Emission Reduction Technology Licensing and Diffusion Under Command-and-Control Regulation

Abstract

In this paper, we discuss a patented emission reduction technology that a monopolistic upstream eco-industry licenses to the polluting firms in a downstream oligopolistic industry, which is subject to command-and-control regulation. We explicitly model the interaction between the outside innovator and the polluting firms, using a non-cooperative game-theoretical framework. We find that full and partial diffusion can both occur in equilibrium, depending on the relationship between environmental regulation stringency and cleanliness improvement of the new technology. Furthermore, we study the impacts of environmental regulation stringency and the improvement in cleanliness on the adoption and the diffusion of the emission reduction technology.

Appendix

1.1 Proofs

Proof of Lemma 1

The first order conditions for each green and brown producer are given by

$$\begin{aligned} \left\{ {\begin{array}{l} a-(m+1)q_g -(n-m)q_b =0 \\ a-mq_g -(n-m+1)q_b =0 \\ \end{array}} \right. \end{aligned}$$

Hence the best response functions:

$$\begin{aligned} q_g^{*} (q_b )= & {} \frac{a-(n-m)q_b }{m+1}\\ q_b^{*} (q_g )= & {} \frac{a-mq_g }{n-m+1} \end{aligned}$$

Solving first order conditions simultaneously and we have $q_g^{*} =q_b^{*} =a/{(n+1)}$. However $q_b^{*} =a/{(n+1)}$ is not feasible since the carbon cap $\kappa <a/{(n+1)}$ requires $q_b <a/{(n+1)}$.

Take the first order derivative of $q_b^{*} (q_g )$ on m:

$$\begin{aligned} \frac{\partial q_b^{*} (q_g )}{\partial m}=\frac{(n+1)}{(n-m+1)^{2}}(\frac{a}{n+1}-q_g ) \end{aligned}$$

Hence if $q_g \ge a/{(n+1)}$, ${\partial q_b^{*} (q_g )}/{\partial m}\le 0$, which means the brown producer’s quantity will decline with mincreases. Therefore, if $q_g \ge a/{(n+1)}$, we have

$$\begin{aligned} q_b^{*} (m;q_g )\ge q_b^{*} (n;q_g )=a-nq_g \ge \frac{a}{n+1} \end{aligned}$$

Hence with the constraint $q_b \le \kappa <a/{(n+1)}$, the optimal output for the brown producer is $q_b^{*} =\kappa$.

Now consider the case $q_g <a/{(n+1)}$. In this case, ${\partial q_b^{*} (q_g )}/{\partial m}>0$, which means the brown producer’s quantity will increase with s increases. Therefore, if $q_g <a/{(n+1)}$, we have

$$\begin{aligned} q_b^{*} (m;q_g )>q_b^{*} (1;q_g )=\frac{a-q_g }{n}>\frac{a}{n+1} \end{aligned}$$

Again with the constraint $q_b \le \kappa <a/{(n+1)}$, the optimal output for the brown producer is $q_b^{*} =\kappa$.

Therefore the optimal output for each brown producer is $q_b^{*} =\kappa$.

Next let us consider the green producers’ production quantity. Since $q_b^{*} =\kappa$, the green producer’s optimal output when there is no carbon emission constraint is

$$\begin{aligned} q_g^{*} (q_b =\kappa )=\frac{a-(n-m)\kappa }{m+1} \end{aligned}$$

And the emission constraint requires $q_g \le \kappa /{(1-g)}$. Hence the optimal output for each green producer is $q_g^{*} =\min \left\{ {\frac{a-(n-m)\kappa }{m+1},\frac{\kappa }{1-g}} \right\}$. $\square$

Proof of Corollary 1

Since $e\in (0,1)$, $\kappa /e>\kappa$ always holds. We only need to check if $\frac{a-(n-m)\kappa }{m+1}>\kappa$.

Since $\left[ {\frac{a-(n-m)\kappa }{m+1}} \right] ^{\prime }_m =\frac{n+1}{(m+1)^{2}}\left[ {\kappa -\frac{a}{n+1}} \right] <0$, we have

$$\begin{aligned} \left[ {\frac{a-(n-m)\kappa }{m+1}} \right]>\frac{a-(n-n)\kappa }{n+1}=\frac{a}{n+1}>\kappa \end{aligned}$$

$\square$

Proof of Lemma 2

In Case A1, from Lemma 1, each licensee’s equilibrium output level is $\kappa /e$, and each non-licensee’s output level is $\kappa$. Given fixed-fee contract F, the profit of a producer that accepts the contract to be a licensee is $\pi _g^{A1} (m,F)=\left[ {a-(n+{(1-e)m}/e)\kappa } \right] \kappa /e-F$, where m is the total number of licensees. The profit of a producer that rejects the contract is $\pi _b^{A1} (m)=\left[ {a-(n+{(1-e)m}/e)\kappa } \right] \kappa$. Hence, if the outside innovator wants m producers to accept its offer to be licensees, it is optimal for the innovator to set the upfront fee F high enough till $\pi _g^{A1} (m,F)=\pi _b^{A1} (m)$. Thus, in Case A1, given that the number of licensees is m, the optimal fixed fee set by the patent holder is

$$\begin{aligned} F^{A1}(m)=\left[ {a-\left( n+\frac{(1-e)m}{e}\right) \kappa } \right] \frac{(1-e)\kappa }{e}. \end{aligned}$$

In Case A2, there exists a threshold $\hat{{m}}\in [0,n]$; thus, the equilibrium output level of each green producer is $\kappa /e$ for $m\le \hat{{m}}$ and ${\left[ {a-(n-m)\kappa } \right] }/{(m+1)}$ for $m\ge \hat{{m}}$. First, we consider the case $m\le \hat{{m}}$. In this case, given fixed-fee contract F, the profit of a producer that accepts the contract to be a licensee is $\left. {\pi _g^{A2} (m,F)} \right| _{m\le \hat{{m}}} =\left[ {a-(n+{(1-e)m}/e)\kappa } \right] \kappa /e-F$. The profit of a producer that rejects the contract is $\left. {\pi _b^{A2} (m)} \right| _{m\le \hat{{m}}} =\left[ {a-(n+{(1-e)m}/e)\kappa } \right] \kappa$. Hence, if the outside innovator wants m producers to accept its offer to be licensees, it is optimal for her to set the upfront fee F high enough till $\left. {\pi _g^{A2} (m,F)} \right| _{m\le \hat{{m}}} =\left. {\pi _b^{A2} (m)} \right| _{m\le \hat{{m}}}$. Thus, in the case of $m\le \hat{{m}}$, the optimal fixed fee set by the patent holder is

$$\begin{aligned} F^{A2}(m)\left| {_{m\le \hat{{m}}} } \right. =\left[ {a-\left( n+\frac{(1-e)m}{e}\right) \kappa } \right] \frac{(1-e)\kappa }{e}. \end{aligned}$$

Now let us consider the case $m\ge \hat{{m}}$. In this case, given the fixed-fee contract F, the profit of a producer that accepts the contract to be a licensee is $\left. {\pi _g^{A2} (m,F)} \right| _{m\ge \hat{{m}}} =\left[ {{\left( {a-(n-m)\kappa } \right) }/{(m+1)}} \right] ^{2}-F$.

The profit of a producer that rejects the fixed-fee contract is $\left. {\pi _b^{A2} (m)} \right| _{m\ge \hat{{m}}} \!=\!\big [ {\left( {a\!-\!(n\!-\!m)\kappa } \right) }/{(m+1)} \big ]\kappa$. If the outside innovator wants m producers to accept the fixed-fee contract, it is optimal for her to set the upfront fee F high enough till $\left. {\pi _g^{A2} (m,F)} \right| _{m\ge \hat{{m}}} =\left. {\pi _b^{A2} (m)} \right| _{m\ge \hat{{m}}}$. Hence, in the case of $m\ge \hat{{m}}$, given that the number of licensees is m, the optimal fixed fee set by the patent holder is

$$\begin{aligned} F^{A2}(m)\left| {_{m\ge \hat{{m}}} } \right. =\frac{\left[ {a-(n-m)\kappa } \right] }{(m+1)}\frac{\left[ {a-(n+1)\kappa } \right] }{(m+1)}. \end{aligned}$$

In Case A3, the equilibrium output of each green producer is ${\left[ {a-(n-m)\kappa } \right] }/{(m+1)}$. Given the fixed-fee contract F, the profit of a producer that accepts the contract to be a licensee is $\pi _g^{A3} (m,F)=\left[ {{\left( {a-(n-m)\kappa } \right) }/{(m+1)}} \right] ^{2}-F$. The profit of a producer that rejects the contract is $\pi _b^{A3} (m)=\left[ {{\left( {a-(n-m)\kappa } \right) }/{(m+1)}} \right] \kappa$. If the outside innovator wants the number of licensees to be m, it is optimal for her to set the upfront fee F high enough till $\pi _g^{A3} (m,F)=\pi _b^{A3} (m)$. Hence, in Case A3, given that the number of licensees is m, the optimal fixed fee set by the patent holder is

$$\begin{aligned} F^{A3}(m)=\frac{\left[ {a-(n-m)\kappa } \right] }{(m+1)}\frac{\left[ {a-(n+1)\kappa } \right] }{(m+1)}. \end{aligned}$$

$\square$

Proof of Theorem 1

From Lemma 2, in Case A1 the optimal fixed fee is the equilibrium price $(\left[ {a-(n+{(1-e)m}/e)\kappa } \right] )$ multiplied by the output difference between each green producer and each brown producer $({(1-e)\kappa }/e=q_g^{*} -q_b^{*} )$. In this case, the increase in m does not diminish the green producers’ competitive advantage in quantity over the brown producers since $q_g^{*} -q_b^{*}$ is independent of m. However, with the increasing number of green producers, the total output will increase, and the equilibrium price for the product will decrease as a result. Thus, the optimal fixed fee set by the outside innovator decreases when the number of licensees increases, that is, ${\partial F^{A1}(m)}/{\partial m}<0$.

Next, let us consider the patent holder’s decision on the number of licensees (i.e., m), which can be formulated as follows:

$$\begin{aligned} \mathop {max}\limits _{m\le n} \;\;\Pi ^{A1}(m)=m\left[ {a-\left( n+\frac{(1-e)m}{e}\right) \kappa } \right] \frac{(1-e)\kappa }{e}, \end{aligned}$$

where $\Pi ^{A1}$ denotes the outside innovator’s profit in Region A1.

The first derivative of the function $\Pi ^{A1}(m)$ is

$$\begin{aligned} \frac{\partial \Pi ^{A1}(m)}{\partial m}=\frac{(1-e)\kappa \left[ {ae-\left( {2(1-e)m+en} \right) \kappa } \right] }{e^{2}} \end{aligned}$$

The second derivative of the function $\Pi ^{A1}(m)$ is

$$\begin{aligned} \frac{\partial ^{2}\Pi ^{A1}(m)}{\partial m^{2}}=-\frac{2(1-e)^{2}\kappa ^{2}}{e^{2}} \end{aligned}$$

Since the second derivative of the function $\Pi ^{A1}(m)$ is negative, using first order condition we have $m^{A1}=\frac{e\left( {a-n\kappa } \right) }{2(1-e)\kappa }$. And $m^{A1}\le n$ iff $e\le \frac{2n\kappa }{a+n\kappa }$. Hence, if $e\le \frac{2n\kappa }{a+n\kappa }$, the optimal number of licensees for the patent holder is $m^{A1{*}}=\frac{e\left( {a-n\kappa } \right) }{2(1-e)\kappa }$, otherwise $m^{A1{*}}=n$.

Let $\hat{{\kappa }}=\frac{(n-1)}{n}\bar{{\kappa }}$, we have $\hat{{\hat{{e}}}}(\kappa )=\kappa /{\bar{{\kappa }}}\ge \frac{2n\kappa }{a+n\kappa }$for $\kappa \ge \hat{{\kappa }}$. Hence, if $\kappa \ge \hat{{\kappa }}$, then $m^{A1{*}}=n$. If $\kappa \le \hat{{\kappa }}$, then

$$\begin{aligned} m^{A1{*}}=\left\{ {\begin{array}{l@{\quad }l} \frac{e\left( {a-n\kappa } \right) }{2(1-e)\kappa },&{} \textit{if}\;\hat{{\hat{{e}}}}(\kappa )\le e\le \frac{2n\kappa }{a+n\kappa } \\ n,&{} \textit{if}\;e\ge \frac{2n\kappa }{a+n\kappa } \\ \end{array}} \right. . \end{aligned}$$

$\square$

Proof of Theorem 2

In Case A2, for $m\le \hat{{m}}$, the optimal fixed fee set by the patent holder is

$$\begin{aligned} F^{A2}(m)\left| {_{m\le \hat{{m}}} } \right. =\left[ {a-\left( n+\frac{(1-e)m}{e}\right) \kappa } \right] \frac{(1-e)\kappa }{e} \end{aligned}$$

where ${(1-e)\kappa }/e=q_g^{*} -q_b^{*}$ is the output difference between each green producer and each brown producer, and $\left[ {a-(n+{(1-e)m}/e)\kappa } \right] =p^{{*}}$ is the equilibrium price for the product. Since the equilibrium price for the product decreases with more producers become licensees, the optimal fixed fee set by the outside innovator will decrease in m, that is, ${\partial F^{A1}(m)}/{\partial m}<0$.

The problem encountered by the outside innovator can be formulated as follows:

$$\begin{aligned} \mathop {max}\limits _{m\le \hat{{m}}} \;\;\Pi ^{A2}(m)=m\left[ {a-\left( n+\frac{(1-e)m}{e}\right) \kappa } \right] \frac{(1-e)\kappa }{e} \end{aligned}$$

where $\Pi ^{A2}$ represents the outside innovator’s profit in Case A2.

For $m\ge \hat{{m}}$, given that the number of licensees is m, the optimal fixed fee set by the patent holder is

$$\begin{aligned} F^{A2}(m)\left| {_{m\ge \hat{{m}}} } \right. =\frac{\left[ {a-(n-m)\kappa } \right] }{(m+1)}\frac{\left[ {a-(n+1)\kappa } \right] }{(m+1)} \end{aligned}$$

where ${\left[ {a-(n+1)\kappa } \right] }/{(m+1)}=q_g^{*} -q_b^{*}$ is the output difference between each green producer and each brown producer, and ${\left[ {a-(n-m)\kappa } \right] }/{(m+1)}=p^{{*}}$ is the equilibrium price for the product. In contrast to the case of $m\le \hat{{m}}$, in the case of $m\ge \hat{{m}}$, the output difference $q_g^{*} -q_b^{*}$ is decreasing with the increasing number of green producers. This implies that the increase in the number of licensees does diminish each green producer’ competitive advantage in quantity over each brown producer.

Thus, for $m\ge \hat{{m}}$, the problem faced by the outside innovator is

$$\begin{aligned} \mathop {max}\limits _{m\ge \hat{{m}}} \;\;\Pi ^{A2}(m)=m\frac{\left[ {a-(n-m)\kappa } \right] \left[ {a-(n+1)\kappa } \right] }{(m+1)^{2}}. \end{aligned}$$

In Case A2, from Lemma 1 the equilibrium output for each green producer is

$$\begin{aligned} q_g^{A2} =\left\{ {\begin{array}{l@{\quad }l} \frac{\kappa }{e},&{} \textit{if}\;\frac{(m+1)\kappa }{(n+1)\bar{{\kappa }}-(n-m)\kappa }\le e\le \hat{{\hat{{e}}}}(\kappa ) \\ \frac{a-(n-m)\kappa }{m+1},&{} \textit{if}\;\hat{{e}}(\kappa )\le e\le \frac{(m+1)\kappa }{(n+1)\bar{{\kappa }}-(n-m)\kappa } \\ \end{array}} \right. \end{aligned}$$

Equivalently, since $\hat{{e}}(\kappa )\le e\le \hat{{\hat{{e}}}}(\kappa )$, the equilibrium output for each green producer is

$$\begin{aligned} q_g^{A2} =\left\{ {\begin{array}{l@{\quad }l} \frac{\kappa }{e},&{} \textit{if}\;m\le \hat{{m}} \\ \frac{a-(n-m)\kappa }{m+1},&{} \textit{if}\;m\ge \hat{{m}} \\ \end{array}} \right. \end{aligned}$$

where $\hat{{m}}=\frac{(n+1)e\bar{{\kappa }}-(1+ne)\kappa }{(1-e)\kappa }$, and it is easily to verify that $0\le \hat{{m}}\le n$ in Case A2.

First let us consider the case $m\le \hat{{m}}$ and the resulting equilibrium output for each green producer is $q_g^{A2} \left| {_{m\le \hat{{m}}} } \right. =\kappa /e$. Given $m\le \hat{{m}}$, the optimal fixed fee for the patent holder is $F^{A2}(m)\left| {_{m\le \hat{{m}}} } \right. =\left[ {a-(n+\frac{(1-e)m}{e})\kappa } \right] \frac{(1-e)\kappa }{e}$ and the problem facing by the outside innovator is

$$\begin{aligned} \mathop {max}\limits _{m\le \hat{{m}}} \;\;\Pi ^{A2}(m)=m\left[ {a-\left( n+\frac{(1-e)m}{e}\right) \kappa } \right] \frac{(1-e)\kappa }{e} \end{aligned}$$

Apply a similar analysis as in the proof of theorem 1 we have the following results:

• (i) if $\kappa \ge \hat{{\kappa }}$, then $m^{A2{*}}\left| {_{m\le \hat{{m}}} } \right. =\hat{{m}}$.

• (ii) if $\kappa \le \hat{{\kappa }}$, then

  $$\begin{aligned} m^{A2{*}}\left| {_{m\le \hat{{m}}} } \right. =\left\{ {\begin{array}{l@{\quad }l} \frac{e\left( {a-n\kappa } \right) }{2(1-e)\kappa },&{} \textit{if}\frac{2\kappa }{a-n\kappa }\le e\le \hat{{\hat{{e}}}}(\kappa ) \\ \hat{{m}},&{} \textit{if}\;\hat{{e}}(\kappa )\le e\le \frac{2\kappa }{a-n\kappa } \\ \end{array}} \right. . \end{aligned}$$

Now let us consider the case $m\ge \hat{{m}}$ and the resulting equilibrium output for each green producer is $q_g^{A2} \left| {_{m\ge \hat{{m}}} } \right. =\frac{a-(n-m)\kappa }{m+1}$. Given $m\ge \hat{{m}}$, the optimal fixed fee set by the patent holder is $F^{A2}(m)\left| {_{m\ge \hat{{m}}} } \right. =\frac{\left[ {a-(n-m)\kappa } \right] \left[ {a-(n+1)\kappa } \right] }{(m+1)^{2}}$ and the problem facing by the outside innovator is

$$\begin{aligned} \mathop {max}\limits _{m\ge \hat{{m}}} \;\;\Pi ^{A2}(m)=m\frac{\left[ {a-(n-m)\kappa } \right] \left[ {a-(n+1)\kappa } \right] }{(m+1)^{2}} \end{aligned}$$

The results are following:

• (i) if $\kappa \ge \hat{{\kappa }}$, then $m^{A2{*}}\left| {_{m\ge \hat{{m}}} } \right. =n$.

• (ii) if $\kappa \le \hat{{\kappa }}$, then

  $$\begin{aligned} m^{A2{*}}\left| {_{m\ge \hat{{m}}} } \right. =\left\{ {\begin{array}{l@{\quad }l} \hat{{m}},&{} \textit{if}\frac{2\kappa }{a-n\kappa }\le e\le \hat{{\hat{{e}}}}(\kappa ) \\ \frac{a-n\kappa }{a-(2+n)\kappa },&{} \textit{if}\;\hat{{e}}(\kappa )\le e\le \frac{2\kappa }{a-n\kappa } \\ \end{array}} \right. . \end{aligned}$$

Based on the above analysis we have:

• (i) if $\kappa \ge \hat{{\kappa }}$, $m^{A2{*}}=n$.

• (ii) if $\kappa \le \hat{{\kappa }}$, then

$$\begin{aligned} m^{A2{*}}=\left\{ {\begin{array}{l@{\quad }l} \frac{e\left( {a-n\kappa } \right) }{2(1-e)\kappa },&{} \textit{if}\;{e}(\kappa )\le e\le \hat{{\hat{{e}}}}(\kappa ) \\ \frac{a-n\kappa }{a-(2+n)\kappa },&{} \textit{if}\;\hat{{e}}(\kappa )\le e\le {e}(\kappa ) \\ \end{array}} \right. . \end{aligned}$$

where${e}(\kappa )=\frac{2\kappa }{a-n\kappa }$. $\square$

Proof of Theorem 3

From Lemma 2, in Case A3, given that the number of licensees is m, the optimal fixed fee set by the patent holder is

$$\begin{aligned} F^{A3}(m)=\frac{\left[ {a-(n-m)\kappa } \right] }{(m+1)}\frac{\left[ {a-(n+1)\kappa } \right] }{(m+1)} \end{aligned}$$

where ${\left[ {a-(n+1)\kappa } \right] }/{(m+1)}=q_g^{*} -q_b^{*}$ is the output difference between each green producer and each brown producer, and ${\left[ {a-(n-m)\kappa } \right] }/{(m+1)}=p^{{*}}$ is the equilibrium price for the product. In this region, the output difference $q_g^{*} -q_b^{*}$ and the equilibrium price $p^{{*}}$ are both decreasing with the increasing number of green producers. Hence, the optimal fixed fee set by the outside innovator will decrease in m, that is, ${\partial F^{A3}(m)}/{\partial m}<0$.

The problem encountered by the outside innovator can be formulated as follows:

$$\begin{aligned} \mathop {max}\limits _m \;\;\Pi ^{A3}(m)=m\frac{\left[ {a-(n-m)\kappa } \right] \left[ {a-(n+1)\kappa } \right] }{(m+1)^{2}} \end{aligned}$$

Let $\Lambda (m)=\frac{m\left[ {a-(n-m)\kappa } \right] }{(m+1)^{2}}$, to maximize $\Pi ^{A3}(m)$ equals to maximize $\Lambda (m)$. The first derivative of the function $\Lambda (m)$ is

$$\begin{aligned} \Lambda '(m)=\frac{a(1-m)+\kappa \left[ {m\left( {2+n} \right) -n} \right] }{\left( {1+m} \right) ^{3}} \end{aligned}$$

Using first order condition we have

$$\begin{aligned} m^{A3}=\frac{a-n\kappa }{a-(2+n)\kappa } \end{aligned}$$

The second derivative of the function $\Lambda (m)$ at $m^{A3}$ is

$$\begin{aligned} {\Lambda }''(m^{A3})=-\frac{\left( {a-\kappa \left( {2+n} \right) } \right) ^{4}}{8\left( {a-\kappa \left( {1+n} \right) } \right) ^{3}}<0 \end{aligned}$$

which means $\Lambda (m)$ is concave at $m^{A3}$.

Firstly, if $\kappa \ge \frac{a}{2+n}$, then ${\Lambda }'(m)=\frac{a(1-m)+\kappa \left[ {m\left( {2+n} \right) -n} \right] }{\left( {1+m} \right) ^{3}}\ge \frac{2a}{\left( {1+m} \right) ^{3}\left( {2+n} \right) }>0$, which means the patent holder’s profit is increasing with the number of licensee increases. Hence the optimal number of licensees for the outside innovator is $m^{A3{*}}=n$. Secondly, if $\kappa <\frac{a}{2+n}$, then we have$m^{A3{*}}=\frac{a-n\kappa }{a-(2+n)\kappa }>0$. We know that $m^{A3}=\frac{a-n\kappa }{a-(2+n)\kappa }\le n$ iff $\kappa \le \hat{{\kappa }}$. And $\hat{{\kappa }}<\frac{a}{2+n}$ holds for sure. Hence in Region A3, the optimal number of licensees for the patent holder is

$$\begin{aligned} m^{A3{*}}=\left\{ {\begin{array}{l@{\quad }l} \frac{a-n\kappa }{a-(2+n)\kappa },&{} \textit{if}\;\kappa \le \hat{{\kappa }} \\ n,&{} { otherwise} \\ \end{array}} \right. . \end{aligned}$$

$\square$