Pricing and Quality Strategies in Crowdfunding with Network Externality

Abstract

To improve the financing performance, optimization models have been developed to analyze the operation mechanism of crowdfunding. However, network externality, which has proven to exist in crowdfunding, was ignored in previous studies. In order to address this gap, we construct an optimization model incorporating network externality and investigate its impact on crowdfunding operation decisions. Our analysis reveals that network externality does not monotonously influence optimal prices, while it always improves the project quality. In addition, we prove that although network externality increases the creator’s profit, it increases the price discrimination level and may hurt consumers, thus reducing social welfare. In extensions, we investigate the effects of demand uncertainty, external financing, and word-of-mouth effect on the optimal decisions. Our work not only provides managerial implications for crowdfunding activities, but also reveals the potential negative effects of network externality for platforms.

Appendix

Proof of Lemma (3.1)

• (1) When $p_{\textrm{c}} \le \lambda q$, then $\Pi$ increases with $p_{\textrm{c}}$. So the optimal price is $p_{\textrm{c}} =\lambda q$ and $\Pi =\lambda q-I$.

• (2) When $q\ge p_{\textrm{c}} >\lambda q$, for a given price $p_{\textrm{c}}$, by solving FOCs, we can get $p_{\textrm{r}} (p_{\textrm{c}} )=\frac{p_{\textrm{c}} -\lambda q}{2(1-\lambda )}$. Substituting $p_{\textrm{r}} (p_{\textrm{c}} )=\frac{p_{c} -\lambda q}{2(1-\lambda )}$ into $\Pi$, we have

  $$\begin{aligned} \Pi (p_{\textrm{r}} (p_{\textrm{c}} ))=\frac{-(3-4\lambda )p_{\textrm{c}}^{2} +(4-6\lambda )qp_{\textrm{c}} +\lambda ^{2} q^{2} }{{ 4(1-}\lambda { )}^{2} { q}}. \end{aligned}$$

  (A. 1)

  If $\lambda >\frac{{ 1}}{{ 2}}$, then $\frac{\partial \Pi (p_{\textrm{r}} (p_{\textrm{c}} ))}{\partial p_{\textrm{c}} } \le 0$, thus the creator will set $p_{\textrm{c}} =\lambda q$, then $\Pi =\lambda q-I$. In this case, no consumers will purchase in the crowdfunding stage. If $\lambda \le \frac{{ 1}}{{ 2}}$, solving FOCs, we can get $p_{\textrm{c}} =\frac{2-3\lambda }{{ 3}-{ 4}\lambda } q$, then $p_{\textrm{r}} =\frac{1-2\lambda }{3-4\lambda } q$. Substituting $p_{\textrm{c}} =\frac{2-3\lambda }{{ 3}-{ 4}\lambda } q$ and $p_{\textrm{r}} =\frac{1-2\lambda }{3-4\lambda } q$ into $\Pi$, we can get $\Pi =\frac{1-\lambda }{3-4\lambda } q-I$.

• (3) When $p_{\textrm{c}} >q$, solving FOCs, then we can get $p_{\textrm{r}} =\frac{q}{2}$. Substituting $p_{\textrm{r}} =\frac{q}{2}$ in $\Pi$, we can get $\Pi =\frac{q}{4} -I$. Through comparisons among the above three cases, we have Lemma (3.1). $\square$

Proof of Proposition (3.2)

From Lemma (3.1), we have: if $\lambda \le \frac{{ 1}}{{ 2}}$, then $\frac{\partial p_{\textrm{c}}^{*} }{\partial \lambda } =-\frac{q}{(3-4\lambda )^{2} }<0,{ \; }\frac{\partial p_{\textrm{r}}^{*} }{\partial \lambda } =\frac{-2q}{(3-4\lambda )^{2} } <0$, and $\frac{\partial \Pi ^{*} }{\partial \lambda } =\frac{q}{(3-4\lambda )^{2} } >0$; if $\lambda >\frac{{ 1}}{{ 2}}$, then $\frac{\partial p_{\textrm{c}}^* }{\partial \lambda } =\lambda >0$ and $\frac{\partial \Pi ^{*} }{\partial \lambda } =q>0$. $\square$

Proof of Proposition (3.4)

Taking the derivative of both sides of the Eq. (8) with respect to $\lambda$, we can obtain

$$\begin{aligned} \frac{\partial \bar{p}_{\textrm{c}} }{\partial \lambda } (1-\frac{\bar{p}_{\textrm{c}} -\lambda q}{(1-\lambda )q} )+\bar{p}_{\textrm{c}} (-\frac{1}{(1-\lambda )q} \frac{\partial \bar{p}_{\textrm{c}} }{\partial \lambda } -\frac{-(1-\lambda )q^{2} +q(\bar{p}_{\textrm{c}} -\lambda q)}{(1-\lambda )^{2} q^{2} } )=0\nonumber \\ \end{aligned}$$

(A. 2)

and

$$\begin{aligned} -\frac{\partial \frac{\bar{p}_{\textrm{c}} -\lambda q}{(1-\lambda )q} }{\partial \lambda } \bar{p}_{\textrm{c}} +(1-\frac{\bar{p}_{\textrm{c}} -\lambda q}{(1-\lambda )q} )\frac{\partial \bar{p}_{\textrm{c}} }{\partial \lambda } =0. \end{aligned}$$

(A. 3)

Using $\frac{2-3\lambda }{{ 3}-{ 4}\lambda } q\ge \bar{p}_{\textrm{c}} \ge \frac{{ 1}}{{ 2}} q$, we have

$$\begin{aligned} \frac{\partial p_{\textrm{c}}^{*} }{\partial \lambda } { =}\frac{\partial \bar{p}_{\textrm{c}} }{\partial \lambda } =\frac{\bar{p}_{\textrm{c}} (\bar{p}_{\textrm{c}} -q)}{(1-\lambda )^{2} q-(1-\lambda )(\bar{p}_{\textrm{c}} -\lambda q)-(1-\lambda )\bar{p}_{\textrm{c}} } \ge 0 \end{aligned}$$

(A. 4)

and

$$\begin{aligned} \frac{\partial \frac{\bar{p}_{\textrm{c}} -\lambda q}{(1-\lambda )q} }{\partial \lambda } { =}\frac{(1-\frac{\bar{p}_{\textrm{c}} -\lambda q}{(1-\lambda )q} )\frac{\partial \bar{p}_{\textrm{c}} }{\partial \lambda } }{{ 2}\bar{p}_{\textrm{c}} } \ge 0. \end{aligned}$$

(A. 5)

Because inequality (A. 5) holds, we can obtain $\Pi ^{*} =\frac{(\bar{p}_{\textrm{c}} -\lambda q)^{2} }{4q(1-\lambda )^{2} } -B$ increases with $\lambda$.

By transforming equation (8), we have

$$\begin{aligned} (q-\bar{p}_{\textrm{c}} )\bar{p}_{\textrm{c}} =(I-B)(1-\lambda )q. \end{aligned}$$

(A. 6)

Taking the derivative of both sides of the Eq. (A. 6) with respect to q, we can obtain

$$\begin{aligned} \bar{p}_{\textrm{c}} +q\frac{\partial \bar{p}_{\textrm{c}} }{\partial q} -2\bar{p}_{\textrm{c}} \frac{\partial \bar{p}_{\textrm{c}} }{\partial q} =(I-B)(1-\lambda )=\frac{(q-\bar{p}_{\textrm{c}} )\bar{p}_{\textrm{c}} }{q}. \end{aligned}$$

(A. 7)

Thus we have $\frac{\partial \bar{p}_{\textrm{c}} }{\partial q} =\frac{\bar{p}_{\textrm{c}}^{2} }{q(2\bar{p}_{\textrm{c}} -q)} >0$. By transforming equation (8), we can get $(q-\bar{p}_{\textrm{c}} )\frac{\bar{p}_{c} }{q} =(I-B)(1-\lambda )$. Because $\frac{\partial (q-\bar{p}_{\textrm{c}} )}{\partial q} =1-\frac{\bar{p}_{\textrm{c}}^{2} }{q(2\bar{p}_{\textrm{c}} -q)} =\frac{\bar{p}_{\textrm{c}}^{2} -q^{2} }{q(2\bar{p}_{\textrm{c}} -q)} <0$, we have $\frac{\bar{p}_{\textrm{c}} }{q}$ increases with q. Therefore, $\Pi ^{*} =\frac{(\frac{\bar{p}_{\textrm{c}} }{q} -\lambda )^{2} q}{4(1-\lambda )^{2} } -B$ increases with q.

Taking the derivative of both sides of the Eq. (8) with respect to $B-I$, we can get

$$\begin{aligned} \frac{\partial \bar{p}_{\textrm{c}} }{\partial B-I} (1-\frac{\bar{p}_{\textrm{c}} -\lambda q}{(1-\lambda )q} )+\bar{p}_{\textrm{c}} (-\frac{1}{(1-\lambda )q} \frac{\partial \bar{p}_{\textrm{c}} }{\partial B-I} )=-1. \end{aligned}$$

(A. 8)

By arranging Eq. (A. 8), we can get

$$\begin{aligned} \frac{\partial p_{\textrm{c}}^{*} }{\partial B-I} { =}\frac{\partial \bar{p}_{\textrm{c}} }{\partial B-I} =\frac{-1}{1-\frac{\bar{p}_{\textrm{c}} -\lambda q}{(1-\lambda )q} -\frac{\bar{p}_{\textrm{c}} }{(1-\lambda )q} }>0,\ \frac{\partial p_{r}^{*} }{\partial B-I} =\frac{1}{2(1-\lambda )} \frac{\partial p_{r}^{*} }{\partial B-I} >0.\nonumber \\ \end{aligned}$$

(A. 9)

Because $\Pi ^{*}$ increases with $\bar{p}_{c}$, and $\bar{p}_{\textrm{c}}$ decreases with I, we have $\Pi ^{*}$ decreases with I. Furthermore,

$$\begin{aligned} \begin{array}{l}{\frac{\partial \Pi ^{*} }{\partial B} =\frac{{ 2}(\bar{p}_{\textrm{c}} -\lambda q)}{4q(1-\lambda )^{2} } \frac{\partial \bar{p}_{\textrm{c}} }{\partial B} -1=\frac{-1}{1-\frac{\bar{p}_{\textrm{c}} -\lambda q}{(1-\lambda )q} -\frac{\bar{p}_{\textrm{c}} }{(1-\lambda )q} } \frac{{ 2}(\bar{p}_{\textrm{c}} -\lambda q)}{4q(1-\lambda )^{2} } -1\ge 0} \\ {\qquad \quad \Leftrightarrow \frac{\bar{p}_{\textrm{c}} -\lambda q}{{ 2}(1-\lambda ){ (2}\bar{p}_{\textrm{c}} -q{ )}} \ge { 1}\Leftrightarrow \frac{2-3\lambda }{{ 3}-{ 4}\lambda } q\ge \bar{p}_{\textrm{c}}. } \end{array}. \end{aligned}$$

(A. 10)

$\square$

Proof of Proposition (3.5)

When $\lambda \ge \frac{1}{2}$, it is obvious that CS is not affected by $\lambda$, B, and I, while increases with q; when $\lambda <\frac{1}{2} \cup B+\frac{2-3\lambda }{(3-4\lambda )^{2} } q\ge I$, we can easily get that CS increases with $\lambda$ and q, while is not influenced by B and I; when $\lambda<\frac{1}{2} \cup B+\frac{2-3\lambda }{(3-4\lambda )^{2} } q<I$, $CS=\frac{{ 5}\bar{p}_{\textrm{c}}^{2} -2\bar{p}_{\textrm{c}} q(4+\lambda )+(4+\lambda ^{2} )q^{2} }{8(1-\lambda )^{2} q}$. The symmetry axis of ${ 5}\bar{\textrm{p}}_{\textrm{c}}^{2} -2\bar{p}_{\textrm{c}} q(4+\lambda )+(4+\lambda ^{2} )q^{2}$ with respect to $\bar{p}_{\textrm{c}}$ is $\frac{4+\lambda }{5} q$ and $\frac{4+\lambda }{5} q>\frac{2-3\lambda }{{ 3}-{ 4}\lambda } q\ge \bar{p}_{c} \ge \frac{{ 1}}{{ 2}} q$. Therefore, CS decreases with $\bar{p}_{\textrm{c}}$. Further, because $\bar{p}_{\textrm{c}}$ increases (decreases) with B (I), we have CS decreases (increases) with B (I). Next, we prove $\frac{\partial CS}{\partial q} <0$ and $\frac{\partial CS}{\partial \lambda } <0$.

$$\begin{aligned}{} & {} \begin{array}{l} {\frac{\partial CS}{\partial q} =\frac{1}{8(1-\lambda )^{2} } (\frac{10\bar{p}_{\textrm{c}} q\frac{\partial \bar{p}_{\textrm{c}} }{\partial q} -5\bar{p}_{\textrm{c}}^{2} }{q^{2} } -2(4+\lambda )\frac{\partial \bar{p}_{\textrm{c}} }{\partial q} +(4+\lambda ^{2} ))} \\ {\;\;\;\;\;\;\;\, =-\frac{1}{8(1-\lambda )^{2} q(2\bar{p}_{\textrm{c}} -q)} ((3+2\lambda )\bar{p}_{\textrm{c}}^{2} -2(4+\lambda ^{2} )q\bar{p}_{\textrm{c}} +(4+\lambda ^{2} )q^{2} )}. \end{array} \end{aligned}$$

(A. 11)

$$\begin{aligned}{} & {} \begin{array}{l} {\frac{\partial CS}{\partial \lambda } =\frac{(10\bar{p}_{\textrm{c}} \frac{\partial \bar{p}_{\textrm{c}} }{\partial \lambda } -2(4+\lambda )q\frac{\partial \bar{p}_{\textrm{c}} }{\partial \lambda } -2\bar{p}_{\textrm{c}} q+2\lambda q^{2} )8(1-\lambda )^{2} q+(5\bar{p}_{\textrm{c}}^{2} -2\bar{p}_{\textrm{c}} q(4+\lambda )+(4+\lambda ^{2} )q^{2} )16(1-\lambda )q}{{ 64}(1-\lambda )^{{ 4}} q^{2} } } \\ {\;\;\;\;\;\;\,\,=\frac{1}{8(1-\lambda )^{3} q} \frac{2(\bar{p}_{\textrm{c}}-q)^{2} (5\bar{p}_{\textrm{c}} -(4+\lambda )q)}{2\bar{p}_{\textrm{c}} -q} }. \end{array} \end{aligned}$$

(A. 12)

When $\bar{p}_{\textrm{c}} \in (\frac{{ 1}}{{ 2}} q,\frac{2-3\lambda }{{ 3}-{ 4}\lambda } q)$, we have $(3+2\lambda )\bar{p}_{\textrm{c}}^{2} -2(4+\lambda ^{2} )q\bar{p}_{\textrm{c}} +(4+\lambda ^{2} )q^{2} >0$ and $5\bar{p}_{\textrm{c}} -(4+\lambda )q<0$ always hold. Therefore, $\frac{\partial CS}{\partial q} <0$, $\frac{\partial CS}{\partial \lambda } <0$. $\square$

Proof of Proposition (3.6)

When $\lambda \ge \frac{1}{2}$ or $\lambda <\frac{1}{2} \cup B+\frac{2-3\lambda }{(3-4\lambda )^{2} } q\ge I$, it is obvious that SW increases with $\lambda$ and q, decreases with I, while is not affected by B. When $\lambda<\frac{1}{2} \cup B+\frac{2-3\lambda }{(3-4\lambda )^{2} } q<I$, we have

$$\begin{aligned} \begin{array}{l} {\frac{\partial SW}{\partial q} =\frac{1}{8(1-\lambda )^{2} } (\frac{-(1-8\lambda )(2\bar{p}_\textrm{c} q\frac{\partial \bar{p}_\textrm{c} }{\partial q} -\bar{p}_\textrm{c}^{2} )}{q^{2} } -14\lambda \frac{\partial \bar{p}_\textrm{c} }{\partial q} +(4+3\lambda ^{2} ))} \\ {\;\;\;\;\;\;\;\;=-\frac{\bar{p}_\textrm{c}^{2} (1+6\lambda )-2\bar{p}_\textrm{cq}(4+3\lambda ^{2} )+(4+3\lambda ^{2} )q^{2} }{(2\bar{p}_\textrm{c}-\textrm{q})q} }\\ {\;\;\;\;\;\;\;\;=\frac{(6 \lambda ^2-6 \lambda +7)\bar{p}_\textrm{c}+(1+5\lambda -6\lambda ^2)(I-B)-(4+3\lambda ^2)q}{(2\bar{p}_\mathrm{c-q})}}. \end{array}\end{aligned}$$

(A. 13)

We now prove whether $\frac{\partial SW}{\partial q}$ is positive or negative. ${\mathop {\lim }\limits _{\bar{p}_{\textrm{c}} \rightarrow \frac{q}{2} }} \frac{\partial SW}{\partial q}={\mathop {\lim }\limits _{\bar{p}_{\textrm{c}} \rightarrow \frac{q}{2} }} -\frac{\bar{p}_\textrm{c}^{2} (1+6\lambda )-2\bar{p}_\textrm{cq}(4+3\lambda ^{2} )+(4+3\lambda ^{2} )q^{2} }{(2\bar{p}_\mathrm{c-q})q}=-\infty <0$, ${\mathop {\lim }\limits _{\bar{p}_{\textrm{c}} \rightarrow \frac{2-3\lambda }{{ 3}-{ 4}\lambda } q}} \frac{\partial SW}{\partial q} ={\mathop {\lim }\limits _{\bar{p}_{\textrm{c}} \rightarrow \frac{2-3\lambda }{{ 3}-{ 4}\lambda } q}} -\frac{\bar{p}_\textrm{c}^{2} (1+6\lambda )-2\bar{p}_\textrm{cq}(4+3\lambda ^{2} )+(4+3\lambda ^{2} )q^{2} }{(2\bar{p}_\mathrm{c-q})q}=\frac{4(1-\lambda )^{2} (2-9\lambda +6\lambda ^{2} )}{3-2\lambda (5-4\lambda )}$. If $\lambda <\frac{1}{12} (9-\sqrt{33} )$, we have ${\mathop {\lim }\limits _{\bar{p}_{\textrm{c}} \rightarrow \frac{2-3\lambda }{{ 3}-{ 4}\lambda } q}} \frac{\partial SW}{\partial q} >0$, therefore, $\frac{\partial SW}{\partial q} >0$ holds in this case. If $\frac{1}{12} (9-\sqrt{33} )<\lambda <\frac{1}{2}$, ${\mathop {\lim }\limits _{\bar{p}_{\textrm{c}} \rightarrow \frac{2-3\lambda }{{ 3}-{ 4}\lambda } q}} \frac{\partial SW}{\partial q} <0$. We can further get when $\bar{p}_\textrm{c}>\frac{(4+3\lambda ^2)q-(1+5\lambda -6\lambda ^2)(I-B)}{6 \lambda ^2-6 \lambda +7}$ ($\bar{p}_\textrm{c}<\frac{(4+3\lambda ^2)q-(1+5\lambda -6\lambda ^2)(I-B)}{6 \lambda ^2-6 \lambda +7}$), $\frac{\partial SW}{\partial q} >0(\frac{\partial SW}{\partial q} <0)$. Substituting $\bar{p}_\textrm{c}=\frac{(4+3\lambda ^2)q-(1+5\lambda -6\lambda ^2)(I-B)}{6 \lambda ^2-6 \lambda +7}$ into $(1-\frac{\bar{p}_{\textrm{c}} -\lambda q}{(1-\lambda )q} )\bar{p}_{\textrm{c}} +B-I$, we have $(1-\frac{\bar{p}_{\textrm{c}} -\lambda q}{(1-\lambda )q} )\bar{p}_{\textrm{c}} +B-I=\frac{-(1-\lambda ) \left( (1+6 \lambda )^2(I-B)^2+12 (I-B) (1-\lambda ) \left( 3 \lambda ^2+4\right) q-3 \left( 3 \lambda ^2+4\right) q^2\right) }{(7-6 (1-\lambda ) \lambda )^2 q}$. We can easily verify that $\frac{-(1-\lambda ) \left( (1+6 \lambda )^2(I-B)^2+12 (I-B) (1-\lambda ) \left( 3 \lambda ^2+4\right) q-3 \left( 3 \lambda ^2+4\right) q^2\right) }{(7-6 (1-\lambda ) \lambda )^2 q}>0$ only when $q>\hat{q}=\frac{\sqrt{(I-B)^2 (7-6 (1-\lambda ) \lambda )^2 \left( 3 \lambda ^2+4\right) }}{\sqrt{3} \left( 3 \lambda ^2+4\right) }+2 (I-B) (1-\lambda )$. Therefore, in this case, if $q>\hat{q}$, we have $\frac{\partial SW}{\partial q} >0$; otherwise, we have $\frac{\partial SW}{\partial q} <0$.

Taking the derivative of SW with respect to $\lambda$, we have

$$\begin{aligned}{} & {} \frac{\partial SW}{\partial \lambda }=\nonumber \\{} & {} \!\quad \frac{(6\lambda q^{2} \!-\!2(1\!-\!8\lambda )\bar{p}_{\textrm{c}} \frac{\partial \bar{p}_{\textrm{c}} }{\partial \lambda } \!+\!8\bar{p}_{\textrm{c}}^{2} \!-\!14\bar{p}_{\textrm{c}} q\!-\!14\lambda q\frac{\partial \bar{p}_{\textrm{c}} }{\partial \lambda } )(1\!-\!\lambda )\!+\!2((4\!+\!3\lambda ^{2} )q^{2} \!-\!(1\!-\!8\lambda )\bar{p}_{\textrm{c}}^{2} -14\lambda \bar{p}_{\textrm{c}} q)}{8(1-\lambda )^{3} q} \nonumber \\{} & {} \!\quad =\frac{1}{8(1-\lambda )^{3} q} \frac{2(\bar{p}_{\textrm{c}} -q)^{2} ({ 7}\bar{p}_{\textrm{c}} -(4+3\lambda )q)}{2\bar{p}_{\textrm{c}} -q} . \end{aligned}$$

(A. 14)

We can obtain ${\mathop {\lim }\limits _{\bar{p}_{\textrm{c}} \rightarrow \frac{q}{2} }} \frac{\partial SW}{\partial \lambda } <0$ and ${\mathop {\lim }\limits _{\bar{p}_{\textrm{c}} \rightarrow \frac{2-3\lambda }{{ 3}-{ 4}\lambda } q}} \frac{\partial SW}{\partial \lambda } { =}\frac{4(1-\lambda )^{2} (2-9\lambda +6\lambda ^{2} )}{3-2\lambda (5-4\lambda )}$. If $\lambda <\frac{{ 1}}{{ 6}}$, ${\mathop {\lim }\limits _{\bar{p}_{\textrm{c}} \rightarrow \frac{2-3\lambda }{{ 3}-{ 4}\lambda } q}} \frac{\partial SW}{\partial \lambda } >0$; if $\frac{{ 1}}{{ 6}}<\lambda <\frac{1}{2}$, ${\mathop {\lim }\limits _{\bar{p}_{\textrm{c}} \rightarrow \frac{2-3\lambda }{{ 3}-{ 4}\lambda } q}} \frac{\partial SW}{\partial \lambda } <0$. Therefore, $\frac{\partial SW}{\partial \lambda } >0$ holds when and only when $\lambda <\frac{1}{6}$ and $\bar{p}_{\textrm{c}} >\frac{(4+3\lambda )q}{{ 7}}$; otherwise, $\frac{\partial SW}{\partial \lambda } <0$. Substituting $\bar{p}_{\textrm{c}} =\frac{(4+3\lambda )q}{{ 7}}$ in $(1-\frac{\bar{p}_{\textrm{c}} -\lambda q}{(1-\lambda )q} )\bar{p}_{\textrm{c}} +B-I$, we can get $(1-\frac{\bar{p}_{\textrm{c}} -\lambda q}{(1-\lambda )q} )\bar{p}_{\textrm{c}} +B-I>0$ when and only when $\lambda >\frac{49 (I-B)}{9 q}-\frac{4}{3}$. Then when $\lambda >\frac{49 (I-B)}{9 q}-\frac{4}{3} (\lambda <\frac{49 (I-B)}{9 q}-\frac{4}{3})$, we have $\bar{p}_{\textrm{c}} >\frac{(4+3\lambda )q}{{ 7}}(\bar{p}_{\textrm{c}} <\frac{(4+3\lambda )q}{{ 7}})$.

We now prove $\frac{\partial SW}{\partial B} <0$. Because $SW=\frac{(4+3\lambda ^{2} )q^{2} -(1-8\lambda )\bar{p}_{\textrm{c}}^{2} -14\lambda \bar{p}_{\textrm{c}} q}{8(1-\lambda )^{2} q} -I$ decreases with $\bar{p}_{\textrm{c}} \in (\frac{{ 1}}{{ 2}} q,\frac{2-3\lambda }{{ 3}-{ 4}\lambda } q)$ and $\frac{\partial \bar{p}_{\textrm{c}} }{\partial B} >0$, we have $\frac{\partial SW}{\partial B} <0$.

We next prove the results about how SW is affected by I.

$$\begin{aligned} \begin{array}{l}{\frac{\partial SW}{\partial I} =\frac{-{ 2}(1-8\lambda )\bar{p}_{\textrm{c}} \frac{\partial \bar{p}_{\textrm{c}} }{\partial I} -14\lambda q\frac{\partial \bar{p}_{\textrm{c}} }{\partial I} }{8(1-\lambda )^{2} q} -1=\frac{2((4+3)\lambda q-7\bar{p}_{\textrm{c}} )}{8(1-\lambda )(2\bar{p}_{\textrm{c}} -q)} }. \end{array}\end{aligned}$$

(A. 15)

If $\lambda \ge \frac{1}{6}$, we have $\frac{(4+3)\lambda }{{ 7}} q\ge \frac{2-3\lambda }{{ 3}-{ 4}\lambda } q$, thus $\frac{\partial SW}{\partial I} >0$ holds when $\bar{p}_{\textrm{c}} \in (\frac{{ 1}}{{ 2}} q,\frac{2-3\lambda }{{ 3}-{ 4}\lambda } q)$; if $\lambda <\frac{1}{6}$, we have $\frac{q}{2}<\frac{(4+3)\lambda }{{ 7}} q<\frac{2-3\lambda }{{ 3}-{ 4}\lambda } q$, thus $\frac{\partial SW}{\partial I} >0$ when $\bar{p}_{\textrm{c}} \in (\frac{{ 1}}{{ 2}} q,\frac{(4+3)\lambda }{{ 7}} q)$ while $\frac{\partial SW}{\partial I} <0$ when $\bar{p}_{\textrm{c}} \in (\frac{(4+3)\lambda }{{ 7}} q, \frac{2-3\lambda }{{ 3}-{ 4}\lambda } q)$. Similarly, when $I<B+\frac{3}{49} (3 \lambda +4) q$ $(I>B+\frac{3}{49} (3 \lambda +4) q)$, we can get $\bar{p}_{\textrm{c}} >\frac{(4+3\lambda )q}{{ 7}}$ $(\bar{p}_{\textrm{c}} <\frac{(4+3\lambda )q}{{ 7}})$. $\square$

Proof of Theorem (3.7)

• (1) If $\lambda >\frac{1}{2}$, $\Pi =\lambda q-aq^{2} -I^{{ 0}}$. By solving $\frac{\partial \Pi }{\partial q} =0$, we get $q^{*} =\frac{\lambda }{2a}$. Substituting $q^{*} =\frac{\lambda }{2a}$ into $\Pi =\lambda q-aq^{2} -I^{{ 0}}$, we have $\Pi ^{*} =\frac{\lambda ^{2} }{4a} -I^{{ 0}}$.

• (2) If $\lambda \le \frac{1}{2}$, (a) when $B+\frac{2-3\lambda }{(3-4\lambda )^{2} } q\ge I=aq^{2} +I^{{ 0}}$ (i.e. $q\le \bar{q}_{{ 1}}$, where $\bar{q}_{{ \textrm{1}}}$ is the positive root of $B+\frac{2-3\lambda }{(3-4\lambda )^{2} } q=aq^{2} +I^{{ 0}}$), $\Pi =\frac{1-\lambda }{3-4\lambda } q-aq^{2} -I^{{ 0}}$. By solving $\frac{\partial \Pi }{\partial q} =0$, we get $q=\frac{\frac{1-\lambda }{3-4\lambda } }{2a}$. Therefore, the optimal quality is $q=\min \{ \frac{\frac{1-\lambda }{3-4\lambda } }{2a},\bar{q}_{{ 1}} \}$. Furthermore, when $B\ge \frac{(5-4\lambda )\lambda ^{2} -1}{4(3-4\lambda )^{3} } +I^{0}$ ($B<\frac{(5-4\lambda )\lambda ^{2} -1}{4(3-4\lambda )^{3} } +I^{{ 0}}$), we have $\frac{\frac{1-\lambda }{3-4\lambda } }{2a} \le \bar{q}_{{\textrm{1}}}$($\frac{\frac{1-\lambda }{3-4\lambda } }{2a} >\bar{q}_{{\textrm{1}}}$).

• (b) When $B+\frac{2-3\lambda }{(3-4\lambda )^{2} } q\le I=aq^{2} +I^{0} \le B+\frac{1}{4(1-\lambda )^{2} } q$, i.e., $\bar{q}_{2} \ge q\ge \bar{q}_{{ 1}}$, we have $\Pi =\frac{(\bar{p}_{\textrm{c}} -\lambda q)^{2} }{4q(1-\lambda )^{2} } -B$. Taking the derivative of both sides of Eq. (8) with respect to q, we can get

  $$\begin{aligned} \left( \frac{\bar{p}_{\textrm{c}} }{(1-\lambda )q^{2} } -\frac{1}{(1-\lambda )q} \frac{\partial \bar{p}_{\textrm{c}} }{\partial q} \right) \bar{p}_{\textrm{c}} +\left( 1-\frac{\bar{p}_{\textrm{c}} -\lambda q}{(1-\lambda )q} \right) \frac{\partial \bar{p}_{\textrm{c}} }{\partial q} =2aq.\qquad \end{aligned}$$

  (A. 16)

  Then $\frac{\partial \bar{p}_{\textrm{c}} }{\partial q} =\frac{2aq-\frac{\bar{p}_{\textrm{c}}^{2} }{(1-\lambda )q^{2} } }{1-\frac{\bar{p}_{\textrm{c}} -\lambda q}{(1-\lambda )q} -\frac{\bar{p}_{\textrm{c}} }{(1-\lambda )q} } =\frac{2aq(1-\lambda )q^{2} -\bar{p}_{\textrm{c}}^{2} }{q(q-2\bar{p}_{\textrm{c}} )}$. Taking the derivative of $\Pi$ with respect to q, we have

  $$\begin{aligned} \frac{\partial \Pi }{\partial q} =\frac{2(\bar{p}_{\textrm{c}} -\lambda q)(\frac{\partial \bar{p}_{\textrm{c}} }{\partial q} -\lambda )\times 4q(1-\lambda )^{2} -4(1-\lambda )^{2} (\bar{p}_{\textrm{c}} -\lambda q)^{2} }{16q^{2} (1-\lambda )^{4} }.\qquad \end{aligned}$$

  (A. 17)

  Substituting $\frac{\partial \bar{p}_{\textrm{c}} }{\partial q}$ in (A. 17), together with (8), we have

  $$\begin{aligned} \frac{\partial \Pi }{\partial q} =0\Leftrightarrow f(q)= & {} -16a^{{ 2}} (1-\lambda )q^{{ 3}} +(3a+4a(1-\lambda )\lambda )q^{2} -\lambda q\nonumber \\{} & {} \quad +(B-I^{{ 0}} )(1-2\lambda )^{2} =0, \end{aligned}$$

  (A. 18)

  then $f'(q)=-{ 48}a^{{ 2}} (1-\lambda )q^{{ 2}} +{ 2}(3a+4a(1-\lambda )\lambda )q-\lambda$. Because $\frac{\lambda }{6a}$ and $\frac{{ 1}}{{ 8}a(1-\lambda )}$ are two positive roots of $f'(q)=0$ and $f'(q)|_{q=0} =-\lambda <0$, we get f(q) decreases in $(0,\frac{\lambda }{6a} )$, increases in $(\frac{\lambda }{6a},\frac{{ 1}}{{ 8}a(1-\lambda )} )$, and decreases in $(\frac{{ 1}}{{ 8}a(1-\lambda )},+\infty )$. Due to $f(0)=(B-I^{{ 0}} )(1-2\lambda )^{2} <0$ and $f(\frac{{ 1}}{{ 8}a(1-\lambda )} )=\frac{(1+64a(B-I^{0} )(1-\lambda )^{2} )(1-2\lambda )^{2} }{64a(1-\lambda )^{2} } >0$, we can get that $\bar{q}_{{ 3}}$ and $\bar{q}_{{ 4}}$ ($\bar{q}_{{ 4}} >\frac{{ 1}}{{ 8}a(1-\lambda )}$ ) are two positives roots of $f(q)=0$. Therefore, $\Pi$ decreases in $(0,\bar{q}_{{ 3}} )$, increases in $(\bar{q}_{{ 3}},\bar{q}_{{ 4}} )$, and decreases in $(\bar{q}_{{ 4}},+\infty )$. Then we have the maximized $\Pi$ achieves at $q=\bar{q}_{{ 4}}$.

Based on (1) and (2), we obtain Theorem (3.7). $\square$

Proof of Proposition (3.8)

When $\lambda >\frac{1}{2}$, $q^{*} =\frac{\lambda }{2a}$. It is obvious that $q^{*}$ increases with $\lambda$, decreases with a, and is not affected by B and $I^{{ 0}}$. When $\lambda \le \frac{1}{2}$, if $B-I^{{ 0}} \ge \frac{n^{2} }{4a} -\frac{mn}{2a}$, $q^{*} =\frac{n}{2a}$. Then we have $\frac{\partial q^{*} }{\partial a} =-\frac{n}{2a^{2} } <0$, $\frac{\partial q^{*} }{\partial \lambda } { =}\frac{1}{2(3-4\lambda )^{2} a} >0$, and $q^{*}$ is not affected by B and $I^{{ 0}}$. When $\lambda \le \frac{1}{2}$ and $B-I^{0} <\frac{n^{2} }{4a} -\frac{mn}{2a}$, taking the derivative of both sides of Eq. (A. 18) with respect to a, we get

$$\begin{aligned}{} & {} {(-48a^{{ 2}} (1-\lambda )(q^{{ *}} )^{2} +2(3a+4a(1-\lambda )\lambda )q^{*} -\lambda )\frac{\partial q^{{ *}} }{\partial a} +(-32a(1-\lambda )(q^{{ *}} )^{{ 3}} } \nonumber \\{} & {} \qquad {+(3+4(1-\lambda )\lambda )(q^{*} )^{2} )=0}. \end{aligned}$$

(A. 19)

Because $q^{*} >\frac{{ 1}}{{ 8}a(1-\lambda )}$, we have $-{ 48}a^{{ 2}} (1-\lambda )(q^*)^{{ 2}} +{ 2}(3a+4a(1-\lambda )\lambda )q^*-\lambda <0$ and $-32a(1-\lambda )(q^{{ *}} )^{{ 3}} +(3+4(1-\lambda )\lambda )(q^{*} )^{2} <0$. Therefore,

$$\begin{aligned} \frac{\partial q^{{ *}} }{\partial a} =-\frac{-32a(1-\lambda )(q^{{ *}} )^{{ 3}} +(3+4(1-\lambda )\lambda )(q^{*} )^{2} }{-48a^{{ 2}} (1-\lambda )(q^{{ *}} )^{2} +2(3a+4a(1-\lambda )\lambda )q^{*} -\lambda } <0.\end{aligned}$$

(A. 20)

Taking the derivative of both sides of Eq. (A. 18) with respect to $\lambda$, we get

$$\begin{aligned} \begin{array}{l}{(-48a^{{ 2}} (1-\lambda )(q^{{ *}} )^{2} +2(3a+4a(1-\lambda )\lambda )q^{*} -\lambda )\frac{\partial q^{{ *}} }{\partial \lambda } +(16a^{{ 2}} (q^{{ *}} )^{{ 3}}} \\ {\qquad +4a(1-2\lambda )(q^{*} )^{2} } {-q^{*} -4(1-2\lambda )(B-I^{0} ))=0.} \end{array}\end{aligned}$$

(A. 21)

Then we have $\frac{\partial q^{{ *}} }{\partial \lambda } =-\frac{16a^{{ 2}} (q^{{ *}} )^{{ 3}} +4a(1-2\lambda )(q^{*} )^{2} -q^{*} -4(1-2\lambda )(B-I^{0} )}{-48a^{{ 2}} (1-\lambda )(q^{{ *}} )^{2} +2(3a+4a(1-\lambda )\lambda )q^{*} -\lambda }$. By solving Eq. (A. 18), we can obtain

$$\begin{aligned} B-I^{{ 0}} { =}\frac{16a^{{ 2}} (1-\lambda )(q^{{ *}} )^{{ 3}} -(3a+4a(1-\lambda )\lambda )(q^{{ *}} )^{2} +\lambda (q^{{ *}} )}{(1-2\lambda )}. \end{aligned}$$

(A. 22)

Substituting $B-I^{{ 0}} { =}\frac{16a^{{ 2}} (1-\lambda )(q^{{ *}} )^{{ 3}} -(3a+4a(1-\lambda )\lambda )(q^{{ *}} )^{2} +\lambda (q^{{ *}} )}{(1-2\lambda )}$ into $\frac{\partial q^{{ *}} }{\partial \lambda }$, we can get

$$\begin{aligned} \frac{\partial q^{{ *}} }{\partial \lambda } =-\frac{q^{{ *}} (4aq^{{ *}} -1)(1+2\lambda -4aq^{{ *}} (3-2\lambda ))}{(-48a^{{ 2}} (1-\lambda )(q^{{ *}} )^{2} +2(3a+4a(1-\lambda )\lambda )q^{*} -\lambda )(1-2\lambda )}. \qquad \end{aligned}$$

(A. 23)

Because $\frac{\frac{1-\lambda }{3-4\lambda } }{2a}>q^{{ *}} >\frac{{ 1}}{{ 8}a(1-\lambda )}$, we have $q^{{ *}} (4aq^{{ *}} -1)(1+2\lambda -4aq^{{ *}} (3-2\lambda ))>0$, then $\frac{\partial q^{{ *}} }{\partial \lambda } >0$.

Finally, taking the derivative of both sides of equation (A. 18) with respect to $B-I^0$, we have

$$\begin{aligned} (-48a^{{ 2}} (1-\lambda )(q^{{ *}} )^{2} +2(3a+4a(1-\lambda )\lambda )q^{*} -\lambda )\frac{\partial q^{{ *}} }{\partial B-I^0} +(1-2\lambda )^{2} =0.\nonumber \\ \end{aligned}$$

(A. 24)

Then $\frac{\partial q^{{ *}} }{\partial B-I} { =}\frac{-(1-2\lambda )^{2} }{-48a^{{ 2}} (1-\lambda )(q^{{ *}} )^{2} +2(3a+4a(1-\lambda )\lambda )q^{*} -\lambda } >0$ holds. $\square$

Proof of Theorem (4.1)

When $\lambda >\frac{{ 1}}{{ 2}}$, the optimal decision under both strategies is $p_{\textrm{c}} =\lambda q$, therefore $p_{\textrm{c}}^{*} =\lambda q$. When $\lambda \le \frac{{ 1}}{{ 2}} \cap \min \{ B+u_{l} \frac{2-3\lambda }{(3-4\lambda )^{2} } q,u_{\textrm{l}} \frac{1-\lambda }{3-4\lambda } q\} \ge I$, because the extreme point of $\Pi _{\textrm{h}}$ is within the feasible area and $\Pi _{\textrm{l}} \le \Pi _{\textrm{h}}$, HP is the optimal strategy.

When $\lambda \le \frac{{ 1}}{{ 2}}$ and $B+u_{\textrm{l}} \frac{2-3\lambda }{(3-4\lambda )^{2} } q<I$, because $\Pi _{\textrm{h}}^{*}$ and $\Pi _{\textrm{l}}^{*}$ are the linear functions of $\beta$, we have $\mathrm \Delta \Pi =\Pi _{\textrm{h}}^{*} -\Pi _{\textrm{l}}^{*}$ is also the linear function of $\beta$. Because $\Delta \Pi { |}_{\beta =0} =\Pi _{h}^{*} \ge 0$, we also have $\Pi _{h} =\Pi _{\textrm{l}}$ when $\beta =1$ and the following formula holds:

$$\begin{aligned}{} & {} { \{ }p_{\textrm{c}} |u_{\textrm{h}} { (}(1-\max \{ \frac{p_{\textrm{c}} -\lambda q}{(1-\lambda )q},0\} )p_{\textrm{c}} \ge I-B\} \nonumber \\{} & {} \qquad \supseteq { \{ }p_{\textrm{c}} |u_{\textrm{l}} { (}(1-\max \{ \frac{p_{\textrm{c}} -\lambda q}{(1-\lambda )q},0\} )p_{\textrm{c}} \ge I-B\}. \end{aligned}$$

(A. 25)

Then we can obtain $\Delta \Pi { |}_{\beta ={ 1}} =\Pi _{\textrm{h}}^{*} -\Pi _{l}^{*} \le 0$. Therefore, there exists a threshold $\bar{\beta }$ such that $\Pi _{\textrm{h}}^{*} \ge \Pi _{\textrm{l}}^{*}$ when $\beta \le \bar{\beta }$ and $\Pi _{\textrm{h}}^{*} <\Pi _{\textrm{l}}^{*}$ when $\beta >\bar{\beta }$. $\square$

Proof of Theorem (4.2)

First, when $\lambda >\frac{{ 1}}{{ 2}}$ and $I\le \lambda q$, the creator will not need loans, so the optimal price is $p_{\textrm{c}}^{*} =\lambda q$.

Second, when $\lambda \le \frac{{ 1}}{{ 2}}$ and $\min \{ B+\frac{2-3\lambda }{(3-4\lambda )^{2} } q,\frac{1-\lambda }{3-4\lambda } q\} \ge I$, because the sum of raised funds in crowdfunding stage and initial funds is enough to cover the cost, the creator will not need loans so the optimal price in this case is the same as that in the case without capital-constraint, i.e., $p_{\textrm{c}}^{*} =\frac{2-3\lambda }{{ 3}-{ 4}\lambda } q,{ \; }p_{\textrm{r}}^{*} =\frac{1-2\lambda }{3-4\lambda } q$.

Third, when $\lambda \le \frac{{ 1}}{{ 2}}$ and $B+\frac{2-3\lambda }{(3-4\lambda )^{2} } q<I$, the profit is

$$\begin{aligned} \Pi =(1-\frac{p_{\textrm{c}} -\lambda q}{(1-\lambda )q} )p_{\textrm{c}} +(\frac{p_{\textrm{c}} -\lambda q}{(1-\lambda )q} -\frac{p_{\textrm{r}} }{q} )p_{\textrm{r}} -I-rA. \end{aligned}$$

(A. 26)

By transforming inequality (22), we obtain $A\ge I-B-(1-\frac{p_{\textrm{c}} -\lambda q}{(1-\lambda )q} )p_{\textrm{c}}$. If $I-B-(1-\frac{p_{\textrm{c}} -\lambda q}{(1-\lambda )q} )p_{\textrm{c}} \le { 0}$, $A=0$ is the optimal loan amounts, i.e., the creator will not need loans. Then the optimal prices are $p_{\textrm{c}} =\bar{p}_{\textrm{c}}$, $p_{\textrm{r}} =\frac{\bar{p}_{\textrm{c}} -\lambda q}{2(1-\lambda )}$, and $\Pi =\frac{(\bar{p}_{\textrm{c}} -\lambda q)^{2} }{4q(1-\lambda )^{2} } -B$; if $I-B-(1-\frac{p_{\textrm{c}} -\lambda q}{(1-\lambda )q} )p_{\textrm{c}} >0$, because $\Pi$ decreases with r, the optimal loan amounts is $A=I-B-(1-\frac{p_{\textrm{c}} -\lambda q}{(1-\lambda )q} )p_{\textrm{c}}$ and

$$\begin{aligned} \Pi =(1-\frac{p_{\textrm{c}} -\lambda q}{(1-\lambda )q} )p_{\textrm{c}} +(\frac{p_{\textrm{c}} -\lambda q}{(1-\lambda )q} -\frac{p_{\textrm{r}} }{q} )p_{\textrm{r}} -I-r(I-B-(1-\frac{p_{\textrm{c}} -\lambda q}{(1-\lambda )q} )p_{\textrm{c}} ).\end{aligned}$$

(A. 27)

By solving $\frac{\partial \Pi }{\partial p_{\textrm{c}} } =0$ and $\frac{\partial \Pi }{\partial p_{\textrm{r}} } =0$, we can get $p_{\textrm{c}} =\frac{q(2+2r(1-\lambda )-3\lambda )}{3+4r(1-\lambda )-4\lambda }$ and $p_{\textrm{r}} =\frac{q(1+r)(1-2\lambda )}{3+4r(1-\lambda )-4\lambda }$. Substituting $p_{\textrm{c}} =\frac{q(2+2r(1-\lambda )-3\lambda )}{3+4r(1-\lambda )-4\lambda }$ and $p_{\textrm{r}} =\frac{q(1+r)(1-2\lambda )}{3+4r(1-\lambda )-4\lambda }$ into $\Pi$, we have

$$\begin{aligned} \Pi =(1+\frac{(1+2r)^{2} }{3+4r(1-\lambda )-4\lambda } )q-I(1+r)+rB. \end{aligned}$$

(A. 28)

Because $\Pi =(1+\frac{(1+2r)^{2} }{3+4r(1-\lambda )-4\lambda } )q-I(1+r)+rB$ decreases with r, we have ${\mathop {\lim }\limits _{r\rightarrow 0}} (1+\frac{(1+2r)^{2} }{3+4r(1-\lambda )-4\lambda } )q-I(1+r)+rB=\frac{1-\lambda }{3-4\lambda } q-I>\frac{(\bar{p}_{\textrm{c}} -\lambda q)^{2} }{4q(1-\lambda )^{2} } -B$ and ${\mathop {\lim }\limits _{r-\rightarrow \infty }} (1+\frac{(1+2r)^{2} }{3+4r(1-\lambda )-4\lambda } )q-I(1+r)+rB<0$. Therefore, there exists a $\bar{r}$ such that when $r\ge \bar{r}$, the creator will not borrow loans and set $p_{\textrm{c}}^{*} =\bar{p}_{\textrm{c}}$ and $p_{\textrm{r}}^{*} =\frac{\bar{p}_{\textrm{c}} -\lambda q}{2(1-\lambda )}$; when $r<\bar{r}$, the creator will borrow loans and set $p_{\textrm{c}}^{*} =\frac{q(2+2r(1-\lambda )-3\lambda )}{3+4r(1-\lambda )-4\lambda }$, $p_{\textrm{r}}^{*} =\frac{q(1+r)(1-2\lambda )}{3+4r(1-\lambda )-4\lambda }$. $\square$

Proof of Theorem (4.3)

When $\lambda >\frac{{ 1}}{{ 2}} (1+e)$, setting $p_{\textrm{c}} =\lambda q$ under HS strategy is better than setting $p_{\textrm{c}} { =} ((1-\lambda )\frac{e}{1+e} +\lambda )q$ under LS strategy; therefore, $p_{\textrm{c}}^{*} =\lambda q$. When $\frac{{ 1}}{{ 2}} (1+e)\ge \lambda >\frac{{ 1}}{{ 2}} (1-e^{2} )$, setting $p_{\textrm{c}} { =}\frac{(1+e-(1-e)\lambda )q}{2(1+e-\lambda )}$ under HS strategy is better than that setting $p_{\textrm{c}} { =}((1-\lambda )\frac{e}{1+e} +\lambda )q$ under LS strategy; therefore, we have $p_{\textrm{c}}^{*} { =}\frac{(1+e-(1-e)\lambda )q}{2(1+e-\lambda )}$ and $p_{\textrm{r}}^{*} =e(1-\bar{\theta })q=\frac{e(1+e)q}{2(1+e-\lambda )}$; when $\lambda \le \frac{{ 1}}{{ 2}} (1-e^{2} )$, setting $p_{\textrm{c}} =\frac{(-2+e^{2} +3\lambda -e(1+\lambda ))}{-3+4\lambda -(2-e)e} q$ under LS strategy is better than setting $p_{\textrm{c}} =((1-\lambda )\frac{e}{1+e} +\lambda )q$ under HS strategy; therefore, we have $p_{\textrm{c}}^{*} { =}\frac{(1+e-(1-e)\lambda )q}{2(1+e-\lambda )}$ and $p_{\textrm{r}}^{*} =\frac{\bar{\theta }+e(1-\bar{\theta })}{2} q=\frac{{ 1}+e-2\lambda }{3+(2-e)e-4\lambda } q$. $\square$

Proof of Proposition (4.4)

When $\lambda >\frac{{ 1}}{{ 2}} (1+e)$, $p_{\textrm{c}}^{*}$ is not influenced by e; when $\frac{{ 1}}{{ 2}} (1+e)\ge \lambda >\frac{{ 1}}{{ 2}} (1-e^{2} )$, we have $\frac{\partial p_{c}^{*} }{\partial e} =\frac{(1-\lambda )\lambda q}{2(1+e-\lambda )^{2} } >0$ and $\frac{\partial p_{\textrm{r}}^{*} }{\partial e} =\frac{q((1+e)^{2} -(1+2e)\lambda )}{2(1+e-\lambda )^{2} } >0$. When $\lambda \le \frac{{ 1}}{{ 2}} (1-e^{2} )$, $\frac{\partial p_{\textrm{c}}^{*} }{\partial e} =-\frac{q((1+e)^{2} -4\lambda )(1-\lambda )}{(-3+4\lambda -(2-e)e)^{2} }$. Therefore, if $\lambda <\min \{ \frac{(1+e)^{2} }{{ 4}},\frac{{ 1}}{{ 2}} (1-e^{2} )\}$, $\frac{\partial p_{\textrm{c}}^{*} }{\partial e} <0$; if $\frac{(1+e)^{2} }{{ 4}} <\lambda \le \frac{{ 1}}{{ 2}} (1-e^{2} )$, $\frac{\partial p_{\textrm{c}}^{*} }{\partial e} >0$; $\frac{\partial p_{\textrm{r}}^{*} }{\partial e} =\frac{q((1+e)^{2} -4e\lambda )}{(-3+4\lambda -(2-e)e)^{2} } >0$ always holds. $\square$