Academic copyright in the publishing game: a contest perspective

Abstract

As scientists are motivated by readership rather than by royalties, one might doubt that academic copyright is required for stimulating research. Consequently, establishing an open access regime is currently intensively being discussed. We contribute to the literature by using a contest-model in which differently talented researchers compete for limited journals space. The contest perspective adds a rent-seeking motive into the publishing game which questions that private incentives for research are always too low due to the positive externalities of scientific progress. In our model, open access always leads to higher social welfare when incentives are too high. When incentives are too low, then open access is only superior if the benefits from larger readership is sufficiently high.

Appendix

Proof of Proposition 1

Part (i). Taking the partial derivatives gives

$$\begin{aligned} \frac{\partial e_{H}^{C}}{\partial \theta }= & {} \frac{\partial e_{L}^{C}}{ \partial \theta }=r^{C}\frac{1-\theta }{\left( 1+\theta \right) ^{3}}<0, \frac{\partial e_{H}^{O}}{\partial \theta }=-\left( 1-g\right) \frac{ \theta -1+g}{\left( 1+\theta -g\right) ^{3}}<0, \\ \frac{\partial e_{L}^{O}}{\partial \theta }= & {} -\left( 1-g\right) ^{2}\frac{ \theta -1+g}{\left( \theta -g+1\right) ^{3}}<0. \end{aligned}$$

Part (ii). Obvious. Part (iii). The ratio of efforts is $\frac{ e_{H}^{*}}{e_{L}^{*}}=\frac{r^{k}}{r^{k}-g_{L}^{k}}$, and thus $e_{H}^{*}=e_{L}^{*}$ for $g=0$ while $e_{H}^{*}>e_{L}^{*}$ for $g>0$. Next,

$$\begin{aligned} \frac{\partial e_{H}^{*}}{\partial g}= & {} -\frac{\left( r^{O}\right) ^{2}\theta \left( r\left( \theta -1\right) +g\right) }{\left( r^{O}\left( \theta +1\right) -g\right) ^{3}}<0, \end{aligned}$$

(19)

$$\begin{aligned} \frac{\partial e_{L}^{*}}{\partial g}= & {} -\frac{2\left( r^{O}\right) ^{2}\theta ^{2}\left( r^{O}-g\right) }{\left( r^{O}\left( \theta +1\right) -g\right) ^{3}}<0. \end{aligned}$$

(20)

Part (iv). From the best response functions in the text, we get

$$\begin{aligned} \frac{\partial e_{H}}{\partial e_{L}}= & {} \frac{1}{2e_{L}\theta ^{2}}\left( \left( e_{L}r^{k}\theta ^{3}\right) ^{0.5}-2e_{L}\theta \right) \\ \frac{\partial e_{L}}{\partial e_{H}}= & {} \frac{1}{2e_{H}}\left( \left( e_{H}\theta \left( r^{k}-g_{L}^{k}\right) \right) ^{0.5}-2e_{H}\theta \right) . \end{aligned}$$

Substituting the equilibrium values into the right hand sides of the two equations gives

$$\begin{aligned} \frac{\partial e_{H}\left( e_{L}^{*}\right) }{\partial e_{L}}= & {} \frac{ r^{k}\left( \theta -1\right) +g_{L}^{k}}{2\theta \left( r^{k}-g_{L}^{k}\right) }>0. \\ \frac{\partial e_{L}\left( e_{H}^{*}\right) }{\partial e_{H}}= & {} -\frac{ \left( r^{k}\left( 1+\theta \right) -g_{L}^{k}\right) ^{2}}{2\left( r^{^{k}}\right) ^{2}\theta \left( r^{k}-g_{L}^{k}\right) }\left( r^{k}\theta \left( r^{k}-g_{L}^{k}\right) \frac{r^{k}\left( \theta -1\right) +g_{L}^{k}}{ \left( r^{k}\left( 1+\theta \right) -g_{L}^{k}\right) ^{2}}\right) <0. \end{aligned}$$

□

Proof of Proposition 2

Recall that we have normalized $r^{O}=1$. Then, $r^{C}<\frac{\left( 1+\theta -g\right) ^{2}}{\left( 1+\theta \right) ^{2}\left( 1-g\right) ^{2}}$ for the low type (part (i)) and $r^{C}< \frac{\left( 1+\theta -g\right) ^{2}}{\left( 1+\theta \right) ^{2}\left( 1-g\right) }$ for the high type (part (ii)) follow immediately from $e_{H}^{*}$ and $e_{L}^{*}$ as given in the text. For the comparative statics in part (i), we have $\frac{\partial \left( RHS1\right) }{ \partial \theta }=\frac{2g\left( 1+\theta -g\right) }{\left( \theta +1\right) ^{3}\left( 1-g\right) ^{2}}>0$, $\frac{\partial \left( RHS1\right) }{\partial g}=-\frac{2\theta \left( 1+\theta -g\right) }{\left( \theta +1\right) ^{2}\left( 1-g\right) ^{3}}<0$. The comparative statics for part (ii) yields $\frac{\partial \left( RHS2\right) }{\partial \theta }= \frac{2g\left( 1+\theta -g\right) }{\left( 1+\theta \right) ^{3}\left( 1-g\right) }>0$, $\frac{\partial \left( RHS2\right) }{dg}=\frac{\left( 1+\theta -g\right) \left( \theta +g-1\right) }{\left( 1+\theta \right) ^{2}\left( 1-g\right) ^{2}}>0$ where RHS1 and RHS2 denote the right hand sides in the inequalities stated in part (i) and part (ii) of the Lemma, respectively. Part (iii). For the low type, $e_{L}^{O}>e_{L}^{C}$ iff $r^{C}<\frac{\left( 1+\theta -g\right) ^{2}}{\left( 1+\theta \right) ^{2}\left( 1-g\right) }$. The RHS of the inequality is increasing in $\theta$, $\frac{\partial \left( \frac{\left( 1+\theta -g\right) ^{2}}{\left( 1+\theta \right) ^{2}\left( 1-g\right) }\right) }{ \partial \theta }=\frac{2g\left( 1+\theta -g\right) }{\left( 1+\theta \right) ^{3}\left( 1-g\right) }>0$ and in g, $\frac{\partial \left( \frac{ \left( 1+\theta -g\right) ^{2}}{\left( 1+\theta \right) ^{2}\left( 1-g\right) }\right) }{\partial g}=\left( \frac{2\theta \left( 1+\theta -g\right) }{\left( \theta +1\right) ^{2}\left( 1-g\right) ^{3}}\right) >0$. For the high type, $e_{H}^{O}>e_{H}^{C}$ iff $r^{C}<\frac{\left( 1+\theta -g\right) ^{2}}{\left( 1+\theta \right) ^{2}\left( 1-g\right) }$. The derivatives of the RHS are $\frac{\partial \left( \frac{\left( 1+\theta -g\right) ^{2}}{\left( 1+\theta \right) ^{2}\left( 1-g\right) }\right) }{ \partial \theta }=\frac{2g\left( 1+\theta -g\right) }{\left( \theta +1\right) ^{3}\left( 1-g\right) }>0$ and $\frac{\partial \left( \frac{\left( 1+\theta -g\right) ^{2}}{\left( 1+\theta \right) ^{2}\left( 1-g\right) } \right) }{\partial g}=\frac{\left( 1+\theta -g\right) \left( \theta -1+g\right) }{\left( 1+\theta \right) ^{2}\left( 1-g\right) ^{2}}>0$ as $\theta >1$.

$$\begin{aligned} Part\ (iv).\frac{\frac{e_{H}^{O}}{e_{L}^{O}}}{\frac{e_{H}^{C}}{ e_{L}^{C}}}=\frac{\frac{\frac{\theta \left( 1-g\right) }{\left( \left( 1+\theta \right) -g\right) ^{2}}}{\frac{\theta \left( 1-g\right) ^{2}}{ \left( \left( 1+\theta \right) -g\right) ^{2}}}}{\frac{\frac{\left( r^{C}\right) ^{2}\theta }{\left( r^{C}\left( 1+\theta \right) \right) ^{2}}}{ \frac{\left( r^{C}\right) ^{2}\theta }{\left( r^{C}\left( 1+\theta \right) \right) ^{2}}}}=\frac{1}{1-g}>1{\text{iff}}\,g>0.\end{aligned}$$

□

Proof of Lemma 1

Part (i). The high type’s optimal effort with closed access is $\frac{\left( r^{C}\right) ^{2}\theta \beta ^{2}}{4}$, while his equilibrium effort is $e_{H}^{C}=\frac{r^{C}\theta }{\left( 1+\theta \right) ^{2}}$. We get $\frac{\left( r^{c}\right) ^{2}\theta \beta ^{2}}{4}-\frac{r^{C}\theta }{\left( 1+\theta \right) ^{2}}>0\Leftrightarrow \beta >\frac{2}{\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) }$ as stated in the Lemma. The low type’s optimal effort with closed access is $\frac{\left( r^{C}\right) ^{2}\beta ^{2}}{4}$, while his equilibrium effort is $e_{L}^{C}=\frac{r^{C}\theta }{\left( 1+\theta \right) ^{2}}$. Thus, $\frac{\left( r^{C}\right) ^{2}\beta ^{2}}{4}-\frac{r^{C}\theta }{\left( 1+\theta \right) ^{2}}>0\Leftrightarrow \beta >\frac{2\theta ^{0..5}}{\left( r^{C}\right) ^{0..5}\left( 1+\theta \right) }$ as stated in the Lemma. Part (ii). The high type’s optimal effort with open access is $\frac{\theta \beta ^{2}}{4}$, while his equilibrium effort is $e_{H}^{O}=\frac{\theta \left( 1-g\right) }{\left( 1+\theta -g\right) ^{2}}$. We get $\frac{\theta \beta ^{2}}{4}-\frac{\theta \left( 1-g\right) }{\left( 1+\theta -g\right) ^{2}}>0\Leftrightarrow \beta >2\frac{\left( 1-g\right) ^{0.5}}{1+\theta -g}$ as stated in the Lemma. The low type’s optimal effort with closed access is $\frac{\beta ^{2}}{4}$, while his equilibrium effort is $e_{L}^{O}=\frac{ \theta \left( 1-g\right) ^{2}}{\left( 1+\theta -g\right) ^{2}}$. Thus, $\frac{\beta ^{2}}{4}-\frac{\theta \left( 1-g\right) ^{2}}{\left( 1+\theta -g\right) ^{2}}>0\Leftrightarrow \beta >\frac{2\theta ^{0.5}\left( 1-g\right) }{1+\theta -g}$ as stated in the Lemma. Part (iii). Efforts are always too low if

$\beta >\max \left( \frac{2}{\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) },\frac{2\theta ^{0.5}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) },\frac{2\left( 1-g\right) ^{0.5}}{1+\theta -g},\frac{2\theta ^{0.5}\left( 1-g\right) }{1+\theta -g}\right)$. Next, (a)$\frac{2\theta ^{0.5}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) }>\frac{2}{\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) }$ as $\theta >1$. (b) Define $\Delta _{1}\equiv \frac{2\theta ^{0.5}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) }-\frac{2\left( 1-g\right) ^{0.5}}{1+\theta -g}$. We show that $\Delta _{1}>0$. As $\frac{\partial \Delta _{1}}{\partial r^{C}}<0$ and $\frac{\partial \Delta _{1}}{\partial g}>0$, we consider the maximum $r^{C}=1$ and the minimum $g=0$ to get $\Delta _{1}=2\frac{\theta ^{0.5}-1}{ \theta +1}>0$. (c) Define $\Delta _{2}\equiv \frac{2\theta ^{0.5}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) }-\frac{2\theta ^{0.5}\left( 1-g\right) }{1+\theta -g}$. We show that $\Delta _{2}\ge 0$. As $\frac{ \partial \Delta _{2}}{\partial r^{C}}<0$ and $\frac{\partial \Delta _{2}}{ \partial g}>0$, we consider $r^{C}=1$ and $g=0$ to get $\Delta _{2}=0$. Part (iv). Efforts are always too high if $\beta <\min \left( \frac{2 }{\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) },\frac{2\theta ^{0.5}}{ \left( r^{C}\right) ^{0.5}\left( 1+\theta \right) },\frac{2\left( 1-g\right) ^{0.5}}{1+\theta -g},\frac{2\theta ^{0.5}\left( 1-g\right) }{1+\theta -g} \right)$. We know that $\frac{2}{\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) }<\frac{2\theta ^{0.5}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) }$. Next, define $\Delta _{3}\equiv \frac{2}{\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) }-\frac{2\left( 1-g\right) ^{0.5}}{1+\theta -g}$. We show that $\Delta _{3}\ge 0$. As $\frac{\partial \Delta _{2}}{ \partial r^{C}}<0$ and $\frac{\partial \Delta _{2}}{\partial g}>0$, we consider $r^{C}=1$ and $g=0$ to get $\Delta _{2}=0$. Thus, we are left with the candidates $\frac{2\left( 1-g\right) ^{0.5}}{1+\theta -g}$ and $\frac{ 2\theta ^{0.5}\left( 1-g\right) }{1+\theta -g}$. Define $R_{1}\equiv \frac{ \frac{2\left( 1-g\right) ^{0.5}}{1+\theta -g}}{\frac{2\theta ^{0.5}\left( 1-g\right) }{1+\theta -g}}=\frac{1}{\theta ^{0.5}\left( 1-g\right) ^{0.5}}$ which is decreasing in g and above 1 iff $g>1-\frac{1}{\theta }$. Then, $\beta <\frac{2\theta ^{0.5}\left( 1-g\right) }{1+\theta -g}$ is the relevant condition and otherwise $\beta <\frac{2\left( 1-g\right) ^{0.5}}{1+\theta -g}$.□

Proof of Proposition 3

Define the ratios under the two regimes as

$$\begin{aligned} \hat{e}_{L}\equiv & {} \frac{e_{L}^{O}}{e_{L}^{C}}=\frac{\frac{\theta \left( 1-g\right) ^{2}}{\left( 1+\theta -g\right) ^{2}}}{\frac{r^{C}\theta }{\left( 1+\theta \right) ^{2}}}=\frac{\left( 1+\theta \right) ^{2}\left( 1-g\right) ^{2}}{r^{C}\left( 1+\theta -g\right) ^{2}} \end{aligned}$$

(21)

$$\begin{aligned} \hat{e}_{H}\equiv & {} \frac{e_{H}^{O}}{e_{H}^{C}}=\frac{\frac{\theta \left( 1-g\right) }{\left( 1+\theta -g\right) ^{2}}}{\frac{r^{C}\theta }{\left( \theta +1\right) ^{2}}}=\frac{\left( 1+\theta \right) ^{2}\left( 1-g\right) }{r^{C}\left( 1+\theta -g\right) ^{2}} \end{aligned}$$

(22)

From $\hat{e}_{L}$ and $\hat{e}_{H}$ we get

$$\begin{aligned} \frac{\partial \hat{e}_{L}}{\partial \theta }= & {} -\frac{2g\left( \theta +1\right) \left( 1-g\right) ^{2}}{r^{C}\left( \theta -g+1\right) ^{3}}<0,\quad \frac{\partial \hat{e}_{H}}{\partial \theta }=-\frac{2g\left( 1+\theta \right) \left( 1-g\right) }{r^{C}\left( \theta -g+1\right) ^{3}}<0 \end{aligned}$$

(23)

$$\begin{aligned} \frac{\partial \hat{e}_{L}}{\partial g}= & {} -\frac{2\theta \left( \theta +1\right) ^{2}\left( 1-g\right) }{r^{C}\left( \theta -g+1\right) ^{3}}<0 , \quad \frac{\partial \hat{e}_{H}}{\partial g}=-\frac{\left( \theta +1\right) ^{2}\left( \theta -1+g\right) }{r^{C}\left( \theta -g+1\right) ^{3} }<0 \end{aligned}$$

(24)

$$\begin{aligned} \frac{\partial \hat{e}_{L}}{\partial r^{C}}= & {} -\frac{\left( 1+\theta \right) ^{2}\left( 1-g\right) ^{2}}{\left( r^{C}\right) ^{2}\left( \theta -g+1\right) ^{2}}<0,\quad \frac{\partial \hat{e}_{H}}{\partial r}=-\frac{ \left( \theta +1\right) ^{2}\left( 1-g\right) }{\left( r^{C}\right) ^{2}\left( \theta -g+1\right) ^{2}}<0. \end{aligned}$$

(25)

Recalling that all efforts are too low by definition of the case considered, the claim follows.□

Proof of Proposition 4

Recall first that we know from Lemma 1 that all efforts are too high if $\beta <\frac{2\theta ^{0.5}\left( 1-g\right) }{ 1+\theta -g}$ for $g\ge 1-\frac{1}{\theta }$ and $\beta <\frac{2\left( 1-g\right) ^{0.5}}{1+\theta -g}$ for $g<1-\frac{1}{\theta }$. Consider first the limit case where $r^{C}\rightarrow r^{O}=1$, so that the only difference between open to closed access is that the low type needs to pay for her submission costs. Then, both efforts are higher under closed access as $\hat{ e}_{L}=\frac{\left( 1-g\right) ^{2}\left( \theta +1\right) ^{2}}{\left( 1+\theta -g\right) ^{2}}<1\forall g>0$ and $\hat{e}_{H}=\frac{\left( 1-g\right) \left( \theta +1\right) ^{2}}{\left( 1+\theta -g\right) ^{2}}<1$ where the last inequality follows from $\frac{\partial \hat{e}_{H}}{\partial \theta }=\frac{-2g\left( 1-g\right) \left( \theta +1\right) }{\left( 1+\theta -g\right) ^{3}}<0$ and $\frac{\partial \hat{e}_{H}}{\partial g}= \frac{\left( \theta +1\right) ^{2}\left( 1-\theta -g\right) }{\left( 1+\theta -g\right) ^{3}}<0$ together with $\frac{e_{H}^{O}}{e_{H}^{C}}=1$ for the minimum values $\theta =1$ and $g=0$. Next, note that only the two efforts matter for the difference in the welfare of the two publishing modes for $r^{C}=1$. And since social welfare is decreasing in efforts by definition of the case considered, welfare is higher for open access if audience is identical. Finally, social welfare for closed access is strictly increasing in $r^{C}$ as

$$\begin{aligned} \frac{\partial SW^{C}}{\partial r^{C}}=\beta \left( \frac{\left( r^{C}\right) ^{0.5}\theta ^{0.5}+r^{C}\theta }{\left( 1+\theta \right) } \right) +\beta r^{C}\left( \frac{\left( r^{C}\right) ^{-0.5}\theta ^{0.5}+2\theta }{2\left( 1+\theta \right) }\right) -\frac{2\theta }{\left( 1+\theta \right) ^{2}}+1>0 \end{aligned}$$

(26)

due to the fact that $\frac{2\theta }{\left( 1+\theta \right) ^{2}}$ is bounded above by 1 for $\theta \ge 1$. Therefore, $SW^{O}-SW^{C}(r^{C}=1)>0\Rightarrow$ $SW^{O}-SW^{C}(r^{C})>0$ $\forall r^{C}<1$.□

Proof of Lemma 2

Part (i). $\frac{\partial e_{H}^{C}}{ \partial \theta _{l}}=-\frac{r}{\theta _{l}^{2}}<0$; $\frac{\partial e_{L}^{C}}{\partial \theta _{l}}=-\frac{2r}{\left( \theta _{l}+1\right) ^{3}} <0$; $\frac{\partial e_{H}^{O}}{\partial \theta _{l}}=-\frac{\left( 1-g\right) \left( g+\theta _{l}-1\right) }{\left( \theta _{l}-g+1\right) ^{3} }<0$; $\frac{\partial e_{L}^{O}}{\partial \theta _{l}}=-2\frac{\left( g-1\right) ^{2}}{\left( \theta _{l}-g+1\right) ^{3}}<0$; $\frac{\partial e_{H}^{O}}{\partial g}=-\frac{\theta _{l}\left( g+\theta _{l}-1\right) }{ \left( \theta _{l}-g+1\right) ^{3}}<0$; $\frac{\partial e_{L}^{O}}{\partial g }=-\frac{2\theta _{l}\left( 1-g\right) }{\left( \theta _{l}-g+1\right) ^{3}} <0$. For part (ii), just recall that $e_{i}^{O}>e_{i}^{C}$, $i=H,L$ for $g\rightarrow 0$ as then only the effect of higher readership exists. Furthermore, $e_{i}^{O}<e_{i}^{C}$, $i=H,L$ for $r^{C}\rightarrow 1$ as then only the effort-reducing effect of the low type’s submission costs exists. Thus, the overall impact depends on which effect dominates, just as in Proposition 2.□

Proof of Lemma 3

Parts (i) and (ii) follow directly from comparing the privately and the socially optimal effort levels.

Part (iii). Efforts are always too low if

$\beta >\max \left( \frac{2\theta _{l}^{0.5}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) },\frac{2\theta _{l}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) },\frac{2\left( \theta _{l}\left( 1-g\right) \right) ^{0.5}}{\left( \theta _{l}+\left( 1-g\right) \right) }, \frac{2\theta _{l}\left( 1-g\right) }{\left( \theta _{l}+\left( 1-g\right) \right) }\right)$. We prove that $\frac{2\theta _{l}^{0.5}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) }>\max \left( \frac{2\theta _{l}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) },\frac{2\left( \theta _{l}\left( 1-g\right) \right) ^{0.5}}{\left( \theta _{l}+\left( 1-g\right) \right) },\frac{2\theta _{l}\left( 1-g\right) }{\left( \theta _{l}+\left( 1-g\right) \right) }\right)$, so that $\beta >\frac{2\theta _{l}^{0.5}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) }$ is the condition for all efforts being always too low:

• (a)
  $$\begin{aligned} \frac{\frac{2\theta _{l}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) }}{\frac{2\theta _{l}^{0.5}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) }}=\theta _{l}^{0.5}>1\,{\text {as}}\,\theta _{l}>1. \end{aligned}$$

  (27)

• (b) Define $R_{1}\equiv \frac{\frac{2\theta _{l}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) }}{\frac{2\left( \theta _{l}\left( 1-g\right) \right) ^{0.5}}{\left( \theta _{l}+\left( 1-g\right) \right) }}$. As we want to show that $R_{1}>1$ and since $R_{1}$ is decreasing in $r^{C}$, we consider the maximum $r^{C}=1$. Then, $R_{1}=\theta _{l}\frac{\left( 1-g\right) +\theta _{l}}{\left( \theta _{l}\left( 1-g\right) \right) ^{0.5}\left( 1+\theta _{l}\right) }$. As $\frac{\partial R_{1}}{\partial g}= \frac{\theta _{l}^{2}\left( \theta _{l}-\left( 1-g\right) \right) }{2\left( \theta _{l}\left( 1-g\right) \right) ^{\frac{3}{2}}\left( 1+\theta _{l}\right) }>0$, we need to consider the minimum g. Then, $R_{1}=1+\frac{ 1+\theta _{l}}{\theta _{l}^{0.5}}>1$.

• (c)
  $$\begin{aligned} \frac{2\theta _{l}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) }-\frac{2\theta _{l}\left( 1-g\right) }{\left( \theta _{l}+\left( 1-g\right) \right) }=2\theta _{l}\frac{\left( \theta _{l}+\left( 1-g\right) \right) -\left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) \left( 1-g\right) }{\left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) \left( \theta _{l}+\left( 1-g\right) \right) }>0\text{:} \end{aligned}$$

  (28)

  The denominator is always positive. The numerator N is decreasing in $r^{C}$ and simplifies to $N=g\theta _{l}>0$ for the maximum $r^{C}=1$.

Parts (a), (b) and (c) together prove that $\beta >\frac{2\theta _{l}}{ \left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) }$ is the critical condition.

Part (iv) Efforts are always too high if

$\beta <\min \left( \frac{2\theta _{l}^{0.5}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) },\frac{2\theta _{l}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) },\frac{2\left( \theta _{l}\left( 1-g\right) \right) ^{0.5}}{\left( \theta _{l}+\left( 1-g\right) \right) }, \frac{2\theta _{l}\left( 1-g\right) }{\left( \theta _{l}+\left( 1-g\right) \right) }\right)$. We know from the proof of part (iii) that $\frac{ 2\theta _{l}^{0.5}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) }< \frac{2\theta _{l}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) }$. Define $R_{2}\equiv \frac{\frac{2\theta _{l}^{0.5}}{\left( r^{C}\right) ^{0.5}\left( 1+\theta _{l}\right) }}{\frac{2\left( \theta _{l}\left( 1-g\right) \right) ^{0.5}}{\left( \theta _{l}+\left( 1-g\right) \right) }}$. We show that $R_{2}\ge 1$. As $\frac{\partial R_{2}}{\partial r^{C}}<0$, we consider the maximum $r^{C}=1$ to get $R_{2}=\frac{\theta _{l}^{0.5}\left( \left( 1-g\right) +\theta _{l}\right) ^{0.5}}{\left( 1+\theta _{l}\right) }$ which is increasing in g. We hence consider $g=0$ which gives $R_{2}=1$. We are hence left with the comparison of $\frac{2\left( \theta _{l}\left( 1-g\right) \right) ^{0.5}}{\left( \theta _{l}+\left( 1-g\right) \right) }$ and $\frac{2\theta _{l}\left( 1-g\right) }{\left( \theta _{l}+\left( 1-g\right) \right) }$ and define $R_{3}\equiv \frac{\frac{2\left( \theta _{l}\left( 1-g\right) \right) ^{0.5}}{\left( \theta _{l}+\left( 1-g\right) \right) }}{\frac{2\theta _{l}\left( 1-g\right) }{\left( \theta _{l}+\left( 1-g\right) \right) }}=\left( \frac{1}{\theta _{l}\left( 1-g\right) }\right) ^{0.5}$ which is increasing in g and above 1 iff $g>1-\frac{1}{\theta _{l}}$. Then, $\beta <\frac{2\theta _{l}\left( 1-g\right) }{\left( \theta _{l}+\left( 1-g\right) \right) }$ is the relevant condition and otherwise $\beta <\frac{2\left( \theta _{l}\left( 1-g\right) \right) ^{0.5}}{\left( \theta _{l}+\left( 1-g\right) \right) }$.□

Proof of Proposition 5

Part (i). Define the ratios of the two types’ efforts under the two regimes as

$$\hat{e}_{L} \equiv {} \frac{e_{L}^{O}}{e_{L}^{C}}=\frac{\frac{\left( 1-g\right) ^{2}}{\left( \theta _{l}+\left( 1-g\right) \right) ^{2}}}{\frac{ r^{C}}{\left( 1+\theta _{l}\right) ^{2}}}=\frac{\left( 1-g\right) ^{2}\left( 1+\theta _{l}\right) ^{2}}{r^{C}\left( \left( 1-g\right) +\theta _{l}\right) ^{2}}$$

(29)

$$\hat{e}_{H}\equiv {} \frac{e_{H}^{O}}{e_{H}^{C}}=\frac{\frac{\theta _{l}\left( 1-g\right) }{\left( \theta _{l}+\left( 1-g\right) \right) ^{2}}}{ \frac{\theta _{l}r^{C}}{\left( 1+\theta _{l}\right) ^{2}}}=\frac{\left( 1-g\right) \left( 1+\theta _{l}\right) ^{2}}{r^{C}\left( \left( 1-g\right) +\theta _{l}\right) ^{2}}$$

(30)

to get

$$\begin{aligned} \frac{\partial \hat{e}_{L}}{\partial \theta _{l}}= & {} -\frac{2g\left( 1-g\right) ^{2}\left( 1+\theta _{l}\right) }{r^{C}\left( \theta _{l}-g+1\right) ^{3}}<0,\quad \frac{\partial \hat{e}_{H}}{\partial \theta _{l}}=-\frac{2g\left( 1-g\right) \left( 1+\theta _{l}\right) }{r^{C}\left( \theta _{l}-g+1\right) ^{3}}<0 \end{aligned}$$

(31)

$$\begin{aligned} \frac{\partial \hat{e}_{L}}{\partial g}= & {} -\frac{2\theta _{l}\left( 1-g\right) \left( 1+\theta _{l}\right) ^{2}}{r^{C}\left( \theta _{l}-g+1\right) ^{3}}<0,\quad \frac{\partial \hat{e}_{H}}{\partial g}=- \frac{\left( 1+\theta _{l}\right) ^{2}\left( g+\theta _{l}-1\right) }{ r^{C}\left( \theta _{l}-g+1\right) ^{3}}<0 \end{aligned}$$

(32)

$$\begin{aligned} \frac{\partial \hat{e}_{L}}{\partial r^{C}}= & {} -\frac{\left( 1-g\right) ^{2}\left( 1+\theta _{l}\right) ^{2}}{\left( r^{C}\right) ^{2}\left( \theta _{l}-g+1\right) ^{2}}<0,\quad \frac{\partial \hat{e}_{H}}{\partial r^{C}}=- \frac{\left( 1-g\right) \left( 1+\theta _{l}\right) ^{2}}{\left( r^{C}\right) ^{2}\left( \theta _{l}-g+1\right) ^{2}}<0. \end{aligned}$$

(33)

Recalling that all efforts are too low by definition of the case considered, the claim follows. Part (ii). Follows exactly the proof of Proposition 4 and is therefore omitted.□

Proof of Lemma 4

Parts (i) and (ii) follow directly from comparing the privately and the socially optimal effort levels.

Part (iii). Efforts are always too low if $\beta >\max \left( \frac{2}{\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) },\frac{2\left( r^{C}\theta \right) ^{0.5}}{lr^{C}\left( 1+\theta \right) },\frac{2\left( 1-g\right) ^{0.5}}{\left( 1+\theta -g\right) },\frac{ 2\theta ^{0.5}\left( 1-g\right) }{l\left( 1+\theta -g\right) }\right)$ (a). $\frac{\frac{2}{\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) }}{\frac{ 2\theta ^{0.5}}{l\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) }}=\frac{l }{\theta ^{0.5}}<1$ as $l<1$ and $\theta >1$. (b) Define $R_{1}\equiv \frac{ \frac{2\theta ^{0.5}}{l\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) }}{ \frac{2\left( 1-g\right) ^{0.5}}{\left( 1+\theta -g\right) }}$. We show that $R_{1}>1$. As $\frac{\partial R_{1}}{\partial r^{C}}<0$ and $\frac{\partial R_{1}}{\partial g}>0$, we consider $r^{C}=1$ and $g=0$ to get $R_{1}=\frac{ \theta ^{0.5}}{l}>1$. (c) Define $R_{2}\equiv \frac{\frac{2\theta ^{0.5}}{ l\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) }}{\frac{2\theta ^{0.5}\left( 1-g\right) }{l\left( 1+\theta -g\right) }}$. We show that $R_{2}\ge 1$. As $\frac{\partial R_{1}}{\partial r^{C}}<0$ and $\frac{ \partial R_{1}}{\partial g}>0$, we consider $r^{C}=1$ and $g=0$ to get $R_{1}=1$. Thus, $\beta >\frac{2\theta ^{0.5}}{l\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) }$ is the critical condition.

Part (iv). Efforts are always too high if $\beta <\min \left( \frac{2}{\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) },\frac{2\left( r^{C}\theta \right) ^{0.5}}{lr^{C}\left( 1+\theta \right) },\frac{2\left( 1-g\right) ^{0.5}}{\left( 1+\theta -g\right) },\frac{ 2\theta ^{0.5}\left( 1-g\right) }{l\left( 1+\theta -g\right) }\right)$. We know from the proof of part (iii) that $\frac{2}{\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) }<\frac{2\theta ^{0.5}}{l\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) }$. Define $R_{3}\equiv \frac{\frac{2}{\left( r^{C}\right) ^{0.5}\left( 1+\theta \right) }}{\frac{2\left( 1-g\right) ^{0.5} }{\left( 1+\theta -g\right) }}$. We show that $R_{3}\ge 1$. As $\frac{ \partial R_{3}}{\partial r^{C}}<0$ and $\frac{\partial R_{3}}{\partial g}>0$, we consider $r^{C}=1$ and $g=0$ which gives $R_{3}=1$. We are hence left with the comparison of $\frac{2\left( 1-g\right) ^{0.5}}{\left( 1+\theta -g\right) }$ and $\frac{2\theta ^{0.5}\left( 1-g\right) }{l\left( 1+\theta -g\right) }$ and define $R_{4}\equiv \frac{\frac{2\left( 1-g\right) ^{0.5}}{ \left( 1+\theta -g\right) }}{\frac{2\theta ^{0.5}\left( 1-g\right) }{l\left( 1+\theta -g\right) }}=\frac{l}{\theta ^{0.5}\left( 1-g\right) ^{0.5}}$ which is increasing in g and $R_{4}>1$ iff $g>1-\frac{l^{2}}{\theta }$. Then, $\beta <\frac{2\theta ^{0.5}\left( 1-g\right) }{l\left( 1+\theta -g\right) }$ is the relevant condition and otherwise $\beta <\frac{l}{\theta ^{0.5}\left( 1-g\right) ^{0.5}}$.□

Proof of Proposition 6

Part (i) . Define the ratios of the two types’ efforts under the two regimes as

$$\begin{aligned} \hat{e}_{L}\equiv & {} \frac{e_{L}^{O}}{e_{L}^{C}}=\frac{\frac{\theta \left( 1-g\right) ^{2}}{\left( 1+\theta -g\right) ^{2}}}{\frac{r^{C}\theta }{\left( 1+\theta \right) ^{2}}}=\frac{\left( 1+\theta \right) ^{2}\left( 1-g\right) ^{2}}{r^{C}\left( 1+\theta -g\right) ^{2}} \end{aligned}$$

(34)

$$\begin{aligned} \hat{e}_{H}\equiv & {} \frac{e_{H}^{O}}{e_{H}^{C}}=\frac{\frac{l\left( 1-g\right) }{\left( l+\left( 1-g\right) \right) ^{2}}}{\frac{lr}{\left( 1+l\right) ^{2}}}=\frac{\left( 1+\theta \right) ^{2}\left( 1-g\right) }{ r^{C}\left( 1+\theta -g\right) ^{2}} \end{aligned}$$

(35)

to get

$$\begin{aligned} \frac{\partial \hat{e}_{L}}{\partial \theta }= & {} -\frac{2g\left( 1+\theta \right) \left( 1-g\right) ^{2}}{r^{C}\left( \theta -g+1\right) ^{3}}<0,\frac{\partial \hat{e}_{H}}{\partial \theta }=-\frac{2g\left( 1+\theta \right) \left( 1-g\right) }{r^{C}\left( \theta -g+1\right) ^{3}}<0 \end{aligned}$$

(36)

$$\begin{aligned} \frac{\partial \hat{e}_{L}}{\partial g}= & {} -\frac{2\theta \left( 1+\theta \right) ^{2}\left( 1-g\right) }{r^{C}\left( \theta -g+1\right) ^{3}}<0 ,\frac{\partial \hat{e}_{H}}{\partial g}=-\frac{\left( 1+\theta \right) ^{2}\left( g+\theta -1\right) }{r^{C}\left( \theta -g+1\right) ^{3}}<0 \end{aligned}$$

(37)

$$\begin{aligned} \frac{\partial \hat{e}_{L}}{\partial r^{C}}= & {} -\frac{\left( 1+\theta \right) ^{2}\left( 1-g\right) ^{2}}{\left( r^{C}\right) ^{2}\left( \theta -g+1\right) ^{2}}<0,\quad \frac{\partial \hat{e}_{H}}{\partial r^{C}}=- \frac{\left( 1+\theta \right) ^{2}\left( 1-g\right) }{\left( r^{C}\right) ^{2}\left( \theta -g+1\right) ^{2}}<0. \end{aligned}$$

(38)

Recalling that all efforts are too low by definition of the case considered, the claim follows. Part (ii). Follows exactly the proof of Proposition 4 and is therefore omitted.□

Proof of Proposition 7

Part (i). From $g_{i}^{*}$ as given in the text, we get $\frac{\partial g_{H}^{*}}{\partial W_{H}}=- \frac{1-g_{L}}{2\sqrt{\left( 1-g_{L}\right) \left( 1+W_{H}\theta -g_{L}\right) }}<0$ and $\frac{\partial g_{L}^{*}}{\partial W_{L}}=- \frac{\theta \left( 1-g_{H}\right) }{2\sqrt{\theta \left( 1-g_{H}\right) \left( W_{L}+\theta \left( 1-g_{H}\right) \right) }}<0$. Part (ii). From $g_{i}^{*}$ as given in the text, we get

$$\begin{aligned} \frac{\partial g_{H}^{*}}{\partial g_{L}}=\frac{\sqrt{\left( 1-g_{L}\right) \left( 1+\theta \left( 1+W_{H}-s\right) -g_{L}\right) }\left( \theta \left( 1+W_{H}-s\right) +2\left( 1-g_{L}\right) -2\sqrt{\left( 1-g_{L}\right) \left( 1+\theta \left( 1+W_{H}-s\right) -g_{L}\right) } \right) }{2\theta \left( 1-g_{L}\right) \left( 1+\theta \left( 1+W_{H}-s\right) -g_{L}\right) } \end{aligned}$$

where the sign depends only on whether $A\equiv \theta \left( 1+W_{H}-s\right) +2\left( 1-g_{L}\right) >B\equiv 2\sqrt{\left( 1-g_{L}\right) \left( 1+\theta \left( 1+W_{H}-s\right) -g_{L}\right) }$. Taking the square on both sides and considering the difference gives $A^{2}-B^{2}=\theta ^{2}\left( 1+W_{H}-s\right) ^{2}>0$. Analogous calculations yield $\frac{\partial g_{L}^{*}}{\partial g_{H}}>0$. Part (iii). We get

$$\begin{aligned} \frac{\partial g_{H}^{*}}{\partial \theta }=\frac{\sqrt{\left( 1-g_{L}\right) \left( 1+\theta \left( 1+W_{H}-s\right) -g_{L}\right) }\left( 2\left( 1-g_{L}\right) +\theta \left( 1+W_{H}-s\right) -2\sqrt{\left( 1-g_{L}\right) \left( 1+\theta \left( 1+W_{H}-s\right) -g_{L}\right) } \right) }{2\theta ^{2}\left( 1+\theta \left( 1+W_{H}-s\right) -g_{L}\right) } \end{aligned}$$

where the sign depends only on whether $A\equiv 2\left( 1-g_{L}\right) +\theta \left( 1+W_{H}-s\right) >B\equiv 2\sqrt{\left( 1-g_{L}\right) \left( 1+\theta \left( 1+W_{H}-s\right) -g_{L}\right) }$. Taking the square on both sides and considering the difference gives $A^{2}-B^{2}=\theta ^{2}\left( 1+W_{H}-s\right) ^{2}>0$. Next,

$$\begin{aligned} \frac{\partial g_{L}^{*}}{\partial \theta }=-\frac{1}{2\theta }\frac{ \sqrt{\theta \left( 1-g_{H}\right) \left( 1+W_{L}-s+\theta \left( 1-g_{H}\right) \right) }\left( 1+W_{L}-s+2\theta \left( 1-g_{H}\right) -2 \sqrt{\theta \left( 1-g_{H}\right) \left( W_{L}-s+\theta -\theta g_{H}+1\right) }\right) }{1+W_{L}-s+\theta \left( 1-g_{H}\right) } \end{aligned}$$

which is negative if and only if $C\equiv 1+W_{L}-s+2\theta \left( 1-g_{H}\right) >D\equiv 2\sqrt{\theta \left( 1-g_{H}\right) \left( W_{L}-s+\theta -\theta g_{H}+1\right) }$. Taking the square on both sides and considering the difference gives $C^{2}-D^{2}=\left( 1+W_{L}-s\right) ^{2}>0$.□