Rent-seeking group contests with one-sided private information

Abstract

We consider a rent-seeking contest where players compete in groups for a prize of given value. One group has private information about its number of members, which can be either small or large. The other groups have possibly different but publicly known sizes. We present an explicit characterization of the groups which are active in the unique equilibrium of the game, and relate the relative magnitude of group efforts to the size of the groups. We compare the decision of each type of the privately informed group to be active in equilibrium to the corresponding decision in a benchmark game with complete information.

Appendix: Proofs

For brevity, the ‘Appendix’ does not give the (straightforward) proof of Lemma 1, the (lengthy) proof of Lemma 3, and details of a number of tedious derivations. They are available upon request from the authors.

1.1 Derivation of (10) and (11)

Considering case $A$, assume $m >2,\,\hat{x}_{A1S}=0,\,\hat{x}_{A1L}=0$ and $\hat{x}_{A}>0$. The FOCs become

$$\begin{aligned}&\displaystyle \frac{V}{\hat{x}_{A} n_S} \le 1 ,&\end{aligned}$$

(21)

$$\begin{aligned}&\displaystyle \frac{V}{\hat{x}_{A} n_L} \le 1 ,&\end{aligned}$$

(22)

$$\begin{aligned}&\displaystyle \frac{(\hat{x}_{A} - \hat{x}_{Ai})V}{(\hat{x}_{A})^2 n_i}=1, \quad \text {if } 2 \le i \le r_{A}+1,&\end{aligned}$$

(23)

$$\begin{aligned}&\displaystyle \frac{V}{\hat{x}_{A} n_i} \le 1, \quad \text {if } r_{A}+1 < i \le m.&\end{aligned}$$

(24)

From (23), we find for $2 \le i \le r_{A}+1$ that

$$\begin{aligned} \hat{x}_{Ai} = \hat{x}_{A}-\frac{\hat{x}_{A}^2n_i}{V}, \end{aligned}$$

(25)

so $\hat{x}_{A} = \sum \nolimits _{j=2}^{r_{A}+1}\hat{x}_{Aj} = \phi _A(r_A),$ i.e., (10). Using this and the definition of $r_A$, we find (11). If (21)–(24) hold, there is a unique solution where $\hat{x}_{A1S}=\hat{x}_{A1L}=0,\,\hat{x}_{Ai}$ can be calculated from (25) for $2 \le i \le r_{A}+1$, and $\hat{x}_{Ai}=0$ for $r_{A}+1 < i \le m$.

1.2 Derivation of (12) and (13)

Examining case $B$, assume $\hat{x}_{B1S}>0,\,\hat{x}_{B1L}=0$ and $\hat{x}_{B}>0$. The FOCs now read

$$\begin{aligned} \frac{\hat{x}_{B} V}{(\hat{x}_{B1S}+\hat{x}_{B})^2 n_S}&=1, \end{aligned}$$

(26)

$$\begin{aligned} \frac{V}{\hat{x}_{B} n_L}&\le 1, \end{aligned}$$

(27)

$$\begin{aligned} p\frac{(\hat{x}_{B1S}+\hat{x}_{B} - \hat{x}_{Bi})V}{(\hat{x}_{B1S}+\hat{x}_{B})^2 n_i}+(1-p)\frac{(\hat{x}_{B}-\hat{x}_{Bi})V}{(\hat{x}_{B})^2 n_i}&= 1,&\text {if } 2 \le i \le r_{B}+1, \end{aligned}$$

(28)

$$\begin{aligned} p\frac{V}{(\hat{x}_{B1S}+\hat{x}_{B}) n_i}+(1-p)\frac{V}{\hat{x}_{B} n_i}&\le 1,&\text {if } r_{B}+1 < i \le m. \end{aligned}$$

(29)

Substituting (26) in (28), we find for $2 \le i \le r_{B}+1$ that

$$\begin{aligned} \hat{x}_{Bi} = \frac{p\sqrt{\hat{x}_{B} V n_s}+(1-p)V-\hat{x}_{B}n_i}{pn_S+(1-p)\frac{V}{\hat{x}_{B}}}. \end{aligned}$$

(30)

Using $\hat{x}_{B} = \sum \nolimits _{j=2}^{r_{B}+1}\hat{x}_{Bj}$, finding $\hat{x}_{B}$ is equivalent to solving the equation

$$\begin{aligned} \left( pn_S+\sum _{j=2}^{r_{B}+1}n_j\right) y^2-\left( r_{B}p\sqrt{Vn_S}\right) y - (1-p)V(r_{B}-1)= 0, \end{aligned}$$

(31)

where $y=\sqrt{\hat{x}_{B}}>0$. We find $\hat{x}_{B} = \phi _B(r_B),$ i.e., (12). Note that $\hat{x}_{Bi} > 0$ if and only if the numerator of (30) is positive, as the denominator is positive. Using this and the definition of $r_B$, we obtain (13). If (26)–(29) hold, there is a unique solution where $\hat{x}_{B1S}$ is defined by (26), $\hat{x}_{B1L}=0,\,\hat{x}_{Bi}$ can be derived from (30) for $2 \le i \le r_{B}+1$, and $\hat{x}_{Bi}=0$ for $r_{B}+1 <i \le m$.

1.3 Derivation of (14) and (15)

Considering case $C$, assume $\hat{x}_{C1S}>0,\,\hat{x}_{C1L}>0$ and $\hat{x}_{C}>0$. The FOCs now equal

$$\begin{aligned} \frac{\hat{x}_{C} V}{(\hat{x}_{C1S}+\hat{x}_{C})^2 n_S} =1, \end{aligned}$$

(32)

$$\begin{aligned} \frac{\hat{x}_{C} V}{(\hat{x}_{C1L}+\hat{x}_{C})^2 n_L} =1,\end{aligned}$$

(33)

$$\begin{aligned} p\frac{(\hat{x}_{C1S}+\hat{x}_{C} - \hat{x}_{Ci})V}{(\hat{x}_{C1S}+\hat{x}_{C})^2 n_i}+(1-p)\frac{(\hat{x}_{C1L}+\hat{x}_{C}-\hat{x}_{Ci})V}{(\hat{x}_{C1L}+\hat{x}_{C})^2 n_i}&= 1,&\text {if } 2 \le i \le r_{C}+1, \end{aligned}$$

(34)

$$\begin{aligned} p\frac{V}{(\hat{x}_{C1S}+\hat{x}_{C}) n_i}+(1-p)\frac{V}{(\hat{x}_{C1L}+\hat{x}_{C}) n_i}&\le 1,&\text {if } r_{C}+1 < i \le m. \end{aligned}$$

(35)

Substituting (32) and (33) in (34), we derive for $2 \le i \le r_{C}+1$ that

$$\begin{aligned} \hat{x}_{Ci} = \frac{p\sqrt{\hat{x}_{C}Vn_S}+(1-p)\sqrt{\hat{x}_{C}Vn_L}-n_i\hat{x}_{C}}{pn_S+(1-p)n_L}. \end{aligned}$$

(36)

Since $\hat{x}_{C} = \sum _{j=2}^{r_{C}+1}\hat{x}_{Cj}$, finding $\hat{x}_{C}$ is equivalent to solving

$$\begin{aligned}&\left( pn_S+ (1-p)n_L\!+\!\sum _{j=2}^{r_{C}+1}n_j\right) y^2\!-\!\left( r_{C}p\sqrt{Vn_S}+r_{C}(1\!-\!p)\sqrt{V n_L}\right) y \!=\! 0, \end{aligned}$$

(37)

where $y=\sqrt{\hat{x}_{C}}>0$. It follows that $\hat{x}_{C} = \phi _C(r_C)$, i.e., (14). Note that $\hat{x}_{Ci} > 0$ if and only if the numerator of (36) is positive, as the denominator is positive. Using this and the definition of $r_C$, we find

$$\begin{aligned}&r_{C} \!=\!\max \left\{ i \in 1,2,\ldots ,m\!-\!1 \,|\, p\sqrt{V n_S}\!+\!(1\!-\!p)\sqrt{V n_L}\!-\!n_{i+1}\sqrt{\phi _{C}(i)}>0\right\} , \end{aligned}$$

(38)

which can be rewritten as (15). If (32)–(35) hold, there is a unique solution where $\hat{x}_{C1S}$ and $\hat{x}_{C1L}$ are found by solving (32) and (33), respectively. The values of $\hat{x}_{Ci}$ for $2 \le i \le r_{C}+1$ can be derived from (36), while $\hat{x}_{Ci}=0$ for $r_{C}+1 < i \le m$.

Proof of Lemma 2

Using (7), we directly obtain (16). Next, from (8), $\phi _{B}(i) < V/n_S$ can be written as

$$\begin{aligned}&i^2p^2+4\left( p+\frac{1}{n_S}\sum _{j=2}^{i+1} n_j\right) (1-p)(i-1)< (2- i)^2p^2\\&\quad +\,4(2- i)\frac{p}{n_S}\sum _{j=2}^{i+1} n_j + 4\left( \frac{1}{n_S}\sum _{j=2}^{i+1} n_j\right) ^2. \end{aligned}$$

Straightforward calculations give (17). Using (8), $\phi _{B}(i) \ge \frac{V}{n_L}$ can be written as

$$\begin{aligned}&i^2p^2n_Sn_L+4n_L \left( pn_S+\sum _{j=2}^{i+1} n_j\right) (1-p)(i-1) \nonumber \\&\quad \ge \left( 2\left( pn_S+\sum _{j=2}^{i+1} n_j\right) -ip\sqrt{n_Sn_L}\right) ^2. \end{aligned}$$

Straightforward manipulations give (18). Finally, using (9), we easily find (19). $\square$

Proof of Lemma 4

For each of the cases $A,\,B$ and $C$, when the associated FOCs hold, a unique solution is defined. If for a case not all four FOCs hold, the case cannot yield an equilibrium. In each case, the third and fourth FOC—i.e., (23) and (24) for case $A$, (28) and (29) for case $B$, and (34) and (35) for case $C$—always hold, as they are implied by the selection of $r_{A}$ in (11), $r_{B}$ in (13), and $r_{C}$ in (15), respectively. This directly follows for the third FOC of each case. We give a proof of this for the fourth FOC in all three cases.

For case $A$, note that (11) implies for $i= r_{A}+1,\ldots ,m-1$ that $n_{i+1} \ge V/\phi _{A}(i)$, which by Lemma 3 gives $n_{i+1} \ge V/\phi _{A}(r_{A})$, which is equivalent to (24). For case $B$, (13) implies for $i = r_{B}+1,\ldots , m-1$ that

$$\begin{aligned} n_{i+1} \ge p\sqrt{\frac{Vn_S}{\phi _{B}(i)}} + (1-p)\frac{V}{\phi _{B}(i)}, \end{aligned}$$

(39)

which by Lemma 3 gives

$$\begin{aligned} n_{i+1} \ge p\sqrt{\frac{Vn_S}{\phi _{B}(r_{B})}} + (1-p)\frac{V}{\phi _{B}(r_{B})}. \end{aligned}$$

(40)

Remark that (40) is equivalent to (29) if you substitute in (26). For case $C$, notice that (38) implies for $i = r_{C}+1,\ldots ,m-1$ that

$$\begin{aligned} n_{i+1} \ge \frac{p\sqrt{Vn_S}+(1-p)\sqrt{Vn_L}}{\sqrt{\phi _{C}(i)}}, \end{aligned}$$

(41)

which by Lemma 3 learns that

$$\begin{aligned} n_{i+1} \ge p\frac{\sqrt{Vn_S}}{\sqrt{\phi _{C}(r_{C})}}+(1-p)\frac{\sqrt{Vn_L}}{\sqrt{\phi _{C}(r_{C})}}. \end{aligned}$$

(42)

We see that (42) is equivalent to (35) if you substitute in (32) and (33).

Finally, considering the two FOCs of each case that do not hold for all parameter values—i.e., (21) and (22) for case $A$, (26) and (27) for case $B$, and (32) and (33) for case $C$—we can easily complete the proof of the lemma. $\square$

Proof of Proposition 1

• Step 1: If we have an equilibrium in case $A$, (16) and part (i) of Lemma 4 give

  $$\begin{aligned} (r_{A}-1)n_S\ge \sum _{j=1}^{r_{A}+1}n_j, \end{aligned}$$

  (43)

  which by (17) implies $\phi _{B}(r_{A}) \ge V/n_S$. Using Lemma 3, this gives $\phi _{B}(r_{B}) \ge V/n_S$, which is inconsistent with an equilibrium for case $B$. Hence, the game cannot have an equilibrium in both case $A$ and case $B$.

• Step 2: When we have an equilibrium in case $C$, we have by Lemma 3 and part (iii) of Lemma 4 that $\phi _{C}(r_{A}) < V/n_L$. By (19), and using $n_S < n_L$, we then have

  $$\begin{aligned}&(r_{A}-1)\left( p\sqrt{n_Sn_L}+(1-p)n_L\right) <pn_S- p\sqrt{n_Sn_L}+\sum _{j=2}^{r_{A}+1}n_j\nonumber \\&\quad \Rightarrow r_{A}\left( pn_S+(1-p)n_L\right) <pn_S+(1-p)n_L+\sum _{j=2}^{r_{A}+1}n_j\nonumber \\&\quad \Rightarrow (r_{A}-1)n_S<\sum _{j=2}^{r_{A}+1}n_j. \end{aligned}$$

  (44)

  Note that (44) implies by (16) that $\phi _{A}(r_{A}) < V/n_S$, which contradicts with an equilibrium for case $A$. Hence, the game cannot have an equilibrium in both case $A$ and case $C$.

• Step 3: When we have an equilibrium in case $C$, we have by Lemma 3 and part (iii) of Lemma 4 that $\phi _{C}(r_{B}) < V/n_L$. Then, (19) yields

  $$\begin{aligned} (r_{B}-1)\left( p\sqrt{n_Sn_L}+(1-p)n_L\right) <pn_S- p\sqrt{n_Sn_L}+\sum _{j=2}^{r_{B}+1}n_j, \end{aligned}$$

  (45)

  which by (18) gives $\phi _{B}(r_{B}) < V/n_L$, implying there is no equilibrium in case $B$. Hence, the game cannot have an equilibrium in both case $B$ and case $C$.

• Step 4: We finally show that the game must have an equilibrium in at least one of the cases $A,\,B$ and $C$. Assume case $B$ does not yield an equilibrium. Then, either $\phi _{B}(r_{B}) \ge V/n_S$ or $\phi _{B}(r_{B}) < V/n_L$. For the case $\phi _{B}(r_{B}) \ge V/n_S$, note that (17) implies

  $$\begin{aligned} (r_{B}-1)n_S\ge \sum _{j=2}^{r_{B}+1}n_j. \end{aligned}$$

  (46)

  Using (7) and Lemma 3, we find that

  $$\begin{aligned} (r_{A}-1)n_S\ge \sum _{j=2}^{r_{A}+1}n_j. \end{aligned}$$

  (47)

  Applying (16) to (47), we know that case $A$ yields an equilibrium. For the case $\phi _{B}(r_{B}) < V/n_L$, Lemma 3 implies that $\phi _{B}(r_{C}) < V/n_L$. From (18) and (19), we know that case $C$ yields an equilibrium.

$\square$

Proof of Proposition 2

The proof follows in a straightforward way from Lemmas 2, 3, 4, and the facts that the small-type group 1 is active in the game with private information if and only if $V/n_S > \phi _A(r_A)$, and group $S$ is active in game $B_S$ if and only if $V/n_S > \phi _S(r_S)$. $\square$

Proof of Proposition 3

Take game $B_L$. Group $L$ in this game will be active if and only if (use 19)

$$\begin{aligned} (r_{L}-1)n_L < \sum _{j=2}^{r_{L}+1}n_j. \end{aligned}$$

(48)

We derive from (19) when the large-type group 1 is active (case $C$) in the private-information game. Note that (15) implies $r_{L}\ge r_{C}$. So, if $r_{L}=1$, then $r_{C}=1$ and group $L$ will always be active in game $B_L$, while the large-type group 1 can be inactive in the game with private information. This gives part (i).

If $r_{L}\ge 2$ and $r_{C}=1$, we compare (48) to (19) with $i = r_{C}=1$ and conclude that both group $L$ in game $B_L$ and the large-type group 1 in the private-information game can be active and inactive, without in one game being (in)active necessarily implying (in)activity for the other. This proves part (ii).

Finally, we examine the situation with $r_{L}\ge 2$ and $r_{C}\ge 2$. Suppose that group $L$ is active in the equilibrium of game $B_L$. By (15), we then have

$$\begin{aligned}&\frac{\sum _{j=2}^{r_{L}+1}n_j}{r_{L}}+\frac{n_L}{r_{L}}-n_{r_{L}+1} > 0\nonumber \\&\quad \Leftrightarrow 1-\frac{r_{L}-1}{\sum _{j=2}^{r_{L}+1}n_j}n_{r_{L}+1}-\frac{n_{r_{L}+1}}{\sum _{j=2}^{r_{L}+1}n_j}+\frac{n_L}{\sum _{j=2}^{r_{L}+1}n_j} > 0\nonumber \\&\quad \Rightarrow 1-\frac{r_{L}-1}{\sum _{j=2}^{r_{L}+1}n_j}n_{r_{L}+1}-\frac{n_{r_{L}+1}}{\sum _{j=2}^{r_{L}+1}n_j}+\frac{1}{r_{L}-1} > 0, \end{aligned}$$

(49)

where (49) is due to (48). Suppose now that $r_{L}>r_{A}$. Then, (11) implies that

$$\begin{aligned} \frac{n_{r_{L}+1}}{\sum _{j=2}^{r_{L}+1}n_j}\ge \frac{1}{r_{L}-1}. \end{aligned}$$

(50)

If we substitute (50) in (49) and compare the result with (11), we obtain $r_{A}\ge r_{L}$, which gives a contradiction. Therefore, we have $r_{A}\ge r_{L} \ge r_{C}$. By Lemma 3 and (48), we now know that

$$\begin{aligned} \frac{1}{n_L}> \frac{r_{L}-1}{\sum _{j=2}^{r_{L}+1}n_j} \Rightarrow \frac{1}{n_L} > \frac{r_{C}-1}{\sum _{j=2}^{r_{C}+1}n_j}. \end{aligned}$$

(51)

Note from (19) that the large-type group 1 in the private-information game is active if and only if

$$\begin{aligned} \sum _{j=2}^{r_{C}+1}n_j&> (r_{C}-1)\left( p\sqrt{n_Sn_L}+(1-p)n_L\right) -pn_S+p\sqrt{n_Sn_L}\nonumber \\&= pn_L\left( r_{C}\sqrt{\frac{n_S}{n_L}}-\frac{n_S}{n_L}\right) +(r_{C}-1)(1-p)n_L. \end{aligned}$$

(52)

Furthermore, because $0 < n_S < n_L$, we have for $r_{C} \ge 2$ that

$$\begin{aligned}&\left( \sqrt{\frac{n_S}{n_L}}-1\right) ^2 > 0\nonumber \\&\quad \Leftrightarrow 2> \frac{1-\frac{n_S}{n_L}}{1-\sqrt{\frac{n_S}{n_L}}}\nonumber \\&\quad \Leftrightarrow r_{C}-1 > r_{C}\sqrt{\frac{n_S}{n_L}}-\frac{n_S}{n_L}. \end{aligned}$$

(53)

By (53) and (51), we know that if group $L$ in game $B_L$ is active in case $r_{C} \ge 2$, then

$$\begin{aligned} \sum _{j=2}^{r_{C}+1}n_j > (r_{C}-1)n_L > pn_L\left( r_{C}\sqrt{\frac{n_S}{n_L}}-\frac{n_S}{n_L}\right) +(r_{C}-1)(1-p)n_L. \end{aligned}$$

(54)

Comparing (54) to (52), we conclude that if group $L$ is active in game $B_L$, the large-type group 1 is also active in the game with private information. Note that the reverse does not hold in general. This proves part (iii). $\square$