Step-by-step group contests with group-specific public-good prizes

Abstract

The achievements reached by group members’ effort, which accompanies the possibility of members’ free-riding, affect the outcomes of competition among groups. In some cases, each achievement has the binary characteristic of “one or nothing.” For example, research groups face the challenge of making a scientific finding or not. The groups compete for a scientific breakthrough by making the related findings in a “step-by-step” manner. One finding could fail because of a mistake caused by a slight lack of effort by one member. Such a characteristic of “one or nothing” motivates group members without any incentive scheme. This study analyzes group contests with group-specific public-good prizes, in which we introduce a step function with the characteristics of “one or nothing” and “step-by-step” as a group impact function. We show the existence of the Nash equilibrium at which no group member free-rides on the others and at which more achievements than those reachable by a single member are reached.

Appendix: Proofs

Proof of Lemma 1

Suppose that member j in $G_i$ on achievement level $\alpha -1$ has an incentive to cover the cost of $m_{\alpha }$ when the other members in $G_i$ expend $\sum _{k=0}^{\alpha -1} m_k$ in total, that is $\frac{\alpha }{\alpha + \beta } v_j^i - m_\alpha \ge \frac{\alpha -1}{\alpha - 1 + \beta } v_j^i \,\iff$$\frac{\beta }{(\alpha - 1 + \beta )(\alpha + \beta )} v_j^i - m_\alpha \ge 0$. Then, suppose that the other members in $G_i$ reduce their total effort by $m_{\alpha -1}$. If j adds $m_{\alpha -1}$ to her expense, $G_i$ can stay on achievement level $\alpha$. If j does not expend any additional cost, $G_i$ steps down to $\alpha - 1$. Then, the difference in j’s payoffs is

$$\begin{aligned}&\left( \frac{\alpha }{\alpha + \beta } v_j^i - (m_\alpha + m_{\alpha -1}) \right) - \left( \frac{\alpha -1}{\alpha - 1 + \beta } v_j^i - m_\alpha \right) \\&\quad = \frac{\beta }{(\alpha - 1 + \beta )(\alpha + \beta )} v_j^i - m_{\alpha -1} \ge \frac{\beta }{(\alpha - 1 + \beta )(\alpha + \beta )} v_j^i - m_\alpha \ge 0, \end{aligned}$$

because of the assumption of $m_1 \le m_2 \le m_3 \le \cdots$. Thus, j has an incentive to pay $m_{\alpha -1}$ for the other members in $G_i$. Furthermore, suppose that the other members in $G_i$ add a reduction in $m_{\alpha -2}$. If j pays $m_{\alpha -2}$ for the others, $G_i$ can stay on achievement level $\alpha$. If j does not, $G_i$ steps down to $\alpha - 1$. Then, the difference in j’s payoffs is

$$\begin{aligned}&\left( \frac{\alpha }{\alpha + \beta } v_j^i - (m_\alpha + m_{\alpha -1} +m_{\alpha -2}) \right) - \left( \frac{\alpha -1}{\alpha - 1 + \beta } v_j^i - (m_\alpha +m_{\alpha -1} )\right) \\&\quad = \frac{\beta }{(\alpha - 1 + \beta )(\alpha + \beta )} v_j^i - m_{\alpha -2} \ge \frac{\beta }{(\alpha - 1 + \beta )(\alpha + \beta )} v_j^i - m_\alpha \ge 0. \end{aligned}$$

Thus, j also has an incentive to pay $m_{\alpha -2}$ for the other members. We repeat this process until j expends all costs of achievement level $\alpha$ for the other members. Then, we have $( \frac{\alpha }{\alpha + \beta } v_j^i - \sum _{k=1}^{\alpha }m_k ) - ( \frac{\alpha -1}{\alpha - 1 + \beta } v_j^i - \sum _{k=2}^{\alpha }m_k ) \ge 0$. Hence, j has an incentive to pay all the costs of achievement level $\alpha$ for the other members in $G_i$.$\square$

Proof of Lemma 2

$\hat{\alpha _i}$ is such that $\frac{\beta }{(\hat{\alpha _i}-1+\beta )(\hat{\alpha _i}+\beta )} v_{n_i}^i - m_{\hat{\alpha _i}} \ge 0$ and $\frac{\beta }{(\hat{\alpha _i}+\beta )(\hat{\alpha _i}+1 +\beta )} v_{n_i}^i - m_{\hat{\alpha _i}+1} < 0$ at each $\beta \ge 1$. From the former inequality, we have

$$\begin{aligned}&\frac{\beta }{m_{\hat{\alpha _i}}} v_{n_i}^i - (\hat{\alpha _i}-1+\beta )(\hat{\alpha _i}+\beta ) \ge 0 \\&\quad \iff \hat{\alpha _i}^2 +(2\beta -1 ) \hat{\alpha _i} + \beta ^2 - \beta - \frac{\beta }{m_{\hat{\alpha _i}}} v_{n_i}^i \le 0 \\&\quad \iff \left( \hat{\alpha _i} - \frac{1-2\beta + \sqrt{1+4\beta v_{n_i}^i / m_{\hat{\alpha _i}} }}{2} \right) \left( \hat{\alpha _i} - \frac{1-2\beta - \sqrt{1+4\beta v_{n_i}^i / m_{\hat{\alpha _i}} }}{2} \right) \le 0 \end{aligned}$$

Noting that $\frac{1-2\beta - \sqrt{1+4\beta v_{n_i}^i / m_{\hat{\alpha _i}} }}{2} < 0$ for $\beta \ge 1$, we have $0 < \hat{\alpha _i} \le \frac{1-2\beta + \sqrt{1+4\beta v_{n_i}^i / m_{\hat{\alpha _i}} }}{2}$. For $\hat{\alpha _i}$ equal to or more than achievement level $t \ge 1$, we need $\frac{1-2\beta + \sqrt{1+4\beta v_{n_i}^i / m_{\hat{\alpha _i}} }}{2} \ge t$. Solving this inequality, we have $v_{n_i}^i \ge \frac{m_t}{\beta } (t + \beta - 1) (t + \beta )$. $\square$

Proof of Lemma 3

(1) Necessity. Suppose that $\varvec{x^*}$ is a Nash equilibrium on a positive achievement level $(\hat{\alpha _i}, \beta )$ or $(\alpha , \beta )$ and $\varvec{x^*}$ does not meet (i) or (ii), that is for some i and j, (i’) $\sum _{j=1}^{n_i} x_j^{i*} > \sum _{k=0}^{t} m_k$ or (i”) $\sum _{j=1}^{n_i} x_j^{i*} < \sum _{k=0}^{t} m_k$, $t \in \{\hat{\alpha _i}, \alpha \}$, or (ii’) $\frac{\beta }{(\hat{\alpha _i}-1+\beta )(\hat{\alpha _i}+\beta )}v_j^i < x_j^{i*},$$\frac{ \hat{\alpha _i} }{ \hat{\alpha _i} + \beta } v_{n_i}^i < x_{n_i}^{i*}$ on $( \hat{\alpha _i}, \beta )$ or $\frac{\beta }{(\alpha -1+\beta )(\alpha +\beta )}v_j^i < x_j^{i*}$ on $(\alpha , \beta )$.

(i’) contradicts the best response of j because decreasing j’s effort that is sufficiently small $\epsilon >0$ increases her payoff by $\epsilon$. (i”) means that $G_i$ is not on t. Thus, these contradict that $\varvec{x^*}$ is a Nash equilibrium on $(t, \beta )$, $t \in \{\hat{\alpha _i}, \alpha \}$.

Noting that the upper limit of each member’s effort is $x_j^i \le \frac{\beta }{(\hat{\alpha _i}- 1+\beta )(\hat{\alpha _i}+\beta )}v_j^i,$$x_{n_i}^i \le \frac{ \hat{\alpha _i} }{ \hat{\alpha _i} + \beta } v_{n_i}^i$ on $( \hat{\alpha _i}, \beta )$ and $x_j^i \le \frac{\beta }{(\alpha -1+\beta )(\alpha +\beta )}v_j^i$ on $(\alpha , \beta )$ for each $j \in G_i$ in the best responses, (ii’) contradicts these limits; in other words, some small $x_j^i < x_j^{i*}$ brings about a better payoff to j. Thus, (i) and (ii) are necessary conditions.

(2) Sufficiency. Suppose that $\varvec{x^*}$, which meets both (i) and (ii), is not a Nash equilibrium. Given the others’ effort, one of the members of $G_i$, say $j\in G_i$, is made better off by deviating from $x_j^{i*}$ to $x_j^i$,

(A): $p_i (X_1^*, \ldots , X_{i-1}^*, X_i^*, X_{i+1}^*, \ldots , X_{N}^*) v_j^i - x_j^{i*} < p_i (X_1^*, \ldots , X_{i-1}^*, X_i, X_{i+1}^*,$$\ldots , X_{N}^*) v_j^i - x_j^{i}$.

First, suppose that $x_j^i > x_j^{i*}$. $G_i$ needs to reach at least one higher achievement level $\alpha + 1$ ($\hat{\alpha _i}+1$ in the case of $\hat{\alpha _i}$) for the increase in j’s effort to obtain $p_i(X_1^*, \ldots , X_{i-1}^*, X_i, X_{i+1}^*, \ldots , X_{N}^*) > p_i(X_1^*, \ldots , X_{i-1}^*, X_i^*, X_{i+1}^*, \ldots , X_{N}^*)$. The condition under which j’s payoff increases by changing her effort is

$$\begin{aligned} \begin{array}{ll} &{} \left( \frac{\alpha +1}{\alpha + 1 + \beta }v_j^i - x_j^i\right) - \left( \frac{\alpha }{\alpha + \beta }v_j^i - x_j^{i*}\right)> 0 \\ &{}\quad \iff \frac{\beta }{(\alpha + 1 +\beta )(\alpha + \beta )}v_j^i > x_j^i - x_j^{i*}. \end{array} \end{aligned}$$

Noting that $\alpha \ge \hat{\alpha _i} + 1$ such that $\frac{\beta }{(\hat{\alpha _i}-1+\beta )(\hat{\alpha _i}+\beta )} v_{n_i}^i - m_{\hat{\alpha _i}} \ge 0$ and $\frac{\beta }{(\hat{\alpha _i}+\beta )(\hat{\alpha _i}+1 +\beta )} v_{n_i}^i - m_{\hat{\alpha _i}+1} < 0$, we have $m_{\alpha +1} \ge m_{\hat{\alpha _i}+1}> \frac{\beta }{(\hat{\alpha _i}+\beta )(\hat{\alpha _i}+1 +\beta )} v_j^i> \frac{\beta }{(\alpha + 1 +\beta )(\alpha + \beta )}v_j^i> x_j^i - x_j^{i*}>0$. This means that $G_i$ cannot reach achievement levels $\alpha +1$ and $\hat{\alpha _i}+1$ by only j’s increase in effort. This contradicts (A).

Second, suppose that $x_j^i < x_j^{i*}$. Then, the achievement level of $G_i$ steps down from $\alpha$ to $\alpha -1$ ($\hat{\alpha _i}-t$, $1 \le t \le \hat{\alpha _i}$ in the case of $\hat{\alpha _i}$) once j decreases her efforts from (i). The condition under which j benefits more by decreasing her effort is

$$\begin{aligned} \begin{array}{ll} &{} \left( \frac{\alpha -1}{\alpha -1 + \beta } v_j^i - x_j^i\right) - \left( \frac{\alpha }{\alpha + \beta }v_j^i - x_j^{i*}\right) > 0 \\ &{}\quad \iff \frac{\beta }{(\alpha -1 +\beta )(\alpha + \beta )}v_j^i < x_j^{i* }- x_j^i \le x_j^{i* }. \end{array} \end{aligned}$$

The last inequality contradicts (ii). At achievement level $\hat{\alpha _i}$, $\frac{t \beta }{(\hat{\alpha _i} - t + \beta )(\hat{\alpha _i}+\beta )}v_j^i < x_j^{i* }- x_j^i \le x_j^{i* } = \sum _{k=0}^{\hat{\alpha _i}}m_k- \sum _{h \ne j}x_h^{i*} \le \sum _{k=0}^{\hat{\alpha _i}} m_k$ is obtained by a similar argument. This inequality also contradicts (ii) or $\frac{ \beta }{(\hat{\alpha _i} - t + \beta )(\hat{\alpha _i} - t + 1+ \beta )} v_{n_i}^i - m_{\hat{\alpha _i}-t+1} \ge 0$ for $1 \le t \le \hat{\alpha _i}$ from the definition of $\hat{\alpha _i}$. Thus, (i) and (ii) are also sufficient conditions. $\square$

Proof of Corollary 1

By using (i) of Lemma 3, the sum of condition (ii) of Lemma 3 of all the members in $G_i$ is the corollary’s condition. $\square$

Proof of Lemma 4

(1) Necessity. Suppose that the condition is not met when there is at least one Nash equilibrium on $(\hat{\alpha _i}, \beta )$ or some $(\alpha , \beta )$. Then, at least one pair of $(\hat{\alpha _i}, \beta )$ or $(\alpha , \beta )$ meets $\frac{\beta }{(\hat{\alpha _i}-1+\beta )(\hat{\alpha _i}+\beta )}\sum _{j=1}^{n_i-1}v_j^i +\frac{\hat{\alpha _i}}{\hat{\alpha _i}+\beta } v_{n_i}^i <\sum _{k=0}^{\hat{\alpha _i}} m_k$ or $\frac{\beta }{(\alpha -1+\beta )(\alpha +\beta )} \sum _{j=1}^{n_i} v_j^i < \sum _{k=0}^{\alpha } m_k$ at the Nash equilibrium. These inequalities contradict Corollary 1. Thus, the conditions are necessary.

(2) Sufficiency. When the condition is met, there is at least one strategy profile $\varvec{x}$ at which we can allocate all members to expend effort levels for their own groups such that $x_j^{i} \le \frac{\beta }{(\hat{\alpha _i}-1+\beta )(\hat{\alpha _i}+\beta )}v_j^i,$$x_{n_i}^{i} \le \frac{\hat{\alpha _i}}{\hat{\alpha _i}+\beta }v_{{n_i}}^i$,$\sum _{j=1}^{n_i} x_j^{i} = \sum _{k=0}^{\hat{\alpha _i}} m_k$ or $x_j^i \le \frac{\beta }{(\alpha -1+\beta )(\alpha +\beta )} v_j^i$, $\sum _{j=1}^{n_i}x_j^i = \sum _{k=0}^{\alpha }m_k$. For example, we allocate some members in each group—from member 1 in ascending order—their full effort level, $x_j^{i} = \frac{\beta }{(\hat{\alpha _i}-1+\beta )(\hat{\alpha _i}+\beta )}v_j^i$ (or $x_j^i = \frac{\beta }{(\alpha - 1 +\beta )(\alpha + \beta )}v_j^i$), allocate only last one member $x_j^{i} \le \frac{\beta }{(\hat{\alpha _i}- 1+\beta )(\hat{\alpha _i}+\beta )}v_j^i$ or $x_{n_i}^{i} \le \frac{\hat{\alpha _i}}{\hat{\alpha _i}+\beta } v_{n_i}^i$(or $x_j^i \le \frac{\beta }{(\alpha - 1 +\beta )(\alpha + \beta )}v_j^i$), and allocate the others $x_j^i = 0$ such that $\sum _{j=1}^{n_i} x_j^i = \sum _{k=0}^{\hat{\alpha _i}} m_k$ (or $\sum _{j=1}^{n_i} x_j^i = \sum _{k=0}^{\alpha } m_k$). From Lemma 3, this $\varvec{x}$ is a Nash equilibrium. Thus, the condition is sufficient. $\square$

Proof of Proposition 2

From Lemma 4, we can allocate each member in each group at least one positive effort level, $x_j^{i*} > 0$, for all i and j, by using Expense Rule A such that the conditions of Lemma 3: $\frac{\beta }{(\hat{\alpha _i}-1+\beta )(\hat{\alpha _i}+\beta )}v_j^i \ge x_j^{i*},$$\frac{\hat{\alpha _i}}{\hat{\alpha _i}+\beta } v_{n_i}^i \ge x_{n_i}^{i*}$, and $\sum _{j=1}^{n_i}x_j^{i*} = \sum _{k=0}^{\hat{\alpha _i}} m_k$, and $\frac{\beta }{(\alpha -1+\beta )(\alpha + \beta )} v_j^i \ge x_j^{i*}$ and $\sum _{j=1}^{n_i}x_j^{i*} = \sum _{k=0}^{\alpha } m_k$. In fact, we multiply both sides of the conditions of Lemma 4 by $\frac{V_j^i}{\sum _{j=1}^{n_i}V_j^i}$. Then, we have the conditions of Lemma 3. Thus, this proposition holds. $\square$

Proof of Theorem 1

Suppose that the condition is met. Since the condition of Lemma 4 is $\frac{1}{N} \sum _{j=1}^{n_i}v_j^i \ge m_1$ at $(\alpha , \beta ) = (1, N -1)$, there is at least one Nash equilibrium at which each member in each group expends a positive effort level on achievement levels $(1,N-1)$, according to Lemma 4 and Proposition 2. $\square$

Proof of Proposition 3

We have this proposition by replacing $(\alpha , \beta )$ with $(t, (N-1)t)$ in the proof of Lemma 4 and by applying Proposition 2. $\square$

Proof of Corollary 2

Suppose that Proposition 3 holds on $(\alpha , \beta ) = (t, (N-1) t)$, $t \ge \hat{\alpha _i} + 1$, $\hat{\alpha _i} \ge 1$ for all i. Noting that $\Delta _{t-1}^t ((N-1)t)=\frac{N-1}{N(Nt-1)}$ on $(\alpha , \beta ) = (t, (N-1) t)$, we compare the benefit and cost on t with those at $t-1$. For all $t \ge \hat{\alpha _i} + 1$,

$$\begin{aligned} \frac{N-1}{N(N(t-1) - 1) } \sum _{j=1}^{n_i} v_j^i> \frac{N-1}{N(Nt - 1)} \sum _{j=1}^{n_i} v_j^i \ge \sum _{k=0}^t m_k > \sum _{k=0}^{t-1} m_k. \end{aligned}$$

(3)

This inequality indicates that Proposition 3 also holds at $(\alpha , \beta ) = (t-1, (N-1)(t-1))$ when that holds at $t \ge \hat{\alpha _i} + 1$. Thus, there is also at least one Nash equilibrium on achievement level $t-1$ if there is at least one Nash equilibrium on t. $\square$

Proof of Lemma 5

$\alpha$ for maximizing $G_i$’s payoff has to meet ${\bar{\Delta }}_{\alpha -1}^\alpha (\beta ) \ge 0$ and ${\bar{\Delta }}_{\alpha }^{\alpha +1} (\beta ) < 0$. On the contrary, from Corollary 1, the achievement level of $G_i$ in the case in which each individual maximizes her payoff meets$\frac{\beta }{(\hat{\alpha _i}-1+\beta )(\hat{\alpha _i}+\beta )} \sum _{j=1}^{n_i-1}v_j^i +\frac{\hat{\alpha _i}}{\hat{\alpha _i}+\beta } v_{n_i}^i\ge \sum _{k=0}^{\hat{\alpha _i}} m_k$ or $\frac{\beta }{(\alpha -1 + \beta )(\alpha + \beta )}\sum _{j=1}^{n_i}v_j^i - \sum _{k=0}^\alpha m_k \ge 0$, $\alpha \ge \hat{\alpha _i}+1$, $\hat{\alpha _i} \ge 1$. Note that $\Delta _{\alpha -1}^{\alpha }(\beta )$ is a strictly decreasing function of $\alpha$ and $\sum _{k=0}^\alpha m_k \ge m_\alpha$ for all $\alpha \ge 1$. Comparing $\frac{\beta }{(\alpha -1 + \beta )(\alpha + \beta )} \sum _{j=1}^{n_i}v_j^i - \sum _{k=0}^\alpha m_k \ge 0$ with $\frac{\beta }{(\alpha +\beta )(\alpha -1+\beta )} \sum _{j=1}^{n_i}v_j^i - m_\alpha \ge 0$, $\alpha$ in the case of maximizing each individual’s payoff is the same level as or less than that in the case of each group maximizing its payoff for any $\beta \ge 1$. $\square$