Real options analysis of climate-change adaptation: investment flexibility and extreme weather events

Abstract

Investments in climate-change adaptation will have to be made while the extent of climate change is uncertain. However, some important sources of uncertainty will fall over time as more climate data become available. This paper investigates the effect on optimal investment decision-making of learning that reduces uncertainty. It develops a simple real options method to value options that are found in many climate-change adaptation contexts. This method modifies a binomial tree model frequently applied to climate-change adaptation problems, incorporating gradual learning using a Bayesian updating process driven by new observations of extreme events. It is used to investigate the timing, scale, or upgradable design of an adaptation project. Recognition that we might have more or different information in the future makes flexibility valuable. The amount of value added by flexibility and the ways in which flexibility should be exploited depend on how fast we learn about climate change. When learning will occur quickly, the value of the option to delay investment is high. When learning will occur slowly, the value of the option to build a small low-risk project instead of a large high-risk one is high. For intermediate cases, the option to build a small project that can be expanded in the future is high. The approach in this paper can support efficient decision-making on adaptation projects by anticipating that we gradually learn about climate change by the recurrence of extreme events.

Appendix: Derivations

If an extreme weather event occurs in the next period then we update the probability that the bad scenario applies to

$$ \begin{array}{@{}rcl@{}} \alpha(i,n+1) &=& \text{Pr}[\text{Bad}|\text{Storm}] \\ &=& \frac{\text{Pr}[\text{Storm}|\text{Bad}] \cdot \text{Pr}[\text{Bad}]}{\text{Pr}[\text{Storm}]} \end{array} $$

$$ \begin{array}{@{}rcl@{}} &=& \frac{\text{Pr}[\text{Storm}|\text{Bad}] \cdot \text{Pr}[\text{Bad}]}{\text{Pr}[\text{Storm}|\text{Good}] \cdot \text{Pr}[\text{Good}] + \text{Pr}[\text{Storm}|\text{Bad}] \cdot \text{Pr}[\text{Bad}]} \\ &=& \frac{\pi_{2} \alpha(i,n)}{\pi_{1}(1- \alpha(i,n)) + \pi_{2} \alpha(i,n)}. \end{array} $$

In contrast, if an event does not occur then our updated probability equals

$$ \begin{array}{@{}rcl@{}} \alpha(i+1,n+1) &=& \text{Pr}[\text{Bad}|\text{No storm}] \\ &=& \frac{\text{Pr}[\text{No storm}|\text{Bad}] \cdot \text{Pr}[\text{Bad}]}{\text{Pr}[\text{No storm}]} \\ &=& \frac{\text{Pr}[\text{No storm}|\text{Bad}] \cdot \text{Pr}[\text{Bad}]}{\text{Pr}[\text{No storm}|\text{Good}] \cdot \text{Pr}[\text{Good}]+\text{Pr}[\text{No storm}|\text{Bad}] \cdot \text{Pr}[\text{Bad}]} \\ &=& \frac{(1-\pi_{2}) \alpha(i,n)}{(1-\pi_{1})(1-\alpha(i,n)) + (1-\pi_{2})\alpha(i,n)}. \end{array} $$

Equivalently, the odds the good scenario applies decrease to

$$ \frac{1-\alpha(i,n+1)}{\alpha(i,n+1)} = \frac{\pi_{1}}{\pi_{2}} \cdot \frac{1-\alpha(i,n)}{\alpha(i,n)} $$

if an event occurs and increase to

$$ \frac{1-\alpha(i+1,n+1)}{\alpha(i+1,n+1)} = \frac{1-\pi_{1}}{1-\pi_{2}} \cdot \frac{1-\alpha(i,n)}{\alpha(i,n)} $$

if one does not occur. That is, we multiply the odds of the good scenario applying by π₁/π₂ each period that an event occurs, and by (1 − π₁)/(1 − π₂) each period one does not occur. It follows that if we initially believe the bad scenario applies with probability α₀ and if i of the next n periods are free of events, we will have revised this probability to the number α(i, n) defined implicitly by

$$ \frac{1-\alpha(i,n)}{\alpha(i,n)} = \left( \frac{1-\pi_{1}}{1-\pi_{2}}\right)^{i} \left( \frac{\pi_{1}}{\pi_{2}}\right)^{n-i} \frac{1-\alpha_{0}}{\alpha_{0}}. $$

Equivalently,

$$ \alpha(i,n) = \frac{1}{1 + \left( \frac{1-\pi_{1}}{1-\pi_{2}}\right)^{i} \left( \frac{\pi_{1}}{\pi_{2}}\right)^{n-i} \frac{1-\alpha_{0}}{\alpha_{0}}}. $$

The probability that an event will occur this period if i of the previous n periods are free of such events therefore equals

$$ \begin{array}{@{}rcl@{}} p(i,n) &=& (1-\alpha(i,n))\pi_{1} + \alpha(i,n) \pi_{2} \\ &=& \frac{\pi_{2} + \pi_{1} \left( \frac{1-\pi_{1}}{1-\pi_{2}}\right)^{i} \left( \frac{\pi_{1}}{\pi_{2}}\right)^{n-i} \frac{1-\alpha_{0}}{\alpha_{0}}}{1 + \left( \frac{1-\pi_{1}}{1-\pi_{2}}\right)^{i} \left( \frac{\pi_{1}}{\pi_{2}}\right)^{n-i} \frac{1-\alpha_{0}}{\alpha_{0}}} \\ &=& \frac{\pi_{2} + \pi_{1} \left( \frac{1-\pi_{1}}{1-\pi_{2}}\right)^{i} \left( \frac{\pi_{1}}{\pi_{2}}\right)^{n-i} \left( \frac{\pi_{2} - \pi_{0}}{\pi_{0}-\pi_{1}}\right) }{1 + \left( \frac{1-\pi_{1}}{1-\pi_{2}}\right)^{i} \left( \frac{\pi_{1}}{\pi_{2}}\right)^{n-i} \left( \frac{\pi_{2} - \pi_{0}}{\pi_{0}-\pi_{1}}\right)}, \end{array} $$

where the relationship between α₀ and the three storm probabilities explains the final step.