Liberalizing trade in environmental goods and services

Abstract

We examine the effects of trade liberalization in environmental goods in a model with one domestic downstream polluting firm and two upstream firms (one domestic, one foreign). The upstream firms offer their technologies to the downstream firm at a flat fee. The domestic government sets the emission tax rate after the outcome of R&D is known. The effect of liberalization on the domestic upstream firm’s R&D incentive is ambiguous. Liberalization usually results in cleaner production, which allows the country to reach higher welfare. However, this increase in welfare is typically achieved at the expense of the environment (a backfire effect).

Appendices

Appendix 1: Conditions for $q_{2}^{s}>0$

Autarky. $q_{d}^{id}$ in (19) is decreasing in $\lambda$ and has an interior minimum in $e_{i}\in \left[ 1/\sqrt{\lambda };1\right]$ given $\lambda$. To make sure that $q_{d}^{id}>0$ for all $e_{i}\in \left[ 1/\sqrt{\lambda };1\right] ,$ we calculate the $\lambda$ where the minimum equals zero. Setting $q_{d}^{id}=0$ and $dq_{d}^{id}/de_{i}=0$ in (19) yields, respectively:

$$\begin{aligned} \frac{\lambda e_{i}^{3}-\lambda e_{i}^{2}+e_{i}+1}{e_{i}(\lambda e_{i}^{2}+1) }= & {} 0 \\ -\lambda ^{2}e_{i}^{4}+4\lambda e_{i}^{2}+1= & {} 0 \end{aligned}$$

The only positive solution for $\lambda$ and $e_{i}$ is $\lambda =\frac{5}{2 }\sqrt{5}+\frac{11}{2}.$ Therefore, $q_{d}^{id}>0$ for all $e_{i}\in \left[ 1/ \sqrt{\lambda };1\right]$ if and only if:

$$\begin{aligned} \lambda <\frac{5}{2}\sqrt{5}+\frac{11}{2}\approx 11.09 \end{aligned}$$

(44)

Free trade. Comparing (19) and (24), we see that $q_{f}^{nf}>q_{d}^{nd}$ by (18). Thus, condition (44) that ensures $q_{d}^{nd}>0$ is also sufficient for $q_{f}^{nf}>0.$

Output $q_{h}^{jh},\ j=f,n,$ in (29) is positive for all values of $e_{j}$ for which the second-order condition holds (which implies that the denominator on the RHS of (29) is positive) if and only if:

$$\begin{aligned} \lim _{e_{j}\downarrow \hat{e}_{j}}\frac{e_{j}\left( 3e_{j}-e_{h}+e_{j}^{3}\lambda -\lambda e_{j}^{2}e_{h}\right) }{2\left( \lambda e_{j}^{4}+3e_{j}^{2}-2e_{h}^{2}\right) }=+\infty \end{aligned}$$

(45)

where $\hat{e}_{j}$ as a function of $e_{h}$ and $\lambda$ is implicitly defined by:

$$\begin{aligned} \lambda e_{j}^{4}+3e_{j}^{2}-2e_{h}^{2}=0 \end{aligned}$$

(46)

The point where the LHS of (45) switches from $+\infty$ to $-\infty$ is where

$$\begin{aligned} 3e_{j}-e_{h}+e_{j}^{3}\lambda -\lambda e_{j}^{2}e_{h}=0 \end{aligned}$$

(47)

and (46) holds. Solving (46) and (47) simultaneously for $\lambda$ and $e_{j},$ we find that the only positive real solution features $\lambda =\frac{1}{2e_{h}^{2}}\left( 3\sqrt{5} +5\right) .$ Then, $q_{h}^{jh}>0$ for all $e_{j}$ if and only if:

$$\begin{aligned} \lambda <\frac{3\sqrt{5}+5}{2e_{h}^{2}}\approx \frac{5.8541}{e_{h}^{2}} \end{aligned}$$

(48)

Appendix 2: The licence fee

In Sect. 4, we introduced the restriction that the licence fee should be decreasing in $e_{1}.$ In this appendix, we discuss the conditions under which this is the case.¹⁵

Fig. 1

The domestic firm’s licence fee $F_{h}^{id}$ under autarky for $\tilde{\alpha }=1$ when the domestic firm has technology $e_{i},\ i=h,n.$

1.1 Autarky

Figure 1 shows the licence fee $F_{h}^{id}$ (given by 20) as a function of $e_{i}$ for different values of $\lambda$ with $\tilde{\alpha }=1$. The condition $dF_{h}^{id}/de_{i}<0$ is binding for $i=n,$ because it is clear from Fig.1 that when $dF_{h}^{nd}/de_{n}<0,$ then $dF_{h}^{hd}/de_{h}<0$ as well, since $e_{h}>e_{n}$. Thus, $e_{n}$ should exceed $\bar{e}_{n},$ where $\bar{e}_{n}$ is defined implicitly by:

$$\begin{aligned} dF_{h}^{nd}(\bar{e}_{n})/de_{n}=0 \end{aligned}$$

(49)

1.2 Free trade

Domestic firm has found the new technology. Comparing $dF_{h}^{nf}/de_{n}$ in (25) to $dF_{h}^{nd}/de_{n}$ in (20) with $i=n,$ we see that qualitatively the only difference lies in the less efficient technology 2 which has $e_{f}<1$ in scenario nf and $e=1$ in nd. At $\bar{e}_{n}$ as defined by (49) we must have $dt^{nd}/de_{n}>0$ by (13). Then, since emissions with the less efficient technology $E_{2}$ are lower in scenario nf than in nd,  $dF_{h}^{nf}(\bar{e}_{n})/de_{n}<0$ and $dF_{h}^{nf}/de_{n}=0$ occurs at an $e_{n}<\bar{e}_{n}$.

Domestic firm has not found the new technology. It can be shown that $F_{f}^{jh}$ in (30)$,j=n,f,$ is first increasing and then decreasing in $e_{j}.$ Then, the condition $dF_{f}^{jh}/de_{j}<0$ is binding for $j=n,$ since when $dF_{f}^{nh}/de_{n}<0,$ then $dF_{h}^{fh}/de_{f}<0$ as well, since $e_{f}>e_{n}$. Thus, $e_{n}$ should exceed $\tilde{e}_{n},$ where $\tilde{e}_{n}$ is defined implicitly by:

$$\begin{aligned} dF_{f}^{nh}(\tilde{e}_{n},e_{h})/de_{n}=0 \end{aligned}$$

(50)

It can be shown that $\tilde{e}_{n}(e_{h})$ is an increasing function of $e_{h}$.

Table 2 Minimum values of $e_n$ from (11)

1.3 Conclusion

We have found two minimum values of $e_{n}$: $\bar{e}_{n}$ in (49) does not depend on $e_{h},$ while $\tilde{e}_{n}$ in (50) is increasing in $e_{h}.$ This means that for low values of $e_{h},$ the binding constraint is $e_{n}>\bar{e}_{n},$ while for higher values of $e_{h}$ it is $e_{n}>\tilde{e}_{n}.$ Table 2 shows how the minimum $e_{n}$ value changes with $e_{h}$ for selected values of $\lambda .$ With $\lambda =3,$ for instance, $\bar{e}_{n}=0.708$ while $\tilde{e}_{n}=0.708$ for $e_{h}=0.779.$ Thus, for $0.708<e_{h}<0.779,$ the binding constraint is $e_{n}> \bar{e}_{n}=0.708.$ For $e_{h}>0.779,$ the binding constraint is $e_{n}> \tilde{e}_{n},$ with $\tilde{e}_{n}$ increasing in $e_{h}.$ For the maximum value of one for $e_{h},$ $\tilde{e}_{n}=0.807.$ For the $\lambda$ values of 3 and 5, the maximum value of $e_{h}$ is one, whereas for higher $\lambda$’s it is constrained by (48).

Appendix 3: Proofs

1.1 Proof of Proposition 1

Let us first collect the expressions for welfare. Substituting (17) and (19) into (16) yields welfare in scenario $id,\ i=h,n$:

$$\begin{aligned} W^{id}=\frac{1}{2\left( \lambda e_{i}^{2}+1\right) }-K_{i} \end{aligned}$$

(51)

Substituting (22) and (23) into (21) gives welfare in scenarios nn and nf as:

$$\begin{aligned} W^{nn}=W^{nf}=\frac{1}{2\left( \lambda e_{n}^{2}+1\right) }-K_{n} \end{aligned}$$

(52)

Substituting (27) and (29) into (26) gives welfare in scenario $jh,\ j=f,n,$ as:

$$\begin{aligned} W^{jh}=\frac{\lambda e_{j}^{4}-2\lambda e_{h}e_{j}^{3}+\lambda e_{h}^{2}e_{j}^{2}+5e_{j}^{2}-2e_{h}e_{j}-e_{h}^{2}}{4\left( \lambda e_{j}^{4}+3e_{j}^{2}-2e_{h}^{2}\right) }-K_{h} \end{aligned}$$

(53)

Before proving the Proposition, we first establish the following two lemmas:

Lemma 1

When the domestic firm has not found the new technology, welfare is higher with free trade than under autarky:$\ W^{jh}>W^{hd}$ with $j=f,n.$

Proof

From (51) with $i=h$ and (53),  it is clear that $W^{hd}=W^{jh}$ for $e_{j}=e_{h}.$ From (53):

$$\begin{aligned} \frac{dW^{jh}}{de_{j}}=\frac{-7e_{j}e_{h}^{2}+2e_{j}^{3}+3e_{j}^{2}e_{h}-2 \lambda e_{j}^{5}-2\lambda e_{j}e_{h}^{4}+6\lambda e_{j}^{2}e_{h}^{3}-2\lambda e_{j}^{3}e_{h}^{2}+\lambda ^{2}e_{j}^{6}e_{h}-\lambda ^{2}e_{j}^{5}e_{h}^{2}}{2\left( 3e_{j}^{2}-2e_{h}^{2}+\lambda e_{j}^{4}\right) ^{2}} \end{aligned}$$

(54)

The sign of $dW^{jh}/de_{j}$ in (54) is the sign of the numerator on the RHS. Defining $a\equiv e_{j}/e_{h},\ b\equiv \lambda e_{j}^{2},$ the sign of the numerator is the sign of:

$$\begin{aligned} \Phi =-7a^{2}+2a^{4}+3a^{3}-2ba^{4}-2b+6ba-2ba^{2}+b^{2}a^{3}-b^{2}a^{2} \end{aligned}$$

(55)

$\Phi$ has a maximum in b for:

$$\begin{aligned} b=b^{*}\equiv \frac{3a-a^{3}-a^{2}-1}{a^{2}(1-a)} \end{aligned}$$

(56)

$b^{*}$ is positive for$\ a\in (\bar{a};1],$ with $\bar{a}\approx 0.414.$ For $a\in \left[ 0;\bar{a}\right] ,$ $\Phi$ reaches its maximum at $b=0$, which from (55) is clearly negative.

Substituting $b=b^{*}$ from (56) into (55), we find the maximum possible value of $\Phi$ given $a\in (0.414;1]$:

$$\begin{aligned} \Phi ^{*}=\frac{1-4a^{4}+6a^{2}-5a}{a^{2}} \end{aligned}$$

Plotting this expression shows that $\Phi ^{*}<0$ for all $a\in (0.414;1]$. Thus, $\Phi <0$ in (55) for all feasible values of a and b,  which means that $dW^{jh}/de_{j}<0$ in (54). This combined with $W^{hd}=W^{jh}$ for $e_{j}=e_{h}$ proves the lemma. $\square$

Lemma 2

In scenario nf with free trade, welfare $W^{nf}$ net of the domestic upstream firm’s net revenue $R_{h}^{nf}$ exceeds welfare $W^{hd}$ in scenario hd under autarky: $W^{nf}-R_{h}^{nf}>W^{hd}.$

Proof

From (52) and (25):

$$\begin{aligned} W^{nf}-R_{h}^{nf}=\frac{1}{2}\frac{\lambda e_{n}^{2}-1}{\left( \lambda e_{n}^{2}+1\right) ^{2}}+\frac{\left( \lambda e_{n}^{3}-\lambda e_{f}e_{n}^{2}+e_{n}+e_{f}\right) ^{2}}{4e_{n}^{2}\left( \lambda e_{n}^{2}+1\right) ^{2}}-K_{n} \end{aligned}$$

(57)

Differentiating (57) with respect to $e_{n}$, we obtain:

$$\begin{aligned} \frac{d\left( W^{nf}-R_{h}^{nf}\right) }{de_{n}}=\frac{\Omega }{ 2e_{n}^{3}\left( \lambda e_{n}^{2}+1\right) ^{3}} \end{aligned}$$

(58)

with

$$\begin{aligned} \Omega \equiv 2a^{2}b\left( 3-b\right) +a\left( b+1\right) \left( b^{2}-4b-1\right) -\left( b-1\right) \left( b^{2}-4b-1\right) \end{aligned}$$

(59)

where $a\equiv e_{n}/e_{f},\ b\equiv \lambda e_{n}^{2}.$ Note that $b<\frac{5 }{2}+\frac{3}{2}\sqrt{5}$ by (48).

The sign of the RHS of (58) is the sign of $\Omega$ which is quadratic in a with a maximum (minimum) for $b>(<)3.$ The highest value of $\Omega$ is then at $\partial \Omega /\partial a=0$ for $b>3$ (if this is an internal maximum) and at either the highest or lowest value of a for $b\le 3$. The highest value of a is 1, for which $\Omega =-2(b+1)<0.$ The lowest value for a is where $dF_{h}^{nf}/de_{n}=0$ from (25). Substituting this into (59), we find $\Omega =-2a^{2}b\left( b+1\right) <0.$ For $b>3,$ the maximum value of $\Omega$ in (59) occurs at:

$$\begin{aligned} a=a^{*}\equiv \frac{\left( b+1\right) \left( b^{2}-4b-1\right) }{4b(b-3)} \end{aligned}$$

Substituting this into (59), the highest possible value of $\Omega$ is:

$$\begin{aligned} \Omega ^{*}=\left( b^{2}-4b-1\right) \left( b^{4}-10b^{3}+24b^{2}-30b-1\right) \end{aligned}$$

We see that $a^{*}>0$ and $\Omega ^{*}<0$ for $b\in \left( 3;2+\sqrt{ 5}\right)$ and $a^{*}<0$ and $\Omega ^{*}>0$ for $b\in \left( 2+ \sqrt{5};\frac{5}{2}+\frac{3}{2}\sqrt{5}\right) .$ Thus, for all values of b for which there is potentially an interior maximum ($a^{*}>0$), $\Omega ^{*}$ is negative. We conclude that $\Omega$ is negative so that the RHS of (58) is negative. The lowest possible value of $(W^{nf}-F_{h}^{nf})$ is thus achieved at the maximum value of $e_{n},$ which is $e_{f}.$ Setting $e_{n}=e_{f}$ in (57), we find from (51):

$$\begin{aligned} W^{nf}-R_{h}^{nf}\ge \frac{1}{2\left( \lambda e_{f}^{2}+1\right) }-K_{n}> \frac{1}{2\left( \lambda e_{h}^{2}+1\right) }-K_{h}=W^{h0} \end{aligned}$$

The inequality follows from (4) and $e_{f}<e_{h}.$ $\square$

We will now prove Proposition 1 by examining each possible combination of R&D decisions in turn.¹⁶

1.1.1 No R&D in autarky; (No R&D, No R&D) with trade

In autarky, welfare is $W^{hd}.$ With trade, welfare is $W^{fh}.$ By Lemma 1, $W^{fh}>W^{hd}.$

1.1.2 No R&D in autarky; (No R&D, R&D) with trade

In autarky, welfare is $W^{hd}.$ With trade, welfare is $W^{nh}$ if the foreign firm’s R&D is successful and $W^{fh}$ if it is not. By Lemma 1, $W^{jh}>W^{hd},\ j=n,f.$

1.1.3 No R&D in autarky; (R&D, R&D) with trade

In autarky, welfare is $W^{hd}.$ With trade, welfare is $W^{nn}-C^{h}=W^{nf}-C^{h}$ if the domestic firm’s R&D is successful and $W^{jh}-R,j=f,n,$ if it is not. Thus, we have:¹⁷

$$\begin{aligned} W^{RR}-W^{N}= & {} p^{h}W^{nf}+\left( 1-p^{h}\right) W^{jh}-W^{hd}-C^{h}> \\> & {} p^{h}\left( W^{nf}-R_{h}^{nf}-W^{hd}\right) +\left( 1-p^{h}\right) \left[ W^{jh}-W^{hd}\right] >0 \end{aligned}$$

The first inequality follows from $C^{h}<C_{2}^{h}$ in (R&D, R&D), with $C_{2}^{h}$ given by (36). The second inequality follows from Lemmas 1 and 2.

1.1.4 R&D in autarky; (No R&D, No R&D) with trade

In autarky, welfare is $W^{nd}-C^{h}$ if R&D by the domestic firm is successful and $W^{hd}-C^{h}$ if it is not. With trade, welfare is $W^{fh}.$ Thus:

$$\begin{aligned} W^{NN}-W^{R}=W^{fh}-p^{h}W^{nd}-\left( 1-p^{h}\right) W^{hd}+C^{h} \end{aligned}$$

Solving for $p^{h}$, we see that expected welfare under free trade is higher than under autarky if and only if inequality (40) holds.

1.1.5 R&D in autarky; (No R&D, R&D) with trade

In autarky, welfare is $W^{nd}-C^{h}$ if R&D by the domestic firm is successful and $W^{hd}-C^{h}$ if it is not. With trade, welfare is $W^{nh}$ if the foreign firm’s R&D is successful and $W^{fh}$ if it is not. Thus:

$$\begin{aligned} W^{NR}-W^{R}=p^{f}W^{nh}+\left( 1-p^{f}\right) W^{fh}-\left[ p^{h}W^{nd}+\left( 1-p^{h}\right) W^{hd}\right] +C^{h} \end{aligned}$$

The RHS is positive if and only if (43) holds.

1.1.6 R&D in autarky; (R&D, R&D) with trade

In autarky, welfare is $W^{nd}-C^{h}$ if R&D by the domestic firm is successful and $W^{hd}-C^{h}$ if it is not. With trade, welfare is $W^{nf}-C^{h}=W^{nn}-C^{h}=W^{nd}-C^{h}$ if the domestic firm’s R&D is successful and $W^{jh}-C^{h},j=f,n,$ if it is not. Thus, we have:

$$\begin{aligned} W^{RR}-W^{R}=(1-p)\left[ W^{jh}-W^{hd}\right] >0 \end{aligned}$$

The inequality follows from Lemma 1.

1.2 Proof of Proposition 2

Let us first collect the expressions for emissions. Emissions in each scenario are given by $e_{1}q_{1}.$ Thus, in scenario $id,\ i=h,n,$ we have from (19):

$$\begin{aligned} E^{id}=\frac{e_{i}}{\lambda e_{i}^{2}+1} \end{aligned}$$

(60)

In scenarios nf and nn,  emissions are, from (23):

$$\begin{aligned} E^{nf}=E^{nn}=\frac{e_{n}}{\lambda e_{n}^{2}+1} \end{aligned}$$

(61)

In scenario $jh,\ j=f,n,$ emissions are, from (29):

$$\begin{aligned} E^{jh}=\frac{e_{j}\left( e_{j}e_{h}+e_{j}^{2}-e_{h}^{2}\right) }{\lambda e_{j}^{4}+3e_{j}^{2}-2e_{h}^{2}} \end{aligned}$$

(62)

Before turning to the Proposition, we first establish:

Lemma 3

When the domestic firm has not found the new technology, emissions are higher with free trade than under autarky:$\ E^{jh}>E^{hd}$ with $j=f,n.$

Proof

From (60) and (62), it is clear that $E^{jh}=E^{hd}$ for $e_{j}=e_{h}.$ From (62):

$$\begin{aligned} \frac{dE^{jh}}{de_{j}}=\frac{-\lambda e_{j}^{6}-2\lambda e_{j}^{5}e_{h}+3\lambda e_{j}^{4}e_{h}^{2}+3e_{j}^{4}-3e_{j}^{2}e_{h}^{2}-4e_{j}e_{h}^{3}+2e_{h}^{4} }{\left( \lambda e_{j}^{4}+3e_{j}^{2}-2e_{h}^{2}\right) ^{2}} \end{aligned}$$

Setting $e_{j}=e_{h}$ yields:

$$\begin{aligned} \left. \frac{dE^{jh}}{de_{j}}\right| _{e_{j}=e_{h}}=\frac{-2e_{h}^{4}}{ \left( \lambda e_{h}^{4}+e_{h}^{2}\right) ^{2}}<0 \end{aligned}$$

Thus, when reducing $e_{j}$ below $e_{h},$ $E^{jh}$ initially rises above $E^{hd}.$ However, for lower values of $e_{j}$, $E^{jh}$ may decline again.

Defining $a\equiv e_{j}/e_{h},\ b\equiv \lambda e_{h}^{2},$ we can write ( 62) as:

$$\begin{aligned} E^{jh}=\frac{e_{j}(a^{2}+a-1)}{ba^{4}+3a^{2}-2} \end{aligned}$$

so that

$$\begin{aligned} E^{jh}-E^{hd}=e_{h}\left[ \frac{(a^{3}+a^{2}-a)}{ba^{4}+3a^{2}-2}-\frac{1}{ b+1}\right] =\frac{e_{h}\left( a^{2}-1\right) \left( a-a^{2}b+ab-2\right) }{ \left( b+1\right) \left( ba^{4}+3a^{2}-2\right) } \end{aligned}$$

The (potentially) positive solutions for $E^{jh}=E^{hd}$ are $e_{j}=e_{h}$ and

$$\begin{aligned} a=\frac{1+b\pm \sqrt{b^{2}-6b+1}}{2} \end{aligned}$$

(63)

There are only real solutions for a when $b^{2}-6b+1\ge 0,$ which is satisfied for $b\le 3-2\sqrt{2}$ and $b\ge 3+2\sqrt{2}.$ The first inequality is irrelevant by (18). In case the second inequality holds, the highest possible value for a is for the maximum value of b given by (48), combined with the “+” sign on the RHS of (63), so that:

$$\begin{aligned} a=\frac{1}{3\sqrt{5}+5}\left( \frac{3}{2}\sqrt{5}+\frac{7}{2}+\sqrt{\left( \frac{3}{2}\sqrt{5}+\frac{5}{2}\right) ^{2}-9\sqrt{5}-14}\right) \approx 0.61834 \end{aligned}$$

(64)

Note that (28) can be written as $ba^{3}+a-2>0.$ Substituting a from (64) and $b=\frac{5}{2}+\frac{3}{2}\sqrt{5}$ from (48), we find $ba^{3}+a-2=0,$ so that (28) is violated. Thus, $E^{jh}=E^{hd}$ cannot hold and pollution is higher with trade than under autarky. $\square$

We will now prove Proposition 2 by examining each possible combination of R&D decisions in turn.¹⁸

1.2.1 No R&D in autarky; (No R&D, No R&D) with trade

In autarky, emissions are $E^{hd}$. With trade, emissions are $E^{fh}.$ By Lemma 3, $E^{fh}>E^{hd}.$

1.2.2 No R&D in autarky; (No R&D, R&D) with trade

In autarky, emissions are $E^{hd}$. With trade, emissions are $E^{nh}$ if the foreign firm’s R&D is successful and $E^{fh}$ if it is not. By Lemma 3, $E^{jh}>E^{hd},\ j=n,f.$

1.2.3 No R&D in autarky; (R&D, R&D) with trade

In autarky, emissions are $E^{hd}$. With trade, emissions are $E^{nn}=E^{nf}$ if the domestic firm’s R&D is successful and $E^{jh},j=f,n,$ if it is not. We know from Sect. 1 that $E^{nn}=E^{nf}>E^{hd}$ and from Lemma 3 that $E^{jh}>E^{hd}$ with $j=f,n.$

1.2.4 R&D in autarky; (No R&D, No R&D) with trade

In autarky, emissions are $E^{nd}$ if R&D is successful and $E^{hd}$ if it is not. With trade, emissions are $E^{fh}$ with $j=f.$ Thus:

$$\begin{aligned} D^{NN}-D^{R}=\frac{1}{2}\lambda (E^{fh})^{2}-\frac{1}{2}\lambda \left[ p^{h}\left( E^{nd}\right) ^{2}+\left( 1-p^{h}\right) \left( E^{hd}\right) ^{2}\right] \end{aligned}$$

Solving for $p^{h},$ we see that the expected pollution damage under free trade is greater than under autarky if and only if (42) holds.

1.2.5 R&D in autarky; (No R&D, R&D) with trade

In autarky, emissions are $E^{nd}$ if R&D is successful and $E^{hd}$ if it is not. With trade, emissions are $E^{nh}$ if the foreign firm’s R&D is successful and $E^{fh}$ if it is not. Thus, we have:

$$\begin{aligned} D^{NR}-D^{R}= & {} \frac{1}{2}\lambda \left[ p^{f}(E^{nh})^{2}+(1-p^{f})\left( E^{fh}\right) ^{2}-p^{h}\left( E^{nd}\right) ^{2}-(1-p^{h})(E^{hd})^{2} \right] \\= & {} \frac{1}{2}\lambda \left[ p^{h}\left[ (E^{nh})^{2}-\left( E^{nd}\right) ^{2}\right] +(1-p^{f})\left[ \left( E^{fh}\right) ^{2}-(E^{hd})^{2}\right] +(p^{f}-p^{h})\left[ (E^{nh})^{2}-(E^{hd})^{2}\right] \right] \end{aligned}$$

By Lemma 3, a sufficient condition for $D^{NR}>D^{R}$ is (41).

1.2.6 R&D in autarky; (R&D, R&D) with trade

In autarky, emissions are $E^{nd}$ if R&D is successful and $E^{hd}$ if it is not. With trade, emissions are $E^{nn}=E^{nf}=E^{nd}$ if the domestic firm’s R&D is successful and $E^{jh},j=f,n,$ if it is not. Thus, we have:

$$\begin{aligned} D^{RR}-D^{R} \\= & {} \frac{1}{2}\lambda \left[ p^{h}\left( E^{nd}\right) ^{2}+p^{f}\left( 1-p^{h}\right) \left( E^{nh}\right) ^{2}+\left( 1-p^{h}\right) (1-p^{f})\left( E^{fh}\right) ^{2} -p^{h}\left( E^{nd}\right) ^{2}-\left( 1-p^{h}\right) \left( E^{hd}\right) ^{2}\right] \\= & {} \frac{1}{2}\lambda (1-p^{h})\left[ p^{f}\left( E^{nh}\right) ^{2}+\left( 1-p^{f}\right) \left( E^{fh}\right) ^{2}-\left( E^{hd}\right) ^{2}\right] >0 \end{aligned}$$

The inequality follows from Lemma 3.