Uncertain discount and hyperbolic preferences

Abstract

This paper studies the interaction between savagean uncertainty and time preferences. We introduce a variation of the discounted subjective expected utility model, where time preferences are state dependent. Before uncertainty is resolved, the individual is unsure about the discount factor that will be used, even when evaluating certain payoffs. The model can account for the present bias and diminishing impatience, even if the future is discounted geometrically. The present bias disappears when the immediate payoff becomes uncertain. Although preferences are not stationary, choices may be time consistent.

Appendix: Proofs

Proof of Theorem 1

We derive I(t) with respect to t (assuming I(t) as function from ${\mathbb {R}}_+\rightarrow {\mathbb {R}}_+$) and we show its derivative is negative.

$$\begin{aligned} \frac{\partial I(t)}{\partial t}=\frac{E_p[\delta (\omega )^t\ln \delta (\omega )] E_p[\delta (\omega )^{t+1}]-E_p[\delta (\omega )^t]E_p[\delta (\omega )^{t+1}\ln \delta (\omega )]}{(E_p[\delta (\omega )^{t+1}])^2} \end{aligned}$$

Since $\delta (\omega )$ are all smaller or equal than one, $E_p[\delta (\omega )^t]\ge E_p[\delta (\omega )^{t+1}]$, hence

$$\begin{aligned}&E_p[\delta (\omega )^t\ln \delta (\omega )] E_p[\delta (\omega )^{t+1}]-E_p[\delta (\omega )^t]E_p[\delta (\omega )^{t+1}\ln \delta (\omega )]\\&\quad \le E_p[\delta (\omega )^{t}](E_p[\delta (\omega )^t\ln \delta (\omega )]-E_p[\delta (\omega )^{t+1}\ln \delta (\omega )]) \end{aligned}$$

the left-hand side of the previous inequality is equal to

$$\begin{aligned} E_p[\delta (\omega )^{t}](E_p[\delta (\omega )^t\ln \delta (\omega )(1-\delta (\omega ))]) \end{aligned}$$

Since $(1-\delta (\omega ))\ge 0$ and $\ln \delta (\omega )\le 0$, the sign of the expression is negative. $\square$

Proof of Theorem 2

Time invariance and time consistency imply stationarity. By time consistency $(z,x)\sim _\tau (z,y)$, if and only if, $x\sim _{\tau +1}y$. By time invariance $x\sim _{\tau +1}y$, if and only if, $x\sim _{\tau }y$.

Stationarity and time consistency imply time invariance. Without loss of generality, let $\tau '>\tau$. By stationarity, $x\sim _\tau y$, if and only if, $(z,x)\sim _{\tau }(z,y)$. By time consistency, $(z,x)\sim _{\tau }(z,y)$, if and only if, $x\sim _{\tau +1}y$. Repeating the same argument until $\tau '$ gives the result.

Time invariance and stationarity imply time consistency. By stationarity $(z,x)\sim _\tau (z,y)$, if and only if, $x\sim _{\tau }y$. By time invariance, $x\sim _{\tau }y$, if and only if, $x\sim _{\tau +1}y$.   

$\square$

Proof of Proposition 1

If $u_\tau \equiv u_{\tau '}$ and $E_{p_\tau }[\delta _\tau (\omega )^{t-\tau }]=E_{p_{\tau '}}[\delta _{\tau '}(\omega )^{t-\tau '}]$, for all $\tau ,\tau '\in {\mathbb {N}}$ time invariance follows immediately. To see the opposite implication, consider the following state-independent acts $h=(x_0,x_1,z,z,z,\ldots )$ and $h'=(y_0,y_1,z,z,z,\ldots )$, who differs in the firs two periods only and assume that $h\sim _\tau h'$. Then by time invariance, $h\sim _\tau h'$, if and only if, $h\sim _{\tau '}h'$. Since $u_\tau \equiv u_{\tau '}$, rescale the utilities to be equal. Then, $u(x_0)+u(x_{1})E_{p_\tau }[\delta _\tau (\omega )]= u(y_0)+u(y_{1})E_{p_\tau }[\delta _\tau (\omega )]$, if and only if, $u(x_0)+u(x_{1})E_{p_{\tau '}}[\delta _\tau (\omega )]= u(y_0)+u(y_{1})E_{p_{\tau '}}[\delta _{\tau '}(\omega )]$, or $u(x_0)-u(y_0)+E_{p_\tau }[\delta _\tau (\omega )](u(x_1)-u(y_1))=u(x_0)-u(y_0)+E_{p_{\tau '}}[\delta _{\tau '}(\omega )](u(x_1)-u(y_1))$ and it follows that $E_{p_{\tau }}[\delta _{\tau }(\omega )]=E_{p_{\tau '}}[\delta _{\tau '}(\omega )]$. $\square$

Proof of Theorem 3

Suppose that $\delta _\tau (\omega )=\delta _{\tau '}(\omega )=\delta (\omega )$ for all $\tau ,\tau '$ and all $\omega \in \varOmega$ and $u_\tau \equiv u_{\tau '}$ for all $\tau ,\tau '$, and we normalize the utilities to be all equal. Then, time consistency holds if

$$\begin{aligned} \sum _{\omega \in \varOmega }p_\tau (\omega )\left[ u(z) +\sum _{s=1}^\infty \delta (\omega )^{s}u(x_{\tau +s})\right] =\sum _{\omega \in \varOmega }p_\tau (\omega )\left[ u(z) +\sum _{s=1}^\infty \delta (\omega )^{s}u(y_{\tau +s})\right] \end{aligned}$$

or

$$\begin{aligned} \sum _{s=1}^\infty u(x_{\tau +s})\left[ \,\sum _{\omega \in \varOmega }p_\tau (\omega ) \delta (\omega )^{s}\right] =\sum _{s=1}^\infty u(y_{\tau +s})\left[ \,\sum _{\omega \in \varOmega }p_\tau (\omega ) \delta (\omega )^{s}\right] \end{aligned}$$

(2)

is equivalent to:

$$\begin{aligned}&\sum _{\omega \in \varOmega }p_{\tau +1}(\omega ) \left[ u(x_{\tau +1})+\sum _{s=1}^\infty \delta (\omega )^{s}u (x_{\tau +1+s})\right] \\&\quad =\sum _{\omega \in \varOmega }p_{\tau +1} (\omega )\left[ u(y_{\tau +1})+\sum _{s=1}^\infty \delta (\omega )^{s}u(y_{\tau +s})\right] \end{aligned}$$

or

$$\begin{aligned} \sum _{s=0}^\infty u(x_{\tau +1+s})\left[ \,\sum _{\omega \in \varOmega }p_{\tau +1}(\omega )\delta (\omega )^{s}\right] =\sum _{s=0}^\infty u(y_{\tau +1+s})\left[ \,\sum _{\omega \in \varOmega }p_{\tau +1}(\omega )\delta (\omega )^{s}\right] \quad \end{aligned}$$

(3)

The left-hand side of Eq. (2) is equal to $u(x_{\tau +1})E_{p_\tau }[\delta (\omega )]+u(x_{\tau +2})E_{p_\tau }[\delta ^2_\omega ]+\cdots$ where the left-hand side of Eq. (3) is equal to $u(x_{\tau +1})+u(x_{\tau +2})E_{p_{\tau +1}}[\delta (\omega )]+\cdots$ Let

$$\begin{aligned} p_{\tau +1}(\omega )=\frac{p_\tau (\omega )\delta (\omega )}{\sum _{\omega '\in \varOmega }p_\tau (\omega ')\delta (\omega ')} \end{aligned}$$

then equality (3) becomes

$$\begin{aligned}&\sum _{s=0}^\infty u(x_{\tau +1+s})\left[ \,\sum _{\omega \in \varOmega }\frac{p_\tau (\omega ) \delta (\omega )^{s+1}}{\sum _{\omega '\in \varOmega }p_\tau (\omega ') \delta ({\omega '})}\right] \\&\quad =\sum _{s=0}^\infty u(y_{\tau +1+s})\left[ \,\sum _{\omega \in \varOmega }\frac{p_\tau (\omega ) \delta (\omega )^{s+1}}{\sum _{\omega '\in \varOmega }p_\tau (\omega ') \delta ({\omega '})}\right] \end{aligned}$$

that corresponds to $u(x_{\tau +1})+u(x_{\tau +2})E_{p_\tau }[\delta (\omega )^{2}]\cdot M+\ldots =u(y_{\tau +1})+u(y_{\tau +2})E_{p_\tau }[\delta (\omega )^{2}]\cdot M+\ldots$, where $M=[E_{p_\tau }[\delta (\omega )]]^{-1}$. Multiplying both sides by $M^{-1}$, we have $u(x_{\tau +1})E_{p_\tau }[\delta (\omega )]+u(x_{\tau +2})E_{p_\tau }[\delta (\omega )^{2}]+\ldots =u(y_{\tau +1})E_{p_\tau }[\delta (\omega )]+u(y_{\tau +2})E_{p_\tau }[\delta (\omega )^{2}]+\ldots$ which is equality (2). $\square$

Proof of Theorem 4

Suppose that $\delta _\tau (\omega )=\delta _{\tau '}(\omega )=\delta (\omega )$ for all $\tau ,\tau '$ and all $\omega \in \varOmega$ and $u_\tau \equiv u_{\tau '}$ for all $\tau ,\tau '$, and we normalize the utilities to be all equal. Then, conditional time consistency holds if

$$\begin{aligned} \sum _{\omega \in E}p_\tau (\omega )\left[ u(z)+\sum _{s=1}^\infty \delta (\omega )^{s}u(x_{\tau +s})\right] =\sum _{\omega \in E}p_\tau (\omega )\left[ u(z)+\sum _{s=1}^\infty \delta (\omega )^{s}u(y_{\tau +s})\right] \end{aligned}$$

or

$$\begin{aligned} \sum _{s=1}^\infty u(x_{\tau +s})\left[ \,\sum _{\omega \in E}p_\tau (\omega )\delta (\omega )^{s}\right] =\sum _{s=1}^\infty u(y_{\tau +s})\left[ \,\sum _{\omega \in E}p_\tau (\omega )\delta (\omega )^{s}\right] \end{aligned}$$

(4)

is equivalent to:

$$\begin{aligned}&\sum _{\omega \in \varOmega }p_{\tau +1,E}(\omega )\left[ u(x_{\tau +1}) +\sum _{s=1}^\infty \delta (\omega )^{s}u(x_{\tau +1+s})\right] \\&\quad =\sum _{\omega \in \varOmega }p_{\tau +1,E}(\omega )\left[ u(y_{\tau +1})+\sum _{s=1}^\infty \delta (\omega )^{s}u(y_{\tau +s})\right] \end{aligned}$$

or

$$\begin{aligned} \sum _{s=0}^\infty u(x_{\tau +1+s})\left[ \,\sum _{\omega \in \varOmega }p_{\tau +1,E}(\omega )\delta (\omega )^{s}\right] =\sum _{s=0}^\infty u(y_{\tau +1+s})\left[ \,\sum _{\omega \in \varOmega }p_{\tau +1,E}(\omega )\delta (\omega )^{s}\right] \end{aligned}$$

(5)

The left-hand side of Eq. (4) is equal to $u(x_{\tau +1})E_{p_\tau }[\delta (\omega )]+u(x_{\tau +2})E_{p_\tau }[\delta ^2_\omega ]+\cdots$ where the left-hand side of Eq. (5) is equal to $u(x_{\tau +1})+u(x_{\tau +2})E_{p_{\tau +1,E}}[\delta (\omega )]+\cdots$ Let

$$\begin{aligned} p_{\tau +1,E}(\omega )=\frac{p_\tau (\omega )\delta (\omega )}{\sum _{\omega '\in E}p_\tau (\omega ')\delta (\omega ')} \end{aligned}$$

if $\omega \in E$ and 0 otherwise. Then equality (5) becomes

$$\begin{aligned}&\sum _{s=0}^\infty u(x_{\tau +1+s})\left[ \,\sum _{\omega \in E}\frac{p_\tau (\omega )\delta (\omega )^{s+1}}{\sum _{\omega '\in E}p_\tau (\omega ')\delta ({\omega '})}\right] \\&\quad =\sum _{s=0}^\infty u(y_{\tau +1+s})\left[ \,\sum _{\omega \in E}\frac{p_\tau (\omega )\delta (\omega )^{s+1}}{\sum _{\omega '\in E}p_\tau (\omega ')\delta ({\omega '})}\right] \end{aligned}$$

that corresponds to $u(x_{\tau +1})+u(x_{\tau +2})[\sum _{\omega \in E}p_\tau (\omega )\delta (\omega )^{2}]\cdot M_E+\ldots =u(y_{\tau +1})+u(y_{\tau +2})[\sum _{\omega \in E}p_\tau (\omega )\delta (\omega )^{2}]\cdot M_E+\cdots$, where $M_E=[\sum _{\omega \in E}p_\tau (\omega )\delta (\omega )]^{-1}$. Multiplying both sides by $M_E^{-1}$, we have $u(x_{\tau +1})[\sum _{\omega \in E}p_\tau (\omega )\delta (\omega )]+u(x_{\tau +2})[\sum _{\omega \in E}p_\tau (\omega )\delta (\omega )^2]+\cdots =u(y_{\tau +1})[\sum _{\omega \in E}p_\tau (\omega )\delta (\omega )]+u(y_{\tau +2})[\sum _{\omega \in E}p_\tau (\omega )\delta (\omega )^2]+\cdots$ which is equality (4). $\square$

Proof of Theorem 1

By standard arguments and the cardinal uniqueness of each $v_\omega (\cdot )$, the Conditional Temporal axiom implies that each conditional preference is represented by a geometrically discounted utility, hence

$$\begin{aligned} x\succcurlyeq _\omega y \Longleftrightarrow \sum _{t=0}^\infty \delta (\omega )^tu_\omega (x_t)\ge \sum _{t=0}^\infty \delta (\omega )^tu_\omega (y_t) \end{aligned}$$

for some $\delta (\omega )\in (0,1]$ and $u_\omega (\cdot ):[0,1]\rightarrow {\mathbb {R}}$. Now take two constant act x, y with $x\succcurlyeq _\omega y$, by Certainty consistency, $x\succcurlyeq _{\omega '} y$ for all $\omega '\in \varOmega$. Then, $u_\omega (x)\frac{1}{1-\delta (\omega )}\ge u_\omega (y)\frac{1}{1-\delta (\omega )}$, if and only if, $u_{\omega '}(x)\frac{1}{1-\delta (\omega ')}\ge u_{\omega '}(y)\frac{1}{1-\delta (\omega ')}$, or $u_{\omega }(x)\ge u_{\omega }(y)$, if and only if, $u_{\omega '}(x)\ge u_{\omega '}(y)$. Since it holds for any x, y, $u_{\omega }(\cdot )$ and $u_{\omega '}(\cdot )$ represent the same preferences over X, by uniqueness up to positive affine transformation of both we can equate all the $u_{\omega }(\cdot )=u(\cdot )$. $\square$

Proof of the Example 2 in Sect. 7. Suppose that $(x,t)\sim (y,t+1)$, then

$$\begin{aligned} V_S(h)=\sum _{\omega \in \varOmega }p(\omega )\ln (\delta ^t_\omega u(x))=\sum _{\omega \in \varOmega }p(\omega )\ln (\delta ^{t+1}_\omega u(y))=V_S(h') \end{aligned}$$

equivalently

$$\begin{aligned}&\sum _{\omega \in \varOmega }p(\omega )(t\ln (\delta _\omega ))+\ln u(x)=\sum _{\omega \in \varOmega }p(\omega )\left( (t+1)\ln (\delta _\omega )\right) +\ln u(y) \\&\quad \Longleftrightarrow t\sum _{\omega \in \varOmega }p(\omega )(\ln (\delta _\omega ))+\ln u(x)=(t+1)\sum _{\omega \in \varOmega }p(\omega )(\ln (\delta _\omega ))+\ln u(y)\\&\quad \Longleftrightarrow \ln u(x)=\sum _{\omega \in \varOmega }p(\omega )(\ln (\delta _\omega ))+\ln u(y)\\&\quad \Longleftrightarrow V_S(g)=V_S(g') \end{aligned}$$

I would like to thank the editor and two anonymous referees for helpful comments.