Global Carbon Budgeting and the Social Cost of Carbon

Abstract

The limited availability of natural resources on the global level relates primarily to the influence of fossil energy use on climate change, rather than to the exhaustion of fossil energy resources. In this chapter we consider a version of the Integrated Assessment climate-economy model with overlapping generations and fossil energy as an input in production. In a laissez-faire equilibrium, individuals do not account in their production plans for the negative external effects of carbon dioxide emission. These effects are internalized by a “social planner” solving the problem of optimal accumulation of global wealth and carbon dioxide concentration in the atmosphere. The carbon budget is defined for this problem as the optimal concentration of carbon dioxide in the long term. The social opportunity cost of fossil fuel consumption supporting this budget is given by the present value of marginal welfare losses of future generations. We examine the effects of welfare discounting on the long-term outcome and, in particular, on the welfare of distant future generations.

Appendices

Appendices

1.1 A.1 Characteristic Eq. (5.16)

The backward-dynamic system corresponding to (5.12), (5.13) is

$$ {W}_{t-1}={\left(\frac{W_t}{\left(1-\alpha -\delta\ \right)A\mathcal{D}\left({Q}_t\right)}\right)}^{\frac{1}{\alpha +\delta }} $$

(5.37)

$$ {Q}_{t-1}={\left(1-\mu \right)}^{-1}\left({Q}_t-\frac{\theta \delta {W}_{t-1}}{\alpha +\delta}\right). $$

(5.38)

Take the derivatives:

$$ \frac{\partial {W}_{t-1}}{\partial {W}_t}=\frac{{W_t}^{-1}}{\alpha +\delta }{\left(\frac{W_t}{\left(1-\alpha -\delta\ \right)A\mathcal{D}\left({Q}_t\right)}\right)}^{\frac{1}{\alpha +\delta }}=\frac{1}{\alpha +\delta}\cdot \frac{W_{t-1}}{W_t} $$

$$ \frac{\partial {W}_{t-1}}{\partial {Q}_t}=-\frac{\mathcal{D}{\left({Q}_t\right)}^{-\frac{1}{\alpha +\delta }-1}}{\alpha +\delta }{\mathcal{D}}^{\prime}\left({Q}_t\right){\left(\frac{W_t}{\left(1-\alpha -\delta\ \right)A}\right)}^{\frac{1}{\alpha +\delta }}=-\frac{1}{\alpha +\delta}\cdot \frac{{\mathcal{D}}^{\prime}\left({Q}_t\right)}{\mathcal{D}\left({Q}_t\right)}\cdot {W}_{t-1} $$

$$ \frac{\partial {Q}_{t-1}}{\partial {W}_t}=-{\left(1-\mu \right)}^{-1}\frac{\theta \delta}{\alpha +\delta}\cdot \frac{\partial {W}_{t-1}}{\partial {W}_t} $$

$$ \frac{\partial {Q}_{t-1}}{\partial {Q}_t}={\left(1-\mu \right)}^{-1}\left(1,-,\frac{\theta \delta}{\alpha +\delta },\cdot, \frac{\partial {W}_{t-1}}{\partial {Q}_t}\right). $$

For the steady state, W_t = W_t − 1 = W^∗, Q_t = Q_t − 1 = Q^∗ and, from (5.14):

$$ {W}^{\ast }=\frac{\mu \left(\alpha +\delta \right)}{\theta \delta}{Q}^{\ast }. $$

Hence, for the steady state we have:

$$ \frac{\partial {W}_{t-1}}{\partial {W}_t}=\frac{1}{\alpha +\delta } $$

$$ \frac{\partial {W}_{t-1}}{\partial {Q}_t}=-\frac{1}{\alpha +\delta}\cdot \frac{{\mathcal{D}}^{\prime}\left({Q}^{\ast}\right)}{\mathcal{D}\left({Q}^{\ast}\right)}\cdot {W}^{\ast }=-\frac{\mu }{\theta \delta}\cdot \frac{{\mathcal{D}}^{\prime}\left({Q}^{\ast}\right){Q}^{\ast }}{\mathcal{D}\left({Q}^{\ast}\right)}=\frac{\mu \varepsilon \left({Q}^{\ast}\right)}{\theta \delta} $$

$$ \frac{\partial {Q}_{t-1}}{\partial {W}_t}=-\frac{\theta \delta}{\left(1-\mu \right){\left(\alpha +\delta \right)}^2} $$

$$ \frac{\partial {Q}_{t-1}}{\partial {Q}_t}={\left(1-\mu \right)}^{-1}\left(1-\frac{\mu \varepsilon \left({Q}^{\ast}\right)}{\alpha +\delta}\right). $$

The characteristic equation for the backward-dynamic system (5.37), (5.38) is

$$ \left|\begin{array}{cc}\frac{1}{\alpha +\delta }-\eta & \frac{\mu \varepsilon \left({Q}^{\ast}\right)}{\theta \delta}\\ {}-\frac{\theta \delta}{\left(1-\mu \right){\left(\alpha +\delta \right)}^2}& {\left(1-\mu \right)}^{-1}\left(1-\frac{\mu \varepsilon \left({Q}^{\ast}\right)}{\alpha +\delta}\right)-\eta \end{array}\right|={\eta}^2-\frac{1-\mu +\alpha +\delta -\mu \varepsilon \left({Q}^{\ast}\right)}{\left(1-\mu \right)\left(\alpha +\delta \right)}\eta +\frac{1}{\left(1-\mu \right)\left(\alpha +\delta \right)}=0 $$

or

$$ \left(1-\mu \right)\left(\alpha +\delta \right){\eta}^2-\left(1+\alpha +\delta -\left(1+\varepsilon \left({Q}^{\ast}\right)\right)\mu \right)\eta +1=0. $$

(5.39)

This equation has two positive real roots if

$$ {\left(1-\mu +\alpha +\delta -\varepsilon \left({Q}^{\ast}\right)\mu \right)}^2>4\left(1-\mu \right)\left(\alpha +\delta \right), $$

which is the case for small ε(Q^∗). The left-hand side of (5.39) is positive for η = 1, because

$$ \left(1-\mu \right)\left(\alpha +\delta \right)>\alpha +\delta -\left(1+\varepsilon \left({Q}^{\ast}\right)\right)\mu $$

(since α + δ < 1 + ε(Q^∗)) implying that the minimal real root of (5.39) is above unity. Consequently, both characteristic roots η₁ and η₂ depicted in Fig. 5.6 are above one, implying that the steady state of the backward-dynamic system (5.37), (5.38) is an unstable node.

1.2 A.2 The First-Order Conditions (5.24)–(5.26)

The Lagrangian for the social planner’s problem (5.18)–(5.22) is

$$ \mathcal{L}=\sum \limits_{t=0}^{\infty }{\beta}^t\left[u\left(\rho {W}_t\right)+{\psi}_t\left(\left(1-\alpha -\delta\ \right)\mathcal{D}\left({Q}_t\right)A\left({\lambda}_t\right){W}_{t-1}^{\alpha +\delta }-{W}_t\right)+{v}_t\left({Q}_t-\left(1-\mu \right){Q}_{t-1}-\theta \left(1-{\lambda}_t\right){W}_{t-1}\right)\right], $$

(5.40)

where we have used equations for output (5.18) and consumption (5.23). Differentiating with respect to λ_t implies

$$ {\psi}_t\left(1-\alpha -\delta\ \right)\mathcal{D}\left({Q}_t\right){A}^{\prime}\left({\lambda}_t\right){W}_{t-1}^{\alpha +\delta }=-{v}_t\theta {W}_{t-1}. $$

Due to (5.18), (5.21), the left-hand side of this equation can be written as ψ_tln^′A(λ_t)W_t, implying (5.24). Differentiating (5.40) with respect to Q_t and taking into account (5.8), (5.21) yields the costate equation:

$$ {v}_t=-{\psi}_t{W}_tl{n}^{\prime}\mathcal{D}\left({Q}_t\right)+\beta \left(1-\mu \right){v}_{t+1}, $$

which is equivalent to (5.25).

Consider the costate Eq. (5.26). Differentiating (5.40) with respect to W_t and taking into account (5.18), (5.21) gives:

$$ \rho {u}^{\prime}\left(\rho {W}_t\right)-{\psi}_t+\beta {\psi}_{t+1}\left(\alpha +\delta\ \right)\frac{W_{t+1}}{W_t}-\beta \theta {v}_{t+1}\left(1-{\lambda}_{t+1}\right)=0. $$

(5.41)

From the first-order condition (5.24), we have:

$$ -\theta {v}_{t+1}={\psi}_{t+1}l{n}^{\prime }A\left({\lambda}_{t+1}\right)\frac{W_{t+1}}{W_t}. $$

Inserting this into (5.41) implies

$$ \rho {u}^{\prime}\left(\rho {W}_t\right)-{\psi}_t+\beta {\psi}_{t+1}\frac{W_{t+1}}{W_t}\left(\alpha +\delta +\left(1-{\lambda}_{t+1}\right)l{n}^{\prime }A\left({\lambda}_{t+1}\right)\right)=0. $$

(5.42)

We have it that

$$ l{n}^{\prime }A\left({\lambda}_{t+1}\right)=\frac{\alpha }{\lambda_{t+1}}-\frac{\delta }{\left(1-{\lambda}_{t+1}\right)}, $$

hence,

$$ \alpha +\delta +\left(1-{\lambda}_{t+1}\right)l{n}^{\prime }A\left({\lambda}_{t+1}\right)=\alpha +\alpha \frac{1-{\lambda}_{t+1}}{\lambda_{t+1}}=\frac{\alpha }{\lambda_{t+1}}\kern0.5em . $$

We can rewrite (5.42) as

$$ \rho {u}^{\prime}\left(\rho {W}_t\right)-{\psi}_t+\beta {\psi}_{t+1}\frac{W_{t+1}}{W_t}\cdot \frac{\alpha }{\lambda_{t+1}}=0. $$

This is equivalent to

$$ {\psi}_t=\rho {u}^{\prime}\left(\rho {W}_t\right)+\beta \left(1-\alpha -\delta\ \right){\psi}_{t+1}{R}_{t+1}, $$

because W_t + 1 = (1 − α − δ )Y_t + 1 from (5.21) and

$$ {R}_{t+1}=\frac{\partial {Y}_{t+1}}{\partial {k}_{ft+1}}=\frac{\alpha {Y}_{t+1}}{k_{ft+1}}=\frac{\alpha {Y}_{t+1}}{\lambda_{t+1}{W}_t}. $$

1.3 A.3 The Steady-State Rule (5.30)

Inserting (5.27) and (5.28) into (5.29) and cancelling ψ out of both sides implies

$$ \left(1-\beta \left(1-\mu \right)\right)\left(1-\lambda \right)l{n}^{\prime }A\left(\lambda \right)=\mu Ql{n}^{\prime}\mathcal{D}(Q). $$

Rearrange terms:

$$ \left(1-\beta \left(1-\mu \right)\right)\left(\alpha \frac{1-\lambda }{\lambda }-\delta \right)=-\mu \varepsilon (Q) $$

$$ \frac{\lambda }{1-\lambda }=\frac{\alpha }{\delta -{\left(1-\beta \left(1-\mu \right)\right)}^{-1}\mu \varepsilon (Q)}\kern0.5em . $$

This yields (5.30).