Relative difference contest success function

Abstract

In this paper, we present a contest success function (CSF), which is homogeneous of degree zero and in which the probability of winning the prize depends on the relative difference of efforts. In a simultaneous game with two players, we present a necessary and sufficient condition for the existence of a pure strategy Nash equilibrium. This equilibrium is unique and interior. This condition does not depend on the size of the valuations as in an absolute difference CSF. We prove that several properties of Nash equilibrium with the Tullock CSF still hold in our framework. Finally, we consider the case of $n$ players, generalize the previous condition and show that this condition is sufficient for the existence of a unique interior Nash equilibrium in pure strategies. For some parameter values of our CSF and when all players are identical, equilibrium entails full rent dissipation for any number of players.

Appendix

Proof of Proposition 1

Let us see first that $(0,0)$ cannot be an equilibrium. In $(0,0)$, the payoff for each player is $V_{i}/2.$ Let $\varepsilon >0$ be sufficiently close to zero. Suppose that player $1$ deviates and increases her effort to $\varepsilon .$ Let $G=(\varepsilon ,0).$ In this case, $f_{1}(G)=\alpha +\beta ,$ and $f_{2}(G)=\alpha -\beta s.$ If $\alpha +\beta \ge 1$, given that $2\alpha +\beta (1-s)=1,$ then $\alpha -\beta s\le 0.$ Thus, $p_{1}=1$. Therefore, player $1$ will win the contest for sure, and her payoff will be $V_{1}-\varepsilon$, which is greater than $V_{1}/2.$ If $f_{1}(G)=\alpha +\beta <1,$ $p_{1}=\alpha +\beta .$ Because $1=2\alpha +\beta (1-s)\le 2\alpha +\beta <2\alpha +2\beta ,$ $\alpha +\beta >1/2.$ Thus, for $\varepsilon$ sufficiently close to zero, $(\alpha +\beta )V_{1}- \varepsilon >V_{1}/2,$ which implies that $G_{1}=\varepsilon$ is a profitable deviation for player 1. Therefore, $(0,0)$ is not an equilibrium.

To prove that there is an equilibrium in which both players are active, consider a simultaneous move auxiliary game with payoffs

$$\begin{aligned} \hat{\pi }_{i}=\left( \alpha +\beta \frac{G_{i}-sG_{j}}{\sum _{k=1}^{2}G_{k}} \right) V_{i}-G_{i},\quad \text { for all }i\in \{1,2\}. \end{aligned}$$

(37)

We show first that, under Assumption (5), the auxiliary game has an interior equilibrium. Second, we show that this equilibrium of the auxiliary game is also an equilibrium of our game.

The first order conditions associated with the auxiliary game are

$$\begin{aligned} \beta \frac{(1+s)G_{j}}{(\sum _{k=1}^{2}G_{k})^{2}}V_{i}-1=0, \quad i\in \{1,2\}\text {.} \end{aligned}$$

(38)

We readily see that the second order conditions hold; thus payoffs of $i$ are concave in $G_{i}.$ Adding up all equations in (38) and setting $X\equiv \sum _{k=1}^{2}G_{k}$, we obtain

$$\begin{aligned} X^{*}=\frac{(1+s)\beta }{\sum _{k=1}^{2}\frac{1}{V_{k}}}. \end{aligned}$$

(39)

Plugging (39) in (38), we find that

$$\begin{aligned} G_{i}^{*}=\frac{\beta (1+s)V_{i}}{Y_{i}}\left( 1-\frac{1}{Y_{i}}\right) , \quad i\in \{1,2\}\text {.} \end{aligned}$$

(40)

Because $Y_{i}>1,$ $G_{i}^{*}>0.$ From the definition of the notional CSF, we have

$$\begin{aligned} f_{i}(G_{1}^{*},G_{2}^{*})=\alpha +\beta \left( 1-\frac{1+s}{Y_{i}}\right) , \quad i\in \{1,2\}\text {.} \end{aligned}$$

(41)

Note that $f_{i}(G_{1}^{*},G_{2}^{*})\ge 0$ if and only if

$$\begin{aligned} Y_{i}(\alpha +\beta )\ge \beta (1+s),\quad i\in \{1,2\}\text {.} \end{aligned}$$

(42)

Under Assumption (5), condition (42) always holds. Furthermore, because $\sum _{k=1}^{2} f_{k}(G_{1}^{*},G_{2}^{*})=1,$ $f_{i}(G_{1}^{*},G_{2}^{*})\le 1$ for all $i\in \{1,2\}$ . Profits in equilibrium amount to

$$\begin{aligned} \pi _{i}^{*}=\frac{V_{i}}{Y_{i}}\left[ (\alpha +\beta )Y_{i}-\beta (1+s)\left( 2-\frac{1 }{Y_{i}}\right) \right] ,i\in \{1,2\}. \end{aligned}$$

(43)

Assumption (5) implies that profits are non-negative; thus $(G_{1}^{*},G_{2}^{*})$ is an equilibrium of the auxiliary game.

Next, we see that $G^{*}=(G_{1}^{*},G_{2}^{*})$ is also an equilibrium of the original game. Clearly, no player has a profitable deviation $G_{i}$ such that $0\le f_{i}(G_{i},G_{-i}^{*})\le 1$ because $G^{*}$ is an equilibrium of the auxiliary game. In particular, $\tilde{G}_{i}$ such that $f_{i}(\tilde{G}_{i},G_{-i}^{*})=1$ is not a profitable deviation. Thus, a profitable deviation $G_{i}^{\prime }$ can only occur when $f_{i}(G_{i}^{\prime },G_{-i}^{*})>1,$ or $f_{i}(G_{i}^{\prime },G_{-i}^{*})<0.$ In the first case, because $f_{i}$ is increasing in $G_{i},$ $G_{i}^{\prime }>\tilde{G}_{i}\,$ and $p_{i}=1.$ The payoff at $(G_{i}^{\prime },G_{-i}^{*})$ is less than the payoff at $(\tilde{G}_{i},G_{-i}^{*}).$ Therefore, $G_{i}^{\prime }\,\$cannot be a profitable deviation. If $f_{i}(G_{i}^{\prime },G_{-i}^{*})<0,$ $p_{i}=0$ and the payoff for player $i$ in this deviation is not larger than zero. Thus, this deviation is not profitable.Finally, we show that there cannot be an equilibrium $(G_{i}^{*},0)$ with $G_{i}^{*}>0$. If this were the case, $\pi _{i}^{*}=(\alpha +\beta )V_{i}-G_{i}^{*}$ so $G_{i}^{*}>0$ cannot maximize the payoff of $i$.

Proof of Proposition 2

We can not have an equilibrium with both players exerting a positive effort because in this case, and given that Assumption (5) does not hold, payoffs would be negative at least for one player. A player will be better off exerting no effort. Thus, if an equilibrium exists, it has to include players exerting zero effort. Suppose $G=(G_{i},0).$ Because Assumption (5) does not hold, it should be the case that $\alpha +\beta >1$ (otherwise Assumption (5) will hold as we prove in the Appendix). Therefore, $f_{i}(G)>1,$ which implies that $f_{j}(G)<0$ for player $j.$ Thus, $p_{i}(G_{i},0)=1.$ However, for any $G_{i}>0,$ player $i$ by decreasing her effort by $\varepsilon$ will be better off. Thus, a pure strategy Nash equilibrium does not exist.

Assumption (23) is implied by $\alpha +\beta \le 1$.

Recall that the sufficient condition is

$$\begin{aligned} Y_{i}(\alpha +\beta )\ge \beta (n-1+s)\left( 2-\frac{n-1}{Y_{i}}\right) \text {, }\quad i\in \{1,..,n\}. \end{aligned}$$

(44)

The minimum of the left hand side of (44) is achieved when

$$\begin{aligned} Y_{i}=\frac{\beta (n-1+s)}{\alpha +\beta }, \end{aligned}$$

and the corresponding value of the left hand side of (44) is

$$\begin{aligned} -\frac{\beta ^{2}(n-1+s)^{2}}{\alpha +\beta }+\beta ((n-1+s)(n-1) \end{aligned}$$

Thus, the the left hand side of (44) is always positive if $\alpha (n-1)\ge s\beta ,$ which considering (17) (i.e., $n\alpha +\beta (1-s)=1$), is $\alpha +\beta \le 1$.

Proof of Proposition 4

Note first that $(0,0,..,0)$ can not be an equilibrium. In $(0,...,0),$ the payoff for each player is $V_{i}/n.$ Let $\varepsilon >0$ be sufficiently close to zero. Suppose that player $i$ deviates and increases her effort such that now $G_{i}=\varepsilon .$ Let $G=(0,..,\varepsilon ,..,0).$ In this case, $f_{i}(G)=\alpha +\beta .$ If $\alpha +\beta \ge 1$, given that $n\alpha +\beta (1-s)=1,$ then $\alpha -(\beta s)/(n-1)\le 0.$ Thus, $p_{i}=1$. Therefore, player $i$ wins the contest for sure, and her payoff will be $V_{i}-\varepsilon ,$ which is greater than $V_{i}/n.$ If $f_{i}(G)=\alpha +\beta <1,$ $p_{i}=\alpha +\beta .$ Because $1=n\alpha +\beta (1-s)\le n\alpha +\beta <n\alpha +n\beta ,$ $\alpha +\beta >1/n,$ for $\varepsilon$ sufficiently close to zero, $(\alpha +\beta )V_{i}-\varepsilon >V_{i}/n.$

To prove that there is an equilibrium where all players are active, consider a simultaneous move auxiliary game with payoffs

$$\begin{aligned} \hat{\pi }_{i}=\left( \alpha +\beta \frac{G_{i}-s\frac{\sum _{j\ne i}G_{j}}{n-1}}{ \sum _{j=1}^{n}G_{j}}\right) V_{i}-G_{i},\quad i\in \{1,..,n\}\text {.} \end{aligned}$$

(45)

We show first that, under assumptions (23) and (24), the auxiliary game has an interior equilibrium. Second, we show that this equilibrium of the auxiliary game is also an equilibrium of our game.

The first order conditions associated with the auxiliary game are

$$\begin{aligned} \beta \frac{\sum _{j\ne i}G_{j}(1+\frac{s}{n-1})}{\left( \sum _{j=1}^{n}G_{j}\right) ^{2}} V_{i}-1=0,\quad i\in \{1,..,n\}. \end{aligned}$$

(46)

We readily see that the second order conditions hold; thus, payoffs of $i$ are concave in $G_{i}$.

Write Eq. (46) as follows:

$$\begin{aligned} \frac{\sum _{j\ne i}G_{j}}{(\sum _{j=1}^{n}G_{j})^{2}}=\frac{1}{\beta V_{i}\left( 1+\frac{s}{n-1}\right) }. \end{aligned}$$

(47)

Adding up the equations in (47) over all players and simplifying we obtain

$$\begin{aligned} \sum _{j=1}^{n}G_{j}=\frac{(n-1)\beta \left( 1+\frac{s}{n-1}\right) }{\sum _{j=1}^{n}\frac{1}{V_{j}}}. \end{aligned}$$

(48)

Realizing that (47) can be written as

$$\begin{aligned} \frac{\sum _{j=1}^{n}G_{j}-G_{i}}{\left( \sum _{j=1}^{n}G_{j}\right) ^{2}}=\frac{1}{\beta V_{i}\left( 1+\frac{s}{n-1}\right) }, \end{aligned}$$

(49)

and considering (48), we obtain

$$\begin{aligned} G_{i}^{*}=\frac{(n-1)\beta \left( 1+\frac{s}{n-1}\right) V_{i}}{Y_{i}}\left( 1-\frac{n-1}{ Y_{i}}\right) ,\quad i\in \{1,..,n\}. \end{aligned}$$

(50)

Because Assumption (24) requires $Y_{i}>n-1,$ for all $i\in \{1,..,n\},G_{i}^{*}\ge 0.$

From the definition of the notional CSF $f_{i}$ , we have

$$\begin{aligned} f_{i}(G^{*})&= \alpha +\beta \frac{G_{i}^{*}\left( 1+\frac{s}{n-1}\right) -s\frac{ \sum _{j=1}^{n}G_{j}^{*}}{n-1}}{\sum _{j=1}^{n}G_{j}^{*}}=\end{aligned}$$

(51)

$$\begin{aligned}&= \alpha +\beta \frac{G_{i}^{*}}{\sum _{j=1}^{n}G_{j}^{*}}\left( 1+\frac{s}{n-1 }\right) -\frac{s\beta }{n-1}. \end{aligned}$$

(52)

Considering (49), we have

$$\begin{aligned} \frac{G_{i}^{*}}{\sum _{j=1}^{n}G_{j}^{*}}=1-\frac{(n-1)}{Y_{i}}, \end{aligned}$$

(53)

and we obtain

$$\begin{aligned} f_{i}(G^{*})&= \alpha +\beta \left( 1-\frac{(n-1)}{Y_{i}}\right) \left( 1+\frac{s}{n-1}\right) - \frac{s\beta }{n-1}\end{aligned}$$

(54)

$$\begin{aligned}&= \alpha +\beta \left( 1-\frac{(n-1)\left( 1+\frac{s}{n-1}\right) }{Y_{i}}\right) . \end{aligned}$$

(55)

Note that $f_{i}(G^{*})\ge 0$ if and only if $(\alpha +\beta )Y_{i}\ge \beta (n-1)(1+\frac{s}{n-1})$. However, given Assumptions (23) and ( 24), this condition always hold. Furthermore, because $\sum _{i=1}^{n}f_{i}(G)=1$, we have $f_{i}(G^{*})\le 1$.

Finally, profits in equilibrium amount to

$$\begin{aligned} \pi _{i}^{*}=\left( \alpha +\beta \left( 1-\frac{(n-1)(1+\frac{s}{n-1})}{Y_{i}} \right) \right) V_{i}-\frac{\left( n-1\right) \beta \left( 1+\frac{s}{n-1}\right) V_{i}}{Y_{i}} \left( 1-\frac{n-1}{Y_{i}}\right) ,\qquad \end{aligned}$$

(56)

which can be rewritten as follows:

$$\begin{aligned} \pi _{i}^{*}&= \left( \alpha +\beta \right) V_{i}-\beta (n-1)\left( 1+\frac{s}{n-1}\right) \frac{ V_{i}}{Y_{i}}-\frac{(n-1)\beta \left( 1+\frac{s}{n-1}\right) V_{i}}{Y_{i}}\left( 1-\frac{n-1}{ Y_{i}}\right) ,\nonumber \\\end{aligned}$$

(57)

$$\begin{aligned} \pi _{i}^{*}&= \frac{V_{i}}{Y_{i}}((\alpha +\beta )Y_{i}-\beta (n-1)(1+ \frac{s}{n-1})-(n-1)\beta \left( 1+\frac{s}{n-1}\right) \left( 1-\frac{n-1}{Y_{i}}\right) ,\nonumber \\\end{aligned}$$

(58)

$$\begin{aligned} \pi _{i}^{*}&= \frac{V_{i}}{Y_{i}}\left( (\alpha +\beta )Y_{i}-\beta (n-1)\left( 1+ \frac{s}{n-1}\right) \left( 2-\frac{n-1}{Y_{i}}\right) \right) . \end{aligned}$$

(59)

Assumption (23) implies that profits are non-negative; thus, $G^{*}=(G_{1}^{*},..,G_{n}^{*})$ is an equilibrium of the auxiliary game. Let us finally see that $G^{*}$ is also an equilibrium of the original game. Clearly, no player has a profitable deviation $G_{i}$ such that $0\le f_{i}(G_{i},G_{-i}^{*})\le 1$ because $G^{*}$ is an equilibrium of the auxiliary game. In particular, $\tilde{G}_{i}$ such that $f_{i}(\tilde{G}_{i},G_{-i}^{*})=1$ is not a profitable deviation. Thus, a profitable deviation $G_{i}^{\prime }$ can only occur when $f_{i}(G_{i}^{\prime },G_{-i}^{*})>1,$ or $f_{i}(G_{i}^{\prime },G_{-i}^{*})<0.$ In the first case, because $f_{i}$ is increasing in $G_{i},$ $G_{i}^{\prime }>G_{i}^{*}.$ Furthermore, because player $i$ has increased her effort, the $f_{j}$ of the other players is reduced; thus, $f_{j}(G_{i}^{\prime },G_{-i}^{*})\le f_{j}(G^{*})\le 1.$ Therefore, by the rationing rule, $p_{i}=1.$ Thus, payoffs at $(G_{i}^{\prime },G_{-i}^{*})$ are less than payoffs at $(\tilde{G} _{i},G_{-i}^{*}).$ Therefore, $G_{i}^{\prime }\,$can not be a profitable deviation. If $f_{i}(G_{i}^{\prime },G_{-i}^{*})<0,$ $p_{i}=0$; then, the payoff for player $i$ in this deviation cannot be greater than zero. Thus, this deviation is not profitable.

Proof of Proposition 5

Suppose that there is a Nash equilibrium with $G_{i}^{*}=0$ for some $i$. Then,

$$\begin{aligned} p_{i}=\alpha -\frac{\beta s}{n-1}=\frac{-\alpha -\beta +1}{n-1}<0. \end{aligned}$$

(60)

Because probabilities add up to one, $p_{i}<0$ for some $i$ implies that some active agent, say $j$, is rationed. However, we already know that in equilibrium no agent can be rationed.

We now tackle the case of $\alpha +\beta \le 1$. A coalition $C$ is a subset of the set of players. Let $c$ be the number of players in coalition $C$. Consider the following condition:

$$\begin{aligned} \text {For any coalition }C\text { and player }i\notin C\text {, }\quad V_{i}\sum _{j\in C}\frac{1}{V_{j}}>c-1. \end{aligned}$$

(61)

Lemma 1 shows that condition (61) is always satisfied whenever Assumption (24) holds.

Lemma 1

If Assumption (24) holds, then $V_{i}\sum _{j\in C}\frac{1}{V_{j}}>c-1$ for any coalition $C$ and player $i\notin C.$

Proof

Suppose that condition (61) does not hold. Then, there is a coalition $C$ and a player $i\notin C$ such that $V_{i}\sum _{j\in C}\frac{1}{V_{j}}\le c-1.$ In particular, for any player $k\notin C$ such that $V_{k}\le V_{i},$ $V_{k}\sum _{j\in C}\frac{1}{V_{j}}\le c-1.$ Let $k_{\min }\notin C$ be such that $V_{k_{\min }}\le V_{k}$ for all $k\notin C.$ Let us see that $Y_{k_{\min }}\le n-1,$ in contradiction with Assumption ( 24). Given that $Y_{k_{\min }}=V_{k_{\min }}\sum _{j=1}^{n} \frac{1}{V_{j}},$ we can rewrite $Y_{k_{\min }}$ as:

$$\begin{aligned} Y_{k_{\min }}=V_{k_{\min }}\sum _{j\in C}\frac{1}{V_{j}}+V_{k_{\min }}\sum _{k\notin C}\frac{1}{V_{k}}. \end{aligned}$$

(62)

Given that $V_{k_{\min }}\sum _{j\in C}\frac{1}{V_{j}}\le c-1,$ and $V_{k_{\min }}\sum _{k\notin C}\frac{1}{V_{k}}\le n-c,$ $Y_{k_{\min }}\le n-1.$ $\square$

Proof of Proposition 6

Suppose we have a Nash equilibrium in which only players $1,2,...,k$ are active. First we show that player $i=k+1,....n$ is not rationed. Indeed

$$\begin{aligned} p_{i}=\alpha -\beta \frac{s}{n-1}=\frac{1-\alpha -\beta }{n-1}\ge 0. \end{aligned}$$

(63)

Thus, player $i$ maximizes, at least in a neighborhood of the Nash equilibrium, over the notional CSF. The FOC of payoff maximization for $i$ is

$$\begin{aligned} \frac{\partial \pi _{i}}{\partial G_{i}}=\beta \frac{\left( 1+\frac{s}{n-1}\right) }{ \sum _{j=1}^{k}G_{j}}V_{i}-1. \end{aligned}$$

(64)

Now computing the expenses made by active players, we see that

$$\begin{aligned} \sum _{j=1}^{k}G_{j}=\beta \frac{(k-1)\left( 1+\frac{s}{n-1}\right) }{\sum _{j=1}^{k}\frac{1 }{V_{j}}}; \end{aligned}$$

(65)

thus,

$$\begin{aligned} \frac{\partial \pi _{i}}{\partial G_{i}}=\frac{V_{i}\sum _{j=1}^{k}\frac{1}{ V_{j}}}{k-1}-1 \end{aligned}$$

(66)

and applying condition (61), we obtain

$$\begin{aligned} \frac{\partial \pi _{i}}{\partial G_{i}}>\frac{k-1}{k-1}-1=0, \end{aligned}$$

(67)

which contradicts that the best reply of $i$ is zero.