Discounting Utility and the Evaluation of Climate Policy

Abstract

This essay discusses the relation between utility discounting and climate policy. Using a cost-benefit model, I show that the planner’s willingness to pay to eliminate a climate event is greater, but less sensitive to discounting, when the event is random instead of deterministic. Examples in an optimizing setting show that policy may be less sensitive to discounting, the more nonlinear is the underlying model. I then explain why, in general, there should be no presumption that the risk of catastrophe swamps discounting in the assessment of climate policy. I conclude by pointing out that intertemporal transfers between the same agent at different points in their life, and transfers between different agents at different points in time, are qualitatively different, and should not be assessed using the same discount rate.

I thank an anonymous reviewer for comments on an earlier draft of this paper. The usual disclaimer applies. Sects. 3.2 and 4 of this paper are adapted from Karp (2009), and Sect. 5 is based on Karp (2013a, b).

Appendix 1

If society pays the premium, thus eliminating risk, the present discounted stream of utility is

$$\begin{aligned} \left. \begin{array}{c} R\left( X;\Delta ,g,\eta \right) \equiv \int _{0}^{\infty }e^{-\rho t}\frac{ (e^{gt}(1-\Delta X))^{1-\eta }-1}{1-\eta }dt= \\ \int _{0}^{\infty }\frac{e^{-\left( \rho +\left( \eta -1\right) g\right) t}((1-\Delta X))^{1-\eta }-e^{-\rho t}}{1-\eta }dt=\frac{1}{1-\eta }\left( \frac{((1-\Delta X))^{1-\eta }}{\left( \rho +\left( \eta -1\right) g\right) }- \frac{1}{\rho }\right) \end{array} \right. \end{aligned}$$

If society does not pay the insurance premium, and the event occurs at time T, the payoff is

$$\begin{aligned} \left. \begin{array}{c} P\left( T;\Delta ,g,\eta \right) \equiv \int _{0}^{T}e^{-\rho t}\frac{ (e^{gt})^{1-\eta }-1}{1-\eta }d\tau +\int _{T}^{\infty }e^{-\rho t}\frac{ (e^{gt}(1-\Delta ))^{1-\eta }-1}{1-\eta }d\tau \\ =\int _{0}^{T}e^{-\rho t}\frac{\left( (e^{gt})^{1-\eta }-1-\left( (e^{gt}(1-\Delta ))^{1-\eta }-1\right) \right) }{1-\eta }d\tau +\int _{0}^{ \infty }e^{-\rho t}\frac{(e^{gt}(1-\Delta ))^{1-\eta }-1}{1-\eta }d\tau \\ =\int _{0}^{T}e^{-\rho t}\frac{e^{-g\left( \eta -1\right) t}\left( 1-\left( 1-\Delta \right) ^{1-\eta }\right) }{1-\eta }d\tau +\int _{0}^{\infty }e^{-\rho t}\frac{(e^{gt}(1-\Delta ))^{1-\eta }-1}{1-\eta }d\tau \\ =\frac{\left( 1-\left( 1-\Delta \right) ^{1-\eta }\right) }{1-\eta } \frac{1-e^{-\left( \rho +\left( \eta -1\right) g\right) T}}{\left( \rho +\left( \eta -1\right) g\right) }+\frac{1}{1-\eta }\left( \frac{\left( 1-\Delta \right) ^{1-\eta }}{\left( \rho +\left( \eta -1\right) g\right) }-\frac{1}{\rho }\right) \end{array} \right. \end{aligned}$$

If the event time T is exponentially distributed with hazard h, then

$$\begin{aligned} Ee^{-\left( \rho +\left( \eta -1\right) g\right) T}=\int _{0}^{\infty }e^{-\left( \rho +\left( \eta -1\right) g\right) t}he^{-ht}dt=\frac{h}{\left( \rho +g\eta -g+h\right) }. \end{aligned}$$

(7)

Using this formula, we have

$$\begin{aligned} EP\left( T;\Delta ,g,\eta \right) =\frac{\left( 1-\left( 1-\Delta \right) ^{1-\eta }\right) }{1-\eta }\frac{1-\frac{h}{\left( \rho +g\eta -g+h\right) }}{\left( \rho +\left( \eta -1\right) g\right) }+\frac{1}{ 1-\eta }\left( \frac{\left( 1-\Delta \right) ^{1-\eta }}{\left( \rho +\left( \eta -1\right) g\right) }-\frac{1}{\rho }\right) \end{aligned}$$

Fig. 11

Maximum premium (as a percent of cost of event) as a function of $\eta$, under certain event time (solid) and random event time (dashed) for $T=200=\frac{1}{h}$, $\Delta =0.4$, $g=0.01$ and $\rho =0.01$

Fig. 12

Maximum premium (as a percent of cost of event) as a function of growth, under certain event time (solid) and random event time (dashed) for $T=200=\frac{1}{h}$, $\Delta =0.4$, $\eta =2$ and $\rho =0.01$

For the certain event time, with $T=\frac{1}{h}$, the maximum premium society would pay, X, is the solution to

$$\begin{aligned} R\left( X;\Delta ,g,\eta \right) =P\left( T;\Delta ,g,\eta \right) \end{aligned}$$

or

$$\begin{aligned} \left. \begin{array}{c} \frac{1}{1-\eta }\left( \frac{((1-\Delta X))^{1-\eta }}{\left( \rho +\left( \eta -1\right) g\right) }-\frac{1}{\rho }\right) = \\ \frac{\left( 1-\left( 1-\Delta \right) ^{1-\eta }\right) }{1-\eta } \frac{1-e^{-\left( \rho +\left( \eta -1\right) g\right) T}}{\left( \rho +\left( \eta -1\right) g\right) }+\frac{1}{1-\eta }\left( \frac{\left( 1-\Delta \right) ^{1-\eta }}{\left( \rho +\left( \eta -1\right) g\right) }-\frac{1}{\rho }\right) . \end{array} \right. \end{aligned}$$

Solving for X

$$\begin{aligned} \left. \begin{array}{c} \left( \frac{((1-\Delta X))^{1-\eta }}{\left( \rho +\left( \eta -1\right) g\right) }\right) =\left( 1-\left( 1-\Delta \right) ^{1-\eta }\right) \frac{1-e^{-\left( \rho +\left( \eta -1\right) g\right) T}}{\left( \rho +\left( \eta -1\right) g\right) }+\left( \frac{\left( 1-\Delta \right) ^{1-\eta }}{\left( \rho +\left( \eta -1\right) g\right) } \right) \Rightarrow \\ (1-\Delta X)=\left( \left( 1-\left( 1-\Delta \right) ^{1-\eta }\right) \left( 1-e^{-\frac{\left( \rho +\left( \eta -1\right) g\right) }{h}}\right) +\left( 1-\Delta \right) ^{1-\eta }\right) ^{\frac{1}{1-\eta }}\Rightarrow \end{array} \right. \end{aligned}$$

Solving for X and setting $y=100X$ gives Eq. 1. I obtain Eq. 2 by taking expectations with respect to T, using Eq. 7, and repeating the steps used to obtain Eq. 1 (Figs. 11 and 12).