Risk-induced discounting

Abstract

We establish a direct connection between time preference and risk about an attribute (health) of the instantaneous utility function. In doing so, we derive a risk-induced discount function that corresponds to a normalized expectation of that attribute. We provide several results characterizing this risk-induced discount function depending on the stochastic properties of the risk, which we model as a discrete Markov process. When it is well-defined, which we refer to as full approximation, the risk-induced discount function coincides with exponential discounting if the Markov process is stationary. However, a slight perturbation of the beliefs can trigger time-inconsistent discounting. When considering non-stationary Markov processes, time-inconsistency also emerges in situations where individuals’ beliefs change in a non-anticipated fashion over time, as exemplified by quasi-hyperbolic discounting. Results are illustrated via several applications.

Appendix

1.1 Proof of Proposition 1

Proof

Under full approximation, $p_{i}^{\prime }P^{k-1}a=\delta _k a_i$ for $k \ge 1$. Therefore, the following system of equations is satisfied:

$$\begin{aligned} (P^{k}-\delta _k I)a = 0 \quad \forall \quad T-t \ge k\ge 1, \end{aligned}$$

(16)

which implies $(\delta _k,a) \in E_r(P^{k})$. Since $(\delta _1,a) \in E_r(P)$ and that each eigenvector has a unique corresponding eigenvalue, then $\delta _k=\delta _1^{k+1}$. Rewriting $\delta \equiv \delta _1$ yields $\delta _{t}(k|i)=p_{i}^{\prime }P^{k-1}a=\delta ^{k+1}$. This completes the proof as any eigenvalue $\delta$ of a transition matrix is such that $0<\delta \le 1$. $\square$

1.2 Proof of Corollary 1

Proof

(1)Using the default state, $p_d^{\prime }a=\delta a_d=\delta$, since $a_d=1$.

(2)–(3) Since $\delta < 1$, then $p_N^{\prime }a=\delta a_N$ is impossible unless $p_{NN}=1$ and $a_N=0$.

(4) Since $P \not =I$ and $\delta < 1$, then the transition probability to at least one higher indexed state must be positive, i.e., $p_{i({i+k})}>0$ for all $1 \le i \le N$ and some $1 \le k \le N-i$.

(5) Since $\delta =1$, then $p_1^{\prime }a=\delta a_1$ and $p_N^{\prime }a=\delta a_N$ are impossible unless $p_{11}=1$ and $p_{NN}=1$. $\square$

1.3 Proof of Proposition 3

Proof

Consider row $p_i$ for some $i \not \in \{d,N \}$ and let $\hat{p}_i$ be a perturbation of $p_i$. Define the $N \times 1$ vector of perturbation $\mathbbm {\epsilon }^{\prime } = [\epsilon _1, \epsilon _2,\ldots \epsilon _N]$ and let $\hat{p}_i=p_i+\epsilon$, where $\hat{p}_{ij} \ge 0$ and $\epsilon ^{\prime }\mathbbm {1}=0$ for all $1 \le j \le N$. Such a perturbation is said to be admissible. Construct the transition matrix $\hat{P}$ from P by substituting row $p_i$ with row $\hat{p_i}$ while leaving all of the other rows the same.

(1) Observe that $\delta _{t}(1|i)=\frac{1}{a_i}\hat{p}_i^{\prime }a=\frac{1}{a_i}(p_i^{\prime }a+\epsilon ^{\prime }a)$. Since $(\delta ,a) \in E_r(P)$ then $\delta _{t}(1|i)=\delta +\frac{1}{a_i}{\epsilon ^{\prime }a}$. Similarly, $\delta _{t}(1|d)={\hat{p}_d^{\prime }a}={{p}_d^{\prime }a} =\delta$. Full approximation is impossible if $\delta _{t}(1|i) \not =\delta _{t}(1|d)$, which occurs when ${\epsilon ^{\prime }a} \not =0$. We can show that the latter condition holds for a suitably chosen perturbation. Pick an index $j_1$, so that $p_{i{j_1}}>0$ and set $\epsilon _{j_1}<0$. Select another index $j_2 \not = j_1$, so that $p_{i{j_2}}<1$ and let $\epsilon _{j_2}=-\epsilon _{j_1}>0$ and assume that $\epsilon _j=0$ for all other j values. By choosing $\epsilon _{j_1}$ small enough, this admissible perturbation exists and by construction ${\epsilon ^{\prime }a}=\epsilon _{j_1}(a_{j_1}-a_{j_2}) \not =0$, since $a_1>a_2> \cdots >a_N \ge 0$.

(2) By Corollary 1 part (4), we can select $j_1>j_2$ and $j_2=i$. Note that the ratio $\frac{\delta _{t}(3|d)}{\delta _{t}(2|d)}=\frac{p_d^{\prime }\hat{P}^{2}a}{p_d^{\prime }\hat{P}a}=\frac{p_d^{\prime }\hat{P}(\hat{P}a)}{p_d^{\prime }\hat{P}a}$ with $\hat{P}a=Pa+(\epsilon ^{\prime }a)\chi _i$, where $\chi _i$ is a characteristic vector taking a value of 1 for row i and 0 otherwise. Since $(\delta ,a) \in E_r(P)$, then $\hat{P}a=\delta a+(\epsilon ^{\prime }a)\chi _i$. Therefore, $\frac{\delta _{t}(3|d)}{\delta _{t}(2|d)}=\frac{\delta p_d^{\prime }\hat{P}a+p_d^{\prime }\hat{P}(\epsilon ^{\prime }a)\chi _i}{p_d^{\prime }\hat{P}a}=\delta +\epsilon ^{\prime }a\frac{p_d^{\prime }\hat{P}\chi _i}{p_d^{\prime }\hat{P}a}$. The term $p_d^{\prime }\hat{P}a$ becomes $p_d^{\prime }Pa+p_{di}\epsilon _{j_1}(a_{j_1}-a_{j_2})=\delta ^2+p_{di}\epsilon _{j_1}(a_{j_1}-a_{j_2})>0$, since $j_1> j_2$ and $\epsilon _{j_1}<0$. Furthermore, the term $p_d^{\prime }\hat{P}\chi _i$ becomes $p_d^{\prime }P\chi _i+p_{di}\epsilon _{i}$, where $\epsilon _{i} \equiv \epsilon _{ij_2}>0$. Since $p_d^{\prime }P\chi _i \ge 0$ and ${\epsilon ^{\prime }a}>0$, then $\frac{\delta _{t}(3|d)}{\delta _{t}(2|d)}>\delta = p_d^{\prime }a=\delta _{t}(1|d)$, which shows that risk-induced discount function in the default state is time-inconsistent. $\square$

1.4 Proof of Proposition 4

Proof

Let $p_d \in \mathbf{S}$. By definition, $p^{\prime }_dP= p_d^{\prime }$ or $(1, p_d) \in E_l(P)$ and therefore

$$\begin{aligned} \delta _{t}(k|d)=p_d^{\prime }P^{k-1}a = p_d^{\prime }a, \quad T-t \ge k \ge 1. \end{aligned}$$

(17)

Letting $\beta =p_d^{\prime }a$ and $\delta =1$ completes the proof. $\square$