Licensing of a New Product Innovation with Risk Averse Agents

Abstract

An outside innovator invents a new product the viability of which is uncertain. The production technology is licensed to a number (strategic choice) of risk-averse potential Cournot producers by means of: an up-front fee, a per-unit royalty, an ad valorem royalty, or a two-part tariff. The incentive to innovate is maximized with a pure up-front fee if potential producers are (or are close to) risk neutral. Otherwise, it is maximized with a combination of up-front fee and ad valorem royalty. Any scheme that contains a royalty component (per-unit or ad valorem) maximizes the innovation diffusion. Irrespective of the magnitude of licensees’ risk aversion, the innovator and consumers are better off, but licensees are worse off with schemes that have an ad valorem royalty component than with a per-unit royalty component. Consumers are best off with pure up-front fee that avoids double marginalization, even though the innovator optimally sells only one license and creates a monopoly. The results remain similar for a risk-averse innovator, but change considerably with a producing innovator.

Appendix

Proof of Proposition 1

Suppose the innovator sells k licenses. Let $q_i$ be the output of licensee i who maximizes his expected utility $\rho \big [(p-c) q_i\big ]^{\frac{1}{\eta }}$. The total Cournot output is $Q(k)= \frac{k\,(a-c)}{(k+1)b}$ and each licensee makes a profit equal to $\frac{(a-c )^2}{(k+1)^2 b}$. A licensee is willing to pay up front X s.t. $U(X)=\rho \bigg (\frac{(a-c )^2}{(k+1)^2 b}\bigg )^{\frac{1}{\eta }}$. Thus $X=\rho ^{\eta }\frac{(a-c )^2}{(k+1)^2 b}$. The total revenue of the innovator is $\rho ^{\eta } \frac{k(a-c )^2}{(k+1)^2 b}$, which is maximized at $k^*=1$. The innovator obtains upfront $\rho ^{\eta } \frac{(a-c)^2}{4b}$ and the licensee obtains $(1-\rho ^{\eta }) \frac{(a-c)^2}{4b}$ with probability $\rho$ and $-\rho ^{\eta } \frac{(a-c)^2}{4b}$ with probability $1-\rho$. The closer to risk neutrality is the licensee, the higher is the innovator’s payoff. The licensee is left with zero profit if he is risk neutral ($\eta =1$). □

Proof of Proposition 2

Given the strategy (k, r) of the innovator, licensee i maximizes $\bigg (p(Q)\,-c-r\bigg )\,q_i$ over $q_i$. The Cournot quantity is

$$\begin{aligned} Q_r(k,r) = {\left\{ \begin{array}{ll} 0 &\quad {\text { if }} \,r \ge a-c \\ \frac{k}{k+1}\left( \frac{a-c-r}{ b}\right) &\quad {\text { if }} \, 0 \le r< a-c \end{array}\right. } \end{aligned}$$

The innovator maximizes his expected revenue $E \Pi ^I_r(k,r) =\rho \, r Q_r(k,r)$ over k and r. The optimal royalty is $r^*=\frac{a-c}{2}$ (independent of k) and the optimal number of licensees is $k^*=N$. □

Proof of Proposition 3 i

Suppose k licensees compete simultaneously in quantities. Each licensee maximizes $q_i\, (p\,(1-v)-c)$ over $q_i$. The Cournot quantity and price are

$$\begin{aligned} Q(k,v)={\left\{ \begin{array}{ll} \frac{k}{k+1}\, \left( \frac{ a\,(1- v) -c }{b\,( 1- v ) }\right) &\quad {\text { if }}\, v\le 1-\frac{c}{a}\\ 0 &\quad {\text { otherwise}} \end{array}\right. }\, {\text { and }}\, p(k,v)= {\left\{ \begin{array}{ll} \frac{a(1-v)+kc}{(k+1)(1-v)} &\quad {\text { if }}\, v\le 1-\frac{c}{a}\\ a &\quad {\text { otherwise}} \end{array}\right. } \end{aligned}$$

(12)

The innovator’s set of relevant strategies is $S= \bigg \{ (k,v)| 0\le v \le 1-\frac{c}{a},\, 0< k \le N\bigg \}$. By (12),

$$\begin{aligned} E \Pi ^{I}_{v}(k,v)=\rho \,v\cdot p(k,v)\cdot Q(k,v)=\dfrac{\rho \,k\,v\,[a\,(1-v)-c][a (1-v)+c\,k]}{(k+1)^2\,b\,(1-v)^2} \end{aligned}$$

(13)

We first show that $k^*=N$. Treating k as a continuous variable we have

$$\begin{aligned} \frac{\partial E \Pi ^{I}_{v} }{\partial k} (k,v)=\dfrac{\rho \,v\,[a(1-v)-c]}{b\,(k+1)^3 \, (1-v)^2} \cdot \bigg \{[2c- a\, (1-v)]\,k+a\,(1-v)\bigg \} \end{aligned}$$

(14)

Since $v < 1-\frac{c}{a}$, by (14)

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{\partial E \Pi ^{I}_{v}}{\partial k}(k,v)> 0 \, &\quad {\text { iff }} \, [2c- a\, (1-v)]\,k+a\,(1-v) > 0\\ \frac{\partial E \Pi ^{I}_{v}}{\partial k}(k,v) = 0 \, &\quad {\text { iff }} \, [2c- a\, (1-v)]\,k+a\,(1-v) = 0 \end{array}\right. } \end{aligned}$$

(15)

Case 1.1 $1-\frac{2\,c}{a}\le v \le 1-\frac{c}{a}$. By (14), $\,\,\frac{\partial E \Pi ^{I}_{v}}{\partial k}(k,v) > 0$, implying $k_{v}^*=N$.

Case 1.2 $0\le v<1-\frac{2\,c}{a}$. In this case, $\,\,\frac{\partial E \Pi ^{I}_{v}}{\partial k}(k,v) > 0$ iff $k < \frac{a(1-v)}{a(1-v)-2c}$, implying $k_{v}^*= \min \bigg (\frac{a(1-v^*)}{a(1-v^*)-2c} , N \bigg )$.

Suppose $k_{v}^*=\frac{a(1-v^*)}{a(1-v^*)-2c}\le N$, equivalently, $v^*\le 1-\frac{2c\,N}{a(N-1)}$ and $1-\frac{2c\,N}{a(N-1)}< 1-\frac{2\,c}{a}$.

Let $g(v) \equiv E \Pi ^{I}_{v}(\frac{a(1-v)}{a(1-v)-2c},v)$. By (13),

$$\begin{aligned} g(v) = E \Pi ^{I}_{v}\bigg (\frac{a(1-v)}{a(1-v)-2c},v\bigg )=\frac{\rho \,a^2\,v}{4\,b} \end{aligned}$$

(16)

g(v) is increasing in v and $v^*\le 1-\frac{2c\,N}{a(N-1)}$, hence $g(v^*) \le G(1-\frac{N\,2c}{a(N-1)})$. Since $(\frac{a(1-v^*)}{a(1-v^*)-2c},v^*)$ maximizes the revenue of the innovator, $g(v^*) = g(1-\frac{2c\,N}{a(N-1)})$ and $E\Pi ^{I}_{v}(\frac{a(1-v^*)}{a(1-v^*)-2c},v^*) = E\Pi ^{I}_{v}(N,1-\frac{2c\,N}{a(N-1)})$. Thus $v^*=1-\frac{2c\,N}{a(N-1)}$ and $k^*_{v}=N$ must hold. □

Lemma 1

$E\Pi ^{I}_{v}(N,v)$ is single peaked in $v \in \left[ 0, 1-\frac{c}{a}\right]$. The maximizer $v^*$ is an interior point.

Proof

$$\begin{aligned} \begin{aligned} \frac{\partial E\Pi ^{I}_{v} }{\partial v} (N,v)&=\frac{-\rho \, N}{ b\,(1-v) ^{3} \left( N+1 \right) ^{2}} \\&\quad \cdot \bigg \{a ^2 \, v^3-3 a ^2\,v^2+ \bigg [3\, a ^2+ \left( N-1 \right) c \, a +N\,c^2 \bigg ] v- \left( a-c \right) \left( Nc+a \right) \bigg \} \end{aligned} \end{aligned}$$

(17)

and it is continuous in v. It is easy to verify that $\frac{\partial E\Pi ^{I}_{v} }{\partial v} (N,0)>0$ and $\frac{\partial E\Pi ^{I}_{v} }{\partial v} (N,1-\frac{c}{a})<0$. Hence $\frac{\partial E\Pi ^{I}_{v} }{\partial v} (N,v)=0$ has a solution in $(0, 1-\frac{c}{a})$. Next

$$\begin{aligned} \frac{\partial ^2 E\Pi ^{I}_{v} }{\partial v^2} (N,v)={\frac{2\rho Nc}{b\,( 1-v ) ^4 \,( N+1 ) ^2}}\bigg \{-[a(N-1)+cN] v +a(N-1)-2cN \bigg \} \end{aligned}$$

$$\begin{aligned}&{\text {If}}\,\, a(N-1)-2cN<0,\,{\text { then }}\, \frac{\partial ^2 E\Pi ^{I}_{v} }{\partial v^2} (N,v)<0\, {\text {for}}\,{\text {all}}\, v \\&{\text {If}}\,\, a(N-1)-2cN \ge 0,\,{\text {then}}\, \frac{\partial ^2 E\Pi ^{I}_{v} }{\partial v^2} (N,v)<0\,{\text {iff}}\, v >\frac{(N-1)a-2cN}{(N-1)a+cN} \end{aligned}$$

For $v \in [0, 1-\frac{c}{a}]$, in the former case, $\frac{\partial E\Pi ^{I}_{v} }{\partial v} (N,v)$ starts positive, monotonically decreasing in v and ends negative. In the latter case, $\frac{\partial E\Pi ^{I}_{v} }{\partial v} (N,v)$ starts positive, first increasing then decreasing in v, and ends negative. Hence the solution to $\frac{\partial E\Pi ^{I}_{v} }{\partial v} (N,v)=0$ is unique and $E\Pi ^{I}_{v}(N,v)$ is single peaked. Its maximizer turns out to be

$$\begin{aligned} v^*&= \frac{B^{\frac{1}{3}}}{3a} - \frac{c \,\left[ N c+ a \left( N -1 \right) \right] }{a B^{\frac{1}{3}} } +1 \end{aligned}$$

(18)

where $B=3c \left( { \sqrt{3}\sqrt{\left[ (a+c)N-a\right] ^3\,c+27\,a^2{N}^2c^2}}-9\, N c\,a \right)$. □

Corollary 2

$\frac{\partial E\Pi _{v}^{I}}{\partial v }(N,v) >0$ implies $v<v^*$, and $\frac{\partial E\Pi _{v}^{I}}{\partial v }(N,v) <0$ implies $v>v^*$.

Given the complicated formula of $E\Pi _{v}^{I}(N,v)$, it is impossible for us to compare its value under different v. However, $\frac{\partial E\Pi _{v}^{I}}{\partial v }(N,v)$ is relatively neat, enabling us to do the comparison.

Proof of Proposition 3 ii

Next we compute the limit of $v^*$ as N grows indefinitely. By (13), for every $v \in [0, 1-\frac{a}{c}]$, $E\Pi ^{I}_{v}(v,N) = \dfrac{\rho \,Nv\,[a\,(1-v)-c] [a (1-v)+c\,N]}{(N+1)^2b\,(1-v)^2 }$,

Hence $\lim \nolimits _{N \rightarrow \infty } E\Pi ^{I}_{v}(v,N) =\frac{cf[a(1-v)-c]}{b\,(1-v)^2} \equiv E\Pi ^{I}_{v}(v,\infty )$. Since $v^* \equiv v^*(N)$ is the maximizer of $E\Pi ^{I}_{v}(v,N)$ over v and $E\Pi ^I_{v}$ is continuous in $[0, 1-\frac{a}{c}]$,

$$\begin{aligned} E\Pi ^I_{v}(v^*(N),N) \ge E\Pi ^I_{v}(v,N),\,{\text {for}}\,{\text {all}}\, v \in \left[ 0, 1-\frac{a}{c}\right] \end{aligned}$$

(19)

The sequence $v^*(N)$ is bounded and therefore has a cluster point. Namely, there exists a subsequence $(N_m)_{m=1}^{\infty }$ and ${\hat{v}}$ s.t. $v^*(N_m ) \rightarrow {\hat{v}}$ as $m \rightarrow \infty$. By (19),

$$\begin{aligned} E\Pi ^I_{v}(v^*(N_m),N_m) \ge E \Pi ^I_{v}(v,N_m), \,\, \forall v \in \left[ 0, 1-\frac{c}{a}\right] \end{aligned}$$

By continuity of $E\Pi _{v}^I$ in v, taking the limit as $m \rightarrow \infty$, we have for all $v \in [0, 1-\frac{c}{a}]$,

$$\begin{aligned} E\Pi ^I_{v}({\hat{v}}, \infty ) \ge E\Pi ^I_{v}(v,\infty ) \end{aligned}$$

proving that ${\hat{v}}$ is a maximizer of $E\Pi ^I_{v}(v,\infty )$ over v. Since $E\Pi ^I_{v}(v,\infty ) =\frac{cf[a(1-v)-c]}{b\,(1-v)^2}$ has a unique maximizer $\frac{a-c}{a+c}$, $\lim \nolimits _{N \rightarrow \infty } v^*(N)$ exists and it is ${\hat{v}} =\frac{a-c}{a+c}$. □

Proof of Proposition 3 iii

Follows from standard Cournot computation. □

Proof of Proposition 4 i

Follows directly from Propositions 1, 2 and 3.□

Proof of Proposition 4 ii

By Proposition 3, the optimal ad valorem price is $p^*_{v}= \dfrac{a(1-v^*)+Nc}{(N+1)(1-v^*)}$. Hence $p^*_{v}>\frac{a+c}{2}=p^*_f$ is equivalent to $v^*>1-\frac{2Nc}{(a+c)(N+1)-2a}$. Since

$$\begin{aligned} \frac{\partial E\Pi ^{I}_{v} }{\partial v}|_{v=1-\frac{2Nc}{(a+c)(N+1)-2a}}&=\frac{\rho \, \left( 3\,Na+cN-a+c \right) \left( a-c \right) }{4\,N \left( 1+N \right) b}>0, \end{aligned}$$

by Corollary 1, $v^*>1-\frac{2Nc}{(a+c)(N+1)-2a}$ and $p^*_{v}>p^*_f$.

Comparing (9) and (7), $p^*_{v}<p^*_r$ iff $v^* < \frac{a-c}{a+c}\equiv {\hat{v}}$. By Corollary 1, since $\dfrac{\partial E \Pi ^{I}_{v} }{\partial v}|_{v={\hat{v}}}=-\dfrac{\rho a(a-c)^2N}{ 4bc \left( N+1 \right) ^{2} } <0$, we have

$$\begin{aligned} v^* < {\hat{v}}=\dfrac{a -c}{a +c} , \end{aligned}$$

(20)

and $p^*_f<p^*_{v}<p^*_r$. By (3), (9) and (7), $p^*_f=\lim \nolimits _{N \rightarrow \infty }p^*_{v}=\lim \nolimits _{N \rightarrow \infty }p^*_r$. □

Proof of Proposition 4 iii

Since $1-\frac{c}{a}-{\hat{v}}= \frac{c(a -c)}{a (a +c)} >0$, by (13)

$$\begin{aligned} E\Pi ^{I^*}_{v}= E\Pi ^{I}_{v}(N,v^*) \ge E\Pi ^{I}_{v}(N,{\hat{v}})=\dfrac{\rho \, (a -c)^2\, N \, [2a+N(a +c)]}{4\,b(N+1)^2\,(a +c)} \end{aligned}$$

It is easy to verify that $a>c$ implies $E\Pi ^{I}_{v}(N,{\hat{v}})>E\Pi ^{I^*}_r$, hence $E\Pi ^{I^*}_{v}> E\Pi ^{I^*}_{r}$.

By (ii), $p^*_{v}> p^*_f= p_M$, where $p_M$ is the monopoly price. Hence $E\Pi ^{I^*}_{v}+N\cdot E\Pi ^{L^*}_{v}=(p^*_{v}-c) Q^*_{v} < \Pi _M= \frac{\rho (a -c)^2}{4\,b}$, where $\Pi _M$ is the expected monopoly profit. By Proposition 3, $E\Pi ^{L}_{v}>0$, hence $E\Pi ^{I^*}_{v}< \frac{\rho (a -c)^2}{4\,b}$.

By (1) and (5), $\Pi ^{I^*}_f < E\Pi ^{I^*}_{r}$ iff $\rho ^{\eta -1} < \frac{N}{N+1}$, or equivalently, $\eta >1 + \ln \frac{ N}{N+1}/\ln \rho \equiv \eta _2$.

By (1) and (11), $\Pi ^{I^*}_f > E\Pi ^{I^*}_{v}$ iff $\rho ^{\eta -1} > \frac{N\,4b}{(N+1)^2 (a-c)^2}\frac{v^*\,[a\,(1-v^*)-c] [a (1-v^*)+c\,N]}{b\,(1-v^*)^2 }$, equivalently, $1 \le \eta < 1+ \ln \bigg (\frac{ N\,4b}{(N+1)^2 (a-c)^2}\frac{v^*\,[a\,(1-v^*)-c] [a (1-v^*)+c\,N]}{b\,(1-v^*)^2 }\bigg ) /\ln \rho \equiv \eta _1$. As N grows indefinitely, $\Pi ^{I^*}_f \le \Pi ^{I^*}_f+E\Pi ^{L^*}_f=\lim \nolimits _{N \rightarrow \infty }E\Pi ^{I^*}_{v}=\lim \nolimits _{N \rightarrow \infty }E\Pi ^{I^*}_{r}=\dfrac{\rho (a -c)^2}{4\,b}$. □

Proof of Proposition 4 iv

By (6) and (10), $E\Pi ^{L^*}_r>E\Pi ^{L^*}_{v}$ iff $\frac{(a -c)^2}{4}>\frac{[a(1-v^*)-c]^2}{1-v^*}$. Equivalently,

$$\begin{aligned} E\Pi ^{L^*}_r>E\Pi ^{L^*}_{v} \,\, {\text { iff}}\,\, J(v^*)\equiv -4a^2 v^{*2}+(7a+c)(a-c)v^*-3(a-c)^2>0 \end{aligned}$$

Let $v_3=\frac{(a-c)(7a+c-\sqrt{a^2+c^2+14ac})}{8a^2}$. It is easy to verify that

$$\begin{aligned} J({\hat{v}})=\dfrac{2c(a-c)^3}{(a+c)^2}>0\,\,{\text {and}} \,\,J(v_3)=0 \end{aligned}$$

Since J(v) is continuous and concave in v, $v_3<v^*<{\hat{v}}$ is a sufficient condition for $J(v^*)>0$. By (20), $v^*<{\hat{v}}$ and we have

$$\begin{aligned}&\dfrac{\partial E \Pi ^{I}_{v} }{\partial v}|_{v=v_3}=\dfrac{\bigg \{h(N)+(a-c)^5+8ac\bigg [3(a-c)^3+2a(N-1)(a-c)^2\bigg ]\bigg \}\cdot 4(a-c)a^2N}{b(N+1)^2\bigg [(a-c)\sqrt{a^2+c^2+14ac}+a^2+6ac+c^2\bigg ]^3}, \end{aligned}$$

where

$$\begin{aligned} h(N)&=16a^2cN \underbrace{\bigg [(a+c)\sqrt{a^2+c^2+14ac}-2c(3a+c) \bigg ]}_{A_4}\\&\quad + \underbrace{\bigg [(a-c)^4+16ac^2(a+c)\bigg ]\sqrt{a^2+c^2+14ac}-128a^2c^3}_{A_5} \end{aligned}$$

Notice that $A_4>0$ iff $(a+c)^2 (a^2+c^2+14ac)>4c^2(3a+c)^2$. Equivalently, $(a-c)^4+ 4c[(a-c)^3+4a(a^2-c^2) ]>0.$ This holds for $a>c$.

On the other hand, $A_5>0$ iff $\bigg [(a-c)^4+16ac^2(a+c)\bigg ]\sqrt{a^2+c^2+14ac}>128a^2c^3$. Equivalently, $16ac^2(a+c)\sqrt{a^2+c^2+14ac}>128a^2c^3,$ which holds for all $a>0, c>0$.

Thus $h(N)>0$ and $\dfrac{\partial E \Pi ^{I}_{v} }{\partial v}|_{v=v_3}>0$. By Corollary 1, $v^*>v_3$. This proves that $E \Pi ^{L^*}_{v}<E \Pi ^{L^*}_{r}$.

By (2) and (6) $E \Pi ^{L^*}_{f}>E \Pi ^{L^*}_{r}$ iff $\eta >1+ \ln \bigg [1-\frac{1}{(N+1)^2} \bigg ]/\ln \rho \equiv \eta _4$. By (2) and (10), $E \Pi ^{L^*}_{f}<E \Pi ^{L^*}_{v}$ iff $1 \le \eta <1+ \ln \bigg \{ 1-\frac{4[a(1-v^*)-c]^2}{(a-c)^2 (N+1)^2 (1-v^*)}\bigg \} /\ln \rho \equiv \eta _3$. $\bigg (\eta _3 <\eta _4$ iff $E\Pi ^{L^*}_r>E\Pi ^{L^*}_{v} \bigg )$. Finally, by (6) and (10), $\lim \nolimits _{N \rightarrow \infty } E\Pi ^{L^*}_{v}(N)= \lim \nolimits _{N \rightarrow \infty }E\Pi ^{L^*}_{r}(N)=0$. □

Proof of Proposition 5

$$\begin{aligned} CS_{v}=\left( \dfrac{N}{N+1}\right) ^2\,\dfrac{ [a(1-v^*)-c]^2}{2b(1-v^*)^2} \end{aligned}$$

(21)

Note that $\frac{\partial CS_{v}}{\partial v}(N,v) <0$. By (20), $v^* <\frac{a-c}{a+c}\equiv {\hat{v}}$ for all $N \ge 1$. Comparing (8) and (21),

$$\begin{aligned} CS_{v} (N,v^*)> CS_{v} (N,{\hat{v}})=\left( \dfrac{N}{N+1}\right) ^2\,\dfrac{(a-c)^2}{8\,b}=CS_r \end{aligned}$$

Comparing (4) and (21), $CS_f>CS_{v} \, {\text {iff}}\,\,\,v^* >1-\dfrac{2Nc}{(N-1)a+(N+1)c}$. Since

$$\begin{aligned} \dfrac{\partial E\Pi _{v}^I}{\partial v} |_{v=1-\frac{2Nc}{(N-1)a+(N+1)c}}={\dfrac{\rho \left( a-c\right) \left[ \left( 3 a+c\right) k-a+c \right] }{4\,k \left( k+1 \right) b }} >0, \end{aligned}$$

by Corollary 1, $v^* >1-\dfrac{2Nc}{(N-1)a+(N+1)c}$ impling $CS_f>CS_{v}$. Also,

$$\begin{aligned} \lim \limits _{N \rightarrow \infty }CS_{v}(N)&= \dfrac{ (a-c)^2}{8b}=\lim \limits _{N \rightarrow \infty }CS_r(N)=CS_f \end{aligned}$$

By Proposition 4, since $p^*_{v}> p^*_f=p_M$, $E\Pi ^{I}_{v}(N,v)+N\cdot E\Pi ^{L}_{v}< \Pi _M=\Pi ^{I^*}_f+\Pi ^{L^*}_f$. We have

$$\begin{aligned} E\Pi ^{I^*}_{r}+N\cdot E\Pi ^{L^*}_{r}&= \dfrac{\rho \, N(N+2)\,(a-c)^2}{4(N+1)^2\,b}\\ E\Pi ^{I^*}_{v}+N\cdot E\Pi ^{L^*}_{v}&=\dfrac{\rho \, N\big [(1-v^*)a-c\big ]\,\big [(1-v^*)(a-c)+Nv^*c\big ]}{(N+1)^2(1-v^*)^2 b} \end{aligned}$$

Let

$$\begin{aligned} t(v) \equiv \bigg \{ \bigg [ (a-c)(N-2)+2Nc \bigg ]\, v - (a-c)(N-2)\bigg \}\bigg [(a+c)\, v-(a-c) \bigg ] \end{aligned}$$

then

$$\begin{aligned} \bigg (E\Pi ^{I^*}_{r}+N\cdot E\Pi ^{L^*}_{r}\bigg )-\bigg ( E\Pi ^{I}_{v}(v)+N\cdot E\Pi ^{L}_{v}(v) \bigg )= \dfrac{\rho \,N\cdot t(v)}{4(N+1)^2(1-v)^2b} \end{aligned}$$

t(v) is quadratic and convex in v, and $t(v)=0$ has two solutions $v_1=\dfrac{(N-2)(a-c)}{(N-2)(a-c)+2Nc}$ and $v_2=\dfrac{a-c }{a+c}={\hat{v}}$. By (17),

$$\begin{aligned} \dfrac{\partial E\Pi _{v}^I}{\partial v} |_{v=v_1}&=\dfrac{\rho (a-c)}{4c N (N+1)^2b}\\&\quad \cdot \bigg \{[2c^2+5ac+a^2]N^2-4(a+c)(a-2c)N+4(a-c)(a-2c) \bigg \} \end{aligned}$$

Let $\alpha (N)= (2c^2+5ac+a^2)N^2-4(a+c)(a-2c)N+4(a-c)(a-2c)$, which is increasing in N for $N\ge 2$ and $\alpha (2)=16c(a+2c)>0$. Also $\alpha (1)=a^2-3ac+18c^2>0$ for all a and c. Hence $\alpha (N) >0$ and $\frac{\partial E\Pi _{v}^I}{\partial v} |_{v=v_1}>0$ for all $N\ge 1$. By Corollary 1, $v_1<v^*$ and by (20), $v^*<v_2$. Since $v_1<v^*<v_2$, $\,\,t(v^*)<0$ and $E\Pi ^{I^*}_{r}+N\cdot E\Pi ^{L^*}_{r}< E\Pi ^{I}_{v}(v^*)+N\cdot E\Pi ^{L}_{v}(v^*)$. □

Proof of Proposition 6

Given (k, r), the total Cournot output is $Q(k,r)= \frac{k}{(k+1)b}(a-c-r)$, for $r<a-c$ if demand is positive. The total royalty payment collected by the innovator is $r\cdot Q(k,r)=\frac{kr(a-c-r)}{(k+1)b}$, which is maximized at $r_{fr}^*= \frac{a-c}{2}$. The net profit of each licensee is $\frac{(a-c)^2}{4b(k+1)^2}$, and his willingness to pay up front is F s.t. $U(F)=\rho \cdot U(\frac{(a-c)^2}{4b(k+1)^2})$. Equivalently, $F= \frac{\rho ^{\eta }(a-c)^2}{4b(k+1)^2}$. The total expected revenue of the innovator is therefore,

$$\begin{aligned} E\Pi ^{I}_{fr}=kF+\rho \cdot r_{fr}^*\cdot Q(k)=\bigg [\frac{ \rho ^{\eta }k}{(k+1)^2} +\frac{\rho k}{k+1} \bigg ] \frac{(a-c)^2}{4b} \end{aligned}$$

It is easily verified that

$$\begin{aligned} \frac{\partial E\Pi ^{I}_{fr} }{\partial k}= \frac{(a-c)^2}{4b(k+1)^3}\bigg [\rho (k+1)-\rho ^{\eta }(k-1) \bigg ]>0 \,\,{\text {for}}\,{\text {all}}\, 0<k \le N \end{aligned}$$

implying $k_{fr}^*=N$ and

$$\begin{aligned} \Pi ^{I^*}_{fr}=\bigg [\frac{ \rho ^{\eta }N}{(N+1)^2} +\frac{\rho N}{N+1} \bigg ] \frac{(a-c)^2}{4b} \end{aligned}$$

(22)

Comparing this with the pure up-front fee scheme (see (1)), we have $\Pi ^{I^*}_{fr}> \Pi ^{I^*}_{f}\,{\text {iff}}\, \eta \ge \frac{\ln \big ( \frac{N^2+N}{N^2+N+1}\big )}{\ln \rho }$. □

Proof of Proposition 7

Given (k, v), each licensee obtains $[(1-v) p(k,v)-c]\, q(k,v)$ and is willing to pay up front F, s.t. $U(F)=\rho U([(1-v) p(k,v)-c] q(k,v))$. Equivalently, $F=\rho ^{\eta } [(1-v) p(k,v)-c]\, q(k,v)$.

The innovator’s expected revenue is

$$\begin{aligned} E\Pi ^{I}_{fv}(k,v)&=kF+ \rho \cdot v\cdot p(k,v)\, Q(k,v)\nonumber \\&=\dfrac{k[(1-v)a-c]}{(1-v)^2(k+1)^2b}\cdot \bigg \{(1-v)[(1-v)a-c] \rho ^{\eta } + [(1-v)a+ck]v \rho \bigg \} \end{aligned}$$

(23)

If $\eta \ge \frac{\ln \big ( \frac{N^2+N}{N^2+N+1}\big )}{\ln \rho }$, by (22), $E\Pi ^{I}_{fv}(N,{\hat{v}})-E\Pi ^{I^*}_{fr}=\dfrac{N(a-c)^3(\rho -\rho ^{\eta })}{4(a+c) b (N+1)^2}>0$. Hence $E\Pi ^{I}_{fv}(N,{\hat{v}}) >E\Pi ^{I^*}_{fr}$ and $E\Pi ^{I^*}_{fv}\ge E\Pi ^{I}_{fv}(N,{\hat{v}}) >E\Pi ^{I^*}_{fr}$. If $\eta =1$, $E\Pi ^{I}_{fv}(k,v)$ is maximized at $v=0$ and $v^*= \frac{\rho ^\eta (a-c)^2}{4b}$. The optimal licensing scheme boils down to pure up-front fee described in Proposition 1. Therefore, there exists ${\tilde{\eta }} \in \bigg (1, \frac{\ln \big ( \frac{N^2+N}{N^2+N+1}\big )}{\ln \rho }\bigg )$ s.t. the optimal scheme consists of positive up-front fee and positive ad valorem royalty if $\eta \ge {\tilde{\eta }}$, and the optimal scheme is pure up-front fee if $1 \le \eta < {\tilde{\eta }}$.

Part (iii) follows from $p_{fv}=p_{v}$. □

Proof of Claim 1

$$\begin{aligned} \begin{aligned} \frac{\partial E\Pi ^{I}_{fv} }{\partial v} (k,v)&=\frac{k}{ b\,(1-v) ^{3} \left( k+1 \right) ^{2}} \\&\quad \cdot \bigg \{a ^2 (\rho - \rho ^{\eta })(1-v)^3+ c \bigg [\rho ^{\eta } c + \left[ a(k-1) +ck\right] \rho \bigg ] (1-v)- 2 \rho c^2 k\bigg \} \end{aligned} \end{aligned}$$

(24)

and it is continuous in v. It is easy to verify that $\frac{\partial E\Pi ^{I}_{v} }{\partial v} (k,0)>0$ and $\frac{\partial E\Pi ^{I}_{v} }{\partial v} (k,1-\frac{c}{a})<0$. Also,

$$\begin{aligned} \frac{\partial ^2 E\Pi ^{I}_{v} }{\partial v^2} (k,v)={\frac{2\rho Nc}{b\,( 1-v ) ^4 \,( N+1 ) ^2}}\bigg \{-[a(N-1)+cN] v +a(N-1)-2cN \bigg \} <0 \end{aligned}$$

For $v \in [0, 1-\frac{c}{a}]$, $\frac{\partial E\Pi ^{I}_{fv} }{\partial v} (k,v)$ starts positive, monotonically decreases in v and ends with a negative value. The solution to $\frac{\partial E\Pi ^{I}_{fv} }{\partial v} (k,v)=0$ is unique and it maximizes $E\Pi ^{I}_{fv}(k,v)$. □

Proof of Proposition 8

As explained in the paragraph exceeding Proposition 8, if the innovator sells licenses, he charges up-front fee only. Suppose the innovator sells $k-1$, $k\ge 2$, licenses, we have

$$\begin{aligned}&\frac{\partial EU(\Pi ^I +(k-1) F) }{\partial k} = -\bigg [ \frac{ (a-c)^2}{b(k+1)^2}\bigg ]^{1/\eta } \cdot \frac{\left( {\rho }^{\eta }k-{\rho }^{\eta }+1\right) ^{\frac{1}{\eta }-1} }{\eta (k^2-1)}\cdot \rho \\&\quad \cdot \bigg \{ \left( k-1 \right) \left[ 2+ \left( k-3\right) {\rho }^{\eta } \right] +(1-\rho ) (k-3)( {\rho }^{\eta }k-{\rho }^{\eta }+1)^{1-\frac{1}{\eta }} (k-1)^ {\frac{1}{\eta }} \bigg \} \end{aligned}$$

Hence the innovator’s payoff is decreasing in k for $k \ge 3$. The optimal k is thus either 2 or 3. Since

$$\begin{aligned} EU(\Pi ^I + F)&=\rho \bigg [\frac{(1+\rho ^\eta )(a-c)^2}{9b} \bigg ]^{ 1/\eta } + (1-\rho ) \bigg [\frac{\rho ^\eta (a-c)^2 }{9b} \bigg ]^{ 1/\eta },\\ EU(\Pi ^I +2 F)&=\rho \bigg [\frac{(1+2\rho ^\eta )(a-c)^2}{16b} \bigg ]^{ 1/\eta } + (1-\rho ) \bigg [\frac{\rho ^\eta (a-c)^2 }{8b} \bigg ]^{ 1/\eta }, \end{aligned}$$

we have

$$\begin{aligned}&EU(\Pi ^I + F)-EU(\Pi ^I +2 F)=\bigg [\frac{(a-c)^2}{b} \bigg ]^{ 1/\eta }\cdot \rho \\&\cdot \quad \bigg \{ \bigg (\frac{1+\rho ^\eta }{9} \bigg )^{ 1/\eta } -\bigg (\frac{1+2\rho ^\eta }{16} \bigg )^{ 1/\eta }- (1-\rho )\bigg [ ( \frac{1}{8} )^{1/\eta }-( \frac{1}{9} )^{1/\eta } \bigg ] \bigg \} \end{aligned}$$

Let $\Delta _1(\rho )= \bigg (\frac{1+\rho ^\eta }{9} \bigg )^{ 1/\eta } -\bigg (\frac{1+2\rho ^\eta }{16} \bigg )^{ 1/\eta }- (1-\rho )\bigg [ \left( \frac{1}{8}\right) ^{1/\eta }-\left( \frac{1}{9} \right) ^{1/\eta } \bigg ]$. Then

$$\begin{aligned} \frac{\partial \Delta _1}{\partial \rho }=\rho ^{\mu -1}\bigg [\frac{1}{9}\bigg (\frac{1+\rho ^\eta }{9} \bigg )^{ \frac{1}{\eta }-1}- \frac{1}{8}\bigg (\frac{1+2\rho ^\eta }{16} \bigg )^{\frac{1}{\eta }-1} \bigg ]+\left( \frac{1}{8} \right) ^{1/\eta }-\left( \frac{1}{9} \right) ^{1/\eta } \end{aligned}$$

Since $\frac{1}{9}\bigg (\frac{1+\rho ^\eta }{9} \bigg )^{ \frac{1}{\eta }-1}- \frac{1}{8}\bigg (\frac{1+2\rho ^\eta }{16} \bigg )^{\frac{1}{\eta }-1} >0$ is equivalent to $2(\frac{1}{8} )^{\frac{1}{\mu -1}}-\left( \frac{1}{9}\right) ^{\frac{1}{\mu -1}}>-2\rho ^\mu \bigg [(\frac{1}{8} )^{\frac{1}{\mu -1}}-\left( \frac{1}{9}\right) ^{\frac{1}{\mu -1}} \bigg ]$, which always holds, we have $\Delta _1(\rho )\ge \Delta _1(0)=2\left( \frac{2}{9}\right) ^{1/\eta }-\left( \frac{1}{16}\right) ^{1/\eta }-\left( \frac{1}{8}\right) ^{1/\eta }>0$. Therefore $EU(\Pi ^I + F)>EU(\Pi ^I +2 F)$.

Finally, $EU(\Pi ^I + F) > EU(\Pi ^M)$ iff $h(\rho )\equiv \rho \bigg (\frac{1+\rho ^\eta }{9} \bigg )^{ 1/\eta } + (1-\rho ) \bigg (\frac{\rho ^\eta }{9} \bigg )^{ 1/\eta }-\rho \cdot \bigg (\frac{1}{4} \bigg )^{1/\eta }>0$. It is easy to verify that $h'(\rho )<0, h(0)>0$ iff $\eta >\frac{\ln 9/4}{\ln 2}$ and $h(1)<0$. Hence there exists a unique $\rho _0$ s.t. $h(\rho _0)=0$ and $EU(\Pi ^I + F) > EU(\Pi ^M)$ iff $\rho <\rho _0$. □

Fig. 1

Innovator’s expected payoff as a function of licensees’ risk aversion magnitude

Fig. 2

Simulation for the optimal $k^*_{fv}$