Risky rents

Abstract

We consider a strategic contest game in which risk-averse agents exert efforts to increase their share of a risky rent. We show that a unique symmetric equilibrium always exists under constant or decreasing absolute risk aversion. We also show that agents exert in general less efforts when they are more risk averse or when the rent is more risky.

Appendix

Proof of Proposition 1

The proof follows from the proof of Proposition 3.1 in Corchón (2007). Rearranging terms, Eq. (5), can be written as follows:

$$\begin{aligned} \frac{\phi ^{\prime }(x_{i})b_{-i}}{\left( \phi (x_{i})+b_{-i}\right) ^{2}}=\frac{E\left[ u^{\prime }\left( \frac{\phi (x_{i})}{\phi (x_{i})+b_{-i}}\widetilde{v}-x_{i}\right) \right] }{E\left[ \widetilde{v}u^{\prime }\left( \frac{\phi (x_{i})}{\phi (x_{i})+b_{-i}}\widetilde{v}-x_{i}\right) \right] }. \end{aligned}$$

Let $x_{m}=\min _{i\in I}x_{i}$ and $x_{M}=\max _{i\in I}x_{i}.$ If the solution is not symmetric, then $x_{m}<x_{M}.$ Since $\phi ^{\prime }(x_{m})b_{-m}>\phi ^{\prime }(x_{M})b_{-M}$ and $\phi (x_{m})+b_{-m}=\phi (x_{M})+b_{-M}$, we have

$$\begin{aligned} \frac{\phi ^{\prime }(x_{m})b_{-m}}{\left( \phi (x_{m})+b_{-m}\right) ^{2}}>\frac{\phi ^{\prime }(x_{M})b_{-M}}{\left( \phi (x_{M})+b_{-M}\right) ^{2}}, \end{aligned}$$

which in turn implies

$$\begin{aligned} \frac{E\left[ \widetilde{v}u^{\prime }\left( \frac{\phi (x_{m})}{\phi (x_{m})+b_{-m}}\widetilde{v}-x_{m}\right) \right] }{E\left[ u^{\prime }\left( \frac{\phi (x_{m})}{\phi (x_{m})+b_{-m}}\widetilde{v}-x_{m}\right) \right] }<\frac{E\left[ \widetilde{v}u^{\prime }\left( \frac{\phi (x_{M})}{\phi (x_{M})+b_{-M}}\widetilde{v}-x_{M}\right) \right] }{E\left[ u^{\prime }\left( \frac{\phi (x_{M})}{\phi (x_{M})+b_{-M}}\widetilde{v}-x_{M}\right) \right] }. \end{aligned}$$

However, this last inequality is not possible if the function $\Psi (x)=\frac{E\left[ \widetilde{v}u^{\prime }\left( \frac{\phi (x)}{B}\widetilde{v}-x\right) \right] }{E\left[ u^{\prime }\left( \frac{\phi (x)}{B}\widetilde{v}-x\right) \right] }$ is decreasing in x for any $B>0$. We have

$$\begin{aligned} \Psi ^{\prime }(x)&=\frac{E\left[ (\frac{\phi ^{\prime }(x)}{B}\widetilde{v}-1)\widetilde{v}u^{\prime \prime }(.)\right] E\left[ u^{\prime }\left( \frac{\phi (x)}{B}\widetilde{v}-x\right) \right] -E\left[ (\frac{\phi ^{\prime }(x)}{B}\widetilde{v}-1)u^{\prime \prime }(.)\right] E\left[ \widetilde{v}u^{\prime }\left( \frac{\phi (x)}{B}\widetilde{v}-x\right) \right] }{E^{2}\left[ u^{\prime }\left( \frac{\phi (x)}{B}\widetilde{v}-x\right) \right] }\\&=\frac{E\left[ (\frac{\phi ^{\prime }(x)}{B}\widetilde{v}-1)\frac{u^{\prime \prime }}{u^{\prime }}(.)\widetilde{v}u^{\prime }(.)\right] }{E\left[ u^{\prime }\left( \frac{\phi (x)}{B}\widetilde{v}-x\right) \right] }-\frac{E\left[ (\frac{\phi ^{\prime }(x)}{B}\widetilde{v}-1)\frac{u^{\prime \prime }}{u^{\prime }}(.)u^{\prime }(.)\right] }{E\left[ u^{\prime }\left( \frac{\phi (x)}{B}\widetilde{v}-x\right) \right] }\times \frac{E\left[ \widetilde{v}u^{\prime }\left( \frac{\phi (x)}{B}\widetilde{v}-x\right) \right] }{E\left[ u^{\prime }\left( \frac{\phi (x)}{B}\widetilde{v}-x\right) \right] } \end{aligned}$$

Similar to the proof of Proposition 6, we now introduce the following probability density function:

$$\begin{aligned} m(v)=\frac{d(v)u^{\prime }\left( \frac{\phi (x)}{B}\widetilde{v}-x\right) }{E\left[ u^{\prime }\left( \frac{\phi (x)}{B}\widetilde{v}-x\right) \right] }, \end{aligned}$$

where d(v) is the probability density function of the random variable $\widetilde{v}$. With this new probability density function, we can rewrite the equality above as

$$\begin{aligned} \Psi ^{\prime }(x)&=\mathrm{COV}\left[ \widetilde{v},\left( \frac{\phi ^{\prime }(x)}{B}\widetilde{v}-1\right) \frac{u^{\prime \prime }\left( \frac{\phi (x)}{B}\widetilde{v}-x\right) }{u^{\prime }\left( \frac{\phi (x)}{B}\widetilde{v}-x\right) }\right] \\&=\mathrm{COV}\left[ \widetilde{v},\frac{d}{dx}\ln \left[ u^{\prime }\left( \frac{\widetilde{v}}{B}\phi (x)-x\right) \right] \right] . \end{aligned}$$

Without loss of generality, we can normalize B to 1. Denoting $H(x,v)=$ $\ln \left[ u^{\prime }(v\phi (x)-x)\right]$, the covariance is thus negative if and only if $\frac{\partial ^{2}H(x,v)}{\partial x\partial v}<0$, which concludes the proof. $\square$