The value of public information in common-value Tullock contests

Abstract

Consider a symmetric common-value Tullock contest with incomplete information in which the players’ cost of effort is the product of a random variable and a deterministic real function of effort, d. We show that the Arrow–Pratt curvature of d,  $R_{d},$ determines the effect on equilibrium efforts and payoffs of the increased flexibility/reduced commitment that more information introduces into the contest: If $R_{d}$ is increasing, then effort decreases (increases) with the level of information when the cost of effort (value) is independent of the state of nature. Moreover, if $R_{d}$ is increasing (decreasing), then the value of public information is nonnegative (nonpositive).

Appendix

Lemma 6.1

A symmetric common-value Tullock contest with complete information in which the players’ cost of effort is a twice differentiable strictly increasing and convex function $c:{\mathbb {R}} _{+}\rightarrow {\mathbb {R}}_{+}$ such that $c(0)=0$ has a unique pure strategy Nash equilibrium. Moreover, this equilibrium is symmetric and interior.

Proof

In the game associated with a symmetric common-value Tullock contest with complete information the set of pure strategies of every player is ${\mathbb {R}}_{+}$, and the payoff function of each player $i\in N$ is $h_{i}:{\mathbb {R}}_{+}^{n}\rightarrow {\mathbb {R}}_{+}$ given for $x\in {\mathbb {R}}_{+}^{n}\backslash \{0\}$ by

$$\begin{aligned} h_{i}(x)=\frac{x_{i}}{\bar{x}}v-c(x_{i}), \end{aligned}$$

where $v>0$ is the players’ common value and $\bar{x}=\sum _{j=1}^{n}x_{j}$, and

$$\begin{aligned} h_{i}(0)=\rho _{i}v-c(0)=\rho _{i}v, \end{aligned}$$

where $\rho \in \Delta ^{n}$ is predetermined. Thus, $h_{i}\left( \cdot ,x_{-i}\right)$ is twice differentiable and concave on ${\mathbb {R}}_{++}$, and

$$\begin{aligned} \frac{\partial h_{i}(x)}{\partial x_{i}}=\frac{\bar{x}-x_{i}}{\bar{x}^{2}} v-c^{\prime }(x_{i}). \end{aligned}$$

Let $x^{*}\in {\mathbb {R}}_{+}^{n}$ be a pure strategy Nash equilibrium. Then, for all $i\in N$, player i’s effort $x_{i}^{*}$ solves the problem

$$\begin{aligned} \max _{x_{i}\in {\mathbb {R}}_{+}}h_{i}\left( x_{i},x_{-i}^{*}\right) , \end{aligned}$$

i.e., $\left( \partial h_{i}(x^{*})/\partial x_{i}\right) x_{i}^{*}=0.$ Clearly, $x^{*}\ne 0;$ since $n\ge 2,$ then $\rho _{i}<1$ for some $i\in N,$ and therefore

$$\begin{aligned} h_{i}(0,0)=\rho _{i}v<v-c(\varepsilon )=h_{i}(\varepsilon ,0) \end{aligned}$$

for $\varepsilon >0$ sufficiently small. Let $k\in N$ be such that $x_{k}^{*}>0.$ Then,

$$\begin{aligned} \frac{\partial h_{i}(0,x_{-i}^{*})}{\partial x_{i}}=\frac{v}{\bar{x} ^{*}}-c^{\prime }(0)>\frac{\bar{x}^{*}-x_{k}^{*}}{\left( \bar{x} ^{*}\right) ^{2}}v-c^{\prime }(x_{k}^{*})=0 \end{aligned}$$

for all $i\in N\backslash \{k\}.$ Thus, $x_{i}^{*}>0$, and therefore $\partial h_{i}(x^{*})/\partial x_{i}=0,$ i.e.,

$$\begin{aligned} \frac{\bar{x}^{*}-x_{i}^{*}}{\left( \bar{x}^{*}\right) ^{2}} v=c^{\prime }(x_{i}^{*}) \end{aligned}$$

for all $i\in N.$ Hence, $x_{i}^{*}=t^{*}>0$ for all $i\in N,$ where $t^{*}$ is the unique solution (recall that $c^{\prime \prime }\ge 0)$ to the equation

$$\begin{aligned} \frac{\left( n-1\right) }{n^{2}t}v=c^{\prime }(t). \end{aligned}$$

(9)

Obviously, the profile $(t^{*},\ldots ,t^{*}),$ where $t^{*}$ is the solution to Eq. (9), is a pure strategy equilibrium. Hence, the contest has a unique pure strategy Nash equilibrium, which is symmetric and interior. $\square$

Lemma 6.2

Let $f:{\mathbb {R}}^{2}\rightarrow {\mathbb {R}}$ be twice differentiable and homogeneous of degree one. Then, f is convex (concave) on ${\mathbb {R}}^{2}$ if and only if $f_{xx}(x,y)\ge 0$ ($f_{xx}\mathrm{(}x,y\mathrm{)}\le 0$) for all $(x,y)\in {\mathbb {R}}^{2}.$

Proof

By Euler’s Theorem,

$$\begin{aligned} f(x,y)=xf_{x}(x,y)+yf_{y}(x,y). \end{aligned}$$

Differentiating with respect to x on both sides on this equation and simplifying yields

$$\begin{aligned} xf_{xx}(x,y)+yf_{yx}(x,y)=0. \end{aligned}$$

(10)

Likewise,

$$\begin{aligned} xf_{xy}(x,y)+yf_{yy}(x,y)=0. \end{aligned}$$

(11)

Hence,

$$\begin{aligned} x^{2}f_{xx}(x,y)=y^{2}f_{yy}(x,y), \end{aligned}$$

and therefore

$$\begin{aligned} f_{xx}(x,y)\lesseqqgtr 0\Leftrightarrow f_{yy}(x,y)\lesseqqgtr 0. \end{aligned}$$

Further, (10) and (11) imply

$$\begin{aligned} f_{xx}(x,y)f_{yy}(x,y)-f_{xy}(x,y)f_{yx}(x,y)=0. \end{aligned}$$

Thus, the eigenvalues of the Hessian matrix of f are nonnegative (nonpositive) when $f_{xx}$ is a nonnegative (nonpositive) function on ${\mathbb {R}}^{2}$. $\square$