Pricing Energy Resources Under Transition to Alternative Energy

Abstract

Resource pricing has specific features in the presence of alternative technologies providing substitutes for conventional non-renewable resources. The sources of alternative energy for fossil fuels include wind, solar and biofuel energy. In this chapter, we consider a model of gradual energy transition for an energy-supplying industry. Exhaustion of a conventional non-renewable resource in this model causes a decline of the relative price of alternative energy. The market share of this energy in the energy mix of consumers is increasing as a result of gradual substitution of renewables for conventional resources. An important property demonstrated below is the Green Paradox: a higher consumer preference for alternative energy implies a higher intensity of the conventional resource extraction in the near term.

Appendices

Appendices

1.1 A.1 Hicksian Demand Functions (4.23)

The Lagrangian for the consumer problem (4.1)–(4.3) is

$$ L={p}_1{y}_1+{p}_2{y}_2-\mu \left({\left({y}_1^{1/\theta }+{\left(\lambda {y}_2\right)}^{1/\theta}\right)}^{\theta }-U\right), $$

where μ is the dual variable for constraint (4.3). The first-order conditions are:

$$ \frac{\partial L}{\partial {y}_1}={p}_1-\mu {y}_1^{1/\theta -1}{\left({y}_1^{1/\theta }+{\left(\lambda {y}_2\right)}^{1/\theta}\right)}^{\theta -1}=0 $$

(4.46)

$$ \frac{\partial L}{\partial {y}_2}={p}_2-\mu {\lambda}^{1/\theta }{y}_2^{1/\theta -1}{\left({y}_1^{1/\theta }+{\left(\lambda {y}_2\right)}^{1/\theta}\right)}^{\theta -1}=0 $$

(4.47)

$$ \frac{\partial L}{\partial \mu }=U-{\left({y}_1^{1/\theta }+{\left(\lambda {y}_2\right)}^{1/\theta}\right)}^{\theta }=0. $$

(4.48)

From (4.46), (4.47), the relative price is:

$$ {p}_2/{p}_1={\lambda}^{1/\theta }{\left({y}_2/{y}_1\right)}^{1/\theta -1}={\lambda}^{1/\theta }{\left({y}_2/{y}_1\right)}^{-1/\sigma }, $$

implying that

$$ {y}_2={y}_1{\left({\lambda}^{1/\theta }{p}_1/{p}_2\right)}^{\sigma }={y}_1{\lambda}^{\sigma -1}{\left({p}_1/{p}_2\right)}^{\sigma }. $$

(4.49)

Insert this into (4.48) and rearrange terms

$$ U={\left({y}_1^{1/\theta }+{\lambda}^{1/\theta }{\left({y}_1{\lambda}^{\sigma -1}{\left({p}_1/{p}_2\right)}^{\sigma}\right)}^{1/\theta}\right)}^{\theta }={\left({y}_1^{1/\theta }+{y_1}^{1/\theta }{\lambda}^{\sigma /\theta }{\left({p}_1/{p}_2\right)}^{\sigma /\theta}\right)}^{\theta }={y}_1{\left(1+{\left(\lambda {p}_1/{p}_2\right)}^{\sigma -1}\right)}^{\theta }={y}_1{\left(1+{\left(\left({p}_2/\lambda \right)/{p}_1\right)}^{1-\sigma}\right)}^{\theta }={y}_1{\left(\frac{p_1^{1-\sigma }+{\left({p}_2/\lambda \right)}^{1-\sigma }}{p_1^{1-\sigma }}\right)}^{\theta }. $$

Consequently,

$$ {y}_1={\left({\left({p}_1^{1-\sigma }+{\left({p}_2/\lambda \right)}^{1-\sigma}\right)}^{\frac{1}{1-\sigma }}/{p}_1\right)}^{\sigma }U={\left(P/{p}_1\right)}^{\sigma }U $$

due to (4.24). Inserting this into (4.49) yields:

$$ {y}_2={\lambda}^{\sigma -1}{\left(P/{p}_2\right)}^{\sigma }U. $$

1.2 A.2 Time Derivatives of the Energy Price Index

Differentiate (4.27):

$$ \dot{P}(t)=r\left(1-\sigma \right){\left({p}_1^0\right)}^{1-\sigma }{e}^{r\left(1-\sigma \right)t}\frac{1}{1-\sigma }{\left({\left({p}_1^0\right)}^{1-\sigma }{e}^{r\left(1-\sigma \right)t}+{\left({p}_2^{\ast }/\lambda \right)}^{1-\sigma}\right)}^{\frac{1}{1-\sigma }-1}= rP(t)\frac{{\left({p}_1^0\right)}^{1-\sigma }{e}^{r\left(1-\sigma \right)t}}{{\left({p}_1^0\right)}^{1-\sigma }{e}^{r\left(1-\sigma \right)t}+{\left({p}_2^{\ast }/\lambda \right)}^{1-\sigma }}= rP(t)\left(1-\frac{{\left({p}_2^{\ast }/\lambda \right)}^{1-\sigma }}{{\left({p}_1^0\right)}^{1-\sigma }{e}^{r\left(1-\sigma \right)t}+{\left({p}_2^{\ast }/\lambda \right)}^{1-\sigma }}\right)= rP(t)\left[1-{\left({p}_2^{\ast }/\lambda \right)}^{1-\sigma }P{(t)}^{\sigma -1}\right]. $$

The second-order time derivative of the price index is

$$ \ddot{P}=r\dot{P}\left(1-\sigma {\left({p}_2^{\ast }/\lambda \right)}^{1-\sigma }P{(t)}^{\sigma -1}\right). $$

The inflection point of P(t) is

$$ \overset{\sim }{P}=\left({p}_2^{\ast }/\lambda \right){\sigma}^{\frac{1}{1-\sigma }}. $$

The initial price index is

$$ {P}_0={\left({\left({p}_1^0\right)}^{1-\sigma }+{\left({p}_2^{\ast }/\lambda \right)}^{1-\sigma}\right)}^{\frac{1}{1-\sigma }}=\left({p}_2^{\ast }/\lambda \right){\left(\frac{{\left({p}_2^{\ast }/\lambda \right)}^{1-\sigma }}{{\left({p}_1^0\right)}^{1-\sigma }+{\left({p}_2^{\ast }/\lambda \right)}^{1-\sigma }}\right)}^{\frac{1}{\sigma -1}}=\left({p}_2^{\ast }/\lambda \right){\psi}^{\frac{1}{\sigma -1}}. $$

Consequently, P(t) is convex-concave if, and only if, $\overset{\sim }{P}\ge {P}_0$ or

$$ \psi \le \frac{1}{\sigma }. $$

1.3 A.3 The Time Path of Capital (4.33)

From (4.30), (4.32):

$$ k(t)=\frac{\lambda^{\sigma -1}U}{a}{\left(\frac{P(t)}{p_2^{\ast }}\right)}^{\sigma }=\overline{k}{\lambda}^{\sigma }{\left(\frac{P(t)}{p_2^{\ast }}\right)}^{\sigma }=\overline{k}{\left(\frac{P(t)}{\left({p}_2^{\ast }/\lambda \right)}\right)}^{\sigma }. $$

(4.50)

From (4.25), (4.27):

$$ P(t)={\left({\left({p}_1^0\right)}^{1-\sigma }{e}^{r\left(1-\sigma \right)t}+{\left({p}_2^{\ast }/\lambda \right)}^{1-\sigma}\right)}^{\frac{1}{1-\sigma }}=\left({p}_2^{\ast }/\lambda \right){\left(1+\frac{1-\psi }{\psi }{e}^{-r\left(\sigma -1\right)t}\right)}^{\frac{1}{1-\sigma }}. $$

Inserting this into (4.50) yields

$$ k(t)=\overline{k}{\left(1+\frac{1-\psi }{\psi }{e}^{-r\left(\sigma -1\right)t}\right)}^{-\theta }. $$

1.4 A.4 The Case σ = 2

Consider function F(ψ):

$$ F\left(\psi \right)=\underset{0}{\overset{\infty }{\int }}{\left(1+\frac{\psi }{1-\psi }{e}^{r\left(\sigma -1\right)t}\right)}^{-\theta } dt. $$

Denote γ = e^{r(σ − 1)t}. Then

$$ t=\frac{ln\gamma}{r\left(\sigma -1\right)}, dt=\frac{d\gamma}{\gamma r\left(\sigma -1\right)} $$

and

$$ F\left(\psi \right)=\frac{1}{r\left(\sigma -1\right)}\underset{1}{\overset{\infty }{\int }}{\gamma}^{-1}{\left(1+\frac{\psi }{1-\psi}\gamma \right)}^{-\theta } d\gamma . $$

For σ = θ = 2, we have:

$$ rF\left(\psi \right)=\underset{1}{\overset{\infty }{\int }}{\gamma}^{-1}{\left(1+\frac{\psi }{1-\psi}\gamma \right)}^{-2} d\gamma =\underset{1}{\overset{\infty }{\int }}\left(\frac{1}{\gamma }-\frac{\psi }{1-\psi +\psi \gamma}-\frac{\psi \left(1-\psi \right)}{{\left(1-\psi +\psi \gamma \right)}^2}\right) d\gamma ={\left.\left( ln\gamma -\mathit{\ln}\left(1+\frac{\psi }{1-\psi}\gamma \right)+\frac{1-\psi }{1-\psi +\psi \gamma}\right)\right|}_1^{\infty }={\left.\left(\mathit{\ln}\frac{\gamma }{1-\psi +\psi \gamma}+\mathit{\ln}\left(1-\psi \right)+\frac{1-\psi }{1-\psi +\psi \gamma}\right)\right|}_1^{\infty }=- ln\psi -\left(1-\psi \right)=\psi - ln\psi -1. $$

implying (4.42).