Financing the capital-constrained online retailer with risk aversion: coordinating strategy analysis

Abstract

This paper considers a capital-constrained online retailer (OR) selling products through an e-commerce platform (EP) who also offers financial services to retailers. During the selling season, the OR exerts an effort to promote market demand through activities like sales promotions, advertising and live-streaming selling events. To investigate the EP-based financing scheme, a game-theoretic model is developed where the EP functions as the leader determining the interest rate and platform usage fee rate, and the OR functions as the follower determining the order quantity and effort level. We explore the impacts of the OR’s risk-aversion and find that when the OR is risk-averse (1) she sets a high effort level, and the EP sets a high usage fee rate; (2) a high risk-averse OR orders less products than low risk-averse OR. We design specific revenue-cost sharing contracts to coordinate the supply chain and demonstrate that the designed contracts are feasible. Moreover, we find that the OR consistently prefers EP financing compared to bank financing.

Appendix

Proof of Lemma 1

In the centralized case, we have $\left(1-\lambda \right)p>w +{c}_{2}>s$. From the second-order derivative of $E\left[{\pi }_{SC}\right]$ with respect to $Q$ and $e$, we have $\frac{{\partial }^{2}E\left[{\pi }_{SC}\right]}{\partial {Q}^{2}}=\left(s-\left(1-\lambda \right)p\right)f\left(Q-e\right)<0$ and $\frac{{\partial }^{2}E\left[{\pi }_{SC}\right]}{\partial {e}^{2}}=-\left(\left(1-\lambda \right)p-s\right)f\left(Q-e\right)-2a<0$. Hence, by solving $\frac{\partial E[{\pi }_{SC}]}{\partial Q}=0$ and $\frac{\partial E[{\pi }_{SC}]}{\partial e}=0$, we derive ${e}^{C}=\frac{p-w-{c}_{2}}{2a}$ and ${Q}^{C}={F}^{-1}\left[\frac{p-w-{c}_{2}}{p-s}\right]+\frac{p-w-{c}_{2}}{2a}$. Q.E.D.

Proof of Proposition 1

Similar to the Proof of Lemma 1, we can easily obtain the results of ${e}^{BN}$ and ${Q}^{BN}$. From the first-order derivative of $E\left[{\pi }_{p}\right]$ on $\lambda$, we have $\frac{dE[{\pi }_{p}]}{d\lambda }=ph\left({Q}^{BN},{e}^{BN}\right)+\lambda p\frac{dh\left({Q}^{BN},{e}^{BN}\right)}{d\lambda }+\left({c}_{1}-{c}_{2}\right)\frac{d{Q}^{BN}}{d\lambda }=ph\left({Q}^{BN},{e}^{BN}\right)-\left(\lambda p+{c}_{1}-{c}_{2}\right)\frac{p}{2a}-\frac{1-F\left({Q}^{BN}-{e}^{BN}\right)}{f\left({Q}^{BN}-{e}^{BN}\right)}\frac{\left(\left(1-\lambda \right)p-s\right)p\left({c}_{1}-{c}_{2}\right)+p\left(w+{c}_{1}-s\right)}{{\left(\left(1-\lambda \right)p-s\right)}^{2}}$. Based on the IGFR assumption, we have $\frac{d\left({Q}^{BN}-{e}^{BN}\right)}{d\lambda }=-\frac{p\left(w+{c}_{1}-s\right)}{f\left({Q}^{BN}-{e}^{BN}\right){\left(\left(1-\lambda \right)p-s\right)}^{2}}<0$ and $\frac{dh\left({Q}^{BN}-{e}^{BN}\right)}{d\lambda }=-\frac{p}{2a}-\frac{1-F\left({Q}^{BN}-{e}^{BN}\right)}{f\left({Q}^{BN}-{e}^{BN}\right)}\frac{p\left(w+{c}_{1}-s\right)}{{\left(\left(1-\lambda \right)p-s\right)}^{2}}<0$. Hence, we have $\frac{{d}^{2}E\left({\pi }_{p}\right)}{d{\lambda }^{2}}<0$ and ${\lambda }^{BN}$ satisfies $\frac{dE\left({\pi }_{p}\right)}{d\lambda }=ph\left({Q}^{BN},{e}^{BN}\right)+\lambda p\frac{dh\left({Q}^{BN},{e}^{BN}\right)}{d\lambda }+\left({c}_{1}-{c}_{2}\right)\frac{d{Q}^{BN}}{d\lambda }=0$. Q.E.D.

Proof of Proposition 2

In scenario BN, we first solve the optimal effort level and order quantity of the OR under revenue-cost sharing contract and have ${e}^{*}=\frac{x\left(1-\lambda \right)p-w\left(1-y\right)-{c}_{1}}{2a(1-y)}$ and ${Q}^{*}={F}^{-1}\left[\frac{w\left(1-y\right)+{c}_{1}-x\left(1-\lambda \right)p}{xs-x\left(1-\lambda \right)p}\right]+\frac{x\left(1-\lambda \right)p-w\left(1-y\right)-{c}_{1}}{2a(1-y)}$. From the results in Lemma 1, we solve the equations ${e}^{*}={e}^{C}$ and ${Q}^{*}={Q}^{C}$,we can obtain that $x=\frac{{c}_{1}\left(p-s\right)}{\left(1-\lambda \right)p\left(p-s\right)-\left(p-{c}_{2}\right)\left(\left(1-\lambda \right)p-s\right)}$ and $y=\frac{\left(1-\lambda \right)p\left(p-s\right)-\left(\left(1-\lambda \right)p-s\right)\left(p+{c}_{1}-{c}_{2}\right)}{\left(1-\lambda \right)p\left(p-s\right)-\left(p-{c}_{2}\right)\left(\left(1-\lambda \right)p-s\right)}$. Q.E.D.

Proof of Corollary 1

From the revenue-cost sharing contract in scenario BN, we have$\frac{\partial {x}^{BN}}{\partial s}=-\frac{{\lambda }^{BN}p{c}_{1}\left(p-{c}_{2}\right)}{{\left(\left(1-{\lambda }^{BN}\right)p\left(p-s\right)-\left(p-{c}_{2}\right)\left(\left(1-{\lambda }^{BN}\right)-s\right)\right)}^{2}}<0$,$\frac{\partial {x}^{BN}}{\partial p}=\frac{{\lambda }^{BN}s{c}_{1}\left(s-{c}_{2}\right)}{{\left(\left(1-{\lambda }^{BN}\right)p\left(p-s\right)-\left(p-{c}_{2}\right)\left(\left(1-{\lambda }^{BN}\right)-s\right)\right)}^{2}}$,$\frac{\partial {x}^{BN}}{\partial {\lambda }^{BN}}=-\frac{\left(p-s\right)p{c}_{1}\left(s-{c}_{2}\right)}{{\left(\left(1-{\lambda }^{BN}\right)p\left(p-s\right)-\left(p-{c}_{2}\right)\left(\left(1-{\lambda }^{BN}\right)-s\right)\right)}^{2}}$,$\frac{\partial {y}^{BN}}{\partial p}=-\frac{{\lambda }^{BN}\left(1-{\lambda }^{BN}\right)ps{c}_{1}}{{\left(\left(1-{\lambda }^{BN}\right)p\left(p-s\right)-\left(p-{c}_{2}\right)\left(\left(1-{\lambda }^{BN}\right)-s\right)\right)}^{2}}<0,\frac{\partial {y}^{BN}}{\partial {\uplambda }^{BN}}=\frac{ps{c}_{1}\left(p-s\right)}{{\left(\left(1-{\uplambda }^{BN}\right)p\left(p-s\right)-\left(p-{c}_{2}\right)\left(\left(1-{\uplambda }^{BN}\right)-s\right)\right)}^{2}}>0$, and$\frac{\partial {y}^{BN}}{\partial s}=\frac{{\lambda }^{BN}\left(1-{\lambda }^{BN}\right){p}^{2}{c}_{1}}{{\left(\left(1-{\lambda }^{BN}\right)p\left(p-s\right)-\left(p-{c}_{2}\right)\left(\left(1-{\lambda }^{BN}\right)-s\right)\right)}^{2}}>0$. Hence, the results in Corollary 1 can be obtained. Q.E.D.

Proof of Proposition 3

In scenario BA, the OR’s objective is to maximize its Mean-CVaR objective function. From the first-order derivative of ${\pi }_{mc}$ with respect to $v$ ,we obtain$\frac{\partial {\pi }_{mc}}{\partial v}=\left(1-t\right)(1-\frac{1}{1-\beta }({\int }_{0}^{\mathit{min}\left(\frac{v+\left(w+{c}_{1}\right)Q+a{e}^{2}-sQ}{\left(1-{\lambda }^{BA}\right)p-s}-e,Q-e\right)}f\left(x\right)dx+{\int }_{Q-e}^{\infty }kf\left(x\right)dx))$. Among them,$k=1$, if$v>\left(\left(1-{\lambda }^{BA}\right)p-s\right)Q-\left(w+{c}_{1}\right)Q-a{e}^{2}$, and$k=0$, if$v\le \left(\left(1-{\lambda }^{BA}\right)p-s\right)Q-\left(w+{c}_{1}\right)Q-a{e}^{2}$. Hence, we have$\frac{\partial {\pi }_{mc}}{\partial v}=\left(1-t\right)(1-\frac{1}{1-\beta }({\int }_{0}^{\mathit{min}\left(\frac{v+\left(w+{c}_{1}\right)Q+a{e}^{2}-sQ}{\left(1-{\lambda }^{BA}\right)p-s}-e,Q-e\right)}f(x)d(x)))$, if$v<\left(1-{\lambda }^{BA}\right)pQ-\left(\omega +{c}_{1}\right)Q-a{e}^{2}$;$\frac{\partial {\pi }_{mc}}{\partial v}=-\left(1-t\right)\frac{\beta }{1-\beta }$, if$v\ge \left(1-{\lambda }^{BA}\right)pQ-\left(w+{c}_{1}\right)Q-a{e}^{2}$.

If$\frac{\partial {\pi }_{mc}}{\partial v}\ge 0$, we have $\beta \le \frac{w+{c}_{1}-s}{t\left(\left(1-{\lambda }^{BA}\right)p-s\right)}$ and the optimal $v$ is ${v}^{*}=\left(\left(1-{\lambda }^{BA}\right)p-s\right)Q-\left(w+{c}_{1}\right)Q-a{e}^{2}$. Substituting ${v}^{*}$ into${\pi }_{mc}$, we have${\pi }_{mc}\left({v}^{*}\right)=tE\left({\pi }_{r}\right)+\left(1-t\right)\left(\left(\left(1-{\lambda }^{BA}\right)p-s\right)Q-\left(w+{c}_{1}\right)Q-a{e}^{2}\!-\!\frac{1}{1\!-\!\beta }{\int }_{0}^{Q-e}\left(\left(1\!-\!{\lambda }^{BA}\right)p\!-\!s\right)\left(x\!+\!e\right)f\left(x\right)dx\right)$. From the second-order derivative, we obtain$\frac{{\partial }^{2}{\pi }_{mc}\left({v}^{*}\right)}{d{Q}^{2}}=-\left(\left(1-{\lambda }^{BA}\right)p-s\right)\frac{1-t\beta }{1-\beta }f\left(Q-e\right)<0$. By solving$\frac{\partial {\pi }_{mc}\left({v}^{*}\right)}{dQ}=0$, we have${Q}_{1}^{BA}={F}^{-1}\left(\frac{\left(\left(1-{\lambda }^{BA}\right)p-\left(w+{c}_{1}\right)\right)\left(1-\beta \right)}{\left(\left(1-{\lambda }^{BA}\right)p-s\right)\left(1-t\beta \right)}\right)+e$.

If $\frac{\partial {\pi }_{mc}}{\partial v}<0$, the results can be derived in the same way. Substituting ${Q}^{BA}$ into ${\pi }_{mc}({v}^{*})$, we obtain that the OR’s optimal effort level on sales is $e^{BA} = \frac{{\left( {1 - \lambda^{BA} } \right)p - w{-}c_{1} }}{2a}$ if $\beta >\frac{w+{c}_{1}-s}{t\left(\left(1-{\lambda }^{BA}\right)p-s\right)}$ or $\beta \le \frac{w+{c}_{1}-s}{t\left(\left(1-{\lambda }^{BA}\right)p-s\right)}$. In sum, we have ${e}^{BA}=\frac{\left(1-{\lambda }^{BA}\right)p-w-{c}_{1}}{2a}$, ${Q}_{1}^{BA}={F}^{-1}\left(\frac{\left(\left(1-{\lambda }^{BA}\right)p-\left(w+{c}_{1}\right)\right)\left(1-\beta \right)}{\left(\left(1-{\lambda }^{BA}\right)p-s\right)\left(1-t\beta \right)}\right)+\frac{\left(1-{\lambda }^{BA}\right)p-w-{c}_{1}}{2a}$ if $\beta \le \frac{w+{c}_{1}-s}{t\left(\left(1-{\lambda }^{BA}\right)p-s\right)}$ and ${Q}_{2}^{BA}={F}^{-1}\left(1+\frac{s-w-{c}_{1}}{t\left(\left(1-{\lambda }^{BA}\right)p-s\right)}\right)+\frac{\left(1-{\lambda }^{BA}\right)p-\omega -{c}_{1}}{2a}$ if $\beta >\frac{w+{c}_{1}-s}{t\left(\left(1-{\lambda }^{BA}\right)p-s\right)}$.

With respect to the results in Proposition 3, if $\beta \le \frac{w+{c}_{1}-s}{t\left(\left(1-{\lambda }^{BA}\right)p-s\right)}$,we have $\frac{\partial E\left({\pi }_{p}\right)}{\partial \lambda }=ph\left({Q}_{1}^{BA},{e}_{1}^{BA}\right)-\lambda p\left(\frac{p}{2a}+\frac{1-F\left({Q}_{1}^{BA}-{e}_{1}^{BA}\right)}{f\left({Q}_{1}^{BA}-{e}_{1}^{BA}\right)}\frac{p\left(w-s\right)}{{\left(\left(1-\lambda \right)p-s\right)}^{2}}\frac{1-\beta }{1-t\beta }\right)$ and $\frac{\partial h\left({Q}_{1}^{BA},{e}_{1}^{BA}\right)}{d\lambda }=-\frac{p}{2a}-\frac{1-F\left({Q}_{1}^{BA}-{e}_{1}^{BA}\right)}{f\left({Q}_{1}^{BA}-{e}_{1}^{BA}\right)}\frac{p\left(w-s\right)}{{\left(\left(1-\lambda \right)p-s\right)}^{2}}\frac{1-\beta }{1-t\beta }<0$. Based on the IGFR assumption and the results that $\frac{\partial \left({Q}_{1}^{BA}-{e}_{1}^{BA}\right)}{d\lambda }=-\frac{1}{f\left({Q}_{1}^{BA}-{e}_{1}^{BA}\right)}\frac{p\left(w-s\right)}{{\left(\left(1-\lambda \right)p-s\right)}^{2}}\frac{1-\beta }{1-t\beta }<0$, we have $\frac{{\partial }^{2}h\left({Q}_{1}^{BA},{e}_{1}^{BA}\right)}{\partial {\lambda }^{2}}<0$, $\frac{{\partial }^{2}{Q}_{1}^{BA}}{\partial {\lambda }^{2}}=\frac{{\partial }^{2}\left({Q}_{1}^{BA}-{e}_{1}^{BA}\right)}{\partial {\lambda }^{2}}<0$, and $\frac{{\partial }^{2}E\left({\pi }_{p}\right)}{\partial {\lambda }^{2}}<0$. Therefore, the optimal usage fee rate satisfies $\frac{dE\left({\pi }_{p}\right)}{d\lambda }=ph\left({Q}_{1}^{BA},{e}_{1}^{BA}\right)+{\lambda }^{BA}p\frac{dh\left({Q}_{1}^{BA},{e}_{1}^{BA}\right)}{d\lambda }+\left({c}_{1}-{c}_{2}\right)\frac{d{Q}_{1}^{BA}}{d\lambda }=0$. If $\beta >\frac{w+{c}_{1}-s}{t\left(\left(1-{\lambda }^{BA}\right)p-s\right)}$, the results can be derived in the same way. Q.E.D.

Proof of Corollary 2

In Proposition 1 and Proposition 3, we show that the optimal usage fee rate satisfies$\frac{dE\left[{\pi }_{p}\right]}{d\lambda }=0$. Let $V\left(\lambda ,t\right)=\frac{dE\left({\pi }_{p}\right)}{d\lambda }=0$,and we have $\frac{\partial \lambda }{\partial t}=-\frac{\frac{\partial V\left(\lambda ,t\right)}{\partial t}}{\frac{\partial V\left(\lambda ,t\right)}{\partial \lambda }}$, in which$\frac{\partial V\left(\lambda ,t\right)}{\partial t}=\frac{\partial V\left(\lambda ,t\right)}{\partial Q}\frac{\partial Q}{\partial t}$. Given a $\lambda$,it is obvious that $\frac{\partial {Q}^{BA}}{\partial t}>0$ and ${e}^{BA}$ is independent of$t$. Note that we have$\frac{\partial h\left(Q,e\right)}{\partial Q}=1-F\left(Q-e\right)>0$,$\frac{dh\left(Q,e\right)}{d\lambda }=-\frac{p}{2a}-\frac{1-F\left(Q-e\right)}{f\left(Q-e\right)}\frac{p\left(w-s\right)}{{\left(\left(1-\lambda \right)p-s\right)}^{2}}$, $\frac{dQ}{d\lambda }=-\frac{p}{2a}-\frac{1-F\left(Q-e\right)}{f\left(Q-e\right)}\frac{p}{\left(1-\lambda \right)p-s}$ and the IGFR assumption, which lead to the results that $\frac{dh\left(Q,e\right)}{d\lambda }$ and $\frac{dQ}{d\lambda }$ increase in$Q$. Hence, we have$\frac{\partial V\left(\lambda ,t\right)}{\partial Q}>0$, $\frac{\partial V\left(\lambda ,t\right)}{\partial t}=\frac{\partial V\left(\lambda ,t\right)}{\partial Q}\frac{\partial Q}{\partial t}>0$ and then$\frac{\partial \lambda }{\partial t}>0$. It is obvious that ${Q}^{BA}={Q}^{BN}$ and ${e}^{BA}={e}^{BN}$ if $t \to 1$. Based on the results that$\frac{\partial \lambda }{\partial t}>0$,we have${\lambda }^{BA}\le {\lambda }^{BN}$. Q.E.D.

Proof of Propositions 4, 6, and 8

Similar to the Proof of Proposition 2, the results are immediate. Q.E.D.

Proof of Proposition 5

Similar to the Proof of Lemma 1, we can easily obtain the results of ${e}^{EN}$ and ${Q}^{EN}$.

By solving the first-order and second-order derivative of $E\left[{\pi }_{p}\right]$ on$\lambda$, we have $\frac{\partial E\left({\pi }_{p}^{EN}\right)}{\partial \lambda }=ph\left({Q}^{EN},{e}^{EN}\right)+p{\lambda }^{EN}\frac{\partial h\left({Q}^{EN},{e}^{EN}\right)}{\partial {\lambda }^{EN}}+\left({c}_{1}-{c}_{2}\right)\frac{\partial {Q}^{EN}}{\partial {\lambda }^{EN}}+\left(w+{c}_{1}\right)r\frac{\partial {Q}^{EN}}{\partial {\lambda }^{EN}}+2a{e}^{EN}r\frac{\partial {e}^{EN}}{\partial {\lambda }^{EN}}$ and$\frac{{\partial }^{2}E\left({\pi }_{p}\right)}{\partial {\lambda }^{2}}=p\frac{\partial h\left({Q}^{EN},{e}^{EN}\right)}{\partial \lambda }+p\left(\frac{\partial {Q}^{EN}}{\partial \lambda }-\frac{\partial {e}^{EN}}{\partial \lambda }\right)\left(1-F\left({Q}^{EN}-{e}^{EN}\right)\right)+p\frac{\partial {e}^{EN}}{\partial \lambda } +{\lambda }^{EN}p\frac{{\partial }^{2}h\left({Q}^{EN},{e}^{EN}\right)}{\partial {\lambda }^{2}}+\left({c}_{1}-{c}_{2}\right)\frac{{\partial }^{2}{Q}^{EN}}{\partial {\lambda }^{2}}+\left(w+{c}_{1}\right)r\frac{{\partial }^{2}{Q}^{EN}}{\partial {\lambda }^{2}}+\frac{r{p}^{2}}{2a{\left(1+r\right)}^{2}}$. Based on the results that $\frac{\partial {Q}^{EN}}{\partial \lambda }=-\frac{p\left(\left(w+{c}_{1}\right)\left(1+r\right)-s\right)}{{\left(\left(1-\lambda \right)p-s\right)}^{2}f\left({Q}^{EN}-{e}^{EN}\right)}-\frac{p}{2a\left(1+r\right)}=-\frac{p\left(1-F\left({Q}^{EN}-{e}^{EN}\right)\right)}{\left(\left(1-\lambda \right)p-s\right)f\left({Q}^{EN}-{e}^{EN}\right)}-\frac{p}{2a\left(1+r\right)}<0$ and$\frac{\partial \left({Q}^{EN}-{e}^{EN}\right)}{\partial \lambda }=-\frac{p\left(\left(w+{c}_{1}\right)\left(1+r\right)-s\right)}{{\left(\left(1-\lambda \right)p-s\right)}^{2}f\left({Q}^{EN}-{e}^{EN}\right)}<0$, we have $\frac{{\partial }^{2}{Q}^{EN}}{\partial {\lambda }^{2}}<0$, $\frac{\partial {e}^{EN}}{\partial \lambda }=-\frac{p}{2a\left(1+r\right)}<0$ and$\frac{\partial h\left({Q}^{EN},{e}^{EN}\right)}{\partial \lambda }=-\frac{p\left(\left(w+{c}_{1}\right)\left(1+r\right)-s\right)\left(1-F\left({Q}^{EN}-{e}^{EN}\right)\right)}{{\left(\left(1-\lambda \right)p-s\right)}^{2}f\left({Q}^{EN}-{e}^{EN}\right)}-\frac{p}{2a\left(1+r\right)}<0$. In addition, based on the results that $p\frac{\partial {e}^{EN}}{\partial \lambda }+\frac{r{p}^{2}}{2a{\left(1+r\right)}^{2}}=-\frac{r{p}^{2}}{2a{\left(1+r\right)}^{2}}<0$ and the IGFR assumption, we have$\frac{{\partial }^{2}E\left({\pi }_{p}\right)}{\partial {\lambda }^{2}}<0$. Therefore, the usage fee rate ${\lambda }^{EN}$ satisfies$\frac{\partial E{\left({\pi }_{p}\right)}^{EN}}{\partial \lambda }=ph\left({Q}^{EN},{e}^{EN}\right)+p{\lambda }^{EN}\frac{\partial h\left({Q}^{EN},{e}^{EN}\right)}{\partial {\lambda }^{EN}}+\left({c}_{1}-{c}_{2}\right)\frac{\partial {Q}^{EN}}{\partial {\lambda }^{EN}}+\left(w+{c}_{1}\right)r\frac{\partial {Q}^{EN}}{\partial {\lambda }^{EN}}+2a{e}^{EN}r\frac{\partial {e}^{EN}}{\partial {\lambda }^{EN}}=0$.

Let$\left(\left(w+{c}_{1}\right)Q+a{e}^{2}-B\right)\left(1+\overline{r }\right)=\left(1-\lambda \right)px+s\left(Q-\left(x+e\right)\right)$, we have $x=L=\frac{\left(\left(w+{c}_{1}\right)Q+a{e}^{2}-B\right)\left(1+\overline{r }\right)-sQ}{\left(1-{\lambda }^{BA}\right)p-s}-e$, and the OR declares bankruptcy if$x<L$. When the capital-constrained OR borrows capital from an EP in a perfectly competitive capital market, the EP can only get the market risk-free interest income by setting a suitable loan interest rate$\overline{r }$. Hence, we have$\mathrm{min}\left[\left(\left(w+{c}_{1}\right)Q+a{e}^{2}-B\right)\left(1+\overline{r }\right),\left(1-\lambda \right)p\mathrm{min}\left(Q+e,x\right)+s{\left(Q-\left(x+e\right)\right)}^{+}\right]=\left(\left(\left(1-{\lambda }^{EN}\right)p-s\right)e+sQ\right)F\left(L\right)+\left(\left(1-{\lambda }^{EN}\right)p-s\right){\int }_{0}^{L}xf\left(x\right)dx+\left(\left(w+{c}_{1}\right)Q+a{e}^{2}-B\right)\left(1+\overline{r }\right)\left(1-F\left(L\right)\right)=\left(\left(w+{c}_{1}\right)Q+a{e}^{2}-B\right)\left(1+r\right)$. Hence, $\overline{r }$ satisfies${A}_{1}+{A}_{2}+{A}_{3}=\left(\left(w+{c}_{1}\right)Q+a{e}^{2}-B\right)\left(1+r\right)$, in which${A}_{1}=\left(\left(\left(1-\uplambda \right)p-s\right){e}^{EN}+s{Q}^{EN}\right)F\left(L\right)$,${A}_{2}=\left(\left(1-\uplambda \right)p-s\right){\int }_{0}^{L}xf\left(x\right)dx$, ${A}_{3}=\left(\left(w+{c}_{1}\right){Q}^{EN}+a{\left({e}^{EN}\right)}^{2}-B\right)\left(1+\overline{r }\right)\left(1-F\left(L\right)\right)$ and$L=\frac{\left(\left(w+{c}_{1}\right){Q}^{EN}+a{\left({e}^{EN}\right)}^{2}-B\right)\left(1+\overline{r }\right)-s{Q}^{EN}}{\left(1-\lambda \right)p-s}-{e}^{EN}$. Q.E.D.

Proof of Corollary 3

We have $\frac{\partial \overline{r}}{\partial r }=\frac{1}{1-F\left(L\right)}$ and $\frac{\partial \overline{r}}{\partial B }=-\frac{1+r-\left(1+\overline{r }\right)\left(1-F\left(L\right)\right)}{\left(\left(w+{c}_{1}\right)Q+a{e}^{2}-B\right)\left(1-F\left(L\right)\right)}$. Then the results can be directly derived. Q.E.D.

Proof of Proposition 7

Similar to the proof of Proposition 3, the results are immediate. Q.E.D.

Proof of Corollary 4

Similar to the Proof of Corollary 6, the results are immediate. Q.E.D.

Proof of Proposition 9

Similar to the proof of Proposition 5, the results are immediate. Q.E.D.

Proof of Corollary 5

From Propositions 5 and 9, if $={r}_{a}$, we substitute ${\lambda }^{B}$ into $\frac{\partial E[{\pi }_{p}^{EN}]}{\partial \lambda }$ and have $\frac{\partial E\left[{\pi }_{p}^{EN}\right]}{\partial \lambda }{|}_{\lambda ={\lambda }^{B}}=\left[\left(w+{c}_{1}\right)r\frac{\partial {Q}^{EN}}{\partial \lambda }+2a{e}^{EN}r\frac{\partial {e}^{EN}}{\partial \lambda }\right]{|}_{\lambda ={\lambda }^{B}}<0$ because $\frac{\partial {Q}^{EN}}{\partial \lambda }<0$ and$\frac{\partial {e}^{EN}}{\partial \lambda }<0$. With the results that $\frac{{\partial }^{2}E\left[{\pi }_{p}^{EN}\right]}{\partial {\lambda }^{2}}<0$ and$\frac{\partial E\left[{\pi }_{r}^{B}\right]}{\partial \lambda }<0$, we have ${\lambda }^{EN}\le {\lambda }^{B}$ and$E\left[{\pi }_{r}^{EN}\right]\ge E\left[{\pi }_{r}^{B}\right]$. Q.E.D.