Supply-Side Climate Policy: On the Role of Exploration and Asymmetric Information

Abstract

In the world economy with interdependent markets for fossil fuel deposits and extracted fossil fuel, a coalition of countries may fight climate change by purchasing fossil fuel deposits for preservation. Harstad (J Polit Econ 120:77–115, 2012) has shown that the coalition’s supply-side climate policy implements the first-best. The present paper focuses on the role exploration and asymmetric information with respect to climate damage plays for the efficiency of unilateral supply-side climate policy. Under the assumption of non-strategic exploration and truthful reporting of climate damage, the deposit policy turns out to be efficient. If exploration is used strategically or the coalition misreports its climate damage, however, the deposit policy becomes inefficient.

Appendix

1.1 Proof of Proposition 1

Conditional on e and g the efficient allocation rules are

$$\begin{aligned}&B'_M (y_M) = B'_N (y_N), \end{aligned}$$

(A1a)

$$\begin{aligned}&B'_M (y_M)- g H'(x_M + x_N) = C'_M (x_M), \, B'_N(x_N) - g H'(x_M + x_N) = C'_N (x_N, e). \end{aligned}$$

(A1b)

The equilibrium of game A is characterized by

$$\begin{aligned} B'_N =p, \quad p= C'_N + g H' \,\, (\text{see } \, (16)). \end{aligned}$$

(A2)

Accounting for (18) in (12) and (13) we obtain

$$\begin{aligned} B'_M =p, \quad p= C'_M + g H'. \end{aligned}$$

(A3)

Comparing (A2) and (A3) with (A1a) and (A1b) establishes Proposition 1(i). The proof of Proposition 1(ii) is straightforward. $\square$

1.2 Parametric Model

Efficiency. For the parametric functions

$$\begin{aligned}&B(y_i) = \alpha y_i - \frac{\beta }{2} y_i^2, \quad C_M(x_M) =\frac{c}{2} x_M^2, \quad C_N(x_N, e) = \frac{c}{2} \left[ x_N+ (\phi - \sigma e) \right] ^2, \\&H (x_M+x_N) = h (x_M +x_N) \end{aligned}$$

(A4)

(with $C_N'= c \left[ x_N + (\phi - \sigma e) \right]$, $\frac{\partial C_N}{\partial e}= - c \sigma \left[ x_N + (\phi - \sigma e) \right]$) we solve

$$\begin{aligned}&\alpha - \beta y_M = \alpha - \beta y_N, \end{aligned}$$

(A5)

$$\begin{aligned}&\alpha - \beta y_M = c x_M + h, \end{aligned}$$

(A6)

$$\begin{aligned}&\alpha - \beta y_N =c \left[ x_N+ (\phi - \sigma e) \right] + h, \end{aligned}$$

(A7)

$$\begin{aligned}&y_M + y_N = x_M + x_N, \end{aligned}$$

(A8)

$$\begin{aligned}&c \sigma \left[ x_N+ (\phi - \sigma e) \right] = 1 \end{aligned}$$

(A9)

to obtain

$$\begin{aligned} y_M^{*}&= y_N^{*} = \frac{(\alpha -h) \sigma -1}{\beta \sigma }, \quad x_M^{*} = \frac{1}{c \sigma } , \quad x_N^{*} = \frac{2c [(\alpha -h) \sigma -1] - \beta }{\beta \sigma }, \\ e^{*}&= \frac{2c [(\alpha - h) \sigma -1] - \beta (2 -c \sigma \phi )}{\beta c \sigma ^2}, \end{aligned}$$

(A10)

which implies

$$\begin{aligned} p^{*} = h + \frac{1}{\sigma }. \end{aligned}$$

(A11)

Stages 1–3 of the games. For the parametric functions it holds

$$\begin{aligned} y_M&= y_N = \frac{\alpha -p}{\beta }, \quad x_M =\frac{p-g h}{c}, \quad x_N = \underline{\xi }, \end{aligned}$$

(A12)

$$\begin{aligned} p&= C'_N(\underline{\xi }, e )+ gh \quad \iff \quad \underline{\xi } = \frac{p- g h}{c} - (\phi - \sigma e) \end{aligned}$$

(A13)

and

$$\begin{aligned} p = C'_N(\overline{\xi }, e ) \quad \iff \quad \overline{\xi } =\frac{p}{c} - (\phi - \sigma e). \end{aligned}$$

(A14)

Inserting (A12) and (A13) into the fuel market clearing condition

$$\begin{aligned} y_M + y_N = x_M +x_N \quad \iff \quad 2 \frac{\alpha -p}{\beta } =\frac{p- g h}{c} + \frac{p-g h}{c} -(\phi - \sigma e) \end{aligned}$$

(A15)

and solving with respect to p we obtain

$$\begin{aligned} p = \hat{P} (e, g):= \frac{2 (\alpha c + \beta g h ) + (\phi - \sigma e) \beta c}{2 (\beta +c)}. \end{aligned}$$

(A16)

Differentiating (A16) with respect to e and g gives $\hat{P}_e=- \frac{ \beta c \sigma }{2 (\beta + c)}$ and $\hat{P}_g= \frac{\beta h}{\beta +c}$.

Inserting (A16) into (A12)–(A15) yields

$$\begin{aligned} \hat{Y}_M (e, g) = \hat{Y}_N (e, g)&= \frac{2 (\alpha - gh ) -(\phi - \sigma e) c}{2 (\beta +c)}, \end{aligned}$$

(A17)

$$\begin{aligned} \hat{X}_M (e, g)&= \frac{2 (\alpha - gh ) + (\phi - \sigma e) \beta }{2 (\beta +c)}, \end{aligned}$$

(A18)

$$\begin{aligned} \overline{\xi } (e, g)&= \frac{2 (\alpha c +\beta gh ) - (\phi - \sigma e) c (\beta +2c)}{2 (\beta +c) c}, \end{aligned}$$

(A19)

$$\begin{aligned} \underline{\xi } (e, g)&= \frac{2 (\alpha - gh ) - (\phi - \sigma e) (\beta +2c)}{2 (\beta +c)}. \end{aligned}$$

(A20)

Next, straightforward calculations lead to

$$\begin{aligned} \theta _{\max } (e, g)&= g h (\overline{\xi } - \underline{\xi }) = \frac{g^2 h^2}{c}, \end{aligned}$$

(A21)

$$\begin{aligned} \theta _{\min } (e, g)&= p (\overline{\xi } - \underline{\xi }) - \frac{c}{2} \left[ \left( \frac{p}{c}\right) ^2 - \left( \frac{p-gh}{c}\right) ^2 \right] = \frac{p gh}{c} - \frac{1}{2 c} \left( 2p g h - g^2 h^2 \right) \\&= \frac{g^2 h^2}{2 c}, \end{aligned}$$

(A22)

$$\begin{aligned} G(e,g)&= \theta _{\max } - \theta _{\min } = \frac{g^2 h^2}{2c}, \end{aligned}$$

(A23)

$$\begin{aligned} \theta (e,g)&= \theta _{\min } + \mu G = (1+\mu ) \frac{g^2 h^2}{2c}. \end{aligned}$$

(A24)

Exploration (GameE). For the parametric functions the first-order condition (25b) turns into

$$\begin{aligned} \frac{{\mathrm {d}} U_N}{{\mathrm {d}} e}&= \underbrace{(p-C'_N)}_{=h} \frac{\partial \underline{\xi }}{\partial e} - (y_N - \underline{\xi } ) \frac{\partial \hat{P}}{\partial e} + \frac{\partial \theta }{\partial e} - \frac{\partial C_N (\underline{\xi }, e)}{\partial e} - 1 \\&= \frac{(\beta +2c) \sigma h}{2(\beta +c)} - \left[ \frac{\alpha -p}{\beta } - \frac{p-h}{c} + (\phi - \sigma e) \right] \left( \frac{\beta c \sigma }{2(\beta +c)} \right) \\&\quad + (p-h) \sigma - 1 =0. \end{aligned}$$

(A25)

Solving $p = \hat{P}(e,1)$ from (A16) and (A25) with respect to p and e we obtain

$$\begin{aligned} \tilde{p} = \frac{2(\beta +c) + (\alpha c + 2 \beta h) \sigma }{3(\beta +c)\sigma }, \quad \tilde{e} = \frac{4c(\alpha \sigma -1) + \beta [(3c\phi +2h) \sigma -4 ]}{3 \beta c \sigma ^2}. \end{aligned}$$

(A26)

Observe that

$$\begin{aligned} p^{*} - \tilde{p} = \frac{\beta +c+(3ch + \beta h -\alpha c) \sigma }{(3 \beta +c)\sigma }, \quad e^{*} - \tilde{e} = - \frac{2[\beta +c+(3ch + \beta h-\alpha c) \sigma ]}{3 \beta c \sigma ^2}. \end{aligned}$$

(A27)

From (A27) we infer

$$\begin{aligned} p^{*} \,\gtreqless \,\tilde{p} \quad\iff & \quad \sigma\, \lesseqgtr \, \frac{\beta +c}{\alpha c-(\beta +3c)h} =: \bar{\sigma }, \\ e^{*} \, \gtreqless \, \tilde{e} \quad\iff & \quad \sigma \, \gtreqless \, \bar{\sigma }. \end{aligned}$$

(A28)

Next, we prove $\sigma < \bar{\sigma }$. From $\phi - \sigma e \ge 0$ and $\tilde{e}=\frac{4c(\alpha \sigma -1) + \beta [(3c\phi +2h) \sigma -4 ]}{3 \beta c \sigma ^2}$ follows $\sigma \le \frac{2(\beta +c)}{2 \alpha c +\beta h}$. Due to³²$\frac{2(\beta +c)}{2 \alpha c + \beta h } < \frac{\beta +c}{\alpha c - ( \beta +3c)h} = \bar{\sigma }$ we get $\sigma \le \bar{\sigma }$. Hence, in view of (A28) we conclude

$$\begin{aligned} p^{*} > \tilde{p}, \quad e^{*} < \tilde{e}. \end{aligned}$$

(A29)

From $y_M=y_N=\frac{\alpha -p}{\beta }$ and $x_M=\frac{p-h}{c}$ we infer

$$\begin{aligned} y_M^{*}=y_N^{*} < \tilde{y}_M=\tilde{y}_N, \quad x_M^{*} > \tilde{x}_M. \end{aligned}$$

(A30)

From $x_N= y_M+y_N-x_M$ we get

$$\begin{aligned} x_N^{*} < \tilde{x}_N. \end{aligned}$$

(A31)

The marginal effects in (25b) have the signs

$$\begin{aligned} (p -C_N') \frac{\partial \underline{\xi }}{\partial e}&= \frac{(\beta +2c) \sigma h}{2(\beta +c)} >0, \end{aligned}$$

(A32)

$$\begin{aligned} - (y_N - \underline{\xi } ) \frac{\partial \hat{P}}{\partial e}&= \underbrace{(y_N - \underline{\xi } )}_{+} \frac{\beta c \sigma }{2(\beta +c)}>0, \end{aligned}$$

(A33)

$$\begin{aligned} \frac{\partial \theta }{\partial e}&= 0. \end{aligned}$$

(A34)

Misreporting (GameG). For the parametric functions the coalition’s first-order condition (28) turns into

$$\begin{aligned} \frac{{\mathrm {d}} U_M}{{\mathrm {d}} g}&= \left[ (g-1) \frac{\partial \hat{X}_M}{\partial g} - \frac{\partial \underline{\xi }}{\partial g} \right] H' - (y_M - x_M) \frac{\partial \hat{P}}{\partial g} - \frac{\partial \theta }{\partial g} \\&= \frac{(2-g) h^2}{\beta +c } - \left( \frac{\alpha - p}{\beta } -\frac{p-gh}{c} \right) \frac{\beta h}{\beta +c} -\frac{(1+\mu ) g h^2}{c} =0. \end{aligned}$$

(A35)

Solving $p = \hat{P}(e ,g)$ from (A16) and (A35) with respect to p and g we obtain

$$\begin{aligned} \tilde{g}&= \frac{[4h + \beta (\phi -\sigma e) ]c}{2 [(1+ \mu ) \beta +(2+\mu ) c] h}, \end{aligned}$$

(A36)

$$\begin{aligned} \tilde{p}&= \frac{[(2+\mu ) (\phi - \sigma e) \beta (\beta +c)+ (2+ \mu ) 2 c \sigma + \beta (4 h+ 2(2+\mu ) \alpha )]c}{ 2 [(1+ \mu ) \beta +(2+\mu ) c] (\beta +c) \sigma }. \end{aligned}$$

(A37)

It can be shown that

$$\begin{aligned} \tilde{g} =1 \quad \iff \quad \mu = \frac{[(\phi - \sigma e) c - 2 h] \beta }{2 (\beta + c) h} =: \bar{\mu } \left( e, \beta , c, h, \sigma \right) . \end{aligned}$$

(A38)

In the sequel we focus on the displacement effects of g and therefore set $e=e^{*}$ to avoid distortions stemming from two different sources, from the inefficiency of g and e. For $\mu =\bar{\mu } \left( e^{*}, \beta , c, h, \sigma \right)$ follows $\tilde{p} = p^{*}$.

Next, straightforward differentiation leads to

$$\begin{aligned} \frac{\partial \tilde{g}}{\partial \mu }&= - \frac{(\beta +c) c [4 h +(\phi - \sigma e^{*}) b]}{ 2[(1+ \mu ) \beta +(2+\mu ) c]^2 h \sigma } <0, \end{aligned}$$

(A39)

$$\begin{aligned} \frac{\partial \tilde{p}}{\partial \mu }&= - \frac{\beta c [4 h +(\phi - \sigma e^{*}) b]}{2 [(1+ \mu ) \beta +(2+\mu ) c]^2 \sigma } <0. \end{aligned}$$

(A40)

Taking advantage of (A39) and (A40) in (A38) we obtain

$$\begin{aligned} \tilde{g} \, \gtreqless \, 1 \quad \iff \quad \tilde{p} \, \gtreqless \, p^{*} \quad \iff \quad \mu \lesseqgtr \bar{\mu } \left( e^{*}, \beta , c, h, \sigma \right) . \end{aligned}$$

(A41)

From $y_M = y_N = \frac{\alpha -p}{b}$ we get

$$\begin{aligned} \tilde{y}_M = \tilde{y}_N \lesseqgtr y_M^{*}=y_N^{*} \quad \iff \quad \mu \lesseqgtr \bar{\mu } \left( e^{*}, \beta , c, h, \sigma \right) . \end{aligned}$$

(A42)

Differentiating $x_M$ and $x_N$ from (A12) with respect to $\mu$ yields

$$\begin{aligned} \frac{\partial x_M}{\partial \mu } = \frac{\partial x_N}{\partial \mu } = \frac{1}{c} \left[ \frac{\partial \tilde{p}}{\partial \mu } - h \frac{\partial \tilde{g}}{\partial \mu } \right] = \frac{[(\phi - \sigma e^{*}) b + 4h]}{2 [(1+ \mu ) \beta +(2+\mu )c]^2 } >0 \end{aligned}$$

(A43)

and hence we get

$$\begin{aligned} \tilde{x}_M \lesseqgtr x_M^{*} \quad \iff \quad \tilde{x}_N \lesseqgtr x_N^{*} \quad \iff \quad \mu \lesseqgtr \bar{\mu } \left( e^{*}, \beta , c, h, \sigma \right) . \end{aligned}$$

(A44)

The marginal effects in (28) have the signs

$$\begin{aligned} \left[ (g-1) \frac{\partial \hat{X}_M}{\partial g} - \frac{\partial \underline{\xi }}{\partial g} \right] H' &= \frac{(2- \tilde{g}) h}{\beta +c } >0, \end{aligned}$$

(A45)

$$\begin{aligned} - (y_M- x_M) \frac{\partial \hat{P}}{\partial g}&= -\underbrace{(y_M-x_M)}_{-} \frac{\beta h}{\beta +c} = \left( \frac{\phi - \sigma e^{*}}{2} \right) \frac{\beta h}{\beta +c} >0, \end{aligned}$$

(A46)

$$\begin{aligned} - \frac{\partial \theta }{\partial g}&= \frac{(1+\mu ) \tilde{g}h^2}{c} <0. \end{aligned}$$

(A47)

Finally, the function $\bar{\mu } \left( e^{*}, \beta , c, h, \sigma \right)$ has the properties

$$\begin{aligned} \bar{\mu }_{\beta }&= \frac{[(\phi - \sigma e^{*}) c - 2 h] c}{ 2 (\beta +c)^2 h}, \end{aligned}$$

(A48)

$$\begin{aligned} \bar{\mu }_{c}&= \frac{[(\phi - \sigma e^{*}) c + 2 h] c}{ 2 (\beta +c)^2 h} >0, \end{aligned}$$

(A49)

$$\begin{aligned} \bar{\mu }_{h}&= -\frac{(\phi - \sigma e^{*}) \beta c}{ 2 (\beta +c)^2 h}<0, \end{aligned}$$

(A50)

$$\begin{aligned} \bar{\mu }_{\sigma }&= - \frac{\beta c e^{*}}{ 2(\beta +c) h}<0. \end{aligned}$$

(A51)

Exploration and misreporting (GameEG).

$$\begin{aligned} \frac{{\mathrm {d}} U_N}{{\mathrm {d}} e}&= \underbrace{(p-C'_N)}_{=g h} \frac{\partial \underline{\xi }}{\partial e} - (y_N - \underline{\xi } ) \frac{\partial \hat{P}}{\partial e} + \frac{\partial \theta }{\partial e} - \frac{\partial C_N (\underline{\xi }, e)}{\partial e} - 1 \\&= \frac{(\beta +2c) \sigma g h}{2(\beta +c)} - \left[ \frac{\alpha -p}{\beta } - \frac{p-h}{c} + (\phi - \sigma e) \right] \left( \frac{\beta c \sigma }{2(\beta +c)} \right) \\&\quad + (p-gh) \sigma - 1 =0, \end{aligned}$$

(A52)

$$\begin{aligned} \frac{{\mathrm {d}} U_M}{{\mathrm {d}} g}&= \left[ (g-1) \frac{\partial \hat{X}_M}{\partial g} - \frac{\partial \underline{\xi }}{\partial g} \right] H' - (y_M - x_M) \frac{\partial \hat{P}}{\partial g} - \frac{\partial \theta }{\partial g} \\&= \frac{(2-g) h^2}{\beta +c } - \left( \frac{\alpha - p}{\beta } -\frac{p-gh}{c} \right) \frac{\beta h}{\beta +c} -\frac{(1+\mu ) g h^2}{c} =0. \end{aligned}$$

(A53)

Solving (A52), (A53) and $p=\hat{P}(e,g)$ with respect to e, g and p yields

$$\begin{aligned} \breve{g}&= \frac{\beta (2h-\sigma ) - 2c(\alpha \sigma -3h\sigma -1)}{3[(1+\mu ) \beta +(2+\mu ) c] h\sigma }, \end{aligned}$$

(A54)

$$\begin{aligned} \breve{e}&= \frac{4c(\alpha \sigma -1) +\beta [(3c\phi +2h) \sigma -4]}{3 \beta c \sigma ^2}, \end{aligned}$$

(A55)

$$\begin{aligned} \breve{p}&= \hat{P}(\breve{e}, \breve{g}) \\&= \frac{\beta ^2(2-h\sigma ) (2+\mu ) +c^2(2+\alpha \sigma ) (2+\mu ) +\beta c[h \sigma (4-\mu ) -\alpha \sigma (1-\mu ) +4(2+\mu )] }{3 (\beta + c) \sigma [(1+\mu ) \beta +(2+\mu ) c]}. \end{aligned}$$

(A56)

Comparing (A54) and (A55) with the efficient allocation yields

$$\begin{aligned} \breve{g} \, \gtreqless \, 1 \quad\iff & \quad \mu \, \gtreqless \, \frac{2[\beta +c -(2\beta h+\alpha c) \sigma ]}{3(\beta +c) h \sigma }=: \breve{\mu } (\beta , c, h, \sigma ), \end{aligned}$$

(A57)

$$\begin{aligned} \breve{e} \, \gtreqless \, e^{*} \quad\iff & \quad \sigma \, \gtreqless \, \frac{\beta +c}{\alpha c -(\beta +3c) h } \equiv \bar{\sigma }. \end{aligned}$$

(A58)

Since (A58) is equivalent to (A28) we get $\sigma < \bar{\sigma }$ and hence $\breve{e} > e^{*}$.