Rent dissipation in large population Tullock contests

Abstract

Tullock contests model rent seeking behavior where agents exert unproductive effort to probabilistically win a fixed prize. Rent dissipation measures the social loss involved in effort exertion in such contests. Tullock contests are characterized by an impact function, which measures how effort impacts success, and a cost of effort function. If these functions are asymmetric and non-linear, then the contest cannot be solved in closed form. Hence, we approximate such contests with a large population contest, for which Nash equilibria and rent dissipation can be explicitly calculated. Rent dissipation is then the ratio of the effort elasticity of impact to the effort elasticity of cost. Greater elasticity of impact incentivizes more exertion of unproductive effort generating higher social loss.

Appendix

1.1 Proofs

Proof of Proposition 2.1

It is well-known that in finite player contests, all players playing ${\hat{x}}=0$ cannot be a Nash equilibrium. Hence, we only need to consider the first payoff in (2). Differentiating that payoff with respect to ${\hat{x}}_i$, we obtain the first order condition

$$\begin{aligned} \frac{\partial {\hat{\pi }}_i}{\partial {\hat{x}}_i}&=\frac{\left( \sum _j{\hat{x}}_j^r\right) r {\hat{x}}_i^{r-1}-{\hat{x}}_i^rr{\hat{x}}_i^{r-1}}{\left( \sum _j{\hat{x}}_j^r\right) ^2}V-\kappa \gamma {\hat{x}}_i^{\gamma -1}\nonumber \\&\quad =\left[ \frac{{\hat{x}}_i^{r-1}}{\sum _j{\hat{x}}_j^r} -\frac{{\hat{x}}_i^{2r-1}}{\left( \sum _j{\hat{x}}_j^r\right) ^2}\right] Vr-\kappa \gamma {\hat{x}}_i^{\gamma -1}\nonumber \\&\quad =0. \end{aligned}$$

(31)

In a symmetric Nash equilibrium, we will have ${\hat{x}}_1^{*}={\hat{x}}_2^{*}=\cdots ={\hat{x}}_n^{*}={\hat{x}}^{*}$. Inserting this condition into (31), we obtain

$$\begin{aligned}&\left( \frac{1}{N{\hat{x}}}-\frac{1}{N^2{\hat{x}}}\right) Vr-\kappa \gamma {\hat{x}}^{\gamma -1}=0\nonumber \\ \Rightarrow&\left( 1-\frac{1}{N}\right) \frac{Vr}{N}=\kappa \gamma {\hat{x}}^\gamma \nonumber \\ \Rightarrow&{\hat{x}}^{*}=\left( \frac{Vr}{\gamma }\right) ^{\frac{1}{\gamma }}\left( \frac{1}{\kappa N}\right) ^{\frac{1}{\gamma }}\left( 1-\frac{1}{N}\right) ^{\frac{1}{\gamma }}. \end{aligned}$$

(32)

$\square$

Proof of Lemma 3.1

Let us consider the first payoff in (7). We can write it equivalently as

$$\begin{aligned}&\frac{\theta _p{\hat{x}}_i^r}{N\left( \sum _{q\in \mathcal {P}}\sum _{j\in q}\theta _q\frac{{\hat{x}}_j^r}{N}\right) }V-\kappa _p{\hat{x}}_i^{\gamma }\nonumber \\&\quad =\frac{1}{N}\left[ \frac{\theta _p{\hat{x}}_i^r}{\sum _{q\in \mathcal {P}}\sum _{j\in q}\theta _q\frac{{\hat{x}}_j^r}{N}}V-N\kappa _p{\hat{x}}_i^{\gamma }\right] \nonumber \\&\quad =\frac{1}{N}\left[ \frac{\theta _p{\hat{x}}_i^r}{\frac{1}{N}\sum _{q\in \mathcal {P}}\sum _{j\in q}\theta _q{\hat{x}}_j^r}V-N\kappa _p{\hat{x}}_i^{\gamma }\right] \nonumber \\&\quad =\frac{1}{N}\left[ \frac{\theta _p\left( \frac{N{\hat{x}}_i}{N}\right) ^r}{\frac{1}{N}\sum _{q\in \mathcal {P}}\sum _{j\in q}\theta _q\left( \frac{N{\hat{x}}_j}{N}\right) ^r}V-N\kappa _p\left( \frac{N{\hat{x}}_i}{N}\right) ^{\gamma }\right] \nonumber \\&\quad =\frac{1}{N}\left[ \frac{\theta _p\left( N{\hat{x}}_i\right) ^r}{\frac{1}{N}\sum _{q\in \mathcal {P}}\sum _{j\in q}\theta _q\left( N{\hat{x}}_j\right) ^r}V-\frac{N}{N^\gamma }\kappa _p\left( N{\hat{x}}_i\right) ^{\gamma }\right] \nonumber \\&\quad =\frac{1}{N}\left[ \frac{\theta _px_i^r}{\frac{1}{N}\sum _{q\in \mathcal {P}}\sum _{j\in q}\theta _q x_j^r}V-k_p x_i^{\gamma }\right] , \end{aligned}$$

(33)

where $x_i=N{\hat{x}}_i$ and $k_p=\frac{N}{N^\gamma }\kappa _p=\frac{\kappa _p}{N^{\gamma -1}}$. Clearly, (33) is strategically equivalent to the payoff

$$\begin{aligned} \frac{\theta _px_i^r}{\frac{1}{N}\sum _{q\in \mathcal {P}}\sum _{j\in q}\theta _q x_j^r}V-k_p x_i^{\gamma }. \end{aligned}$$

This is the first payoff in (8). Applying the same transformations and logic to the second part of the payoff in (7), we obtain the second payoff in (8). $\square$

Proof

The Nash equilibrium (21) follows from the fact that (19) has a unique solution. The assumption $\left( \frac{\theta _p Vr}{k_p\alpha ^{*}\gamma }\right) ^{\frac{1}{\gamma -r}}<{\bar{x}}$ implies $b_p(\alpha ^{*})=\left( \frac{\theta _p Vr}{k_p\alpha ^{*}\gamma }\right) ^{\frac{1}{\gamma -r}}$. The value of $\alpha _p^{*}$ in (22) then follows from inserting (20) into $b_p(\alpha ^{*})$. By (9), $A(\mu ^{*})=\sum _{p\in \mathcal {P}}m_p\theta _pb_p^r(\alpha ^{*})$. The fact that $\alpha ^{*}$ is the unique solution to (19) then implies that $A(\mu ^{*})=\alpha ^{*}>0$.¹⁷$\square$

Proof of Proposition 5.3

Recall from Proposition 5.2 that at the Nash equilibrium $\mu ^{*}$, every agent of type p plays strategy $\alpha ^{*}_p$ as specified in (20). Therefore, using (10) and (20) we can write the Nash equilibrium payoff a type p agent as

$$\begin{aligned} F_{\alpha ^{*}_p,p}(\mu ^{*})&=\frac{\theta _p(\alpha _p^{*})^r}{\sum _{q\in \mathcal {P}} m_q\theta _q(\alpha _q^{*})^r}V-k_p(\alpha _p^{*})^{\gamma }\nonumber \\&\quad =\frac{\theta _p^{\frac{\gamma }{\gamma -r}}k_p^{\frac{-r}{\gamma -r}}}{\sum _{q\in \mathcal {P}} m_q\theta _q^{\frac{\gamma }{\gamma -r}}k_q^{\frac{-r}{\gamma -r}}}V\left( 1-\frac{r}{\gamma }\right) . \end{aligned}$$

(34)

Using (34) and the fact that the Nash equilibrium $\mu ^{*}$ is all in monomorphic population states, we calculate the aggregate payoff (16) in F at $\mu ^{*}$ to be

$$\begin{aligned} {\bar{F}}(\mu ^{*})&=\sum _{p\in \mathcal {P}}\int _{\mathcal {S}}F_{x,p}(\mu ^{*})\mu _p^{*}(dx)\\&\quad =\sum _{p\in \mathcal {P}}m_pF_{\alpha ^{*}_p,p}(\mu ^{*})\\&\quad =V\left( 1-\frac{r}{\gamma }\right) . \end{aligned}$$

$\square$

Proof of Proposition 7.1

Since $r=\gamma$, we have $F_{x,p}(\mu )=\left( \frac{\theta _p}{A(\mu )}V-k_p\right) x^r$. As we want $A(\mu )>0$, we cannot have $\frac{\theta _p}{A(\mu )}V-k_p<0$ in equilibrium for all p. That would mean all agents will play $x=0$.

Suppose in equilibrium, we have $\frac{\theta _p}{A(\mu )}V-k_p>0$ for some p. Then all agents of that type will play ${\bar{x}}$ so that $A(\mu )\ge m_p\theta _p{\bar{x}}^r$. We assume that ${\bar{x}}$ is sufficiently large that in that case, $\frac{\theta _p}{m_p\theta _p{\bar{x}}^r}V-k_p<0$, which would imply $\frac{\theta _p}{A(\mu )}V-k_p<0$ for that type p.

Therefore, it must be that at any Nash equilibrium $\mu$, $\frac{\theta _p}{A(\mu )}V-k_p=0$ for some p. For such p, equilibrium payoff would be zero for all agents even if that agent plays $x>0$.

But if $\frac{\theta _p}{A(\mu )}V-k_p=0$ for some p, then it must be that at the Nash equilibrium $\mu$, $A(\mu )=\frac{V\theta _p}{k_p}$. Now consider $q\in \mathcal {P}$ such that $\frac{k_q}{\theta _q}>\frac{k_p}{\theta _p}$. But then, we would have $F_{x,q}(\mu )=\left( \frac{\theta _q}{A(\mu )}V-k_q\right) x\le 0$ so that for such types of players, the best response to $\mu$ is 0.

Hence, if a type p agent plays $x>0$ at a Nash equilibrium, it must mean that p minimizes $\frac{k_q}{\theta _q}$ for all q such that $\theta _q>0$. Thus, $p\in \mathcal {P}^{*}$. Therefore, the set of Nash equilibria of F is NE(F) as described in (30), which is clearly a convex set. As all agents either play $x=0$ or if they play $x>0$, then $\frac{\theta _p}{A(\mu )}V-k_p=0$, it must be that all agents obtain zero equilibrium payoff. $\square$

1.2 Further details of numerical simulations in Sect. 6

The common parameter values given to us are $\{V,r,\gamma \}=\{10,0.5,1\}$. There are three types of players with the number of players of type p being $N_p$, $p\in \{1,2,3\}$. Let ${\hat{x}}^{*}_p$ be the Nash equilibrium effort level of a type p contestant. Under these conditions, it is not difficult to show that the first order condition that determines the Nash equilibrium in the finite player asymmetric contest (7) is

$$\begin{aligned} \frac{\theta _1 (0.5)(10)}{\sqrt{{\hat{x}}^{*}_p}\sum _{q=1}^3N_q\theta _q\sqrt{{\hat{x}}^{*}_q}} -\frac{\theta _p^2(0.5)(10)}{\left( \sum _{q=1}^3\theta _qN_q\sqrt{{\hat{x}}^{*}_q}\right) ^2}=\kappa _p. \end{aligned}$$

(35)

Thus, we obtain three FOCs, which can be numerically solved. That solution would provide the Nash equilibrium effort levels for each of the three types under the given parameter values.

In Table 1, the parameter values in the impact and cost functions are $\{\theta _1,\theta _2,\theta _3\}=\{1,1,1\}$ and $\{\kappa _1,\kappa _2,\kappa _3\}=\{1,2,3\}$. Applying these parameters to (35), we obtain the Nash equilibrium effort levels for different types, which we present below for different values of N. The Nash equilibrium determines the type specific equilibrium payoffs ${\hat{\pi }}_p^{*}$ through (7), which in turn generate the values in Table 1. It may be verified that $\sum _{p=1}^3 N_p{\hat{\pi }}_p^{*}$ will give the aggregate payoff $\bar{{\hat{\pi }}}^{*}$ listed in the second column of Table 1.

• 1. $N_1=N_2=N_3=2$. The Nash equilibrium effort levels of the three types are ${\hat{x}}_1^{*}=0.9458$, ${\hat{x}}_2^{*}=0.31$ and ${\hat{x}}_3^{*}=0.1521$. The associated equilibrium payoff levels are ${\hat{\pi }}_1^{*}=1.5876$, ${\hat{\pi }}_2^{*}=0.8304$ and ${\hat{\pi }}_3^{*}=0.5597$.

• 2. $N_1=N_2=N_3=4$. The Nash equilibrium effort levels of the three types are ${\hat{x}}_1^{*}=0.5712$, ${\hat{x}}_2^{*}=0.1636$ and ${\hat{x}}_3^{*}=0.0763$. The associated equilibrium payoff levels are ${\hat{\pi }}_1^{*}=0.7442$, ${\hat{\pi }}_2^{*}=0.3768$ and ${\hat{\pi }}_3^{*}=0.2519$.

• 3. $N_1=N_2=N_3=10$. The Nash equilibrium effort levels of the three types are ${\hat{x}}_1^{*}=0.2544$, ${\hat{x}}_2^{*}=0.0672$ and ${\hat{x}}_3^{*}=0.0304$. The associated equilibrium payoff levels are ${\hat{\pi }}_1^{*}=0.2833$, ${\hat{\pi }}_2^{*}=0.142$ and ${\hat{\pi }}_3^{*}=0.0947$.

• 4. $N_1=N_2=N_3=20$. The Nash equilibrium effort levels of the three types are ${\hat{x}}_1^{*}=0.1317$, ${\hat{x}}_2^{*}=0.0338$ and ${\hat{x}}_3^{*}=0.0152$. The associated equilibrium payoff levels are ${\hat{\pi }}_1^{*}=0.1391$, ${\hat{\pi }}_2^{*}=0.0696$ and ${\hat{\pi }}_3^{*}=0.0464$.

• 5. $N_1=N_2=N_3=30$. The Nash equilibrium effort levels of the three types are ${\hat{x}}_1^{*}=0.0888$, ${\hat{x}}_2^{*}=0.0226$ and ${\hat{x}}_3^{*}=0.0101$. The associated equilibrium payoff levels are ${\hat{\pi }}_1^{*}=0.0921$, ${\hat{\pi }}_2^{*}=0.0461$ and ${\hat{\pi }}_3^{*}=0.0307$.

• 6. $N_1=N_2=N_3=40$. The Nash equilibrium effort levels of the three types are ${\hat{x}}_1^{*}=0.067$, ${\hat{x}}_2^{*}=0.0169$ and ${\hat{x}}_3^{*}=0.0076$. The associated equilibrium payoff levels are ${\hat{\pi }}_1^{*}=0.0688$, ${\hat{\pi }}_2^{*}=0.0344$ and ${\hat{\pi }}_3^{*}=0.0229$.

• 7. $N_1=N_2=N_3=50$. The Nash equilibrium effort levels of the three types are ${\hat{x}}_1^{*}=0.0538$, ${\hat{x}}_2^{*}=0.0136$ and ${\hat{x}}_3^{*}=0.0061$. The associated equilibrium payoff levels are ${\hat{\pi }}_1^{*}=0.055$, ${\hat{\pi }}_2^{*}=0.0275$ and ${\hat{\pi }}_3^{*}=0.0183$.

In Table 2, the impact and cost parameters are $\{\theta _1,\theta _2,\theta _3\}=\{1.2,1.8,2.3\}$ and $\{\kappa _1,\kappa _2,\kappa _3\}=\{1,2,3\}$. A similar procedure as in the above details for Table 1 generates the Nash equilibrium effort and payoff levels under these parameters. Details are presented below.

• 1. $N_1=N_2=N_3=2$. The Nash equilibrium effort levels of the three types are ${\hat{x}}_1^{*}=0.6447$, ${\hat{x}}_2^{*}=0.3492$ and ${\hat{x}}_3^{*}=0.2461$. The associated equilibrium payoff levels are ${\hat{\pi }}_1^{*}=0.8759$, ${\hat{\pi }}_2^{*}=0.9803$ and ${\hat{\pi }}_3^{*}=1.0624$.

• 2. $N_1=N_2=N_3=4$. The Nash equilibrium effort levels of the three types are ${\hat{x}}_1^{*}=0.3483$, ${\hat{x}}_2^{*}=0.1923$ and ${\hat{x}}_3^{*}=0.1375$. The associated equilibrium payoff levels are ${\hat{\pi }}_1^{*}=0.405$, ${\hat{\pi }}_2^{*}=0.455$ and ${\hat{\pi }}_3^{*}=0.4947$.

• 3. $N_1=N_2=N_3=10$. The Nash equilibrium effort levels of the three types are ${\hat{x}}_1^{*}=0.1453$, ${\hat{x}}_2^{*}=0.0811$ and ${\hat{x}}_3^{*}=0.0585$. The associated equilibrium payoff levels are ${\hat{\pi }}_1^{*}=0.1543$, ${\hat{\pi }}_2^{*}=0.1736$ and ${\hat{\pi }}_3^{*}=0.1889$.

• 4. $N_1=N_2=N_3=20$. The Nash equilibrium effort levels of the three types are ${\hat{x}}_1^{*}=0.0737$, ${\hat{x}}_2^{*}=0.0412$ and ${\hat{x}}_3^{*}=0.0299$. The associated equilibrium payoff levels are ${\hat{\pi }}_1^{*}=0.0759$, ${\hat{\pi }}_2^{*}=0.0854$ and ${\hat{\pi }}_3^{*}=0.0929$.

• 5. $N_1=N_2=N_3=30$. The Nash equilibrium effort levels of the three types are ${\hat{x}}_1^{*}=0.0493$, ${\hat{x}}_2^{*}=0.0277$ and ${\hat{x}}_3^{*}=0.02$. The associated equilibrium payoff levels are ${\hat{\pi }}_1^{*}=0.0503$, ${\hat{\pi }}_2^{*}=0.0566$ and ${\hat{\pi }}_3^{*}=0.0616$.

• 6. $N_1=N_2=N_3=40$. The Nash equilibrium effort levels of the three types are ${\hat{x}}_1^{*}=0.0371$, ${\hat{x}}_2^{*}=0.0208$ and ${\hat{x}}_3^{*}=0.0151$. The associated equilibrium payoff levels are ${\hat{\pi }}_1^{*}=0.0376$, ${\hat{\pi }}_2^{*}=0.0423$ and ${\hat{\pi }}_3^{*}=0.0461$.

• 7. $N_1=N_2=N_3=50$. The Nash equilibrium effort levels of the three types are ${\hat{x}}_1^{*}=0.0297$, ${\hat{x}}_2^{*}=0.0167$ and ${\hat{x}}_3^{*}=0.0121$. The associated equilibrium payoff levels are ${\hat{\pi }}_1^{*}=0.03$, ${\hat{\pi }}_2^{*}=0.0338$ and ${\hat{\pi }}_3^{*}=0.0368$.