Why uncertainty matters: discounting under intertemporal risk aversion and ambiguity

Abstract

Uncertainty has an almost negligible impact on project value in the standard economic model. I show that a comprehensive evaluation of uncertainty and uncertainty attitude changes this picture fundamentally. The illustration of this result relies on the discount rate, which is the crucial determinant in balancing immediate costs against future benefits, and the single most important determinant of optimal mitigation policies in the integrated assessment of climate change. First, the paper removes an implicit assumption of (intertemporal or intrinsic) risk neutrality from the standard economic model. Second, the paper introduces aversion to non-risk uncertainty (ambiguity). I show a close formal similarity between the model of intertemporal risk aversion, which is a reformulation of the widespread Epstein–Zin–Weil model, and a recent model of smooth ambiguity aversion. I merge the models, achieving a threefold disentanglement between risk aversion, ambiguity aversion, and the propensity to smooth consumption over time.

Appendices

Appendix 1

Notation: The calculations in the appendix make frequent use of the abbreviations $\rho =1-\eta$ and $\alpha =1-{{\mathrm{RRA}}}$, characterizing the exponents of the isoelastic aggregators.

Proof of Proposition 1

The first part of the proof calculates the marginal value of an additional certain unit of consumption in the second period ($\mathrm{d}x_2$), expressed in terms of marginal first-period consumption units ($\mathrm{d}x_1$). This value derives from the marginal trade-off that leaves aggregate welfare

$$\begin{aligned} U(x_1,p)=\frac{x_1^\rho }{\rho } + \beta \frac{1}{\rho } \left[ {\hbox {E}}_{p} x_2^\alpha \, \right] ^{\frac{\rho }{\alpha }} \end{aligned}$$

unchanged:

$$\begin{aligned} d U(x_1,p)&= x_1^{\rho -1} \mathrm{d}x_1 + \beta \frac{1}{\alpha } \left[ {\hbox {E}}_{p} x_2^{\alpha } \right] ^{\frac{\rho }{\alpha }-1} \, {\hbox {E}}_{p} \alpha x_2^{\alpha -1} \mathrm{d}x_2 \mathop {=}\limits ^{!} 0\\&\Rightarrow x_1^{\rho -1} \mathrm{d}x_1 = - \beta \left[ {\hbox {E}}_{p} x_2^{\alpha } \, \right] ^{\frac{\rho }{\alpha }-1} \, {\hbox {E}}_{p} x_2^{\alpha -1} \mathrm{d}x_2\\&\Rightarrow \frac{\mathrm{d}x_1}{\mathrm{d}x_2} = - \beta \left[ {\hbox {E}}_{p} \left( \frac{x_2}{x_1}\right) ^{\alpha } \, \right] ^{\frac{\rho }{\alpha }-1} \, {\hbox {E}}_{p} \left( \frac{x_2}{x_1}\right) ^{\alpha -1}\\&\Rightarrow \frac{\mathrm{d}x_1}{\mathrm{d}x_2} = - \beta \left[ {\hbox {E}}_{p} e^{\alpha \ln {\frac{x_2}{x_1}}} \, \right] ^{\frac{\rho }{\alpha }-1} \, {\hbox {E}}_{p} e^{(\alpha -1)\ln {\frac{x_2}{x_1}}}\\&\Rightarrow \frac{\mathrm{d}x_1}{\mathrm{d}x_2} = - \beta \left[ e^{\alpha \mu + \alpha ^2 \frac{\sigma ^2}{2}} \, \right] ^{\frac{\rho }{\alpha }-1} \, e^{(\alpha -1)\mu + (1-\alpha )^2\frac{\sigma ^2}{2}}\\&\Rightarrow \frac{\mathrm{d}x_1}{\mathrm{d}x_2} = - \beta e^{\rho \mu + \alpha \rho \frac{\sigma ^2}{2} -\alpha \mu - \alpha ^2 \frac{\sigma ^2}{2}} \, e^{(\alpha -1)\mu + (1-\alpha )^2\frac{\sigma ^2}{2}}\\&\Rightarrow \frac{\mathrm{d}x_1}{\mathrm{d}x_2} = - \beta e^{(\rho -1) \mu + (\alpha \rho + 1-2\alpha ) \frac{\sigma ^2}{2}}\\&\Rightarrow \frac{\mathrm{d}x_1}{\mathrm{d}x_2} = - \beta e^{(\rho -1) \mu + (\alpha \rho + 1-2\alpha ) \frac{\sigma ^2}{2}}. \end{aligned}$$

The second part of the proof translates the relation into rates by defining the social discount rate $r=-\ln \frac{\mathrm{d}x_1}{-\mathrm{d}x_2}\,\left( =-\ln \frac{\mathrm{d}x_2}{-\mathrm{d}x_1}|_{\bar{U}}\right)$, the rate of pure time preference $\delta =-\ln \beta$, and $\eta =1-\rho \,\left( =\frac{1}{\sigma }\right)$. Further below, I make use of the relation $1=\frac{1-\eta }{\rho }$.

$$\begin{aligned}&\Rightarrow r = \delta + \left( 1-\rho \right) \mu - \left( \alpha \left( \rho -1\right) + 1-\alpha \right) \frac{\sigma ^2}{2} \end{aligned}$$

(16)

$$\begin{aligned}&\Rightarrow r = \delta + \eta \mu - \eta ^2 \frac{\sigma ^2}{2} + \left( \eta ^2 + \alpha \left( \eta +1\right) - 1 \right) \frac{\sigma ^2}{2} \nonumber \\&\Rightarrow r = \delta + \eta \mu - \eta ^2 \frac{\sigma ^2}{2} + \left( \eta ^2 + \frac{\alpha }{\rho }\left( 1-\eta \right) \left( \eta +1\right) - 1 \right) \frac{\sigma ^2}{2} \nonumber \\&\Rightarrow r = \delta + \eta \mu - \eta ^2 \frac{\sigma ^2}{2} + \left( \eta ^2 + \frac{\alpha }{\rho }\left( 1-\eta ^2\right) - 1 \right) \frac{\sigma ^2}{2}\nonumber \\&\Rightarrow r = \delta + \eta \mu - \eta ^2 \frac{\sigma ^2}{2} - \left( 1-\frac{\alpha }{\rho }\right) \left( 1-\eta ^2\right) \frac{\sigma ^2}{2}\nonumber \\&\Rightarrow r = \delta + \eta \mu - \eta ^2 \frac{\sigma ^2}{2} - {{\mathrm{RIRA}}}\, |1-\eta ^2| \frac{\sigma ^2}{2}. \end{aligned}$$

(17)

$\square$

Proof of Proposition 2

For the isoelastic specification and with the definition,

$$\begin{aligned} U_2(\epsilon ) = f^{-1} \left[ {\hbox {E}}_{p(x_2,y)} f\circ u(x_2+\epsilon y) \right] =\frac{1}{\rho } \left[ {\hbox {E}}_{p(x_2,y)} (x_2+\epsilon y)^\alpha \, \right] ^{\frac{\rho }{\alpha }} \end{aligned}$$

Eq. (9) translates into

$$\begin{aligned} x_1^{\rho -1} \, \mathrm{d}x_1+\beta \left. \frac{\mathrm{d}}{\mathrm{d}\epsilon } U_2(\epsilon ) \right| _{\epsilon =0}\mathrm{d}\epsilon \mathop {=}\limits ^{!} 0 \end{aligned}$$

(18)

In order to calculate $\left. \frac{\mathrm{d}}{\mathrm{d}\epsilon } U_2(\epsilon ) \right| _{\epsilon =0}\mathrm{d}\epsilon$ the following definition is useful.

$$\begin{aligned} V_\epsilon (a,b)= {\hbox {E}}_{p(x_2,y)} (x_2+\epsilon y)^a y^b. \end{aligned}$$

(19)

Then,

$$\begin{aligned} \left. \frac{\mathrm{d}}{\mathrm{d}\epsilon } U_2(\epsilon ) \right| _{\epsilon =0}&= \left. \frac{1}{\alpha }V_\epsilon (\alpha ,0)^{\frac{\rho }{\alpha }-1} \alpha V_\epsilon (\alpha -1,1) \right| _{\epsilon =0}\nonumber \\&= V_0(\alpha ,0)^{\frac{\rho }{\alpha }-1} V_0(\alpha -1,1) \end{aligned}$$

(20)

where equality between the first and the second lines follow from Lebesgue’s dominated convergence theorem. Analogously to step 1 in the proof of Proposition 1, I calculate with $z=\ln y$

$$\begin{aligned} V_0(\alpha ,0)&= x_1^{\alpha } \, {\hbox {E}}_{p(x_2,y)} \Big (\frac{x_2}{x_1}\Big )^\alpha =x_1^{\alpha } \, {\hbox {E}}_{p(g,z)} e^{\alpha g}\nonumber \\&= x_1^{\alpha } \, \int \limits _{-\infty }^{\infty } \int \limits _{-\infty }^{\infty } e^{\alpha g} \frac{e^{-\frac{1}{2(1-\kappa ^2)} \left[ \left( \frac{g-\mu _g}{\sigma _g}\right) ^2 + \left( \frac{z-\mu _y}{\sigma _y}\right) ^2 - 2\kappa \left( \frac{g-\mu _g}{\sigma _g}\right) \left( \frac{z-\mu _y}{\sigma _y}\right) \right] }}{ 2 \pi \sigma _g \sigma _y \sqrt{1-\rho ^2}} \mathrm{d}g \ \mathrm{d}z\qquad \quad \end{aligned}$$

(21)

$$\begin{aligned}&= x_1^{\alpha } \, e^{\alpha \mu _g+ \alpha ^2 \frac{\sigma _g^2}{2}}. \end{aligned}$$

(22)

Similarly,

$$\begin{aligned} V_0(\alpha -1,1)&= x_1^{\alpha -1} \, {\hbox {E}}_{p(x,y)} \Big (\frac{x_2}{x_1}\Big )^{\alpha -1} y = x_1^{\alpha -1} \, {\hbox {E}}_{p(g,z)} e^{(\alpha -1)g+z}\\&= x_1^{\alpha -1} \, e^{-(1-\alpha ) \left[ \mu _g - (1-\alpha ) \frac{\sigma _g^2}{2} + \kappa \sigma _g\sigma _y \right] +\mu _y +\frac{\sigma _y^2}{2}} \end{aligned}$$

so that

$$\begin{aligned} \left. \frac{\mathrm{d}}{\mathrm{d}\epsilon } U_2(\epsilon ) \right| _{\epsilon =0}&= x_1^{\rho -\alpha +\alpha -1} e^{(\alpha \mu _g+ \alpha ^2 \frac{\sigma _g^2}{2})({\frac{\rho }{\alpha }-1})} e^{-(1-\alpha ) \left[ \mu _g - (1-\alpha ) \frac{\sigma _g^2}{2} + \kappa \sigma _g\sigma _y \right] +\mu _y +\frac{\sigma _y^2}{2}}\nonumber \\&= x_1^{\rho -1} e^{(\rho -1) \mu _g+ \left[ \alpha (\rho -1)+(1-\alpha )\right] \frac{\sigma _g^2}{2} -(1-\alpha ) \kappa \sigma _g\sigma _y +\mu _y +\frac{\sigma _y^2}{2}}. \end{aligned}$$

(23)

Substituting the result into Eq. (18) and solving for the discount rate yields

$$\begin{aligned} r = \ln \frac{\mathrm{d}\epsilon }{-\mathrm{d}x_1}&= \delta + (1-\rho ) \mu _g - [\alpha (\rho -1) + 1-\alpha ] \frac{\sigma ^2}{2}\\&+ (1-\alpha ) \kappa \sigma _g\sigma _y - \left( \mu _y +\frac{\sigma _y^2}{2}\right) . \end{aligned}$$

The first line corresponds to Eq. (16) and, thus, Eq. (17), yielding the risk-free discount rate under intertemporal risk aversion. Moreover, the random variable $y$ was assumed to yield an expected value (project payoff) of unity, which implies

$$\begin{aligned} {\hbox {E}}_{p(x,y)} y = e^{\mu _y +\frac{\sigma _y^2}{2}} \mathop {=}\limits ^{!} 1 \qquad \Rightarrow \quad \mu _y + \frac{\sigma _y^2}{2}=0, \end{aligned}$$

eliminating the last bracket. Finally, $1-\alpha$ has to be expressed in terms of $\eta$ (capturing the effects of the standard model) and ${{\mathrm{RIRA}}}$ (capturing the additional effects of intertemporal risk averison). I find for $\rho >0$ that

$$\begin{aligned} 1-\alpha = 1-(1-\eta )(1-{{\mathrm{RIRA}}}) = \eta + (1-\eta ) {{\mathrm{RIRA}}}\end{aligned}$$

and for $\rho <0$ that

$$\begin{aligned} 1-\alpha = 1-(1-\eta )(1+{{\mathrm{RIRA}}}) = \eta - (1-\eta ) {{\mathrm{RIRA}}}. \end{aligned}$$

In both cases, this yields

$$\begin{aligned} 1-\alpha = \eta + |1-\eta | {{\mathrm{RIRA}}}, \end{aligned}$$

(24)

which gives rise to the form stated in the proposition.

Proof of Corollary 1

In case 1 of the risk-free discount rate, Eq. (8) translates $r_{50} = 50 \delta$ into the condition $\eta \, 50\mu \mathop {=}\limits ^{!} \eta ^2 \frac{\sigma ^2}{2} + {{\mathrm{RIRA}}}\, |1-\eta ^2| \frac{\sigma ^2}{2}$, which results in the stated equation for $\sigma$. Similarly in case 2, Eq. (10), $\sigma =\sigma _g=\sigma _y$, and $\eta \, 50\mu _g \mathop {=}\limits ^{!} \eta ^2 \frac{\sigma _g^2}{2} + {{\mathrm{RIRA}}}\, |1-\eta ^2| \frac{\sigma _g^2}{2}- \eta \kappa \ \sigma _g \sigma _y - |1-\eta | {{\mathrm{RIRA}}}\kappa \ \sigma _g \sigma _y$ yield the result. Without the condition $\sigma =\sigma _g=\sigma _y$, the same reasoning gives statement 3 of the corollary.

Proof of Proposition 3

Define for the isoelastic specification

$$\begin{aligned} U_2^a(\epsilon )&= \varPhi ^{-1} \left\{ \int \limits _\varTheta \varPhi \left[ f^{-1}{\hbox {E}}_{p_\theta (x_2,y)} f\circ u(x_2+\epsilon y) \right] \mathrm{d}\pi (\theta ) \right\} \\&= \frac{1}{\rho }\left\{ \int \limits _\varTheta \left[ {\hbox {E}}_{p_\theta (x_2,y)} (x_2+\epsilon y)^\alpha \right] ^{\frac{\rho }{\alpha }\varphi } \mathrm{d}\pi (\theta ) \right\} ^{\frac{1}{\varphi }}. \end{aligned}$$

I have to solve once more the equation

$$\begin{aligned} \mathrm{d}V(x_1,p,\pi )= x_1^{\rho -1} \, \mathrm{d}x_1+\beta \left. \frac{\mathrm{d}}{\mathrm{d}\epsilon } U^a_2(\epsilon ) \right| _{\epsilon =0}\mathrm{d}\epsilon \mathop {=}\limits ^{!} 0 \end{aligned}$$

(25)

for $r=\ln \frac{\mathrm{d}\epsilon }{-\mathrm{d}x_1}$. Making once more use of the definition

$$\begin{aligned} V_\epsilon (a,b)= {\hbox {E}}_{p_\theta (x_2,y)} (x_2+\epsilon y)^a y^b \,, \end{aligned}$$

where $\theta$ replaces $\mu _g$ in $p_{(x,y)}$ of Eqs. (19) and (21), I find

$$\begin{aligned} \left. \frac{\mathrm{d}}{\mathrm{d}\epsilon } U_2^a(\epsilon ) \right| _{\epsilon =0}&= \frac{1}{\rho }\frac{1}{\varphi }\left\{ \int \limits _\varTheta V_\epsilon (\alpha ,0)^{\frac{\rho }{\alpha }\varphi } \mathrm{d}\pi (\theta ) \right\} ^{\frac{1}{\varphi }-1}\nonumber \\&\left. \left\{ \int \limits _\varTheta {\frac{\rho }{\alpha }\varphi } V_\epsilon (\alpha ,0)^{\frac{\rho }{\alpha }\varphi -1} \alpha V_\epsilon (\alpha -1,1) \mathrm{d}\pi (\theta ) \right\} \right| _{\epsilon =0}\nonumber \\&= \left\{ \int \limits _\varTheta V_0(\alpha ,0)^{\frac{\rho \varphi }{\alpha }} \mathrm{d}\pi (\theta ) \right\} ^{\frac{1}{\varphi }-1}\nonumber \\&\int \limits _\varTheta V_0(\alpha ,0)^{\frac{\rho \varphi }{\alpha }-1} V_0(\alpha -1,1) \mathrm{d}\pi (\theta ). \end{aligned}$$

(26)

With the help of Eq. (22), the expression $\{ \cdot \}$ calculates to

$$\begin{aligned} \int \limits _\varTheta x_1^{\alpha (\frac{\rho \varphi }{\alpha })} \, e^{(\alpha \theta + \alpha ^2 \frac{\sigma _g^2}{2})(\frac{\rho \varphi }{\alpha })} \mathrm{d}\pi (\theta )&= x_1^{\rho \varphi } e^{\rho \varphi \alpha \frac{\sigma _g^2}{2}}\int \limits _\varTheta e^{\rho \varphi \theta } \frac{e^{-\frac{1}{2} \left( \frac{\theta -\mu _g}{\sigma _g}\right) ^2}}{ \sqrt{2 \pi } \sigma _g } \mathrm{d}\theta \\&= x_1^{\rho \varphi } e^{\rho \varphi \alpha \frac{\sigma _g^2}{2}} e^{\rho \varphi \mu _g+\rho ^2\varphi ^2\frac{\tau _g^2}{2}}. \end{aligned}$$

Acknowledging the equality of Eqs. (20) and (23) and their similarity to the second integrand in Eq. (26) (for $\rho \leftrightarrow \rho \varphi$), this second integral becomes

$$\begin{aligned}&\int \limits _\varTheta V_0(\alpha ,0)^{\frac{\rho \varphi }{\alpha }-1} V_0(\alpha -1,1) \mathrm{d}\pi (\theta ) \\&\quad =\int \limits _\varTheta x_1^{\rho \varphi -1} e^{(\rho \varphi -1) \theta + \left[ \alpha (\rho \varphi -1)+(1-\alpha )\right] \frac{\sigma _g^2}{2} -(1-\alpha ) \kappa \sigma _g\sigma _y +\mu _y +\frac{\sigma _y^2}{2}} \mathrm{d}\pi (\theta ) \\&\quad =x_1^{\rho \varphi -1} e^{\left[ \alpha (\rho \varphi -1)+(1-\alpha )\right] \frac{\sigma _g^2}{2} -(1-\alpha ) \kappa \sigma _g\sigma _y +\mu _y +\frac{\sigma _y^2}{2}} \int \limits _\varTheta e^{(\rho \varphi -1) \theta } \mathrm{d}\pi (\theta ) \\&\quad =x_1^{\rho \varphi -1} e^{\left[ \alpha (\rho \varphi -1)+(1-\alpha )\right] \frac{\sigma _g^2}{2} -(1-\alpha ) \kappa \sigma _g\sigma _y +\mu _y +\frac{\sigma _y^2}{2}} e^{(\rho \varphi -1)\mu _g + (\rho \varphi -1)^2\frac{\tau _g^2}{2}}. \end{aligned}$$

Substituting these results back into Eq. (26) delivers

$$\begin{aligned} \left. \frac{\mathrm{d}}{\mathrm{d}\epsilon } U_2^a\left( \epsilon \right) \right| _{\epsilon =0}&= x_1^{\rho \varphi \left( \frac{1}{\varphi }-1\right) } e^{\left( \rho \varphi \alpha \frac{\sigma _g^2}{2}\right) \left( \frac{1}{\varphi }-1\right) } e^{\left( \rho \varphi \mu _g+\rho ^2\varphi ^2\frac{\tau _g^2}{2}\right) \left( \frac{1}{\varphi }-1\right) }\\&x_1^{\rho \varphi -1} e^{\left[ \alpha \left( \rho \varphi -1\right) +\left( 1-\alpha \right) \right] \frac{\sigma _g^2}{2} -\left( 1-\alpha \right) \kappa \sigma _g\sigma _y +\mu _y +\frac{\sigma _y^2}{2}} e^{\left( \rho \varphi -1\right) \mu _g + \left( \rho \varphi -1\right) ^2\frac{\tau _g^2}{2}}\\&= x_1^{\rho -1} e^{\left[ \alpha \left( \rho -1\right) +\left( 1-\alpha \right) \right] \frac{\sigma _g^2}{2}+\left( \rho -1\right) \mu _g +\left[ \rho \varphi \left( \rho -1\right) +1-\rho \varphi \right] \frac{\tau _g^2}{2}-\left( 1-\alpha \right) \kappa \sigma _g\sigma _y +\mu _y +\frac{\sigma _y^2}{2}}. \end{aligned}$$

Substituting this result into Eq. (25) and solving for $r=\ln \frac{\mathrm{d}\epsilon }{-\mathrm{d}x_1}$ yields analogously to the proof of Proposition 2, the discount rate

$$\begin{aligned} r&= \delta + \eta \mu _g - \eta ^2 \frac{\sigma _g^2}{2} - {{\mathrm{RIRA}}}\, |1-\eta ^2| \frac{\sigma _g^2}{2} + \ \eta \, \kappa \ \sigma _g \sigma _y \\&+ |1-\eta | {{\mathrm{RIRA}}}\ \kappa \ \sigma _g \sigma _y - [1-2\rho \varphi +\rho ^2\varphi ]\frac{\tau _g^2}{2}. \end{aligned}$$

The last term can be rearranged to the form

$$\begin{aligned}{}[1-2\rho \varphi +\rho ^2\varphi ]\frac{\tau _g^2}{2}&= [(1-\varphi )+\varphi (1-\rho )-\varphi \rho (1-\rho )]\frac{\tau _g^2}{2}\\&= [(1-\varphi )+(1-\rho )^2+(\varphi -1)(1-\rho )^2]\frac{\tau _g^2}{2}\\&= [\eta ^2+(1-\varphi )(1-\eta ^2)]\frac{\tau _g^2}{2} =\eta ^2 \frac{\tau _g^2}{2} + {{\mathrm{RAA}}}|1-\eta ^2| \frac{\tau _g^2}{2}, \end{aligned}$$

completing the proof. $\square$

Proof of Proposition 4

Up to Eq. (26), the proof is identical to that of Proposition 3. In the next step, in $V_0(\alpha ,0)^{\frac{\rho \varphi }{\alpha }}$ the ambiguity parameter $\theta$ replaces $\kappa$ instead of $\mu _g$. Thus, the first integral in Eq. (26) becomes

$$\begin{aligned} \int \limits _\varTheta x_1^{\alpha (\frac{\rho \varphi }{\alpha })} \, e^{(\alpha \mu _g+ \alpha ^2 \frac{\sigma _g^2}{2})(\frac{\rho \varphi }{\alpha })} \mathrm{d}\pi (\theta )&= x_1^{\rho \varphi } e^{\rho \varphi \mu _g+\rho \varphi \alpha \frac{\sigma _g^2}{2}} \int \limits _{-1}^{1} \frac{1}{2} \mathrm{d}\theta \\&= x_1^{\rho \varphi } e^{\rho \varphi \mu _g+\rho \varphi \alpha \frac{\sigma _g^2}{2}}. \end{aligned}$$

For the integrand of the second integral in Eq. (26), I find

$$\begin{aligned} V_0(\alpha -1,1)= x_1^{\alpha -1} e^{(\alpha -1) \mu _g+(\alpha -1)^2 \frac{\sigma _g^2}{2}+(\alpha -1) \theta \sigma _g \sigma _y+\mu _y+\frac{\sigma _y^2}{2}} \end{aligned}$$

delivering the integral

$$\begin{aligned}&\int \limits _\varTheta V_0(\alpha ,0)^{\frac{\rho \varphi }{\alpha }-1} V_0(\alpha -1,1) \mathrm{d}\pi (\theta ) \\&\quad =\int \limits _\varTheta x_1^{\rho \varphi -\alpha } e^{\rho \varphi \mu _g+\rho \varphi \alpha \frac{\sigma _g^2}{2}-\alpha \mu _g-\alpha ^2 \frac{\sigma _g^2}{2}}\\&\quad x_1^{\alpha -1} e^{(\alpha -1) \mu _g+(\alpha -1)^2 \frac{\sigma _g^2}{2}+(\alpha -1) \theta \sigma _g \sigma _y+\mu _y+\frac{\sigma _y^2}{2}} \mathrm{d}\pi (\theta ) \\&\quad =x_1^{\rho \varphi -1} e^{(\rho \varphi -1) \mu _g+(\rho \varphi \alpha -2\alpha -1) \frac{\sigma _g^2}{2} +\mu _y+\frac{\sigma _y^2}{2}} \int \limits _{-1}^{1} e^{(\alpha -1) \theta \sigma _g \sigma _y} \frac{1}{2} \mathrm{d}\theta \\&\quad =x_1^{\rho \varphi -1} e^{(\rho \varphi -1) \mu _g+(\rho \varphi \alpha -2\alpha -1) \frac{\sigma _g^2}{2} +\mu _y+\frac{\sigma _y^2}{2}} \quad \frac{\sinh [(\alpha -1) \sigma _g \sigma _y]}{(\alpha -1)\sigma _g\sigma _y}. \end{aligned}$$

Substituting these results back into Eq. (26) returns the second-period welfare change in $\epsilon$:

$$\begin{aligned} \left. \frac{\mathrm{d}}{\mathrm{d}\epsilon } U_2^a(\epsilon )\right| _{\epsilon =0}&= x_1^{\rho \varphi \left( \frac{1}{\varphi }-1\right) } e^{\left( \rho \varphi \mu _g+\rho \varphi \alpha \frac{\sigma _g^2}{2}\right) \left( \frac{1}{\varphi }-1\right) }\\&x_1^{\rho \varphi -1} e^{(\rho \varphi -1) \mu _g+(\rho \varphi \alpha -2\alpha -1) \frac{\sigma _g^2}{2} +\mu _y+\frac{\sigma _y^2}{2}} \quad \frac{\sinh [(\alpha -1) \sigma _g \sigma _y]}{(\alpha -1)\sigma _g\sigma _y}\\&= x_1^{\rho -1} e^{(\rho -1) \mu _g +(\rho \alpha -2\alpha -1) \frac{\sigma _g^2}{2} +\mu _y+\frac{\sigma _y^2}{2}} \quad \frac{\sinh [(\alpha -1) \sigma _g \sigma _y]}{(\alpha -1)\sigma _g\sigma _y}. \end{aligned}$$

Substituting this result into Eq. (25) and solving for $r=\ln \frac{\mathrm{d}\epsilon }{-\mathrm{d}x_1}$ yields analogously to the proof of Proposition 2, the discount rate

$$\begin{aligned} r =&\delta + \eta \mu _g - \eta ^2 \frac{\sigma _g^2}{2} - {{\mathrm{RIRA}}}\, |1-\eta ^2| \frac{\sigma _g^2}{2} - \ln \left[ \frac{\sinh [(\alpha -1) \sigma _g \sigma _y]}{(\alpha -1)\sigma _g\sigma _y} \right] . \end{aligned}$$

By symmetry of the hyperbolic sine, the sign of $(\alpha -1)$ can be flipped simultaneously in the numerator and the denominator. Using Eq. (24) to substitute for $(1-\alpha )$ then yields the result stated in the proposition. $\square$

Appendix 2

The following proposition formalizes how intertemporal risk aversion, defined in the sense of Eq. (4), translates into the curvature of the function $f$ in a preference representation of the form (3).³⁴

Proposition 5

Let preferences over $X\times P$ be represented by Eq. (3) with a continuous function $u:X\rightarrow {\mathbb R}$ and a strictly increasing and continuous function $f:U\rightarrow {\mathbb R}$, where $U=u(X)$ and $\beta =1$.

• (a) The corresponding decision maker is (weakly) intertemporal risk averse [loving], if and only if the function $f$ is concave [convex].

• (b) The corresponding decision maker is intertemporal risk neutral, if and only if there exist $a,b \in {\mathbb R}$ such that $f(z)=a z + b$. An intertemporal risk neutral decision maker maximizes intertemporally additive expected utility (Eq. (1)).

Proof of Proposition 5

(a) Sufficiency of axiom (4): The premise of axiom (4) translates with $\beta =1$ into the representation (3) as

$$\begin{aligned} \displaystyle \left( x^*,x^*\right)&\sim \displaystyle \left( x_1,x_2\right) \nonumber \\ \Leftrightarrow \displaystyle u(x^*) + u(x^*)&= \displaystyle u(x_1) + u(x_2)\\ \Leftrightarrow \displaystyle u(x^*)&= \displaystyle \frac{1}{2} u(x_1) + \frac{1}{2} u(x_2)\nonumber \end{aligned}$$

(27)

Writing the implication of the axiom in terms of representation (3) yields

$$\begin{aligned} \begin{array}{llll} &{} (x^*,x^*) &{} \succ &{} \displaystyle \left( x^*,\frac{1}{2} x_1 + \frac{1}{2} x_2\right) \\ \Leftrightarrow &{} u(x^*) + &{} \ge &{} \displaystyle f^{-1} \left( \frac{1}{2} f\circ u(x_1) + \frac{1}{2} f\circ u(x_2)\right) . \end{array} \end{aligned}$$

(28)

Combining Eqs. (27) and (28) returns

$$\begin{aligned} \begin{array}{lccc}&\displaystyle \frac{1}{2} u(x_1) + \frac{1}{2} u(x_2)&\ge \displaystyle f^{-1} \left( \frac{1}{2} f\circ u(x_1) + \frac{1}{2} f\circ u(x_2)\right) , \end{array} \end{aligned}$$

(29)

which for an increasing [decreasing] version of $f$ is equivalent to

$$\begin{aligned} \begin{array}{lccc} \Leftrightarrow&\displaystyle f \left( \frac{1}{2} u(x_1) + \frac{1}{2} u(x_2) \right)&>\; [<]\;&\displaystyle \frac{1}{2} f\circ u(x_1) + \frac{1}{2} f\circ u(x_2). \end{array} \end{aligned}$$

Defining $z_i=u(x_i)$, the equation becomes

$$\begin{aligned} \begin{array}{lccc} \Leftrightarrow&\displaystyle f \left( \frac{1}{2} z_1 + \frac{1}{2} z_2 \right)&\ge \; [\le ]\;&\displaystyle \frac{1}{2} f (z_1) + \frac{1}{2} f (z_2). \end{array} \end{aligned}$$

(30)

Because preferences are assumed to be representable in the form (3), there exists a certainty equivalent $x^*$ to all lotteries $\frac{1}{2} x_1 + \frac{1}{2} x_2$ with $x_1,x_2 \in X$. Taking $x^*$ to be the certainty equivalent, the premise and, thus, Eq. (30) have to hold for all $z_1,z_2 \in u(X)$. Therefore, $f$ has to be concave [convex] on $U(x)$ ((Hardy et al. 1964, 75)).

Necessity of axiom 4 The necessity is seen to hold by going backward through the proof of sufficiency above. Strict concavity [convexity] of $f$ with $f$ increasing [decreasing] implies that Eq. (30) and, thus, Eq. (29) have to hold for $z_1,z_2 \in u(X)$. The premise corresponding to (27) guarantees that Eq. (29) implies Eq. (28) which yields the implication in condition (4). Replacing $\succeq$ by $\preceq$ and $\ge$ by $\le$ in the proof above implies that the decision maker is intertemporal risk averse, if and only if $f$ is convex [for an increasing version of $f$ and concave for $f$ decreasing].

(b) The decision maker is intertemporal risk neutral, if and only if $f$ is concave and convex on $u(X)$, which is equivalent to $f$ being linear.³⁵ However, a linear function $f$ cancels out in representation (3) and makes it identical to the intertemporally additive expected utility standard representation (1).

Appendix 3

1.1 Relation to Proposition 1 and Gollier (2002)

Equation (14) in Gollier (2002) is

$$\begin{aligned} r=\delta + \eta \ {\hbox {E}}\ \tilde{g} - {{\mathrm{RRA}}}(1+\eta ) \frac{\text{ Var }(\tilde{g})}{2} \ , \end{aligned}$$

where $\delta =\frac{1}{\beta }-1,\,\tilde{g}=\frac{\tilde{x}_2}{x_1}-1$, and $\beta$ denotes Gollier’s utility discount factor. The tilde ($\tilde{}$) marks random variables. In contrast, Eq. (16) for the isoelastic normal case translates into

$$\begin{aligned} r = \delta + \eta \mu - ((1-{{\mathrm{RRA}}}) (-\eta ) + {{\mathrm{RRA}}}) \frac{\sigma ^2}{2} = \delta + \eta \mu - {{\mathrm{RRA}}}(1+\eta ) \frac{\sigma ^2}{2} + \eta \frac{\sigma ^2}{2}. \end{aligned}$$

Thus, Gollier’s approximate formula underestimates the social discount rate by the term $\eta \frac{\sigma ^2}{2}$. Much of the difference can be traced back to the first step of his derivation, where the Arrow–Pratt approximation of the certainty equivalent (or equivalently the certainty-equivalent growth rate) loses a term proportional to $\frac{\sigma ^2}{2}$ in a setting with isoelastic preferences and a normal growth rate.

In a setting with CARA utility and a normal distribution of the growth rate Gollier’s approximation does better. I assume that Gollier’s growth rate $\tilde{g}=\frac{\tilde{x}_2}{x_1}-1\sim N(\mu ,\sigma ^2)$, which is equivalent to a distribution of second-period consumption $\tilde{x}_2 \sim N\big (x_1(1+\mu ),x_1^2\sigma ^2\big ) \equiv N(\mu _x,\sigma _x^2)$. Moreover, the assumption of CARA aggregators translates into $u(x)=-\frac{\exp (-A_u x)}{A_u}$ for utility and $v(x)=-\frac{\exp (-A_v x)}{A_v}$ for Arrow–Pratt risk aversion, adopting Gollier’s notation where $A_u$ is absolute aversion to intertemporal substitution and $A_v$ is absolute aversion to risk in the Arrow–Pratt sense. Then, the resulting exact discount rate becomes

$$\begin{aligned} r&= \frac{1}{\beta } \exp \left( A_u \left[ \mu _x - x_1 - A_v \frac{\sigma _x^2}{2} \right] \right) \end{aligned}$$

(31)

$$\begin{aligned}&\approx \delta + A_u \ (\mu _x - x_1) - A_u \ A_v \frac{\sigma _x^2}{2} = \delta + A_u \ x_1 \ \mu - A_u \ A_v \ x_1^2 \frac{\sigma ^2}{2}. \end{aligned}$$

(32)

The second line is equivalent to Gollier’s equation (12), before replacing his absolute aversion measures with relative aversion measures. Thus, in the CARA-normal case, the quality of his approximation is given by the quality of the approximation going from Eqs. (31) to (32). For this approximation to hold, risk has to be moderate and expected absolute growth has to be small. Small absolute growth is equivalent to a low expected growth rate or a low present consumption level.