Lobbying for and Against Subsidizing Green Energy

Abstract

We consider a small open economy that operates a carbon emission trading scheme and subsidizes green energy. Taking cap-and-trade as given, we seek to explain the subsidy as the outcome of a trilateral tug of war between the green lobby, the brown lobby and the consumer lobby. With parametric functions we fully solve the competitive economic equilibrium and the lobbying Nash equilibrium. The rate of the green subsidy results from complementary or opposing political pressures of the three interest groups. If the brown lobby is stronger than the green one, our main results are (i) that the outcome of the three-party lobbying game is a green tax, if preferences are not green, and (ii) that green consumer preferences are necessary but not sufficient for generating a green subsidy in the lobbying game.

Appendices

Appendix 1: Social Optimum

For an interior solution, the FOCs of solving (7) are:

$$\begin{aligned} \begin{array}{lll} {\mathcal L}_e = \lambda _b B'(e) - \lambda _e - \lambda _t p_e =0, &{} {\mathcal L}_b = \lambda _z - \lambda _b =0, &{} {\mathcal L}_x = - \lambda _x + \lambda _t =0, \\ {\mathcal L}_g =n \gamma + \lambda _z - \lambda _g =0, &{} {\mathcal L}_{x_d} =n - n \lambda _t =0, &{} {\mathcal L}_z = -\lambda _z + \lambda _\psi =0, \\ {\mathcal L}_{k_g} = \lambda _g G'(k_g) - \lambda _k =0, &{} {\mathcal L}_{z_d} = n V'(z_d) - n \lambda _\psi =0, &{} {\mathcal L}_{k_x} = \lambda _x - \lambda _k =0. \\ \end{array}\quad \end{aligned}$$

(39)

These FOCs imply $\lambda _k= \lambda _t = \lambda _x=1$, $\lambda _b = \lambda _z$, $\lambda _g = \lambda _z + n \gamma$ and give rise to the equations $\lambda _z = V'(z_d)$, $(\lambda _z + n \gamma ) G'(k_g)=1$ and $\lambda _z B'(e) = \lambda _e + p_e$, which imply (8), in turn.

Appendix 2: Proofs

1.1 Proof of Proposition 2

• (i) From profit maximization of the green industry [$\pi _g'=0$, see also (11)], follows

  $$\begin{aligned} k_g = (p_z+s)^2 \mu , \quad g = 2 \mu (p_z+s) \quad \text{ and } \quad \pi _g= \mu (p_z+s)^2. \end{aligned}$$

  (40)

  Inserting $z_{d}= c - p_z$ and $g$ from (40) in the equilibrium condition of the domestic energy market $n z_d = \bar{b} +g$ we obtain after rearrangement of terms

  $$\begin{aligned} \hat{p}_z = \bar{\kappa } - (1 - \kappa ) s \quad \text{ and } \quad \hat{p}_z + s = \bar{\kappa } + \kappa s, \end{aligned}$$

  (41)

  where $\kappa := \frac{n}{n+2\mu } \in ]0, 1[, \bar{\kappa } := \kappa (c - \frac{\bar{b} }{n} )>0$ and $\bar{b}:= \bar{e}^\beta$. The ’hat’ indicates that $\hat{p}_z$ is the equilibrium price. Observe that $\bar{\kappa } = c - \frac{\bar{b} }{n} = \hat{p}_z + \frac{\hat{g} }{n}>0$. In view of (40) and (41) we obtain the equilibrium profit (14).

• (ii) From profit maximization of the brown industry [$\pi _b'=0$, see also (10)] follows

  $$\begin{aligned} t = \beta p_z \bar{e}^{\beta - 1} - p_e>0 \end{aligned}$$

  (42)

  taking into account that the emission cap $\bar{e}$ is strictly binding. Hence for given $p_z$ the maximum profit is

  $$\begin{aligned} \pi _b= (1-\beta ) \bar{b} p_z. \end{aligned}$$

  (43)

  In view of (43) and (41) we obtain the equilibrium profit (15).

• (iii) Next, we consider the utility of consumer group $x$:

  $$\begin{aligned} \hat{u}_x = c \hat{z}_d - \frac{\hat{z}_d^2}{2} + \hat{y}_x -p_z \hat{z}_d + \gamma \hat{g} = \left[ (c - \hat{p}_z) \hat{z}_d - \frac{\hat{z}_d^2}{2} \right] + \frac{\bar{k} + \hat{t} \bar{e} - s \hat{g}}{n} + \gamma \hat{g}. \nonumber \\ \end{aligned}$$

  (44)

Combine the term in square brackets with $z_d = c -p_z$ and (41) to get

$$\begin{aligned} (c- \hat{p}_z) \hat{z}_d - \frac{\hat{z}_d^2}{2} = \frac{(c - \hat{p}_z)^2}{2} = \frac{\left[ c - \bar{\kappa } + (1-\kappa ) s \right] ^2}{2}. \end{aligned}$$

(45)

Insert (40)

$$\begin{aligned} \bar{k} + \hat{t} \bar{e} - s \hat{g} = (\bar{k} + \beta \bar{b} \bar{\kappa } - p_e \bar{e}) - \left[ 2 \bar{\kappa } \mu + (1- \kappa ) \beta \bar{b} \right] s - 2 \kappa \mu s^2 \end{aligned}$$

(46)

to obtain the equilibrium utility

$$\begin{aligned} \hat{U}^x(s)&= \frac{\left[ c - \bar{\kappa } + (1-\kappa ) s \right] ^2}{2} + \frac{\bar{k} + \beta \bar{b} \bar{\kappa } -p_e \bar{e}- \left[ 2 \bar{\kappa } \mu + (1- \kappa ) \beta \bar{b} \right] s - 2 \kappa \mu s^2}{n}\nonumber \\&+ 2 \gamma \mu (\bar{\kappa } + \kappa s) = 2 \kappa \mu \left[ \bar{w}_x + (\gamma + w_{x1}) s - w_{x2} s^2 \right] , \end{aligned}$$

(47)

where $\bar{w}_x:= \frac{(c - \bar{\kappa })^2}{4 \kappa \mu } + \frac{\bar{\kappa } + \beta \bar{b} \bar{\kappa } - p_e \bar{e}}{2 \kappa \mu n} + \frac{\gamma \bar{\kappa }}{\kappa }$, $w_{x1} := \frac{(1- \kappa ) (c - \bar{\kappa }) }{2 \kappa \mu } - \frac{ 2 \bar{\kappa } \mu + (1- \kappa ) \beta \bar{b}}{2 \kappa \mu n}$ and $w_{x2}:= \frac{1}{n} - \frac{(1-\kappa )^2}{4 \kappa \mu }$. Maximization of $\hat{U}^x(s)$ with respect to $s$ leads to the consumer group’s favorite subsidy rate (17).

The sign of $w_{x1}$ and $w_{x2}$: To prove $w_{x2} >0$ we insert $\kappa = \frac{n}{n+2 \mu }$ in $w_{x2}= \frac{1}{n} - \frac{(1-\kappa )^2}{4 \kappa \mu }$ and rearrange terms to get $w_{x2} = \frac{n+ \mu }{ n(n+ 2 \mu )}$.

Next, we prove

$$\begin{aligned} w_{x1} \gtreqless 0 \quad \Longleftrightarrow \quad n \lesseqgtr \frac{(2-\beta ) \bar{b}}{2c} + \sqrt{\frac{(2-\beta )^2 \bar{b}^2}{4 c^2} + \frac{2(1-\beta )\bar{b}\mu }{c}}=: \tilde{n}. \end{aligned}$$

(48)

Verify that

$$\begin{aligned} w_{x1}&:= \frac{(1-\kappa ) (c - \bar{\kappa })}{2 \kappa \mu } - \frac{2 \bar{\kappa } \mu + (1-\kappa )\beta \bar{b}}{2 \kappa \mu n} \gtreqless 0 \nonumber \\&\iff (1-\kappa ) \left[ (c - \bar{\kappa }) n - \beta \bar{b} \right] - 2 \bar{\kappa } \mu \gtreqless 0 \nonumber \\&\iff \frac{1}{n + 2 \mu } \left[ (c - \bar{\kappa }) n - \beta \bar{b} \right] - \bar{\kappa } \gtreqless 0 \nonumber \\&\iff \left[ (c - \bar{\kappa }) n - \beta \bar{b} \right] - (n+2\mu ) \bar{\kappa } \gtreqless 0 \nonumber \\&\iff \left[ (c - \bar{\kappa }) n - \beta \bar{b} \right] - c n + \bar{b} \gtreqless 0 \nonumber \\&\iff - \bar{\kappa } n +(1-\beta ) \bar{b} \gtreqless 0 \nonumber \\&\iff - \frac{n}{n+2 \mu } (c n - \bar{b}) + (1-\beta ) \bar{b} \gtreqless 0 \nonumber \\&\iff - c n^2 + (2-\beta ) \bar{b}n+2(1-\beta ) \bar{b} \mu \gtreqless 0 \nonumber \\&\iff -n^2 + \frac{(2-\beta ) \bar{b}}{c} n + \frac{2(1-\beta ) \bar{b} \mu }{c} \gtreqless 0. \end{aligned}$$

(49)

The last equivalence establishes (48).

• (iv) For sufficiently large $n$ the function $\hat{U}^g$ is progressively increasing in $s$: The quadratic term in $\hat{U}^g$ is: $- 2 \mu \kappa w_{x2} s^2 + \frac{1}{n_g} \mu \kappa ^2 s^2$ which is equivalent to $\left( - 2 w_{x2} + \frac{\kappa }{n_g}\right) \kappa \mu s^2$. Insert $w_{x2} = \frac{n+c\mu }{n(n+2\mu )}$ and $\kappa = \frac{n}{n+2\mu }$ to obtain

  $$\begin{aligned} - 2 w_{x2} + \frac{\kappa }{n_g}>0 \quad&\iff \quad -2 \frac{n+ \mu }{n(n+2\mu ) }+ \frac{n}{n_g(n+2 \mu )} >0 \\&\iff \quad - 2n -2 \mu +\frac{n^2}{n_g}>0. \end{aligned}$$

1.2 Proof of (19)

The sum of utilities of all consumers is given by

$$\begin{aligned} \sum \limits _j \hat{u}_j = n \left( c \hat{z}_d - \frac{\hat{z}_d^2}{2} + \gamma g \right) + \sum \limits _jx_{jd}. \end{aligned}$$

(50)

Making use of the trade balance (3) and $x = k_x =\bar{k}- k_g$ in (50) we obtain

$$\begin{aligned} \sum \limits _j \hat{u}_j = n \left( c \hat{z}_d - \frac{\hat{z}_d^2}{2} + \gamma g \right) + \bar{k} - k_g - p_e e. \end{aligned}$$

(51)

Next, we insert $\hat{z}_d = c - \hat{p}_z$, (40) and (41) in (51) to obtain

$$\begin{aligned} \frac{\sum \nolimits _j \hat{u}_j}{n}&= c \hat{z}_d - \frac{\hat{z}_d^2}{2} + \gamma g + \frac{ \bar{k} - k_g - p_e e}{n} = c \left( \bar{\xi } + (1-\kappa ) s \right) \nonumber \\&-\, \frac{1}{2} \left( \bar{\xi } + (1-\kappa ) s \right) ^2 + \gamma 2 \mu (\bar{\kappa } + \kappa s) + \frac{ \bar{k} - \mu \left( \bar{\kappa } + \kappa s\right) ^2 - p_e e}{n}, \end{aligned}$$

(52)

where $\bar{\xi } : = c - \bar{\kappa }$. Further rearrangements of (52) yield

$$\begin{aligned} \frac{\sum \nolimits _j \hat{u}_j}{n}&= c \bar{\xi } - \frac{ \xi ^2}{2} + 2 \gamma \mu \bar{\kappa } + \frac{\bar{k} - \mu \bar{\kappa }^2 -p_e e}{n}\nonumber \\&+ \left( c (1-\kappa ) - \bar{\xi } (1- \kappa ) + 2 \gamma \mu \kappa - \frac{2 \mu \bar{\kappa } \kappa }{n} \right) s -\, \left( \frac{ (1-\kappa )^2}{2} + \frac{\mu \kappa ^2}{n} \right) s^2 \nonumber \\ \end{aligned}$$

(53)

Next, observe that

$$\begin{aligned} c (1-\kappa ) - \bar{\xi } (1- \kappa ) + 2 \gamma \mu \kappa - \frac{2 \mu \bar{\kappa } \kappa }{n} = \bar{\kappa } \left( 1-\kappa - \frac{2 \mu \kappa }{n} \right) + 2 \gamma \mu \kappa = 2 \gamma \mu \kappa \end{aligned}$$

(54)

and that

$$\begin{aligned} \frac{ (1-\kappa )^2}{2} + \frac{\mu \kappa ^2}{n} = \frac{(2 \mu )^2}{2 (n +2 \mu )^2} + \frac{\mu n^2}{n (n +2 \mu )^2}= \frac{\mu }{n+2\mu } = \frac{\mu \kappa }{n}. \end{aligned}$$

(55)

Making use of (54) and (55) in (53) establishes

$$\begin{aligned} \frac{\sum \nolimits _j \hat{u}_j}{n} = \nu + 2 \gamma \mu \kappa s - \left( \frac{\mu \kappa }{n} \right) s^2, \end{aligned}$$

(56)

where $\nu := c \bar{\xi } - \frac{ \xi ^2}{2} + 2 \gamma \mu \bar{\kappa } + \frac{\bar{k} - \mu \bar{\kappa }^2 -p_e e}{n}$.

1.3 Proof of Proposition 3

(ii) Rewrite (28) to get

$$\begin{aligned} \rho ^N_g = \frac{a_g \mu \kappa ^2}{1-a^2_g \kappa ^2 \mu } \left[ \frac{\bar{\kappa }}{\kappa } + S\left( \rho ^N_b, 0, \rho ^N_x\right) \right] . \end{aligned}$$

Observe that the producer price $p_z + s = \bar{\kappa } + \kappa s$ from (41) has to be strictly positive. Then we immediately get $\frac{\bar{\kappa }}{\kappa } + s > 0$ which ensures $\rho ^{N}_g > 0$ and hence we obtain $S (\rho ^N_b, \rho ^N_g, \rho ^N_x) > S(\rho ^N_b, 0, \rho ^N_x)$.

(i) I. Proof of:

$$\begin{aligned} S\left( \rho ^N_b, \rho _g^N, \rho _x^{N}\right) \gtreqless S\left( \rho ^N_b, \rho _g^N, 0\right) \quad \Longleftrightarrow \quad s^*_x (\gamma ) \gtreqless S \left( \rho ^N_b, \rho _g^N, 0\right) . \end{aligned}$$

From (29) we obtain

$$\begin{aligned} \rho ^N_x = \frac{a_x \kappa \mu n_x}{1+2a^2_x \kappa \mu w_{x2}} \left[ s^*_x (\gamma ) - S\left( \rho ^N_b, \rho ^N_g, 0\right) \right] \end{aligned}$$

(57)

where

$$\begin{aligned} {\mathrm {sign}} \,a_x = {\mathrm {sign}} \,\left[ s^*_x (\gamma ) - S(\rho ^N_b, \rho _g^N, 0)\right] \end{aligned}$$

(58)

From (57) and (58) we infer

$$\begin{aligned} s^*_x (\gamma ) \gtreqless S\left( \rho ^N_b, \rho _g^N, 0\right) \quad \Longrightarrow \quad a_x \gtreqless 0 \quad \Longrightarrow \quad S\left( \rho ^N_b, \rho _g^N, \rho ^{N}_x\right) \gtreqless S\left( \rho ^N_b, \rho _g^N, 0\right) \end{aligned}$$

and

$$\begin{aligned} S\left( \rho ^N_b, \rho _g^N, \rho ^{N}_x\right) \gtreqless S\left( \rho ^N_b, \rho _g^N, 0\right) \quad \Longrightarrow \quad a_x \gtreqless 0 \quad \Longrightarrow \quad s^*_x (\gamma ) \gtreqless S\left( \rho ^N_b, \rho _g^N, 0\right) . \end{aligned}$$

II. Proof of:

$$\begin{aligned} s^*_x (\gamma ) \gtreqless S\left( \rho ^N_b, \rho _g^N, \rho ^{N}_g\right) \quad \Longleftrightarrow \quad s^*_x (\gamma ) \gtreqless S\left( \rho ^N_b, \rho _g^N, 0\right) . \end{aligned}$$

For $\rho ^{N}_x > 0$ the equation $\Pi ^x_{r_x} = 0$ from (26) reads

$$\begin{aligned} \rho ^{N}_x = 2 \kappa \mu n_x w_{x2} \left\{ a_x \left[ s^*_x (\gamma ) - S\left( \rho ^N_b, \rho _g^N, \rho ^{N}_x\right) \right] \right\} . \end{aligned}$$

(59)

Comparing (57) and (59) immediately yields

$$\begin{aligned} s^*_x (\gamma ) \gtreqless S\left( \rho ^N_b, \rho _g^N, \rho ^{N}_x\right) \quad \Longleftrightarrow \quad s^*_x (\gamma ) \gtreqless S\left( \rho ^N_b, \rho _g^N, 0\right) . \end{aligned}$$

1.4 Proof of Proposition 4:

• (i) From (29) with $a_g = 0$ we obtain

  $$\begin{aligned} \rho ^{BX}_x = \frac{a_x \kappa \mu n_x 2 w_{x2}}{1+2a^2_x \kappa \mu w_{x2}} \left[ s^*_x (\gamma ) - S \left( \rho ^N_b, 0, 0\right) \right] , \end{aligned}$$

  (60)

  where

  $$\begin{aligned} {\mathrm {sign}} \,a_x = {\mathrm {sign}} \,\left[ s^*_x (\gamma ) - S\left( \rho ^N_b, 0, 0\right) \right] . \end{aligned}$$

  (61)

  Next, $\Pi ^x_{r_x} = 0$ from (26) implies

  $$\begin{aligned} \rho ^{BX}_x = 2 \kappa \mu n_x w_{x2} \left\{ a_x \left[ s^*_x (\gamma ) - S\left( \rho ^N_b, 0, \rho ^{BX}_x\right) \right] \right\} . \end{aligned}$$

  (62)

  (60)–(62) estabish

  $$\begin{aligned} s^*_x (\gamma ) \gtreqless S\left( \rho ^N_b, 0, \rho ^{BX}_x\right) \gtreqless S\left( \rho ^N_b, 0, 0\right) \quad \iff \quad s^*_x (\gamma ) \gtreqless S\left( \rho ^N_b, 0, 0\right) . \end{aligned}$$

• (ii) For $a_x = 0$, (28) turns into

  $$\begin{aligned} \rho ^{BG}_g = \frac{a_g \mu \kappa ^2}{1- a^2_g \kappa ^2 \mu } \left( \frac{\bar{\kappa }}{\kappa } - a_b \rho ^N_b \right) = \frac{a_g \mu \kappa ^2}{1-a^2_g \kappa ^2 \mu } \left[ \frac{\bar{\kappa }}{\kappa } + S\left( \rho ^N_b, 0, 0\right) \right] . \end{aligned}$$

  (63)

  Due to $\frac{\bar{\kappa }}{\kappa } + s > 0$, we get $\rho ^{BG}_g > 0$ which implies $S(\rho ^N_b, \rho ^{BG}_g, 0) > S(\rho ^N_b, 0, 0)$.

• (iii) For $a_b= 0$, we rewrite (28) as

  $$\begin{aligned} \rho ^{GX}_g = \frac{a_g \mu \kappa ^2}{1-a^2_g \kappa ^2 \mu } \left[ \frac{\bar{\kappa }}{\kappa } + S\left( 0, 0, \rho _x^{GX}\right) \right] . \end{aligned}$$

  (64)

  Due to $\frac{\bar{\kappa }}{\kappa } + s > 0$, we get $\rho ^{GX}_g > 0$ and hence $S(0, \rho ^{GX}_g, \rho _x^{GX}) > S(0, 0, \rho _x^{GX})$.

From (29) with $a_b = 0$ we have

$$\begin{aligned} \rho ^{GX}_x = \frac{a_x \kappa \mu n_x}{1 + 2 a^2_x \kappa \mu w_{x2}} \left[ s^*_x (\gamma ) - S\left( 0, \rho ^{GX}_g, 0\right) \right] \end{aligned}$$

(65)

where

$$\begin{aligned} {\mathrm {sign}} \,a_x = {\mathrm {sign}} \,\left[ s^*_x (\gamma ) - S\left( 0, \rho ^{GX}_g, 0\right) \right] . \end{aligned}$$

(66)

Next, the equation $\Pi ^x_{r_x}=0$ from (26) can be written as

$$\begin{aligned} \rho ^{GX}_x = \kappa \mu n_x 2 w_{x2} \left\{ a_x \left[ s^*_x (\gamma ) - S\left( 0, \rho ^{GX}_g, \rho ^{GX}_x\right) \right] \right\} . \end{aligned}$$

(67)

From (65)–(67) we infer

$$\begin{aligned} s^*_x (\gamma ) \gtreqless S\left( 0, \rho ^{GX}_g, \rho ^{GX}_x\right) \gtreqless S\left( 0, \rho ^{GX}_g, 0\right) \quad \Longleftrightarrow \quad s^*_x \gtreqless S\left( 0, \rho ^{GX}_g, 0\right) . \end{aligned}$$

Appendix 3: Comparative Statics of the Nash Equilibrium

Differentiation of $\rho ^N_b = \frac{a_b (1-\beta ) \bar{e}^{\beta } (1-\kappa )}{2}$ yields

$$\begin{aligned} \frac{\partial \rho ^N_b}{\partial a_b}&= \frac{(1-\beta )\bar{e}^{\beta } (1-\kappa )}{2} > 0, \\ \frac{\partial \rho ^N_b}{\partial a_g}&= \frac{\partial \rho ^N_b}{\partial a_x} = \frac{\partial g^N_b}{\partial \gamma } = 0, \\ \frac{\partial \rho ^N_b}{\partial \bar{e}}&= \frac{a_b (1-\beta ) \beta \bar{e}^{\beta - 1} (1-\kappa )}{2} > 0. \end{aligned}$$

Differentiation of

$$\begin{aligned} \rho ^N_x = \frac{a_x n_x\kappa \mu \left[ (w_{x1}+ \gamma ) \left( 1- a_g^2 \kappa ^2 \mu \right) + a_b^2 w_{x2} (1-\kappa ) (1-\beta ) \bar{b} - 2 a_g^2 \kappa w_{x2} \mu \bar{\kappa } \right] }{1- a_g^2\kappa ^2 \mu + 2 a_x^2 n_x \kappa w_{x2} \mu } \end{aligned}$$

yields

$$\begin{aligned} \frac{\partial \rho ^N_x}{\partial a_b}&= \frac{2a_b a_x n_x \bar{b} \mu (1-\kappa ) \kappa (1-\beta ) w_{x2} }{1-a^2_g \kappa ^2 \mu + 2 a^2_x n_x \kappa w_{x2} \mu } > 0, \\ \frac{\partial \rho ^N_x}{\partial \gamma }&= \frac{ a_b^2 a_x n_x (1-\kappa )(1-\beta ) \kappa w_{x2} \mu }{1-a^2_g \kappa ^2 \mu + 2 a^2_x n_x \kappa w_{x2} \mu } > 0, \\ \frac{\partial \rho ^N_x}{\partial \bar{b}}&= \frac{a_x n_x\kappa \mu \left[ \left( 1- a_g^2 \kappa ^2 \mu \right) \frac{\partial w_{x1}}{\partial \bar{b}} + a_b^2 w_{x2} (1-\kappa ) (1-\beta ) \ - 2 a_g^2 \kappa w_{x2} \mu \frac{ \partial \bar{\kappa }}{\partial \bar{b}} \right] }{1- a_g^2\kappa ^2 \mu + 2 a_x^2 n_x \kappa w_{x2} \mu } > 0, \end{aligned}$$

where

$$\begin{aligned} \frac{\partial \bar{\kappa }}{\partial \bar{b}}&= - \frac{\kappa c}{n}<0, \\ \frac{\partial w_{x1}}{\partial \bar{b}}&= -\frac{(1-\kappa ) \beta }{2 \kappa \mu n } - \left( \frac{1-\kappa }{2c \kappa \mu } +\frac{2 \mu }{ 2 \kappa \mu n} \right) \cdot \underbrace{\frac{\partial \bar{\kappa }}{\partial \bar{b}}}_{= - \frac{\kappa c}{n}} = \frac{(1-\beta ) c}{n} + \frac{1-\kappa }{2 \mu n}>0. \end{aligned}$$

Differentiating

$$\begin{aligned} \rho ^N_g = \frac{a_g \kappa \mu \left[ a_x^2 n_x \kappa ^2 \mu (w_{x1}+\gamma ) + (1+ 2 a_x^2 n_x \kappa w_{x2} \mu ) \bar{\kappa } -0.5 a_b^2 \kappa (1-\kappa ) (1-\beta ) \bar{b} \right] }{1- a_g^2\kappa ^2 \mu + 2 a_x^2 n_x \kappa w_{x2} \mu } \end{aligned}$$

we obtain

$$\begin{aligned} \frac{\partial \rho _g^N}{\partial a_b}&= - \frac{a_b a_g \bar{b} (1-\kappa ) (1-\beta ) \kappa ^2 \mu }{1-a^2_g \kappa ^2 \mu + 2 a^2_x n_x \kappa w_{x2} \mu } < 0, \\ \frac{\partial \rho _g^N}{\partial \gamma }&= \frac{a_g a^2_x n_x \kappa ^3 \mu ^2}{1-a^2_g \kappa ^2 \mu + 2 a^2_x n_x \kappa w_{x2} \mu } > 0, \end{aligned}$$

Finally, differentiation of

$$\begin{aligned} S\left( \rho ^N_b, \rho ^N_g, \rho ^N_x\right)&= \frac{a_x^2 n_x \kappa \mu (w_{x1} +\gamma )+ a_g^2 \kappa \mu \bar{\kappa } - a_b \rho _b^N}{1- a_g^2\kappa ^2 \mu + 2 a_x^2 n_x \kappa w_{x2} \mu } \\&= \frac{\kappa \mu \left[ a_x^2 n_x (w_{x1} +\gamma )+ a_g^2 \bar{\kappa }\right] - 0.5 a_b^2 (1-\kappa ) (1-\beta ) \bar{b}}{1- a_g^2\kappa ^2 \mu + 2 a_x^2 n_x \kappa w_{x2} \mu } \end{aligned}$$

leads to

$$\begin{aligned} \frac{\partial S\left( \rho ^N_b, \rho ^N_g, \rho ^N_x\right) }{\partial a_b}&= -\frac{a_b \bar{b} (1-\kappa ) (1-\beta ) }{1-a^2_g \kappa ^2 \mu + 2a_{x}^{2}\, n_x \kappa w_{x2} \mu } <0,\\ \frac{\partial S\left( \rho ^N_b, \rho ^N_g, \rho ^N_x\right) }{\partial a_g}&= \frac{2 a_g \kappa \mu \left[ \left( a_x^2 n_x \kappa \mu (w_{x1} + \gamma ) - a_b \rho _b^N\right) \kappa \right] + \left( 1+ 2 a_x^2 n_x \kappa w_{x2} \mu \right) \bar{\kappa }}{\left( 1-a^2_g \kappa ^2 \mu + 2 a_{x}^{2}\, n_x \kappa w_{x2} \mu \right) ^2},\\ \frac{\partial S\left( \rho ^N_b, \rho ^N_g, \rho ^N_x\right) }{\partial a_x}&= \frac{2 a_x n_x \kappa \mu \left[ w_{x1} +\gamma + 2 a_b w_{x2} \rho _b^N - a_g^2 \kappa \mu \left[ (w_{x1}+ \gamma ) \kappa + 2 w_{x2} \bar{\kappa } \right] \right] }{\left( 1-a^2_g \kappa ^2 \mu + 2 a_{x}^{2}\, n_x \kappa w_{x2} \mu \right) ^2},\\ \frac{\partial S\left( \rho ^N_b, \rho ^N_g, \rho ^N_x\right) }{\partial \gamma }&= \frac{a^2_x \kappa n_x \mu }{1-a^2_g \kappa ^2 \mu + 2a_{x}^{2}\, n_x \kappa w_{x2} \mu } > 0. \end{aligned}$$