Appendix
1.1 Proof of Lemma 1
Conditions \(FO{C}_{1}\) and \(FO{C}_{2}\) imply
$$\int x{f}_{h}\left(x,\theta ,h\left({a}_{1},{a}_{2}\right)\right)dx=\frac{{c}_{1}^{\prime}\left({a}_{1}\right)}{{h}_{1}^{\prime}\left(a\right)}+\frac{{c}_{2}^{\prime}\left({a}_{2}\right)}{{h}_{2}^{\prime}\left(a\right)},$$
implying
$${R}_{h}^{\mathrm{^{\prime}}}\left(\theta ,h\left({a}_{1},{a}_{2}\right)\right)=\frac{{c}_{1}^{\mathrm{^{\prime}}}\left({a}_{1}\right)}{{h}_{1}^{\mathrm{^{\prime}}}\left(a\right)}+\frac{{c}_{2}^{\mathrm{^{\prime}}}\left({a}_{2}\right)}{{h}_{2}^{\mathrm{^{\prime}}}\left(a\right)}.$$
Hence, problem becomes
$$\begin{array}{l}\underset{{s}_{i}\in \mathcal{S},{\widehat{a}}_{i}\in \mathcal{A}}{{\text{max}}}{E}_{\theta }\left(R\left(\theta ,h\left({\widehat{a}}_{1}\left(\theta \right),{\widehat{a}}_{2}\left(\theta \right)\right)\right)-{c}_{1}\left({\widehat{a}}_{1}\left(\theta \right)\right)-{c}_{2}\left({\widehat{a}}_{2}\left(\theta \right)\right)\right)\\ {\text{ s.t. }} {R}_{h}^{\prime}\left(\theta ,h\left(\widehat{a}\left(\theta \right)\right)\right)=\frac{{c}_{1}^{\mathrm{^{\prime}}}\left({\widehat{a}}_{1}\left(\theta \right)\right)}{{h}_{1}^{\mathrm{^{\prime}}}\left(\widehat{a}\left(\theta \right)\right)}+\frac{{c}_{2}^{\mathrm{^{\prime}}}\left({\widehat{a}}_{2}\left(\theta \right)\right)}{{h}_{2}^{\mathrm{^{\prime}}}\left(\widehat{a}\left(\theta \right)\right)},\\ \quad {FOC}_{2}, \, SO{C}_{1}, \, SO{C}_{2},\\ \quad {s}_{1}\left(x\right)+{s}_{2}\left(x\right)=x,\text{ for all }x\in {\mathbb{R}}_{+}.\end{array}$$
Since the objective function is unrelated to the revenue-sharing scheme, this problem can be solved in two steps. First, we find a solution \(\widehat{a}\left(\theta \right)\) from the following problem:
$$\begin{array}{l}\underset{{\widehat{a}}_{1},{\widehat{a}}_{2}\in \mathcal{A}}{{\text{max}}}{E}_{\theta }\left(R\left(\theta ,h\left({\widehat{a}}_{1}\left(\theta \right),{\widehat{a}}_{2}\left(\theta \right)\right)\right)-{c}_{1}\left({\widehat{a}}_{1}\left(\theta \right)\right)-{c}_{2}\left({\widehat{a}}_{2}\left(\theta \right)\right)\right)\\ \text{ s.t. } {R}_{h}^{\prime}\left(\theta ,h\left(\widehat{a}\left(\theta \right)\right)\right)=\frac{{c}_{1}^{\mathrm{^{\prime}}}\left({\widehat{a}}_{1}\left(\theta \right)\right)}{{h}_{1}^{\mathrm{^{\prime}}}\left(\widehat{a}\left(\theta \right)\right)}+\frac{{c}_{2}^{\mathrm{^{\prime}}}\left({\widehat{a}}_{2}\left(\theta \right)\right)}{{h}_{2}^{\mathrm{^{\prime}}}\left(\widehat{a}\left(\theta \right)\right)}.\end{array}$$
By the Pontryagin maximum principle, this problem can be further changed to the problem. This problem is not related to the revenue-sharing scheme.
Second, given \(\left({\widehat{a}}_{1},{\widehat{a}}_{2}\right)\) from the above, we look for a revenue-sharing scheme \({s}_{i}(x)\) that satisfies \(FO{C}_{2},SO{C}_{1}\) and \(SO{C}_{2}.\) We can first find a sharing scheme \({s}_{2}\left(x\right)\) that satisfies \(FO{C}_{2}\):
$$\int {s}_{2}\left(x\right){f}_{h}\left(x;\theta ,h\left({\widehat{a}}_{1}\left(\theta \right),{\widehat{a}}_{2}\left(\theta \right)\right)\right)dx=\frac{{c}_{2}^{\mathrm{^{\prime}}}\left({\widehat{a}}_{2}\left(\theta \right)\right)}{{h}_{2}^{\mathrm{^{\prime}}}\left(\widehat{a}\left(\theta \right)\right)},$$
(20)
and then define \({s}_{1}\left(x\right)\) by the condition \({s}_{1}\left(x\right)+{s}_{2}\left(x\right)=x\). If \({s}_{2}\left(x\right)\) can be a linear function, as in Case 1, verifying \(SO{C}_{1}\) and \(SO{C}_{2}\) will be easy.
1.2 Proof of Lemma 2
The objective function can be written as
$$U\equiv {r}_{1}{\text{ln}}\left(\int {s}_{1}\left(x\right)f\left(x,h\left(a\right)\right)dx-{c}_{1}\left({a}_{1}\right)\right)+{r}_{2}{\text{ln}}\left(\int {s}_{2}\left(x\right)f\left(x,h\left(a\right)\right)dx-{c}_{2}\left({a}_{2}\right)\right).$$
Then, problem can be written as
$$\begin{array}{l}\underset{{a}_{1},{a}_{2}\in {\mathbb{A}}}{{\text{max}}} \underset{{s}_{i}\in \mathcal{S}}{{\text{max}}} U\\ \text{ s.t. } {s}_{1}\left(x\right)+{s}_{2}\left(x\right)=x,\text{ for all }x\in {\mathbb{R}}_{+}.\end{array}$$
(21)
Given \(\left({a}_{1},{a}_{2}\right)\), consider problem:
$$\begin{array}{l}\underset{{s}_{i}\in \mathcal{S}}{{\text{max}}} U\\ \text{ s.t. } {s}_{1}\left(x\right)+{s}_{2}\left(x\right)=x,\text{ for all }x\in {\mathbb{R}}_{+}.\end{array}$$
(22)
The FOC isFootnote 6
$$\begin{array}{c}0=\frac{\partial U}{\partial {s}_{1}}=\frac{ {r}_{1}f\left(x,h\left(a\right)\right)}{\int {s}_{1}\left(x\right)f\left(x,h\left(a\right)\right)dx-{c}_{1}\left({a}_{1}\right)}-\frac{{r}_{2}f\left(x,h\left(a\right)\right)}{\int \left(x-{s}_{1}\left(x\right)\right)f\left(x,h\left(a\right)\right)dx-{c}_{2}\left({a}_{2}\right)},\end{array}$$
implying
$${r}_{1}\left[R\left(h\left(a\right)\right)-\int {s}_{1}\left(x\right)f\left(x,h\left(a\right)\right)dx-{c}_{2}\left({a}_{2}\right)\right]={r}_{2}\left[\int {s}_{1}\left(x\right)f\left(x,h\left(a\right)\right)dx-{c}_{1}\left({a}_{1}\right)\right],$$
implying
$$\int {s}_{1}\left(x\right)f\left(x,h\left(a\right)\right)dx=\frac{{r}_{1}R\left(h\left(a\right)\right)+{r}_{2}{c}_{1}\left({a}_{1}\right)-{r}_{1}{c}_{2}\left({a}_{2}\right)}{{r}_{1}+{r}_{2}}.$$
(23)
The SOC is satisfied:
$$\frac{{\partial }^{2}U}{\partial {s}_{1}^{2}}=-\frac{ {r}_{1}{\left[f\left(x,h\left(a\right)\right)\right]}^{2}}{{\left[\int {s}_{1}\left(x\right)f\left(x,h\left(a\right)\right)dx-{c}_{1}\left({a}_{1}\right)\right]}^{2}}-\frac{{r}_{2}{\left[f\left(x,h\left(a\right)\right)\right]}^{2}}{{\left[\int \left(x-{s}_{1}\left(x\right)\right)f\left(x,h\left(a\right)\right)dx-{c}_{2}\left({a}_{2}\right)\right]}^{2}}<0.$$
That is, given\(\left({a}_{1},{a}_{2}\right)\), any admissible \({s}_{1}\left(\cdot \right)\) satisfying is optimal. Consider a simple linear sharing scheme\({s}_{1}\left(x\right)={\alpha }_{1}x\). Then, by, we have
$${\alpha }_{1}=\frac{{r}_{1}R\left(h\left(a\right)\right)+{r}_{2}{c}_{1}\left({a}_{1}\right)-{r}_{1}{c}_{2}\left({a}_{2}\right)}{\left({r}_{1}+{r}_{2}\right)R\left(h\left(a\right)\right)}.$$
That is, equation has a simple solution of the form \({s}_{1}\left(x\right)={\alpha }_{1}x\). Then,
$${\alpha }_{2}=1-{\alpha }_{1}=\frac{{r}_{2}R\left(h\left(a\right)\right)-{r}_{2}{c}_{1}\left({a}_{1}\right)+{r}_{1}{c}_{2}\left({a}_{2}\right)}{\left({r}_{1}+{r}_{2}\right)R\left(h\left(a\right)\right)},$$
implying \({s}_{2}\left(x\right)={\alpha }_{2}x.\) Then, the objective function of is
$$\begin{array}{l}{\left({\alpha }_{1}R\left(h\left(a\right)\right)-{c}_{1}\left({a}_{1}\right)\right)}^{{r}_{1}}{\left({\alpha }_{2}R\left(h\left(a\right)\right)-{c}_{2}\left({a}_{2}\right)\right)}^{{r}_{2}}\\ ={\left(\frac{{r}_{1}R\left(h\left(a\right)\right)-{r}_{1}{c}_{1}\left({a}_{1}\right)-{r}_{1}{c}_{2}\left({a}_{2}\right)}{{r}_{1}+{r}_{2}}\right)}^{{r}_{1}}{\left(\frac{{r}_{2}R\left(h\left(a\right)\right)-{r}_{2}{c}_{1}\left({a}_{1}\right)-{r}_{2}{c}_{2}\left({a}_{2}\right)}{{r}_{1}+{r}_{2}}\right)}^{{r}_{2}}\\ =\frac{{r}_{1}^{{r}_{1}}{r}_{2}^{{r}_{2}}}{{\left({r}_{1}+{r}_{2}\right)}^{{r}_{1}+{r}_{2}}}{\left[R\left(h\left(a\right)\right)-{c}_{1}\left({a}_{1}\right)-{c}_{2}\left({a}_{2}\right)\right]}^{{r}_{1}+{r}_{2}}.\end{array}$$
Then, problem becomes
$$\underset{{a}_{1},{a}_{2}\in {\mathbb{A}}}{{\text{max}}} R\left(h\left(a\right)\right)-{c}_{1}\left({a}_{1}\right)-{c}_{2}\left({a}_{2}\right).$$
1.3 Derivation of case 1
We have
$$R\left(\theta ,h\right)=E\left(\widetilde{A}\theta h\right)=\theta h, \, R\left(h\right)={E}_{\theta }\left(R\left(\widetilde{\theta },h\right)\right)=\overline{\theta }h.$$
Then, for a limited contract, problem becomes
$$\begin{array}{l}\underset{ {a}_{1,}{a}_{2}\in A}{\text{max}} \theta h\left({a}_{1},{a}_{2}\right)-\frac{1}{2}{a}_{1}^{2}-\frac{1}{2}{a}_{2}^{2}\\ \text{ s.t. } \theta =\frac{{a}_{1}}{{\mu }_{1}}+\frac{{a}_{2}}{{\mu }_{2}}.\end{array}$$
The solution is
$${a}_{1}=\frac{{\mu }_{1}^{3}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta , {a}_{2}=\frac{{\mu }_{2}^{3}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta .$$
Then,
$$\begin{array}{l}{W}_{l}^{S}={E}_{\theta }\left(R\left(\theta ,h\left({a}_{1},{a}_{2}\right)\right)-{c}_{1}\left({a}_{1}\right)-{c}_{2}\left({a}_{2}\right)\right)\\ ={E}_{\theta }\left(\theta \left(\frac{{\mu }_{1}^{4}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta +\frac{{\mu }_{2}^{4}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta \right)-\frac{1}{2}{\left(\frac{{\mu }_{1}^{3}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta \right)}^{2}-\frac{1}{2}{\left(\frac{{\mu }_{2}^{3}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta \right)}^{2}\right)\\ =\left(\frac{{\mu }_{1}^{4}+{\mu }_{2}^{4}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}-\frac{{\mu }_{1}^{6}+{\mu }_{2}^{6}}{2{\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}^{2}}\right){E}_{\theta }\left({\theta }^{2}\right) \\ =\frac{{\sigma }_{\theta }^{2}+{\overline{\theta }}^{2}}{2{\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}^{2}}\left({\mu }_{1}^{6}+{\mu }_{2}^{6}+2{\mu }_{1}^{2}{\mu }_{2}^{2}\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)\right)\\ =\frac{{\sigma }_{\theta }^{2}+{\overline{\theta }}^{2}}{2}\frac{{\mu }_{1}^{4}+{\mu }_{2}^{4}+{\mu }_{1}^{2}{\mu }_{2}^{2}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}.\end{array}$$
We also have
$$\frac{{c}_{2}^{\mathrm{^{\prime}}}\left({a}_{2}\right)}{{h}_{2}^{\mathrm{^{\prime}}}\left(a\right)}=\frac{{a}_{2}}{{\mu }_{2}}=\frac{{\mu }_{2}^{2}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta .$$
Let \({s}_{2}\left(x\right)={\alpha }_{2}x\), where \({\alpha }_{2}\) is a constant. Then,
$$\int {s}_{2}\left(x\right){f}_{h}\left(x,\theta ,h\left({a}_{1},{a}_{2}\right)\right)dx={\alpha }_{2}{R}_{h}^{\mathrm{^{\prime}}}\left(\theta ,h\left(a\right)\right)={\alpha }_{2}\theta .$$
Hence, if\({\alpha }_{2}=\frac{{\mu }_{2}^{2}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\), then equation is satisfied for any\(\theta\). That is, there is a linear sharing scheme satisfying equation. We also have
$$h={\mu }_{1}{a}_{1}+{\mu }_{2}{a}_{2}=\frac{{\mu }_{1}^{4}+{\mu }_{2}^{4}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta .$$
For a comprehensive contract, problem becomes
$${W}_{c}^{S}=\underset{ {a}_{1,}{a}_{2}\in {\mathbb{A}}}{\text{max}} \overline{\theta }h\left({a}_{1},{a}_{2}\right)-\frac{1}{2}{a}_{1}^{2}-\frac{1}{2}{a}_{2}^{2}.$$
The FOCs are
$$\overline{\theta }{\mu }_{1}={a}_{1}, \overline{\theta }{\mu }_{2}={a}_{2}.$$
Then,
$${W}_{c}^{S}=\overline{\theta }\left({\mu }_{1}^{2}\overline{\theta }+{\mu }_{2}^{2}\overline{\theta }\right)-\frac{1}{2}{\left(\overline{\theta }{\mu }_{1}\right)}^{2}-\frac{1}{2}{\left(\overline{\theta }{\mu }_{2}\right)}^{2}=\frac{{\mu }_{1}^{2}+{\mu }_{2}^{2}}{2}{\overline{\theta }}^{2},$$
and
$$h\left(a\right)={\mu }_{1}{a}_{1}+{\mu }_{2}{a}_{2}=\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)\overline{\theta },$$
and
$$\begin{array}{l}{\alpha }_{1}=\frac{{r}_{1}\overline{\theta }h\left(a\right)+\frac{{r}_{2}}{2}{a}_{1}^{2}-\frac{{r}_{1}}{2}{a}_{2}^{2}}{({r}_{1}+{r}_{2})\overline{\theta }h\left(a\right)}=\frac{{r}_{1}\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right){\overline{\theta }}^{2}+\frac{{r}_{2}}{2}{\mu }_{1}^{2}{\overline{\theta }}^{2}-\frac{{r}_{1}}{2}{\mu }_{2}^{2}{\overline{\theta }}^{2}}{\left({r}_{1}+{r}_{2}\right)\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right){\overline{\theta }}^{2}}=\frac{\left(2{r}_{1}+{r}_{2}\right){\mu }_{1}^{2}+{r}_{1}{\mu }_{2}^{2}}{2\left({r}_{1}+{r}_{2}\right)\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)},\\ =\frac{\left({r}_{1}+{r}_{2}\right){\mu }_{1}^{2}+{r}_{1}({\mu }_{1}^{2}+{\mu }_{2}^{2})}{2\left({r}_{1}+{r}_{2}\right)\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}=\frac{1}{2}\left(\frac{{\mu }_{1}^{2}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}+\frac{{r}_{1}}{{r}_{1}+{r}_{2}}\right),\end{array}$$
and symmetrically,
$${\alpha }_{2}=\frac{1}{2}\left(\frac{{\mu }_{2}^{2}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}+\frac{{r}_{2}}{{r}_{1}+{r}_{2}}\right).$$
1.4 Proof of Proposition 1
We have \({W}_{l}^{S}>{W}_{c}^{S}\) iff
$$\frac{{\sigma }_{\theta }^{2}+{\overline{\theta }}^{2}}{2}\frac{{\mu }_{1}^{6}-{\mu }_{2}^{6}}{{\mu }_{1}^{4}-{\mu }_{2}^{4}}>\frac{{\mu }_{1}^{2}+{\mu }_{2}^{2}}{2}{\overline{\theta }}^{2},$$
or
$${\left(\frac{{\sigma }_{\theta }}{\overline{\theta }}\right)}^{2}+1>\frac{\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)\left({\mu }_{1}^{4}-{\mu }_{2}^{4}\right)}{{\mu }_{1}^{6}-{\mu }_{2}^{6}},$$
or
$${\left(\frac{{\sigma }_{\theta }}{\overline{\theta }}\right)}^{2}>\frac{{\mu }_{1}^{4}{\mu }_{2}^{2}-{\mu }_{1}^{2}{\mu }_{2}^{4}}{{\mu }_{1}^{6}-{\mu }_{2}^{6}}=\frac{{r}_{\mu }^{2}-{r}_{\mu }^{4}}{1-{r}_{\mu }^{6}}=\frac{{r}_{\mu }^{2}}{1+{r}_{\mu }^{2}+{r}_{\mu }^{4}}.$$
The right-hand side of the above inequality is increasing when \({r}_{\mu }<1\) and decreasing when \({r}_{\mu }>1\) and reaches maximum at \({r}_{\mu }=1\).
1.5 Proof of Lemma 3
By \(FO{C}_{1}\) and \(FO{C}_{2}\), we have
$${R}_{h}^{\mathrm{^{\prime}}}\left(\theta ,h\left({a}_{1},{a}_{2}\right)\right)=\frac{{c}_{1}^{\mathrm{^{\prime}}}\left({a}_{1}\right)}{{h}_{1}^{\mathrm{^{\prime}}}\left(a\right)}+\frac{{c}_{2}^{\mathrm{^{\prime}}}\left({a}_{2}\right)}{{h}_{2}^{\mathrm{^{\prime}}}\left(a\right)}.$$
By \(FO{C}_{1}\) and \(FO{C}_{3}\), we have
$$\left({h}_{1}^{\mathrm{^{\prime}}}{a}_{1}^{\mathrm{^{\prime}}}\left(\theta \right)+{h}_{2}^{\mathrm{^{\prime}}}{a}_{2}^{\mathrm{^{\prime}}}\left(\theta \right)\right)\frac{{c}_{1}^{\mathrm{^{\prime}}}\left({a}_{1}\right)}{{h}_{1}^{\mathrm{^{\prime}}}\left(a\right)}={c}_{1}^{\mathrm{^{\prime}}}\left({a}_{1}\left(\theta \right)\right){a}_{1}^{\mathrm{^{\prime}}}\left(\theta \right),$$
implying
$${h}_{1}^{\mathrm{^{\prime}}}{a}_{1}^{\mathrm{^{\prime}}}\left(\theta \right)+{h}_{2}^{\mathrm{^{\prime}}}{a}_{2}^{\mathrm{^{\prime}}}\left(\theta \right)={h}_{1}^{\mathrm{^{\prime}}}\left(a\right){a}_{1}^{\mathrm{^{\prime}}}\left(\theta \right),$$
implying
$${a}_{2}^{\mathrm{^{\prime}}}\left(\theta \right)=0,$$
for all \(\theta\). Then, problem becomes
$$\begin{aligned}&\underset{{s}_{i}\in \mathcal{S},{\widehat{a}}_{i}\in \mathcal{A}}{{\text{max}}}{E}_{\theta }\left(R\left(\theta ,h\left(\widehat{a}\left(\theta \right)\right)\right)-{c}_{1}\left({\widehat{a}}_{1}\left(\theta \right)\right)-{c}_{2}\left({\widehat{a}}_{2}\left(\theta \right)\right)\right)\\ & \quad{\text{s.t}}. {R}_{h}^{\mathrm{^{\prime}}}\left(\theta ,h\left(\widehat{a}\left(\theta \right)\right)\right)=\frac{{c}_{1}^{\mathrm{^{\prime}}}\left({\widehat{a}}_{1}\left(\theta \right)\right)}{{h}_{1}^{\mathrm{^{\prime}}}\left(\widehat{a}\left(\theta \right)\right)}+\frac{{c}_{2}^{\mathrm{^{\prime}}}\left({\widehat{a}}_{2}\left(\theta \right)\right)}{{h}_{2}^{\mathrm{^{\prime}}}\left(\widehat{a}\left(\theta \right)\right)},\\ & \quad {\widehat{a}}_{2}^{\mathrm{^{\prime}}}\left(\theta \right)=0,\\ & \quad {FOC}_{2},SO{C}_{1}, \, SO{C}_{2}, \, SO{C}_{3},\\ & \quad {s}_{1}\left(x\right)+{s}_{2}\left(x\right)=x,\text{ for all }x\in {\mathbb{R}}_{+}.\end{aligned}$$
This problem can be solved in two steps. We first solve for \(\left({a}_{1}\left(\theta \right),{a}_{2}\left(\theta \right)\right)\) from, and then solve for a revenue-sharing scheme \({s}_{i}\left(x\right)\) from the remaining conditions.
1.6 Derivation of Case 2
We have
$$R\left(\theta ,h\right)=E\left(\widetilde{A}\theta h\right)=\theta h, \, R\left(h\right)={E}_{\theta }\left(R\left(\widetilde{\theta },h\right)\right)=\overline{\theta }h.$$
Then, with a limited contract, problem becomes
$$\begin{aligned}&\quad \underset{ {a}_{1,}{a}_{2}\in \mathcal{A}}{\text{max}} {E}_{\theta }\left(\theta h\left({a}_{1}(\theta ),{a}_{2}\left(\theta \right)\right)-\frac{1}{2}{a}_{1}^{2}\left(\theta \right)-\frac{1}{2}{a}_{2}^{2}\left(\theta \right)\right)\\ \text{ s.t. } &\theta =\frac{{a}_{1}\left(\theta \right)}{{\mu }_{1}}+\frac{{a}_{2}\left(\theta \right)}{{\mu }_{2}},\\ & \quad {a}_{2}^{\prime}\left(\theta \right)=0.\end{aligned}$$
By the first condition, we have
$$1=\frac{{a}_{1}^{\mathrm{^{\prime}}}}{{\mu }_{1}}+\frac{{a}_{2}^{\mathrm{^{\prime}}}}{{\mu }_{2}}=\frac{{a}_{1}^{\mathrm{^{\prime}}}}{{\mu }_{1}},$$
implying \({a}_{1}^{\prime}={\mu }_{1}\), implying
$${a}_{1}\left(\theta \right)={\mu }_{1}\theta +{\overline{a} }_{1},$$
(24)
where \({\overline{a} }_{1}\) is a constant. By the second condition, we have
$${a}_{2}\left(\theta \right)={\overline{a} }_{2},$$
(25)
where \({\overline{a} }_{2}\) is a constant. Then, by the first condition in the problem,
$$\theta =\frac{{\mu }_{1}\theta +{\overline{a} }_{1}}{{\mu }_{1}}+\frac{{\overline{a} }_{2}}{{\mu }_{2}},$$
implying
$${\overline{a} }_{2}=-\frac{{\mu }_{2}}{{\mu }_{1}}{\overline{a} }_{1},$$
(26)
implying
$$0=\frac{{\overline{a} }_{1}}{{\mu }_{1}}+\frac{{\overline{a} }_{2}}{{\mu }_{2}}.$$
Since \({a}_{1},{a}_{2}\ge 0\), this implies \({\overline{a} }_{1}={\overline{a} }_{2}=0\). Hence, the solution is
$${a}_{1}\left(\theta \right)={\mu }_{1}\theta , {a}_{2}\left(\theta \right)=0.$$
This implies
$$h={\mu }_{1}^{2}\theta .$$
Then,
$$\begin{array}{c}{W}_{l}^{A}={E}_{\theta }\left(R\left(\theta ,h\left({a}_{1},{a}_{2}\right)\right)-{c}_{1}\left({a}_{1}\right)-{c}_{2}\left({a}_{2}\right)\right)={E}_{\theta }\left(\theta \left({\mu }_{1}^{2}\theta \right)-\frac{1}{2}{\left({\mu }_{1}\theta \right)}^{2}\right)=\frac{1}{2}{\mu }_{1}^{2}{E}_{\theta }\left({\theta }^{2}\right).\end{array}$$
Further, we have
$$\frac{{c}_{2}^{\mathrm{^{\prime}}}\left({a}_{2}\right)}{{h}_{2}^{\mathrm{^{\prime}}}\left(a\right)}=\frac{{a}_{2}}{{\mu }_{2}}=0.$$
Let \({s}_{2}\left(x\right)={s}_{0}+{\alpha }_{2}x,\) where \({s}_{0}\) and \({\alpha }_{2}\) are two constants. Then, equation implies \({\alpha }_{2}=0\) and \({s}_{0}\) can be any value. That is, there is a linear contract satisfying equation.
1.7 Proof of Proposition 2
We have \({W}_{l}^{A}>{W}_{c}^{A}\) iff
$$\frac{1}{2}{\mu }_{1}^{2}\left({\sigma }_{\theta }^{2}+{\overline{\theta }}^{2}\right)>\frac{{\mu }_{1}^{2}+{\mu }_{2}^{2}}{2}{\overline{\theta }}^{2},$$
or \({r}_{\sigma }>{r}_{\mu }.\)
With a limited contract, firm 1’s expected standard is \(E\left({a}_{1}\right)={\mu }_{1}\overline{\theta }.\) With a comprehensive contract, firm 1’s standard \({a}_{1}={\mu }_{1}\overline{\theta }\). Hence, the two standards are equal ex-ante.
1.8 Proof of Proposition 3
It is obvious that \({W}_{l}^{S}>{W}_{l}^{A}\) and \({W}_{c}^{S}={W}_{c}^{A}\).
1.9 Proof of Lemma 4
By \(FO{C}_{1}\) and \(FO{C}_{2}\), we have
$${R}_{h}^{\mathrm{^{\prime}}}\left(\theta ,h\left(a\right)\right)=\frac{{c}_{1}^{\mathrm{^{\prime}}}\left({a}_{1}\right)}{{h}_{1}^{\mathrm{^{\prime}}}\left(a\right)}+\frac{{c}_{2}^{\mathrm{^{\prime}}}\left({a}_{2}\right)}{{h}_{2}^{\mathrm{^{\prime}}}\left(a\right)}.$$
By \(FO{C}_{3}\) and \(FO{C}_{4}\), we have
$$\iint xf\left(x,\theta ,h\left(\widehat{a}\left(\theta \right)\right)\right)dx{f}_{g}^{\mathrm{^{\prime}}}\left(\theta ;g\left(b\right)\right)d\theta =\frac{{C}_{1}^{\mathrm{^{\prime}}}\left({b}_{1}\right)}{{g}_{1}^{\mathrm{^{\prime}}}\left(b\right)}+\frac{{C}_{2}^{\mathrm{^{\prime}}}\left({b}_{2}\right)}{{g}_{2}^{\mathrm{^{\prime}}}\left(b\right)},$$
implying
$$\int R\left(\theta ,h\left(\widehat{a}\left(\theta \right)\right)\right){f}_{g}^{\mathrm{^{\prime}}}\left(\theta ;g\left(b\right)\right)d\theta =\frac{{C}_{1}^{\mathrm{^{\prime}}}\left({b}_{1}\right)}{{g}_{1}^{\mathrm{^{\prime}}}\left(b\right)}+\frac{{C}_{2}^{\mathrm{^{\prime}}}\left({b}_{2}\right)}{{g}_{2}^{\mathrm{^{\prime}}}\left(b\right)},$$
implying
$${R}_{g}^{\mathrm{^{\prime}}}\left(h\left(\widehat{a}\right),g\left({b}_{1},{b}_{2}\right)\right)=\frac{{C}_{1}^{\mathrm{^{\prime}}}\left({b}_{1}\right)}{{g}_{1}^{\mathrm{^{\prime}}}\left(b\right)}+\frac{{C}_{2}^{\mathrm{^{\prime}}}\left({b}_{2}\right)}{{g}_{2}^{\mathrm{^{\prime}}}\left(b\right)}.$$
Then, problem becomes
$$\begin{aligned}&\underset{{s}_{i}\in S,{\widehat{a}}_{i}\in \mathcal{A},{b}_{i}\in {\mathbb{B}}}{{\text{max}}}\int \left(R\left(\theta ,h\left(\widehat{a}\left(\theta \right)\right)\right)-{c}_{1}\left({\widehat{a}}_{1}\left(\theta \right)\right)-{c}_{2}\left({\widehat{a}}_{2}\left(\theta \right)\right)\right)f\left(\theta ;g\left({b}_{1},{b}_{2}\right)\right)d\theta -{C}_{1}\left({b}_{1}\right)-{C}_{2}\left({b}_{2}\right)\\ &{\text{ s.t. }} {R}_{h}^{\mathrm{^{\prime}}}\left(\theta ,h\left(a\right)\right)=\frac{{c}_{1}^{\mathrm{^{\prime}}}\left({a}_{1}\right)}{{h}_{1}^{\mathrm{^{\prime}}}\left(a\right)}+\frac{{c}_{2}^{\mathrm{^{\prime}}}\left({a}_{2}\right)}{{h}_{2}^{\mathrm{^{\prime}}}\left(a\right)},\\ & \quad {R}_{g}^{\mathrm{^{\prime}}}\left(h\left(\widehat{a}\right),g\left({b}_{1},{b}_{2}\right)\right)=\frac{{C}_{1}^{\mathrm{^{\prime}}}\left({b}_{1}\right)}{{g}_{1}^{\mathrm{^{\prime}}}\left(b\right)}+\frac{{C}_{2}^{\mathrm{^{\prime}}}\left({b}_{2}\right)}{{g}_{2}^{\mathrm{^{\prime}}}\left(b\right)},\\ & \quad FO{C}_{2}, FO{C}_{4}, SO{C}_{1}, SO{C}_{2}, SO{C}_{3}, SO{C}_{4},\\ & \quad {s}_{1}\left(x\right)+{s}_{2}\left(x\right)=x,\text{ for all }x\in {\mathbb{R}}_{+}.\end{aligned}$$
(27)
This problem can be solved in two steps. First, solve for \(\left({\widehat{a}}_{1}\left(\theta \right),{\widehat{a}}_{2}\left(\theta \right),{b}_{1},{b}_{2}\right)\) from
$$\begin{array}{l}\underset{{\widehat{a}}_{i}\in \mathcal{A},{b}_{i}\in {\mathbb{B}}}{{\text{max}}}\int \left(R\left(\theta ,h\left(\widehat{a}\left(\theta \right)\right)\right)-{c}_{1}\left({\widehat{a}}_{1}\right)-{c}_{2}\left({\widehat{a}}_{2}\right)\right)f\left(\theta ;g\left({b}_{1},{b}_{2}\right)\right)d\theta -{C}_{1}\left({b}_{1}\right)-{C}_{2}\left({b}_{2}\right)\\ {\text{ s. t.}} {R}_{h}^{\mathrm{^{\prime}}}\left(\theta ,h\left(a\right)\right)=\frac{{c}_{1}^{\mathrm{^{\prime}}}\left({a}_{1}\right)}{{h}_{1}^{\mathrm{^{\prime}}}\left(a\right)}+\frac{{c}_{2}^{\mathrm{^{\prime}}}\left({a}_{2}\right)}{{h}_{2}^{\mathrm{^{\prime}}}\left(a\right)},\\ {R}_{g}^{\mathrm{^{\prime}}}\left(h\left(\widehat{a}\right),g\left({b}_{1},{b}_{2}\right)\right)=\frac{{C}_{1}^{\mathrm{^{\prime}}}\left({b}_{1}\right)}{{g}_{1}^{\mathrm{^{\prime}}}\left(b\right)}+\frac{{C}_{2}^{\mathrm{^{\prime}}}\left({b}_{2}\right)}{{g}_{2}^{\mathrm{^{\prime}}}\left(b\right)}.\end{array}$$
This problem can be further divided into two steps: solve for \(\left({\widehat{a}}_{1}\left(\theta \right),{\widehat{a}}_{2}\left(\theta \right)\right)\) from
$$\begin{array}{l}\underset{{a}_{i}\in {\mathbb{A}}}{{\text{max}}} R\left(\theta ,h\left({a}_{1},{a}_{2}\right)\right)-{c}_{1}\left({a}_{1}\right)-{c}_{2}\left({a}_{2}\right)\\ {\text{s.t. }} {R}_{h}^{\mathrm{^{\prime}}}\left(\theta ,h\left(a\right)\right)=\frac{{c}_{1}^{\mathrm{^{\prime}}}\left({a}_{1}\right)}{{h}_{1}^{\mathrm{^{\prime}}}\left(a\right)}+\frac{{c}_{2}^{\mathrm{^{\prime}}}\left({a}_{2}\right)}{{h}_{2}^{\mathrm{^{\prime}}}\left(a\right)},\end{array}$$
and then, given \(\left({\widehat{a}}_{1}\left(\theta \right),{\widehat{a}}_{2}\left(\theta \right)\right)\), solve for \(\left({b}_{1},{b}_{2}\right)\) from
$$\begin{array}{l}\underset{{b}_{i}\in {\mathbb{B}}}{{\text{max}}} \int \left(R\left(\theta ,h\left(\widehat{a}\left(\theta \right)\right)\right)-{c}_{1}\left({\widehat{a}}_{1}\left(\theta \right)\right)-{c}_{2}\left({\widehat{a}}_{2}\left(\theta \right)\right)\right)f\left(\theta ;g\left({b}_{1},{b}_{2}\right)\right)d\theta -{C}_{1}\left({b}_{1}\right)-{C}_{2}\left({b}_{2}\right)\\ {\text{s.t. }} {R}_{g}^{\mathrm{^{\prime}}}\left(h\left(\widehat{a}\right),g\left({b}_{1},{b}_{2}\right)\right)=\frac{{C}_{1}^{\mathrm{^{\prime}}}\left({b}_{1}\right)}{{g}_{1}^{\mathrm{^{\prime}}}\left(b\right)}+\frac{{C}_{2}^{\mathrm{^{\prime}}}\left({b}_{2}\right)}{{g}_{2}^{\mathrm{^{\prime}}}\left(b\right)}.\end{array}$$
Second, find a sharing scheme satisfying \(FO{C}_{2}\) and \(FO{C}_{4}\). We can then further verify the SOCs. There are typically many sharing schemes that can satisfy \(FO{C}_{2}\) and \(FO{C}_{4}\). We should try to find a simple sharing scheme so that the SOCs are easy to verify.
1.10 Proof of Lemma 5
The objective function of can be written as
$$\begin{aligned}U &\equiv {r}_{1}{\text{ln}}\left(\iint {s}_{1}\left(x\right)f\left(x;\theta ,h\left(a\right)\right)dxf\left(\theta ;g\left(b\right)\right)d\theta -{c}_{1}\left({a}_{1}\right)-{C}_{1}\left({b}_{1}\right)\right)\\ & \quad +{r}_{2}{\text{ln}}\left(\iint {s}_{2}\left(x\right)f\left(x;\theta ,h\left(a\right)\right)dxf\left(\theta ;g\left(b\right)\right)d\theta -{c}_{2}\left({a}_{2}\right)-{C}_{2}\left({b}_{2}\right)\right).\end{aligned}$$
Then, the problem can be written as
$$\begin{array}{l}\underset{{a}_{i}\in {\mathbb{A}}, \, {b}_{i}\in {\mathbb{B}}}{{\text{max}}} \underset{{s}_{i}\in \mathcal{S}}{{\text{max}}} U\\ \text{s.t. } {s}_{1}\left(x\right)+{s}_{2}\left(x\right)=x,\text{ for all }x\in {\mathbb{R}}_{+}.\end{array}$$
(28)
Given \(\left({a}_{1},{a}_{2},{b}_{1},{b}_{2}\right)\), consider problem
$$\begin{array}{l}\underset{{s}_{i}\in \mathcal{S}}{{\text{max}}} U\\ \text{ s.t. } {s}_{1}\left(x\right)+{s}_{2}\left(x\right)=x,\text{ for all }x\in {\mathbb{R}}_{+}.\end{array}$$
(29)
The FOC is
$$\begin{aligned}0=\frac{\partial U}{\partial {s}_{1}} & =\frac{ {r}_{1}\int f\left(x;\theta ,h\left(a\right)\right)f\left(\theta ;g\left(b\right)\right)d\theta }{\iint {s}_{1}\left(x\right)f\left(x;\theta ,h\left(a\right)\right)dxf\left(\theta ;g\left(b\right)\right)d\theta -{c}_{1}\left({a}_{1}\right)-{C}_{1}\left({b}_{1}\right)}\\ &\quad -\frac{{r}_{2}\int f\left(x;\theta ,h\left(a\right)\right)f\left(\theta ;g\left(b\right)\right)d\theta }{\iint {s}_{2}\left(x\right)f\left(x;\theta ,h\left(a\right)\right)dxf\left(\theta ;g\left(b\right)\right)d\theta -{c}_{2}\left({a}_{2}\right)-{C}_{2}\left({b}_{2}\right)},\end{aligned}$$
implying
$$\begin{array}{l}{r}_{1}\left[R\left(h\left(a\right),g\left(b\right)\right)-\iint {s}_{1}\left(x\right)f\left(x;\theta ,h\left(a\right)\right)dxf\left(\theta ;g\left(b\right)\right)d\theta -{c}_{2}\left({a}_{2}\right)-{C}_{2}\left({b}_{2}\right)\right]\\ \quad ={r}_{2}\left[\iint {s}_{1}\left(x\right)f\left(x;\theta ,h\left(a\right)\right)dxf\left(\theta ;g\left(b\right)\right)d\theta -{c}_{1}\left({a}_{1}\right)-{C}_{1}\left({b}_{1}\right)\right],\end{array}$$
implying
$$\iint {s}_{1}\left(x\right)f\left(x;\theta ,h\left(a\right)\right)dxf\left(\theta ;g\left(b\right)\right)d\theta =\frac{{r}_{1}R\left(h\left(a\right),g\left(b\right)\right)+{r}_{2}{c}_{1}\left({a}_{1}\right)+{r}_{2}{C}_{1}\left({b}_{1}\right)-{r}_{1}{c}_{2}\left({a}_{2}\right)-{r}_{1}{C}_{2}\left({b}_{2}\right)}{{r}_{1}+{r}_{2}}.$$
(30)
That is, given\(\left({a}_{1},{a}_{2},{b}_{1},{b}_{2}\right)\), any admissible \({s}_{1}\left(\cdot \right)\) satisfying the above equation is optimal. Consider\({s}_{1}\left(x\right)={\alpha }_{1}x\). Then, by, we have
$${\alpha }_{1}=\frac{{r}_{1}R\left(h\left(a\right),g\left(b\right)\right)+{r}_{2}{c}_{1}\left({a}_{1}\right)+{r}_{2}{C}_{1}\left({b}_{1}\right)-{r}_{1}{c}_{2}\left({a}_{2}\right)-{r}_{1}{C}_{2}\left({b}_{2}\right)}{\left({r}_{1}+{r}_{2}\right)R\left(h\left(a\right),g\left(b\right)\right)}.$$
That is, we find a simple sharing scheme \({s}_{1}\left(x\right)={\alpha }_{1}x\) that satisfies. Then,
$${\alpha }_{2}=1-{\alpha }_{1}=\frac{{r}_{2}R\left(h\left(a\right),g\left(b\right)\right)-{r}_{2}{c}_{1}\left({a}_{1}\right)-{r}_{2}{C}_{1}\left({b}_{1}\right)+{r}_{1}{c}_{2}\left({a}_{2}\right)+{r}_{1}{C}_{2}\left({b}_{2}\right)}{\left({r}_{1}+{r}_{2}\right)R\left(h\left(a\right),g\left(b\right)\right)}.$$
Then,
$$\begin{array}{l}W\left(a,b\right)={\left({\alpha }_{1}R\left(h\left(a\right),g\left(b\right)\right)-{c}_{1}\left({a}_{1}\right)-{C}_{1}\left({b}_{1}\right)\right)}^{{r}_{1}}{\left({\alpha }_{2}R\left(h\left(a\right),g\left(b\right)\right)-{c}_{2}\left({a}_{2}\right)-{C}_{2}\left({b}_{2}\right)\right)}^{{r}_{2}}\\ ={\left(\frac{{r}_{1}R\left(h\left(a\right),g\left(b\right)\right)+{r}_{2}{c}_{1}\left({a}_{1}\right)+{r}_{2}{C}_{1}\left({b}_{1}\right)-{r}_{1}{c}_{2}\left({a}_{2}\right)-{r}_{1}{C}_{2}\left({b}_{2}\right)}{{r}_{1}+{r}_{2}}-{c}_{1}\left({a}_{1}\right)-{C}_{1}\left({b}_{1}\right)\right)}^{{r}_{1}}\\ \cdot {\left(\frac{{r}_{2}R\left(h\left(a\right),g\left(b\right)\right)-{r}_{2}{c}_{1}\left({a}_{1}\right)-{r}_{2}{C}_{1}\left({b}_{1}\right)+{r}_{1}{c}_{2}\left({a}_{2}\right)+{r}_{1}{C}_{2}\left({b}_{2}\right)}{{r}_{1}+{r}_{2}}-{c}_{2}\left({a}_{2}\right)-{C}_{2}\left({b}_{2}\right)\right)}^{{r}_{2}}\\ ={\left({r}_{1}\frac{R\left(h\left(a\right),g\left(b\right)\right)-{c}_{1}\left({a}_{1}\right)-{c}_{2}\left({a}_{2}\right)-{C}_{1}\left({b}_{1}\right)-{C}_{2}\left({b}_{2}\right)}{{r}_{1}+{r}_{2}}\right)}^{{r}_{1}}\\ \cdot {\left({r}_{2}\frac{R\left(h\left(a\right),g\left(b\right)\right)-{c}_{1}\left({a}_{1}\right)-{c}_{2}\left({a}_{2}\right)-{C}_{1}\left({b}_{1}\right)-{C}_{2}\left({b}_{2}\right)}{{r}_{1}+{r}_{2}}\right)}^{{r}_{2}}.\end{array}$$
Then, problem becomes
$$\underset{{a}_{i}\in {\mathbb{A}},{b}_{i}\in {\mathbb{B}}}{{\text{max}}} R\left(h\left(a\right),g\left(b\right)\right)-{c}_{1}\left({a}_{1}\right)-{c}_{2}\left({a}_{2}\right)-{C}_{1}\left({b}_{1}\right)-{C}_{2}\left({b}_{2}\right).$$
1.11 Derivation of Case 3A
We have
$$R\left(\theta ,h\right)=E\left(\widetilde{A}\theta h\right)=\theta h,$$
and
$$R\left(h\left(\widehat{a}\right),g\right)={E}_{\theta }\left(R\left(\theta ,h(\widehat{a}\left(\theta \right)\right)\right)={E}_{\theta }\left(\theta h(\widehat{a}\left(\theta \right)\right).$$
Then, problem becomes
$$\begin{array}{l}\underset{ {a}_{1,}{a}_{2}\in {\mathbb{A}}}{\text{max}} \theta h\left({a}_{1},{a}_{2}\right)-\frac{1}{2}{a}_{1}^{2}-\frac{1}{2}{a}_{2}^{2}\\ \text{ s.t. } \theta =\frac{{a}_{1}}{{\mu }_{1}}+\frac{{a}_{2}}{{\mu }_{2}}.\end{array}$$
The solution is
$${\widehat{a}}_{1}\left(\theta \right)=\frac{{\mu }_{1}^{3}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta , {\widehat{a}}_{2}\left(\theta \right)=\frac{{\mu }_{2}^{3}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta .$$
Then,
$$\begin{array}{l}{U}_{l}^{Q}={E}_{\theta }\left(R\left(\theta ,h\left(\widehat{a}\left(\theta \right)\right)\right)-{c}_{1}\left({\widehat{a}}_{1}\left(\theta \right)\right)-{c}_{2}\left({\widehat{a}}_{2}\left(\theta \right)\right)\right)\\ ={E}_{\theta }\left(\theta \left(\frac{{\mu }_{1}^{4}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta +\frac{{\mu }_{2}^{4}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta \right)-\frac{1}{2}{\left(\frac{{\mu }_{1}^{3}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta \right)}^{2}-\frac{1}{2}{\left(\frac{{\mu }_{2}^{3}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta \right)}^{2}\right)\\ =\frac{{E}_{\theta }\left({\theta }^{2}\right)}{2{\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}^{2}}\left[2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)\left({\mu }_{1}^{4}+{\mu }_{2}^{4}\right)-{\mu }_{1}^{6}-{\mu }_{2}^{6}\right]\\ =\frac{{\mu }_{1}^{4}+{\mu }_{2}^{4}+{\mu }_{1}^{2}{\mu }_{2}^{2}}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}{E}_{\theta }\left({\theta }^{2}\right).\end{array}$$
We have
$$E\left({\widetilde{\theta }}^{2}\right)=g\left(b\right)E\left({\widetilde{B}}^{2}\right)=g\left(b\right)\left({\sigma }_{B}^{2}+{\overline{B} }^{2}\right),$$
and
$$\begin{aligned}R\left(h\left(\widehat{a}\right),g\right)&={E}_{\theta }\left[\theta h\left(\widehat{a}\left(\theta \right)\right)\right]={E}_{\theta }\left[\theta \left(\frac{{\mu }_{1}^{4}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta +\frac{{\mu }_{2}^{4}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta \right)\right]\\ \quad & =\frac{{\mu }_{1}^{4}+{\mu }_{2}^{4}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}{E}_{\theta }\left({\theta }^{2}\right)=\frac{{\mu }_{1}^{4}+{\mu }_{2}^{4}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\left({\sigma }_{B}^{2}+{\overline{B} }^{2}\right)g.\end{aligned}$$
Then, problem becomes
$$\begin{array}{l}{W}_{l}^{Q}=\underset{{b}_{i}\in {\mathbb{B}}}{{\text{max}}} \frac{{\mu }_{1}^{4}+{\mu }_{2}^{4}+{\mu }_{1}^{2}{\mu }_{2}^{2}}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}\left({\sigma }_{B}^{2}+{\overline{B} }^{2}\right)\left({\beta }_{1}{b}_{1}+{\beta }_{2}{b}_{2}\right)-\frac{1}{2}{b}_{1}^{2}-\frac{1}{2}{b}_{2}^{2}\\ \text{ s.t. } \frac{{\mu }_{1}^{4}+{\mu }_{2}^{4}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\left({\sigma }_{B}^{2}+{\overline{B} }^{2}\right)=\frac{{b}_{1}}{{\beta }_{1}}+\frac{{b}_{2}}{{\beta }_{2}}.\end{array}$$
The solution is
$$\begin{array}{c}{b}_{1}={\beta }_{1}\frac{\left({\mu }_{1}^{4}+{\mu }_{2}^{4}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)+{\mu }_{1}^{2}{\mu }_{2}^{2}\left({\beta }_{1}^{2}-{\beta }_{2}^{2}\right)}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}\left({\sigma }_{B}^{2}+{\overline{B} }^{2}\right),\\ {b}_{2}={\beta }_{2}\frac{\left({\mu }_{1}^{4}+{\mu }_{2}^{4}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)+{\mu }_{1}^{2}{\mu }_{2}^{2}\left({\beta }_{2}^{2}-{\beta }_{1}^{2}\right)}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}\left({\sigma }_{B}^{2}+{\overline{B} }^{2}\right).\end{array}$$
Then,
$$\begin{array}{c}{\beta }_{1}{b}_{1}+{\beta }_{2}{b}_{2}=\frac{\left({\mu }_{1}^{4}+{\mu }_{2}^{4}\right){\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}^{2}+{\mu }_{1}^{2}{\mu }_{2}^{2}{\left({\beta }_{1}^{2}-{\beta }_{2}^{2}\right)}^{2}}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}\left({\sigma }_{B}^{2}+{\overline{B} }^{2}\right).\end{array}$$
Then,
$$\overline{\theta }=\overline{B}\sqrt{g }=\overline{B }\sqrt{\frac{\left({\mu }_{1}^{4}+{\mu }_{2}^{4}\right){\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}^{2}+{\mu }_{1}^{2}{\mu }_{2}^{2}{\left({\beta }_{1}^{2}-{\beta }_{2}^{2}\right)}^{2}}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}\left({\sigma }_{B}^{2}+{\overline{B} }^{2}\right)} ,$$
and
$$\begin{aligned}{W}_{l}^{Q}& =\frac{{\mu }_{1}^{4}+{\mu }_{1}^{2}{\mu }_{2}^{2}+{\mu }_{2}^{4}}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}\frac{\left({\mu }_{1}^{4}+{\mu }_{2}^{4}\right){\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}^{2}+{\mu }_{1}^{2}{\mu }_{2}^{2}{\left({\beta }_{1}^{2}-{\beta }_{2}^{2}\right)}^{2}}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}{\left({\sigma }_{B}^{2}+{\overline{B} }^{2}\right)}^{2}\\ &\quad -\frac{1}{2}{\left({\beta }_{1}\frac{\left({\mu }_{1}^{4}+{\mu }_{2}^{4}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)+{\mu }_{1}^{2}{\mu }_{2}^{2}\left({\beta }_{1}^{2}-{\beta }_{2}^{2}\right)}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}\right)}^{2}{\left({\sigma }_{B}^{2}+{\overline{B} }^{2}\right)}^{2}\\ & \quad -\frac{1}{2}{\left({\beta }_{2}\frac{\left({\mu }_{1}^{4}+{\mu }_{2}^{4}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)+{\mu }_{1}^{2}{\mu }_{2}^{2}\left({\beta }_{2}^{2}-{\beta }_{1}^{2}\right)}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}\right)}^{2}{\left({\sigma }_{B}^{2}+{\overline{B} }^{2}\right)}^{2}.\end{aligned}$$
Since
$$\begin{aligned}& 2\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)\left({\mu }_{1}^{4}+{\mu }_{1}^{2}{\mu }_{2}^{2}+{\mu }_{2}^{4}\right)\left(\left({\mu }_{1}^{4}+{\mu }_{2}^{4}\right){\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}^{2}+{\mu }_{1}^{2}{\mu }_{2}^{2}{\left({\beta }_{1}^{2}-{\beta }_{2}^{2}\right)}^{2}\right)\\ & \quad -{\beta }_{1}^{2}{\left[\left({\mu }_{1}^{4}+{\mu }_{2}^{4}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)+{\mu }_{1}^{2}{\mu }_{2}^{2}\left({\beta }_{1}^{2}-{\beta }_{2}^{2}\right)\right]}^{2}\\ & \quad -{\beta }_{2}^{2}{\left[\left({\mu }_{1}^{4}+{\mu }_{2}^{4}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)+{\mu }_{1}^{2}{\mu }_{2}^{2}\left({\beta }_{2}^{2}-{\beta }_{1}^{2}\right)\right]}^{2}\\ & \quad =\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)\left({\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}^{2}{\left({\mu }_{1}^{4}+{\mu }_{1}^{2}{\mu }_{2}^{2}+{\mu }_{2}^{4}\right)}^{2}-4{\beta }_{1}^{2}{\beta }_{2}^{2}{\mu }_{1}^{4}{\mu }_{2}^{4}\right),\end{aligned}$$
we have
$${W}_{l}^{Q}=\frac{{\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}^{2}{\left({\mu }_{1}^{4}+{\mu }_{1}^{2}{\mu }_{2}^{2}+{\mu }_{2}^{4}\right)}^{2}-4{\beta }_{1}^{2}{\beta }_{2}^{2}{\mu }_{1}^{4}{\mu }_{2}^{4}}{8\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right){\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}^{2}}{\left({\sigma }_{B}^{2}+{\overline{B} }^{2}\right)}^{2}.$$
For a comprehensive contract, consider problem:
$${W}_{c}^{Q}=\underset{{a}_{i}\in {\mathbb{A}},{b}_{i}\in {\mathbb{B}}}{{\text{max}}} R\left(h\left(a\right),g\left(b\right)\right)-\frac{1}{2}{a}_{1}^{2}-\frac{1}{2}{a}_{2}^{2}-\frac{1}{2}{b}_{1}^{2}-\frac{1}{2}{b}_{2}^{2}.$$
We have
$$R\left(\theta ,h\right)=E\left(\widetilde{A}\theta h\right)=\theta h,$$
and
$$R\left(g,h\right)={E}_{\theta }\left(\theta h\right)=E\left(\widetilde{B}\sqrt{g}h\right)=\overline{B}\sqrt{g }h.$$
Then, the problem becomes
$${W}_{c}^{Q}=\underset{{a}_{i}\in {\mathbb{A}},{b}_{i}\in {\mathbb{B}}}{{\text{max}}} \overline{B}\sqrt{{\beta }_{1}{b}_{1}+{\beta }_{2}{b}_{2}}\left({\mu }_{1}{a}_{1}+{\mu }_{2}{a}_{2}\right)-\frac{1}{2}{a}_{1}^{2}-\frac{1}{2}{a}_{2}^{2}-\frac{1}{2}{b}_{1}^{2}-\frac{1}{2}{b}_{2}^{2}.$$
The FOCs are
$$\begin{array}{l}{\mu }_{1}\overline{B}\sqrt{{\beta }_{1}{b}_{1}+{\beta }_{2}{b}_{2}}={a}_{1},\\ {\mu }_{2}\overline{B}\sqrt{{\beta }_{1}{b}_{1}+{\beta }_{2}{b}_{2}}={a}_{2},\\ {\beta }_{1}\overline{B}\frac{{\mu }_{1}{a}_{1}+{\mu }_{2}{a}_{2}}{2\sqrt{{\beta }_{1}{b}_{1}+{\beta }_{2}{b}_{2}}}={b}_{1},\\ {\beta }_{2}\overline{B}\frac{{\mu }_{1}{a}_{1}+{\mu }_{2}{a}_{2}}{2\sqrt{{\beta }_{1}{b}_{1}+{\beta }_{2}{b}_{2}}}={b}_{2},\end{array}$$
implying
$${b}_{i}={\beta }_{i}{\overline{B} }^{2}\frac{\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)\sqrt{{\beta }_{1}{b}_{1}+{\beta }_{2}{b}_{2}}}{2\sqrt{{\beta }_{1}{b}_{1}+{\beta }_{2}{b}_{2}}}=\frac{{\beta }_{i}}{2}{\overline{B} }^{2}\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right).$$
Then,
$${a}_{i}={\mu }_{i}\overline{B}\sqrt{\frac{{\beta }_{1}^{2}}{2}{\overline{B} }^{2}\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)+\frac{{\beta }_{2}^{2}}{2}{\overline{B} }^{2}\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}={\mu }_{i}{\overline{B} }^{2}\sqrt{\frac{({\beta }_{1}^{2}+{\beta }_{2}^{2})\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}{2}}.$$
Then,
$$\overline{\theta }=\overline{B}\sqrt{g }={\overline{B} }^{2}\sqrt{\frac{({\beta }_{1}^{2}+{\beta }_{2}^{2})\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}{2}} ,$$
and
$$\begin{array}{l}R\left(g,h\right)={\overline{B} }^{2}\sqrt{\frac{\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}{2}}\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right){\overline{B} }^{2}\sqrt{\frac{({\beta }_{1}^{2}+{\beta }_{2}^{2})\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}{2}}\\ \end{array}=\frac{1}{2}{\overline{B} }^{4}\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right){\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}^{2},$$
and
$$\begin{array}{c}{W}_{c}^{Q}=\frac{{\overline{B} }^{4}}{8}\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right){\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}^{2}.\end{array}$$
1.12 Proof of Proposition 4
For part (a), we isolate the effect of the disparity between \({\beta }_{1}\) and \({\beta }_{2}\) by assuming \({\beta }_{1}={\beta }_{2}\equiv \beta\). Then,
$${W}_{c}^{Q}=\frac{1}{4}{\beta }^{2}{\overline{B} }^{4}{\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}^{2},$$
and
$$\begin{array}{c}{W}_{l}^{Q}=\frac{1}{4}\left({\mu }_{1}^{4}+{\mu }_{2}^{4}\right){\beta }^{2}{\left({\sigma }_{B}^{2}+{\overline{B} }^{2}\right)}^{2}.\end{array}$$
Then, \({W}_{l}^{Q}>{W}_{c}^{Q}\) iff
$${r}_{B}^{2}>\frac{1+{r}_{\mu }^{2}}{\sqrt{1+{r}_{\mu }^{4}}}-1.$$
(31)
Consider
$$\begin{array}{c}\frac{\partial }{\partial z}\left(\frac{{\left(1+z\right)}^{2}}{1+{z}^{2}}\right)=\frac{2\left(1+z\right)(1-z)}{{\left(1+{z}^{2}\right)}^{2}}.\end{array}$$
This implies that the right-hand side of is increasing when \({r}_{\mu }<1\) and decreasing when \({r}_{\mu }>1\), reaching its maximum at \({r}_{\mu }=1\).
For part (b), we isolate the effect of the disparity between \({\mu }_{1}\) and \({\mu }_{2}\) by assuming \({\mu }_{1}={\mu }_{2}\equiv \mu\). Then,
$${W}_{c}^{Q}={\mu }^{4}{\overline{B} }^{4}\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)/2,$$
and
$${W}_{l}^{Q}=\frac{9\left({\beta }_{1}^{4}+{\beta }_{2}^{4}\right)+14{\beta }_{1}^{2}{\beta }_{2}^{2}}{32\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}{\mu }^{4}{\left({\sigma }_{B}^{2}+{\overline{B} }^{2}\right)}^{2}.$$
Then, \({W}_{l}^{Q}>{W}_{c}^{Q}\) iff
$${r}_{B}^{2}>\frac{4\left(1+{r}_{\beta }^{2}\right)}{\sqrt{9\left(1+{r}_{\beta }^{4}\right)+14{r}_{\beta }^{2}}}-1.$$
(32)
Consider
$$\frac{\partial }{\partial z}\left(\frac{{\left(1+z\right)}^{2}}{9\left(1+{z}^{2}\right)+14z}\right)=\frac{4(1-{z}^{2})}{{\left[9\left(1+{z}^{2}\right)+14z\right]}^{2}}.$$
This implies that the right-hand side of is increasing when \({r}_{\beta }<1\) and decreasing when \({r}_{\beta }>1\), reaching maximum at \({r}_{\beta }=1\).
1.13 Derivation of Case 3B
We have
$$R\left(\theta ,h\right)=E\left(\widetilde{A}\theta h\right)=\theta h,$$
and
$$R\left(h\left(\widehat{a}\right),g\right)={E}_{\theta }\left(R\left(\theta ,h(\widehat{a}\left(\theta \right)\right)\right)={E}_{\theta }\left(\theta h(\widehat{a}\left(\theta \right)\right).$$
Then, problem becomes
$$\begin{array}{l}\underset{ {a}_{1,}{a}_{2}\in {\mathbb{A}}}{\text{max}} \theta h\left({a}_{1},{a}_{2}\right)-\frac{1}{2}{a}_{1}^{2}-\frac{1}{2}{a}_{2}^{2}\\ \text{ s.t. } \theta =\frac{{a}_{1}}{{\mu }_{1}}+\frac{{a}_{2}}{{\mu }_{2}}.\end{array}$$
The solution is
$${\widehat{a}}_{1}\left(\theta \right)=\frac{{\mu }_{1}^{3}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta , \,{\widehat{a}}_{2}\left(\theta \right)=\frac{{\mu }_{2}^{3}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\theta .$$
Then,
$$\begin{array}{c}{U}_{l}^{Q}={E}_{\theta }\left(R\left(\theta ,h\left(\widehat{a}\left(\theta \right)\right)\right)-{c}_{1}\left({\widehat{a}}_{1}\left(\theta \right)\right)-{c}_{2}\left({\widehat{a}}_{2}\left(\theta \right)\right)\right)=\frac{{\mu }_{1}^{4}+{\mu }_{2}^{4}+{\mu }_{1}^{2}{\mu }_{2}^{2}}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}{E}_{\theta }\left({\theta }^{2}\right).\end{array}$$
We have
$$E\left({\widetilde{\theta }}^{2}\right)=g\left(b\right)E\left({\widetilde{\theta }}_{0}^{2}\right)=g\left(b\right)\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right),$$
and
$$\begin{array}{c}R\left(h\left(\widehat{a}\right),g\right)={E}_{\theta }\left[\theta h\left(\widehat{a}\left(\theta \right)\right)\right]=\frac{{\mu }_{1}^{4}+{\mu }_{2}^{4}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}{E}_{\theta }\left({\theta }^{2}\right)=\frac{{\mu }_{1}^{4}+{\mu }_{2}^{4}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right)g.\end{array}$$
Then, problem becomes
$$\begin{array}{l}{W}_{l}^{Q}=\underset{{b}_{i}\in {\mathbb{B}}}{{\text{max}}} \frac{{\mu }_{1}^{4}+{\mu }_{2}^{4}+{\mu }_{1}^{2}{\mu }_{2}^{2}}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right)\left(1+{\beta }_{1}{b}_{1}+{\beta }_{2}{b}_{2}\right)-\frac{1}{2}{b}_{1}^{2}-\frac{1}{2}{b}_{2}^{2}\\ \text{ s.t. } \frac{{\mu }_{1}^{4}+{\mu }_{2}^{4}}{{\mu }_{1}^{2}+{\mu }_{2}^{2}}\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right)=\frac{{b}_{1}}{{\beta }_{1}}+\frac{{b}_{2}}{{\beta }_{2}}.\end{array}$$
The solution is
$$\begin{array}{l}{b}_{1}={\beta }_{1}\frac{\left({\mu }_{1}^{4}+{\mu }_{2}^{4}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)+{\mu }_{1}^{2}{\mu }_{2}^{2}\left({\beta }_{1}^{2}-{\beta }_{2}^{2}\right)}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right),\\ {b}_{2}={\beta }_{2}\frac{\left({\mu }_{1}^{4}+{\mu }_{2}^{4}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)+{\mu }_{1}^{2}{\mu }_{2}^{2}\left({\beta }_{2}^{2}-{\beta }_{1}^{2}\right)}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right).\end{array}$$
Then,
$$\begin{array}{c}{\beta }_{1}{b}_{1}+{\beta }_{2}{b}_{2}=\frac{\left({\mu }_{1}^{4}+{\mu }_{2}^{4}\right){\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}^{2}+{\mu }_{1}^{2}{\mu }_{2}^{2}{\left({\beta }_{1}^{2}-{\beta }_{2}^{2}\right)}^{2}}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right).\end{array}$$
Then,
$$\overline{\theta }={\overline{\theta }}_{0}\sqrt{g}={\overline{\theta }}_{0}\sqrt{1+\frac{\left({\mu }_{1}^{4}+{\mu }_{2}^{4}\right){\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}^{2}+{\mu }_{1}^{2}{\mu }_{2}^{2}{\left({\beta }_{1}^{2}-{\beta }_{2}^{2}\right)}^{2}}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right)} ,$$
and
$${W}_{l}^{Q}=\frac{{\mu }_{1}^{4}+{\mu }_{1}^{2}{\mu }_{2}^{2}+{\mu }_{2}^{4}}{2\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right)+\frac{{\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}^{2}{\left({\mu }_{1}^{4}+{\mu }_{1}^{2}{\mu }_{2}^{2}+{\mu }_{2}^{4}\right)}^{2}-4{\beta }_{1}^{2}{\beta }_{2}^{2}{\mu }_{1}^{4}{\mu }_{2}^{4}}{8\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right){\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}^{2}}{\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right)}^{2}.$$
For a comprehensive contract, consider problem:
$${W}_{c}^{Q}=\underset{{a}_{i}\in {\mathbb{A}},{b}_{i}\in {\mathbb{B}}}{{\text{max}}} R\left(h\left(a\right),g\left(b\right)\right)-\frac{1}{2}{a}_{1}^{2}-\frac{1}{2}{a}_{2}^{2}-\frac{1}{2}{b}_{1}^{2}-\frac{1}{2}{b}_{2}^{2}.$$
We have
$$R\left(\theta ,h\right)=E\left(\widetilde{A}\theta h\right)=\theta h,$$
and
$$R\left(g,h\right)={E}_{\theta }\left(\theta h\right)=E\left({\widetilde{\theta }}_{0}\sqrt{g}h\right)={\overline{\theta }}_{0}\sqrt{g}h.$$
Then, the problem becomes
$${W}_{c}^{Q}=\underset{{a}_{i}\in {\mathbb{A}},{b}_{i}\in {\mathbb{B}}}{{\text{max}}} {\overline{\theta }}_{0}\sqrt{1+{\beta }_{1}{b}_{1}+{\beta }_{2}{b}_{2}}\left({\mu }_{1}{a}_{1}+{\mu }_{2}{a}_{2}\right)-\frac{1}{2}{a}_{1}^{2}-\frac{1}{2}{a}_{2}^{2}-\frac{1}{2}{b}_{1}^{2}-\frac{1}{2}{b}_{2}^{2}.$$
The FOCs are
$$\begin{array}{l}{\mu }_{1}{\overline{\theta }}_{0}\sqrt{1+{\beta }_{1}{b}_{1}+{\beta }_{2}{b}_{2}}={a}_{1},\\ {\mu }_{2}{\overline{\theta }}_{0}\sqrt{1+{\beta }_{1}{b}_{1}+{\beta }_{2}{b}_{2}}={a}_{2},\\ {\beta }_{1}{\overline{\theta }}_{0}\frac{{\mu }_{1}{a}_{1}+{\mu }_{2}{a}_{2}}{2\sqrt{1+{\beta }_{1}{b}_{1}+{\beta }_{2}{b}_{2}}}={b}_{1},\\ {\beta }_{2}{\overline{\theta }}_{0}\frac{{\mu }_{1}{a}_{1}+{\mu }_{2}{a}_{2}}{2\sqrt{1+{\beta }_{1}{b}_{1}+{\beta }_{2}{b}_{2}}}={b}_{2},\end{array}$$
implying
$${b}_{i}=\frac{{\beta }_{i}}{2}{\overline{\theta }}_{0}^{2}\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right).$$
Then,
$${a}_{{\text{i}}}={\mu }_{i}{\overline{\theta }}_{0}\sqrt{1+\frac{1}{2}({\beta }_{1}^{2}+{\beta }_{2}^{2})\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right){\overline{\theta }}_{0}^{2}}.$$
Then,
$$\overline{\theta }={\overline{\theta }}_{0}\sqrt{g}={\overline{\theta }}_{0}\sqrt{1+\frac{1}{2}({\beta }_{1}^{2}+{\beta }_{2}^{2})\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right){\overline{\theta }}_{0}^{2}} ,$$
and
$$\begin{array}{c}R\left(g,h\right)=\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right){\overline{\theta }}_{0}^{2}+\frac{1}{2}\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right){\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}^{2}{\overline{\theta }}_{0}^{4},\end{array}$$
and
$$\begin{array}{c}{W}_{c}^{Q}=\frac{1}{2}\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right){\overline{\theta }}_{0}^{2}+\frac{1}{8}\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right){\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}^{2}{\overline{\theta }}_{0}^{4}.\end{array}$$
1.14 Proof of Proposition 5
For part (a), since
$${\beta }_{1}^{2}+{\beta }_{2}^{2}>2{\beta }_{1}{\beta }_{2}$$
and
$${\mu }_{1}^{4}+{\mu }_{1}^{2}{\mu }_{2}^{2}+{\mu }_{2}^{4}>{\mu }_{1}^{4}+{\mu }_{2}^{4}>2{\mu }_{1}^{2}{\mu }_{2}^{2},$$
we have
$${\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}^{2}{\left({\mu }_{1}^{4}+{\mu }_{1}^{2}{\mu }_{2}^{2}+{\mu }_{2}^{4}\right)}^{2}>4{\beta }_{1}^{2}{\beta }_{2}^{2}4{\mu }_{1}^{4}{\mu }_{2}^{4},$$
implying \({W}_{l}^{Q}>{W}_{l}^{S}\). We also obviously have \({W}_{c}^{Q}>{W}_{c}^{S}\).
For part (b), consider inequality \({W}_{l}^{Q}>{W}_{c}^{Q}\), or
$$\begin{aligned}&\frac{4}{{\beta }_{1}^{2}+{\beta }_{2}^{2}}\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right)\frac{{\mu }_{1}^{6}-{\mu }_{2}^{6}}{{\mu }_{1}^{4}-{\mu }_{2}^{4}}+\frac{{\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}^{2}{\left({\mu }_{1}^{4}+{\mu }_{1}^{2}{\mu }_{2}^{2}+{\mu }_{2}^{4}\right)}^{2}-4{\beta }_{1}^{2}{\beta }_{2}^{2}{\mu }_{1}^{4}{\mu }_{2}^{4}}{{\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}^{2}{\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}^{2}}{\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right)}^{2}\\ & \quad >\frac{4}{{\beta }_{1}^{2}+{\beta }_{2}^{2}}\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right){\overline{\theta }}_{0}^{2}+{\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}^{2}{\overline{\theta }}_{0}^{4},\end{aligned}$$
or
$$\begin{aligned}&\frac{4}{{\beta }_{1}^{2}+{\beta }_{2}^{2}}\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right)\frac{{\mu }_{1}^{6}-{\mu }_{2}^{6}}{{\mu }_{1}^{4}-{\mu }_{2}^{4}}+{\left(\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right)\frac{{\mu }_{1}^{6}-{\mu }_{2}^{6}}{{\mu }_{1}^{4}-{\mu }_{2}^{4}}\right)}^{2}-\frac{4{\beta }_{1}^{2}{\beta }_{2}^{2}{\mu }_{1}^{4}{\mu }_{2}^{4}}{{\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}^{2}{\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}^{2}}{\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right)}^{2}\\ & \quad >{\left(\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right){\overline{\theta }}_{0}^{2}+\frac{2}{{\beta }_{1}^{2}+{\beta }_{2}^{2}}\right)}^{2}-{\left(\frac{2}{{\beta }_{1}^{2}+{\beta }_{2}^{2}}\right)}^{2},\end{aligned}$$
or
$$\begin{aligned}&{\left(\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right)\frac{{\mu }_{1}^{6}-{\mu }_{2}^{6}}{{\mu }_{1}^{4}-{\mu }_{2}^{4}}+\frac{2}{{\beta }_{1}^{2}+{\beta }_{2}^{2}}\right)}^{2}-{\left(\frac{2}{{\beta }_{1}^{2}+{\beta }_{2}^{2}}\right)}^{2}-\frac{4{\gamma }_{1}^{2}{\gamma }_{2}^{2}{\mu }_{1}^{4}{\mu }_{2}^{4}}{{\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}^{2}{\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}^{2}}{\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right)}^{2}\\ & \quad >{\left(\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right){\overline{\theta }}_{0}^{2}+\frac{2}{{\beta }_{1}^{2}+{\beta }_{2}^{2}}\right)}^{2}-{\left(\frac{2}{{\beta }_{1}^{2}+{\beta }_{2}^{2}}\right)}^{2},\end{aligned}$$
or
$$\begin{aligned}&{\left(\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right)\frac{{\mu }_{1}^{6}-{\mu }_{2}^{6}}{{\mu }_{1}^{4}-{\mu }_{2}^{4}}+\frac{2}{{\beta }_{1}^{2}+{\beta }_{2}^{2}}\right)}^{2}-\frac{4{\gamma }_{1}^{2}{\gamma }_{2}^{2}{\mu }_{1}^{4}{\mu }_{2}^{4}}{{\left({\beta }_{1}^{2}+{\beta }_{2}^{2}\right)}^{2}{\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right)}^{2}}{\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right)}^{2}\\ & \quad >{\left(\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right){\overline{\theta }}_{0}^{2}+\frac{2}{{\beta }_{1}^{2}+{\beta }_{2}^{2}}\right)}^{2}.\end{aligned}$$
(33)
We also find that condition is equivalent to
$${\left(\left({\sigma }_{{\theta }_{0}}^{2}+{\overline{\theta }}_{0}^{2}\right)\frac{{\mu }_{1}^{6}-{\mu }_{2}^{6}}{{\mu }_{1}^{4}-{\mu }_{2}^{4}}+\frac{2}{{\beta }_{1}^{2}+{\beta }_{2}^{2}}\right)}^{2}>{\left(\left({\mu }_{1}^{2}+{\mu }_{2}^{2}\right){\overline{\theta }}_{0}^{2}+\frac{2}{{\beta }_{1}^{2}+{\beta }_{2}^{2}}\right)}^{2}.$$
Hence, if holds, then holds. That is, if \({W}_{l}^{Q}>{W}_{c}^{Q}\), then \({W}_{l}^{S}>{W}_{c}^{S}\). This is equivalent to: if \({W}_{l}^{S}<{W}_{c}^{S}\), then \({W}_{l}^{Q}<{W}_{c}^{Q}\).
