-1)}{2(1+v^{b})^{2}}\!. \end{aligned}$$

Since \(v^{b}<v^{a}=1\) (decreasing marginal values) this difference is negative and therefore \(TE_{1}<TE_{2}.\) \(\square \)

1.3 The proof of Proposition 3

If the budget constraint is nonrestrictive both of the restrictions in the maximization problem (6) are nonrestrictive such that

$$\begin{aligned} x_{i}^{b}+{\widetilde{x}}_{i}^{a}&< w \\ x_{i}^{b}+{\widetilde{x}}_{i}^{b}&< w. \end{aligned}$$

The first-order conditions of the maximization problems in the second stage (4) and (5) are

$$\begin{aligned} v^{a}\frac{{\widetilde{x}}_{j}^{b}}{\left({\widetilde{x}}_{i}^{a}+{\widetilde{x}} _{j}^{b}\right)^{2}}-1&= 0 \\ v^{b}\frac{{\widetilde{x}}_{j}^{a}}{\left({\widetilde{x}}_{i}^{b}+{\widetilde{x}} _{j}^{a}\right)^{2}}-1&= 0. \end{aligned}$$

Because of the symmetry we denote

$$\begin{aligned} {\widetilde{x}}_{i}^{a}&= {\widetilde{x}}_{j}^{a}={\widetilde{x}}^{a} \\ {\widetilde{x}}_{i}^{b}&= {\widetilde{x}}_{j}^{b}={\widetilde{x}}^{b}\!. \end{aligned}$$

The solution of the above two first-order conditions is

$$\begin{aligned} {\widetilde{x}}^{a}&= \frac{v^{b}(v^{a})^{2}}{(v^{a}+v^{b})^{2}}\nonumber \\ {\widetilde{x}}^{b}&= \frac{v^{a}(v^{b})^{2}}{(v^{a}+v^{b})^{2}}. \end{aligned}$$
(22)

The first-order condition of the maximization problem in the first stage (Eq. (6)) is

$$\begin{aligned} \left[v^{b}+v^{a}\frac{{\widetilde{x}}_{i}^{a}}{{\widetilde{x}}_{i}^{a}+ {\widetilde{x}}_{j}^{b}}-{\widetilde{x}}_{i}^{a}-v^{b}\frac{{\widetilde{x}}_{i}^{b} }{{\widetilde{x}}_{i}^{b}+{\widetilde{x}}_{j}^{a}}+{\widetilde{x}}_{i}^{b}\right]\frac{ x_{j}^{b}}{\left(x_{i}^{b}+x_{j}^{b}\right)^{2}}=1, \end{aligned}$$

where \({\widetilde{x}}_{i}^{a},{\widetilde{x}}_{j}^{a},{\widetilde{x}}_{i}^{b},{\widetilde{x}}_{j}^{b}\) are given by (22). Because of the symmetry we denote

$$\begin{aligned} x_{i}^{b}=x_{j}^{b}=x^{b}\!. \end{aligned}$$

Then the solution of the above first-order condition is

$$\begin{aligned} x^{b}=\frac{2(v^{b})^{2}v^{a}+v^{b}(v^{a})^{2}+(v^{a})^{3}}{4(v^{a}+v^{b})^{2}}. \end{aligned}$$

By using the normalization (\(v^{a}=1\)) we obtain

$$\begin{aligned} {\widetilde{x}}^{a}-{\widetilde{x}}^{b}=\frac{v^{b}(1-v^{b})}{(v^{b}+1)^{2}}. \end{aligned}$$

Since \(0<v^{b}<1,\) the difference \({\widetilde{x}}^{a}-{\widetilde{x}}^{b}\) is always positive. Furthermore,

$$\begin{aligned} x^{b}-{\widetilde{x}}^{b}=\frac{-2(v^{b})^{2}+v^{b}+1}{4(v^{b}+1)^{2}}. \end{aligned}$$

Since the expression \(-2(v^{b})^{2}+v^{b}+1\) is positive for all \(0<v^{b}<1,\) the difference \(x^{b}-{\widetilde{x}}^{b}\) is always positive. We also have

$$\begin{aligned} x^{b}-{\widetilde{x}}^{a}=\frac{2(v^{b})^{2}-3v^{b}+1}{4(v^{b}+1)^{2}}. \end{aligned}$$

Since the expression \(2(v^{b})^{2}-3v^{b}+1\) is positive for all \(0<v^{b}<0.5 \) and negative for all \(0.5<v^{b}<1\) we obtain that the difference \(x^{b}-{\widetilde{x}}^{a}\) is positive for all \(0<v^{b}<0.5\) and is negative for all \(0.5<v^{b}<1.\) The relations between a player’s allocations of effort is therefore

$$\begin{aligned} x^{b}&\ge {\widetilde{x}}^{a}>{\widetilde{x}}^{b}\quad \text{ if}\; 0<v^{b}\le 0.5\\ {\widetilde{x}}^{a}&\ge x^{b}>{\widetilde{x}}^{b}\quad \text{ if}\; 0.5<v^{b}<1. \end{aligned}$$

Now we examine the conditions under which the budget constraint is nonrestrictive. If the restrictions are nonrestrictive we have

$$\begin{aligned} x^{b}+{\widetilde{x}}^{a}&= \frac{2(v^{b})^{2}+5v^{b}+1}{4(v^{b}+1)^{2}}<w \\ x^{b}+{\widetilde{x}}^{b}&= \frac{6(v^{b})^{2}+v^{b}+1}{4(v^{b}+1)^{2}}<w. \end{aligned}$$

Since \(x^{b}+{\widetilde{x}}^{a}>x^{b}+{\widetilde{x}}^{b}\) we obtain that the constraints are nonrestrictive iff \(w>x^{b}+{\widetilde{x}}^{a}.\) Thus, the condition that implies nonrestrictive budget constraints is

$$\begin{aligned} w>\frac{2(v^{b})^{2}+5v^{b}+1}{4(v^{b}+1)^{2}}. \end{aligned}$$

\(\square \)

1.4 The proof of Proposition 4

We proved in Proposition 3 that if \(v^{a}\) is normalized to be 1, the budget constraint is nonrestrictive if

$$\begin{aligned} w>\frac{2(v^{b})^{2}+5v^{b}+1}{4(v^{b}+1)^{2}}. \end{aligned}$$

In this case the total effort in the first stage of the contest is

$$\begin{aligned} TE_{1}=2x^{b}=\frac{2(v^{b})^{2}+v^{b}+1}{2(1+v^{b})^{2}} \end{aligned}$$

and the total effort in the second stage of the contest is

$$\begin{aligned} TE_{2}={\widetilde{x}}^{a}+{\widetilde{x}}^{b}=\frac{v^{b}}{1+v^{b}}. \end{aligned}$$

The difference between the total efforts in both stages when the budget constraint is nonrestrictive is

$$\begin{aligned} TE_{1}-TE_{2}=\frac{1-v^{b}}{2(1+v^{b})^{2}}. \end{aligned}$$

Since \(v^{b}<v^{a}=1\) (increasing marginal values) this difference is positive and therefore \(TE_{1}>TE_{2}.\) \(\square \)

1.5 The proof of Proposition 5

(1) Assume first that the players have decreasing marginal values. If the budget constraint is restrictive both of the restrictions in the maximization problem (3) are restrictive such that

$$\begin{aligned} x_{i}^{a}+{\widetilde{x}}_{i}^{b}&= w \\ x_{i}^{a}+{\widetilde{x}}_{i}^{a}&= w. \end{aligned}$$

Player \(i\)’s maximization problem in the first stage is then

$$\begin{aligned}&\max _{x_{i}^{a}}\left(v^{a}+v^{b}\frac{w-x_{i}^{a}}{w-x_{i}^{a}+w-x_{j}^{a}} -\left(w-x_{i}^{a}\right)\right)\frac{x_{i}^{a}}{x_{i}^{a}+x_{j}^{a}}\\&\quad +\left(v^{a}\frac{w-x_{i}^{a}}{w-x_{i}^{a}+w-x_{j}^{a}}-\left(w-x_{i}^{a}\right)\right) \left(1-\frac{x_{i}^{a}}{x_{i}^{a}+x_{j}^{a}}\right)-x_{i}^{a}\!. \end{aligned}$$

Therefore the first-order condition is

$$\begin{aligned}&\left[v^{a}+v^{b}\frac{w-x_{i}^{a}}{w-x_{i}^{a}+w-x_{j}^{a}}-\left(w-x_{i}^{a}\right)-v^{a} \frac{w-x_{i}^{a}}{w-x_{i}^{a}+w-x_{j}^{a}}+\left(w-x_{i}^{a}\right)\right]\\&\quad \times \frac{x_{j}^{a}}{ \left(x_{i}^{a}+x_{j}^{a}\right)^{2}}+(v^{b}-v^{a})\frac{-\left(w-x_{j}^{a}\right)}{\left(w-x_{i}^{a}+w-x_{j}^{a}\right)^{2}}\frac{ x_{i}^{a}}{x_{i}^{a}+x_{j}^{a}}\\&\quad +v^{a}\frac{-\left(w-x_{j}^{a}\right)}{ \left(w-x_{i}^{a}+w-x_{j}^{a}\right)^{2}} =0. \end{aligned}$$

Because of the symmetry we denote

$$\begin{aligned} x_{i}^{a}&= x_{j}^{a}=x^{a} \\ {\widetilde{x}}_{i}^{a}&= {\widetilde{x}}_{j}^{a}={\widetilde{x}}^{a} \\ {\widetilde{x}}_{i}^{b}&= {\widetilde{x}}_{j}^{b}={\widetilde{x}}^{b}. \end{aligned}$$

Then, the solution of the above first-order condition is:

$$\begin{aligned} x^{a}=\frac{w}{2} \end{aligned}$$

and then by our assumption

$$\begin{aligned} {\widetilde{x}}^{a}={\widetilde{x}}^{b}=w-x^{a}=\frac{w}{2}. \end{aligned}$$

In the second stage, player \(i\)’s maximization problems are given by (1) and (2). The first-order conditions of these maximization problems are

$$\begin{aligned}&v^{b}\frac{{\widetilde{x}}_{j}^{a}}{\left({\widetilde{x}}_{i}^{b}+{\widetilde{x}} _{j}^{a}\right)^{2}}-1 \\&v^{a}\frac{{\widetilde{x}}_{j}^{b}}{\left({\widetilde{x}}_{i}^{a}+{\widetilde{x}} _{j}^{b}\right)^{2}}-1. \end{aligned}$$

In order that both constraints will be restrictive these first-order conditions of the maximization problems in the second stage should be positive. Thus, both constraints are restrictive iff

$$\begin{aligned} \frac{v^{b}}{2w}-1&> 0 \Rightarrow w<\frac{v^{b}}{2}. \end{aligned}$$

In this case the total effort in the first stage of the contest is

$$\begin{aligned} TE_{1}=2x^{a}=w \end{aligned}$$

and the total effort in the second stage is

$$\begin{aligned} TE_{2}={\widetilde{x}}^{a}+{\widetilde{x}}^{b}=w. \end{aligned}$$

Therefore

$$\begin{aligned} TE_{1}=TE_{2}. \end{aligned}$$

(2) Assume now that the players have increasing marginal values. When the budget constraint is restrictive both of the restrictions in the maximization problem (6) are restrictive such that

$$\begin{aligned} x_{i}^{b}+{\widetilde{x}}_{i}^{a}&= w \\ x_{i}^{b}+{\widetilde{x}}_{i}^{b}&= w. \end{aligned}$$

Player \(i\)’s maximization problem in the first stage is then

$$\begin{aligned}&\max _{x_{i}^{b}}\left(v^{b}+v^{a}\frac{w-x_{i}^{b}}{w-x_{i}^{b}+w-x_{j}^{b}} -\left(w-x_{i}^{b}\right)\right)\frac{x_{i}^{b}}{x_{i}^{b}+x_{j}^{b}}\\&\quad +\left(v^{b}\frac{w-x_{i}^{b} }{w-x_{i}^{b}+w-x_{j}^{b}}-\left(w-x_{i}^{b}\right)\right)\left(1-\frac{x_{i}^{b}}{ x_{i}^{b}+x_{j}^{b}}\right)-x_{i}^{b}. \end{aligned}$$

The first-order condition is

$$\begin{aligned}&\left[v^{b}+v^{a}\frac{w-x_{i}^{b}}{w-x_{i}^{b}+w-x_{j}^{b}}-\left(w-x_{i}^{b}\right)-v^{b} \frac{w-x_{i}^{b}}{w-x_{i}^{b}+w-x_{j}^{b}}+\left(w-x_{i}^{b}\right)\right]\\&\quad \times \frac{x_{j}^{b}}{ \left(x_{i}^{b}+x_{j}^{b}\right)^{2}}+(v^{a}-v^{b})\frac{-\left(w-x_{j}^{b}\right)}{\left(w-x_{i}^{b}+w-x_{j}^{b}\right)^{2}}\frac{ x_{i}^{b}}{x_{i}^{b}+x_{j}^{b}}\\&\quad +v^{b}\frac{-\left(w-x_{j}^{b}\right)}{ \left(w-x_{i}^{b}+w-x_{j}^{b}\right)^{2}}=0. \end{aligned}$$

Because of the symmetry we denote

$$\begin{aligned} x_{i}^{b}&= x_{j}^{b}=x^{b} \\ {\widetilde{x}}_{i}^{b}&= {\widetilde{x}}_{j}^{b}={\widetilde{x}}^{b} \\ {\widetilde{x}}_{i}^{a}&= {\widetilde{x}}_{j}^{a}={\widetilde{x}}^{a}\!. \end{aligned}$$

The solution of the above first-order condition is

$$\begin{aligned} x^{b}=\frac{w}{2} \end{aligned}$$

and then by our assumption

$$\begin{aligned} {\widetilde{x}}^{a}={\widetilde{x}}^{b}=w-x^{b}=\frac{w}{2}. \end{aligned}$$

In the second stage, player \(i\)’s maximization problems are given by (4) and (5). The first-order conditions of these maximization problems are

$$\begin{aligned}&v^{a}\frac{{\widetilde{x}}_{j}^{b}}{\left({\widetilde{x}}_{i}^{a}+{\widetilde{x}} _{j}^{b}\right)^{2}}-1 \\&v^{b}\frac{{\widetilde{x}}_{j}^{a}}{\left({\widetilde{x}}_{i}^{b}+{\widetilde{x}} _{j}^{a}\right)^{2}}-1. \end{aligned}$$

In order that both constraints will be restrictive these first order conditions of the maximization problems in the second stage should be positive. Thus, both constraints are restrictive iff

$$\begin{aligned} \frac{v^{b}}{2w}-1&> 0\Rightarrow w<\frac{v^{b}}{2}. \end{aligned}$$

In this case the total effort in the first stage of the contest is

$$\begin{aligned} TE_{1}=2x^{b}=w \end{aligned}$$

and the total effort in the second stage is

$$\begin{aligned} TE_{2}={\widetilde{x}}^{a}+{\widetilde{x}}^{b}=w. \end{aligned}$$

Therefore

$$\begin{aligned} TE_{1}=TE_{2}. \end{aligned}$$

\(\square \)

1.6 The proof of Proposition 6

If the budget constraint is partially restrictive only the second restriction in the maximization problem (3) is restrictive such that

$$\begin{aligned} x_{i}^{a}+{\widetilde{x}}_{i}^{b}&< w \\ x_{i}^{a}+{\widetilde{x}}_{i}^{a}&= w. \end{aligned}$$

Thus, if player \(i\) does not win in the first stage his effort in the second stage is \({\widetilde{x}}_{i}^{a}=w-x_{i}^{a}.\) If, on the other hand, he wins in the first stage his maximization problem in the second stage is

$$\begin{aligned} \max _{{\widetilde{x}}_{i}^{b}}v^{b}\frac{{\widetilde{x}}_{i}^{b}}{{\widetilde{x}} _{i}^{b}+{\widetilde{x}}_{j}^{a}}-{\widetilde{x}}_{i}^{b}. \end{aligned}$$

The first-order condition of this maximization problem is

$$\begin{aligned} v^{b}\frac{{\widetilde{x}}_{j}^{a}}{\left({\widetilde{x}}_{i}^{b}+{\widetilde{x}}_{j}^{a}\right)^{2}}-1=0. \end{aligned}$$
(23)

Player \(i\)’s maximization problem in the first stage is then

$$\begin{aligned}&\max _{x_{i}^{a}}\left(v^{a}+v^{b}\frac{{\widetilde{x}}_{i}^{b}}{{\widetilde{x}} _{i}^{b}+w-x_{j}^{a}}-{\widetilde{x}}_{i}^{b}\right)\frac{x_{i}^{a}}{ x_{i}^{a}+x_{j}^{a}}\\&\quad +\left(v^{a}\frac{w-x_{i}^{a}}{w-x_{i}^{a}+{\widetilde{x}} _{j}^{b}}-\left(w-x_{i}^{a}\right)\right) \left(1-\frac{x_{i}^{a}}{x_{i}^{a}+x_{j}^{a}}\right)-x_{i}^{a}. \end{aligned}$$

Therefore the first-order condition is

$$\begin{aligned}&\left[v^{a}+v^{b}\frac{{\widetilde{x}}_{i}^{b}}{{\widetilde{x}}_{i}^{b}+w-x_{j}^{a}} -{\widetilde{x}}_{i}^{b}-v^{a}\frac{w-x_{i}^{a}}{w-x_{i}^{a}+{\widetilde{x}} _{j}^{b}}+\left(w-x_{i}^{a}\right)\right]\frac{x_{j}^{a}}{\left(x_{i}^{a}+x_{j}^{a}\right)^{2}} \nonumber \\&+\left(v^{a}\frac{-{\widetilde{x}}_{j}^{b}}{\left(w-x_{i}^{a}+{\widetilde{x}} _{j}^{b}\right)^{2}}+1\right)\left(1-\frac{x_{i}^{a}}{x_{i}^{a}+x_{j}^{a}}\right)-1=0. \end{aligned}$$
(24)

Because of the symmetry we denote

$$\begin{aligned} x_{i}^{a}&= x_{j}^{a}=x^{a} \\ {\widetilde{x}}_{i}^{a}&= {\widetilde{x}}_{j}^{a}={\widetilde{x}}^{a} \\ {\widetilde{x}}_{i}^{b}&= {\widetilde{x}}_{j}^{b}={\widetilde{x}}^{b}\!. \end{aligned}$$

The solution of the first-order conditions (when \(v^{a}=1\)) from both stages (24) and (23) implies that the equilibrium effort in the first stage \(x^{a}\) is determined by the following equation

$$\begin{aligned}&\left[1+v^{b}+2w-2x^{a}-\sqrt{w-x^{a}}\left(2\sqrt{v^{b}}+\frac{1}{\sqrt{v^{b}}}\right)\right]v^{b}(w-x^{a}) \\&=\left[\sqrt{v^{b}(w-x^{a})}-(w-x^{a})(v^{b}+1)\right]2x^{a}+v^{b}(w-x^{a})4x^{a}\!, \end{aligned}$$

where

$$\begin{aligned} {\widetilde{x}}^{a}&= w-x^{a} \\ {\widetilde{x}}^{b}&= \sqrt{v^{b}(w-x^{a})}-w+x^{a}\!. \end{aligned}$$

According to Propositions 1 and 5, the budget constraint is partially restrictive iff

$$\begin{aligned} \frac{v^{b}}{2}<w<\frac{(v^{b})^{3}+(v^{b})^{2}+6v^{b}}{4(v^{b}+1)^{2}}. \end{aligned}$$

\(\square \)

1.7 The proof of Proposition 7

If the budget constraint is partially restrictive only the first restriction in the maximization problem (6) is restrictive such that

$$\begin{aligned} x_{i}^{b}+{\widetilde{x}}_{i}^{a}&= w \\ x_{i}^{b}+{\widetilde{x}}_{i}^{b}&< w. \end{aligned}$$

Thus if player \(i\) wins in the first stage his effort in the second stage is \({\widetilde{x}}_{i}^{a}=w-x_{i}^{b}.\) If, on the other hand, he does not win in the first stage his maximization problem in the second stage is

$$\begin{aligned} \max _{{\widetilde{x}}_{i}^{b}}v^{b}\frac{{\widetilde{x}}_{i}^{b}}{{\widetilde{x}} _{i}^{b}+{\widetilde{x}}_{j}^{a}}-{\widetilde{x}}_{i}^{b}. \end{aligned}$$

The first order of this maximization problem is

$$\begin{aligned} v^{b}\frac{{\widetilde{x}}_{j}^{a}}{\left({\widetilde{x}}_{i}^{b}+{\widetilde{x}} _{j}^{a}\right)^{2}}-1=0. \end{aligned}$$
(25)

Player \(i\)’s maximization problem in the first stage is then

$$\begin{aligned}&\max _{x_{i}^{b}}\left(v^{b}+v^{a}\frac{w-x_{i}^{b}}{w-x_{i}^{b}+{\widetilde{x}} _{j}^{b}}-\left(w-x_{i}^{b}\right)\right)\frac{x_{i}^{b}}{x_{i}^{b}+x_{j}^{b}}\\&\quad +\left(v^{b}\frac{ {\widetilde{x}}_{i}^{b}}{{\widetilde{x}}_{i}^{b}+w-x_{j}^{b}}-{\widetilde{x}} _{i}^{b}\right) \left(1-\frac{x_{i}^{b}}{x_{i}^{b}+x_{j}^{b}}\right)-x_{i}^{b}. \end{aligned}$$

The first-order-condition is

$$\begin{aligned}&\left[v^{b}+v^{a}\frac{w-x_{i}^{b}}{w-x_{i}^{b}+{\widetilde{x}}_{j}^{b}} -\left(w-x_{i}^{b}\right)-v^{b}\frac{{\widetilde{x}}_{i}^{b}}{{\widetilde{x}} _{i}^{b}+w-x_{j}^{b}}+{\widetilde{x}}_{i}^{b}\right]\frac{x_{j}^{b}}{ \left(x_{i}^{b}+x_{j}^{b}\right)^{2}} \nonumber \\&\quad +\left(v^{a}\frac{-{\widetilde{x}}_{j}^{b}}{\left(w-x_{i}^{b}+{\widetilde{x}} _{j}^{b}\right)^{2}}+1\right)\frac{x_{i}^{b}}{x_{i}^{b}+x_{j}^{b}}-1 =0. \end{aligned}$$
(26)

Because of the symmetry we denote

$$\begin{aligned} x_{i}^{b}&= x_{j}^{b}=x^{b} \\ {\widetilde{x}}_{i}^{a}&= {\widetilde{x}}_{j}^{a}={\widetilde{x}}^{a} \\ {\widetilde{x}}_{i}^{b}&= {\widetilde{x}}_{j}^{b}={\widetilde{x}}^{b}. \end{aligned}$$

The solution of the first-order conditions (when \(v^{a}=1\)) from both stages (26) and (25) implies that the equilibrium effort in the first stage \(x^{b}\) is determined by the following equation

$$\begin{aligned}&\frac{\sqrt{w-x^{b}}}{\sqrt{v^{b}}}+2\sqrt{v^{b}(w-x^{b})}-2w-2x^{b} \\&=\frac{\left[\sqrt{v^{b}}-(v^{b}+1)\sqrt{w-x^{b}}\right]2x^{b}}{v^{b}\sqrt{w-x^{b}}}\!, \end{aligned}$$

where

$$\begin{aligned} {\widetilde{x}}^{a}&= w-x^{b} \\ {\widetilde{x}}^{b}&= \sqrt{v^{b}(w-x^{b})}-w+x^{b}\!. \end{aligned}$$

According to Propositions 3 and 5 the budget constraint is partially restrictive iff

$$\begin{aligned} \frac{v^{b}}{2}<w<\frac{2(v^{b})^{2}+5v^{b}+1}{4(v^{b}+1)^{2}}. \end{aligned}$$

\(\square \)

1.8 The proof of Theorem 1

If the budget constraint is restrictive all of the four restrictions in the maximization problems (13) and (14) are restrictive such that

$$\begin{aligned} x^{a}+{\widetilde{x}}^{b}&= w_{1} \\ x^{a}+{\widetilde{x}}^{a}&= w_{1} \\ y^{c}+{\widetilde{y}}^{d}&= w_{2} \\ y^{c}+{\widetilde{y}}^{c}&= w_{2}. \end{aligned}$$

Thus we denote

$$\begin{aligned} {\widetilde{x}}&= {\widetilde{x}}^{b}={\widetilde{x}}^{a} \\ {\widetilde{y}}&= {\widetilde{y}}^{d}={\widetilde{y}}^{c} \\ x&= x^{a} \\ y&= y^{c}. \end{aligned}$$

Then the first-order condition of player 1’s maximization problem (15) is

$$\begin{aligned}&\left((v^{a}-v^{b})\frac{-r(w_{1}-x)^{r-1}(w_{2}-y)^{r}}{((w_{1}-x)^{r}+(w_{2}-y)^{r})^{2}}\right)\frac{(y)^{r}}{(x)^{r}+(y)^{r}}\nonumber \\&\quad +\left((v^{a}-v^{b})\frac{(w_{1}-x)^{r}}{(w_{1}-x)^{r}+(w_{2}-y)^{r}}-v^{a}\right) \frac{-r(x)^{r-1}(y)^{r}}{((x)^{r}+(y)^{r})^{2}} \nonumber \\&\quad +v^{b}\frac{-r(w_{1}-x)^{r-1}(w_{2}-y)^{r}}{((w_{1}-x)^{r}+(w_{2}-y)^{r})^{2}}=0. \end{aligned}$$
(27)

Similarly, the first-order condition of player 2’s maximization problem (16) is

$$\begin{aligned}&\left((v^{c}-v^{d})\frac{-r(w_{2}-y)^{r-1}(w_{1}-x)^{r}}{\left((w_{1}-x)^{r}+(w_{2}-y)^{r}\right)^{2}}\right)\frac{(x)^{r}}{(x)^{r}+(y)^{r}} \nonumber \\&\quad +\left((v^{c}-v^{d})\frac{(w_{2}-y)^{r}}{(w_{1}-x)^{r}+(w_{2}-y)^{r}}-v^{c}\right) \frac{-r(y)^{r-1}(x)^{r}}{\left((x)^{r}+(y)^{r}\right)^{2}} \nonumber \\&\quad +v^{d}\frac{-r(w_{2}-y)^{r-1}(w_{1}-x)^{r}}{\left((w_{1}-x)^{r}+(w_{2}-y\right)^{r})^{2}}=0. \end{aligned}$$
(28)

Thus, it can be verified that the solution of the above first-order conditions (27) and (28) is

$$\begin{aligned} x&= {\widetilde{x}}=\frac{w_{1}}{2} \\ y&= {\widetilde{y}}=\frac{w_{2}}{2}. \end{aligned}$$

The budget constraints are restrictive if the first-order conditions of the maximization problems in the second stage (9), (10), (11) and (12) are positive. Thus,

$$\begin{aligned} v^{a}\frac{({\widetilde{y}})^{r}r({\widetilde{x}})^{r-1}}{(({\widetilde{x}})^{r}+( {\widetilde{y}})^{r})^{2}}-1&> 0 \\ v^{b}\frac{({\widetilde{y}})^{r}r({\widetilde{x}})^{r-1}}{(({\widetilde{x}})^{r}+( {\widetilde{y}})^{r})^{2}}-1&> 0 \\ v^{c}\frac{({\widetilde{x}})^{r}r({\widetilde{y}})^{r-1}}{(({\widetilde{x}})^{r}+( {\widetilde{y}})^{r})^{2}}-1&> 0 \\ v^{d}\frac{({\widetilde{x}})^{r}r({\widetilde{y}})^{r-1}}{(({\widetilde{x}})^{r}+( {\widetilde{y}})^{r})^{2}}-1&> 0. \end{aligned}$$

This happens when

$$\begin{aligned} \frac{2v^{a}(w_{2})^{r}r(w_{1})^{r-1}}{((w_{1})^{r}+(w_{2})^{r})^{2}}&> 1 \\ \frac{2v^{b}(w_{2})^{r}r(w_{1})^{r-1}}{((w_{1})^{r}+(w_{2})^{r})^{2}}&> 1 \\ \frac{2v^{c}(w_{1})^{r}r(w_{2})^{r-1}}{((w_{1})^{r}+(w_{2})^{r})^{2}}&> 1 \\ \frac{2v^{d}(w_{1})^{r}r(w_{2})^{r-1}}{((w_{1})^{r}+(w_{2})^{r})^{2}}&> 1. \end{aligned}$$

In this case the total effort in the first stage of the contest is

$$\begin{aligned} TE_{1}=x^{a}+y^{c}=\frac{w_{1}+w_{2}}{2}. \end{aligned}$$

The total effort in the second stage of the contest is

$$\begin{aligned} TE_{2}={\widetilde{x}}^{a}+{\widetilde{y}}^{d}={\widetilde{x}}^{b}+{\widetilde{y}}^{c}=\frac{w_{1}+w_{2}}{2}. \end{aligned}$$

Therefore

$$\begin{aligned} TE_{1}=TE_{2}. \end{aligned}$$

\(\square \)

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Megidish, R., Sela, A. Sequential contests with synergy and budget constraints. Soc Choice Welf 42, 215–243 (2014). https://doi.org/10.1007/s00355-013-0723-5

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