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Hyperbolic discounting in an intergenerational model with altruistic parents

Abstract

Hyperbolic utility discounting has emerged as a leading alternative to exponential discounting because it can explain time-inconsistent behaviors. Intuitively, hyperbolic discounting should play a crucial role in intergenerational models characterized by intertemporal trade-offs. In this paper, we incorporate hyperbolic discounting into a dynamic model in which parents are altruistic towards their children. Agents live for four periods and choose levels of consumption, fertility, investment in their children’s human capital, and bequests to maximize discounted utility. In the steady state, hyperbolic discounting decreases fertility, increases human capital investment, and shifts consumption towards younger ages. These effects are more pronounced in the time-consistent problem (in which agents cannot commit to a course of action) than in the commitment problem, which can be interpreted as realized and intended actions, respectively. The preference-based discrepancy between intended fertility and realized fertility has important implications for the empirical literature that compares the two.

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Notes

  1. It is relatively easier to place the education decision after the fertility decision is a static model, such as that in Nakagawa et al. (forthcoming).

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Acknowledgement

We would like to thank the editor, Alessandro Cigno, and three anonymous reviewers for their important comments. We also want to thank Otávio Bartalotti, Elizabeth Hoffman, Joshua Rosenbloom, Sunanda Roy, Quinn Weninger, John Winters, Wendong Zhang, and participants in the Graduate Students Seminar at Iowa State University for helpful comments. We thank Bryan Jackson for proofreading the manuscript. The authors are responsible for all remaining errors.

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Correspondence to Minghao Li.

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Appendix

Appendix

Appendix 1. The proof of Proposition 1

Proof. Let λ be the Lagrangian multiplier associated with the budget constraint. Optimal consumption choices satisfy

$$ \lambda ={u}^{\prime}\left({c}_1\right)=\beta \delta R{u}^{\prime}\left({c}_2\right)=\beta {\delta}^2{R}^2{u}^{\prime}\left({c}_3\right) $$
(8)

The optimal human capital investment, fertility, and bequest conditions read

$$ {\displaystyle \begin{array}{c}\frac{\lambda n}{R}=\beta {\delta}^2\varPhi (n)\frac{\partial {V}_1^c\left(e^{\prime },b^{\prime}\right)}{\partial e^{\prime }}\\ {}\lambda \left[\frac{\left(1+g\right)h(e)x+e^{\prime }}{R}+\frac{b^{\prime }}{R^2}\right]=\beta {\delta}^2\varPhi^{\prime }(n){V}_1^c\left(e^{\prime },b^{\prime}\right)\\ {}\frac{\lambda n}{R^2}=\beta {\delta}^2\varPhi (n)\frac{\partial {V}_1^c\left(e^{\prime },b^{\prime}\right)}{\partial b^{\prime }}\end{array}} $$

From the Envelope Theorem (ET thereafter), we have

$$ {\displaystyle \begin{array}{c}\frac{\partial {{\mathrm{V}}_1^{\mathrm{c}}}_1\left(e,b\right)}{\partial e}=\lambda \left[1+\frac{\left(1+g\right)\left(1- nx\right)}{R}\right]h^{\prime }(e)\\ {}\frac{\partial {\mathrm{V}}_1^{\mathrm{c}}\left(e,b\right)}{\partial b}=\lambda \end{array}} $$

In the steady state, the equation above and the bequest condition decide fertility

$$ \beta {\delta}^2{R}^2\frac{\varPhi \left({n}^c\right)}{n^c}=1 $$

Appendix 2. The proof of Proposition 2

We use backward induction to solve the time-consistent problem: first, Self 3 chooses c3 and b′ given e, a3 and n; then, Self 2 chooses c2, n, e, and a3 given a2; at last, Self 1 chooses c1 and a2 given e and b. The three problems are described as follows.

Self 3 solves the following maximization problem:

$$ {V}_3^t\left({e}^{\prime },{a}_3,n\right)=\underset{c_3,b^{\prime }}{\max}\left\{u\left({c}_3\right)+\varPhi (n){V}_1^t\left({e}^{\prime },b^{\prime}\right)\right\} $$

subject to

$$ {c}_3={a}_3- nb^{\prime } $$

The solutions to Self 1’s problem are denoted by \( {c}_3={c}_3^t\left({e}^{\prime },{a}_3,n\right),b={b}^t\left({e}^{\prime },{a}_3,n\right) \).

Self 2 solves the following maximization problem:

$$ {V}_2^t\left(e,{a}_2\right)=\underset{c_2,{a}_3,n,e^{\prime }}{\max}\left\{u\left({c}_2\right)+\beta \delta {V}_3^t\left({e}^{\prime },{a}_3,n\right)\right\} $$

subject to

$$ {c}_2={a}_2+\left(1+g\right)h(e)\left(1- nx\right)-n{e}^{\prime }-\frac{a_3}{R} $$

The solutions to Self 2’s problem are denoted by \( {c}_2={c}_2^t\left(e,{a}_2\right),{a}_3={a}_3^t\left(e,{a}_2\right),n={n}^t\left(e,{a}_2\right) \).

Self 1 solves the following maximization problem:

$$ {V}_1^t\left(e,b\right)={\max}_{c_1,{a}_2}\left\{u\left({c}_1\right)+\beta \delta {V}_2^t\left(e,{a}_2\right)\right\} $$

subject to

$$ {c}_1=h(e)+b-\frac{a_2}{R} $$

The solution to Self 1’s problem is denoted by \( {c}_1={c}_1^t\left(e,b\right),{a}_2={a}_2^t\left(e,b\right) \).

Self 3’s first-order condition (FOC thereafter) with respect to bequests is

$$ u^{\prime}\left({c}_3\right)n=\varPhi (n)\frac{\partial {V}_1^t\left(e^{\prime },b^{\prime}\right)}{\partial b^{\prime }} $$

Self 3’s ET conditions are

$$ {\displaystyle \begin{array}{c}\frac{\partial {V}_3^t\left({e}^{\prime },{a}_3,n\right)}{\partial {e}^{\prime }}=\varPhi (n)\frac{\partial {V}_1^t\left(e^{\prime },b^{\prime}\right)}{\partial e^{\prime }}\\ {}\frac{\partial {V}_3^t\left({e}^{\prime },{a}_3,n\right)}{\partial {a}_3}=u^{\prime}\left({c}_3\right)\\ {}\frac{\partial {V}_3^t\left({e}^{\prime },{a}_3,n\right)}{\partial n}=-{u}^{\prime}\left({c}_3\right){b}^{\prime }+\varPhi^{\prime }(n){V}_1^t\left(e^{\prime },b^{\prime}\right)\end{array}} $$

Self 2’s FOCs with respect to e, a3, n are

$$ {\displaystyle \begin{array}{c}u^{\prime}\left({c}_2\right)n=\beta \delta \frac{\partial {V}_3^t\left({e}^{\prime },{a}_3,n\right)}{\partial e^{\prime }}\\ {}\frac{u^{\prime}\left({c}_2\right)}{R}=\beta \delta \frac{\partial {V}_3^t\left({e}^{\prime },{a}_3,n\right)}{\partial {a}_3}\\ {}{u}^{\prime}\left({c}_2\right)\left(1+g\right)h(e)x=\frac{\partial {V}_3^t\left({e}^{\prime },{a}_3,n\right)}{\partial n}\end{array}} $$

Self 2’s ET conditions are

$$ {\displaystyle \begin{array}{c}\frac{\partial {V}_2^t\left(e,{a}_2\right)}{\partial e}={u}^{\prime}\left({c}_2\right)\left(1+g\right)h^{\prime }(e)\left(1- nx\right)\\ {}\frac{\partial {V}_2^t\left(e,{a}_2\right)}{\partial {a}_2}={u}^{\prime}\left({c}_2\right)\end{array}} $$

Self 1’s FOC with respect to a2 and ET conditions read

$$ {\displaystyle \begin{array}{c}\frac{u^{\prime}\left({c}_1\right)}{R}=\beta \delta \frac{\partial {V}_2^t\left(e,{a}_2\right)}{\partial {a}_2}\\ {}\frac{\partial {V}_1^t\left(e,b\right)}{\partial e}={u}^{\prime}\left({c}_1\right){h}^{\prime }(e)+\beta \delta \frac{\partial {V}_2^t\left(e,{a}_2\right)}{\partial e}\\ {}\frac{\partial {V}_1^t\left(e,b\right)}{\partial b}={u}^{\prime}\left({c}_1\right)\end{array}} $$

From the FOCs and ET conditions with respect to a2 and a3, we have the following equation to describe consumption:

$$ {u}^{\prime}\left({c}_1\right)=\beta \delta R{u}^{\prime}\left({c}_2\right)={\left(\beta \delta R\right)}^2{u}^{\prime}\left({c}_3\right) $$
(9)

In the steady state, we also have

$$ {u}^{\prime}\left({c}_3\right)n=\varPhi (n)\frac{\partial {V}_1^t\left(e,b\right)}{\partial b}=\varPhi (n){u}^{\prime}\left({c}_1\right) $$

Combining the two equations above to eliminate consumption, we have

$$ {\left(\beta \delta R\right)}^2\frac{\varPhi \left({n}^t\right)}{n^t}=1 $$

Appendix 3. The proof of Proposition 3

In the commitment problem, combining the FOC and ET conditions with respect to human capital from A1 and using the fact that \( {V}_1^c\left(e,b\right)={V}_1^c\left(e^{\prime },b^{\prime}\right) \) in the steady state, we have

$$ \beta {\delta}^2\varPhi (n)\left[1+\frac{\left(1+g\right)\left(1- nx\right)}{R^2}\right]h^{\prime }(e)=\frac{n}{R} $$

Rearranging the equation and using the result of Proposition 1, we have

$$ \left[\frac{1}{R}+\frac{\left(1+g\right)\left(1-{n}^cx\right)}{R^2}\right]h^{\prime}\left({e}^c\right)=1 $$

In the time-consistent problem, combining Self 2’s FOC and all three ET conditions with respect to education,

$$ {u}^{\prime}\left({c}_2\right)=\beta \delta \varPhi (n)\left[{u}^{\prime}\left({c}_1\right){h}^{\prime }(e)+\beta \delta {u}^{\prime}\left({c}_2\right)\left(1+g\right)h^{\prime }(e)\left(1- nx\right)\right] $$

Plugging Eq. (8) into the equation above to eliminate consumption and using the result of Proposition 2, we have

$$ \left[\frac{1}{R}+\frac{\left(1+g\right)\left(1-{n}^tx\right)}{R^2}\right]h^{\prime}\left({e}^t\right)=1 $$

Appendix 4. The proof of Proposition 4

In both the commitment problem and the time-consistent problem, a reduction in β will lead to an increase in Φ(n)/n, and therefore will reduce n because Φ(n) is strictly increasing and concave. Hyperbolic discounting means that β < 1. So, the existence of hyperbolic discounting will reduce fertility in the steady states of both problems. From Eq. (7), we can see a negative relationship between fertility and human capital. Consequently, the existence of hyperbolic discounting will increase the steady-state human capital in both problems.

Dividing Eq. (6) by Eq. (5), we have

$$ \beta \frac{\varPhi \left({n}^t\right)}{n^t}=\frac{\varPhi \left({n}^c\right)}{n^c}\Rightarrow \frac{\varPhi \left({n}^t\right)}{n^t}>\frac{\varPhi \left({n}^c\right)}{n^c}\Rightarrow {n}^t<{n}^c $$

Combining Eq. (7) with the inequality above, we have

$$ {n}^t<{n}^c\Rightarrow 1-{n}^t>1-{n}^c\Rightarrow {h}^{\prime}\left({e}^t\right)<{h}^{\prime}\left({e}^c\right)\Rightarrow {e}^t>{e}^c $$

When β = 1, we have an exponential discounting problem with solutions ne and ee satisfying \( {\delta}^2{R}^2\frac{\varPhi \left({n}^e\right)}{n^e}=1 \) and \( \left[\frac{1}{R}+\frac{\left(1+g\right)\left(1-{n}^ex\right)}{R^2}\right]h^{\prime}\left({e}^e\right)=1 \). So, we have

$$ \beta \frac{\varPhi \left({n}^c\right)}{n^c}=\frac{\varPhi \left({n}^e\right)}{n^e}\Rightarrow \frac{\varPhi \left({n}^c\right)}{n^c}>\frac{\varPhi \left({n}^e\right)}{n^e}\Rightarrow {n}^c<{n}^e $$

Similarly, we have ec > ee.

Appendix 5. The proof of Proposition 5

From Appendix 4, we have \( \beta \frac{\varPhi \left({n}^t\right)}{n^t}=\frac{\varPhi \left({n}^c\right)}{n^c} \). Plugging \( \varPhi (n)=\frac{n^{1-\epsilon }}{1-\epsilon } \) into the equation, we have \( {n}^t={n}^c{\beta}^{\frac{1}{\epsilon }} \) which can be used to estimate \( {\beta}^{\frac{1}{\epsilon }} \) as a whole coefficient with nt and nc. Taking logs on both sides, we have \( \log \left({n}^t\right)=\log \left({n}^c\right)+\frac{1}{\epsilon}\log \left(\beta \right) \) which can be used to estimate β with nt, nc, and ϵ, or be used to estimate ϵ with nt, nc, and β.

Appendix 6. The proof of Proposition 6

Define \( \theta (c)=\frac{u^{\prime }(c)c}{u(c)},\theta (n)=\frac{\varPhi^{\prime }(n)n}{\varPhi (n)} \). From Eq. (8) and the budget constraint (4), the steady-state consumption in three periods in the commitment problem can be expressed as

$$ {c}_1^c={\alpha}_1{I}^c,{c}_2^c={\alpha}_2{I}^c,{c}_3^c={\alpha}_3{I}^c $$
(10)

where α1, α2, α3 are defined as in Proposition 6. \( {I}^c={c}_1^c+\frac{c_2^c}{R}+\frac{c_3^c}{R^2} \). We need to show the specific form of Ic. For simplicity, I will not show the superscript “c” until the end of the proof. Substituting the three equations above into Eq. (1), the maximization problem becomes

$$ {V}_1\left(e,b\right)={\mathit{\max}}_{c_1,n,{e}^{\prime },{b}^{\prime }}\left\{{\left({\alpha}_1\right)}^{-\gamma}\frac{I^{1-\gamma }}{1-\gamma }+\beta {\delta}^2\varPhi (n){V}_1\left(e^{\prime },b^{\prime}\right)\right\}={\mathit{\max}}_{c_1,n,{e}^{\prime },{b}^{\prime }}\left\{\frac{u\left({c}_1\right)}{\alpha_1}+\beta {\delta}^2\varPhi (n){V}_1\left(e^{\prime },b^{\prime}\right)\right\} $$
(11)

We have three new FOCs, which are given as follows:

$$ {\displaystyle \begin{array}{c}\frac{\partial {V}_1\left(e,b\right)}{\partial b}={u}^{\prime}\left({c}_1\right)\\ {}{u}^{\prime}\left({c}_1\right)\frac{n}{R^2}=\beta {\delta}^2\varPhi (n)\frac{\partial {V}_1\left(e^{\prime },b^{\prime}\right)}{\partial b^{\prime }}\\ {}{u}^{\prime}\left({c}_1\right)\left(\frac{\left(1+g\right)h(e)x+e^{\prime }}{R}+\frac{b^{\prime }}{R^2}\right)=\beta {\delta}^2\varPhi^{\prime }(n){V}_1\left(e^{\prime },b^{\prime}\right)\end{array}} $$

From the first two FOCs, we have

$$ {u}^{\prime}\left({c}_1\right)=\beta {\delta}^2{R}^2\frac{\varPhi (n)}{n}{u}^{\prime}\left({c}_1^{\prime}\right) $$

From the third FOC, we have

$$ \varPhi (n){V}_1\left(e^{\prime },b^{\prime}\right)={u}^{\prime}\left({c}_1\right)\left(\frac{\left(1+g\right)h(e)x+e^{\prime }}{R}+\frac{b^{\prime }}{R^2}\right)\frac{\varPhi (n)}{\beta {\delta}^2\varPhi^{\prime }(n)} $$

Rearranging the equation above with the definition of θ(c) and θ(n),

$$ \varPhi (n){V}_1\left(e^{\prime },b^{\prime}\right)=\frac{u\left({c}_1\right)}{c_1}\left(\frac{\left(1+g\right)h(e) nx+ ne^{\prime }}{R}+\frac{nb^{\prime }}{R^2}\right)\frac{\theta \left({c}_1\right)}{\beta {\delta}^2\theta (n)} $$

Using Eq. (10) to rewrite the budget constraint (4),

$$ \frac{c_1}{\alpha_1}={c}_1+\frac{c_2}{R}+\frac{c_3}{R^2}=\left[h(e)+b+\frac{\left(1+g\right)h(e)}{R}\right]-\left[\frac{\left(1+g\right)h(e) nx+ ne^{\prime }}{R}+\frac{nb^{\prime }}{R^2}\right] $$
(12)

Combining the two equations above, we have

$$ \varPhi (n){V}_1\left(e^{\prime },b^{\prime}\right)=\frac{u\left({c}_1\right)}{c_1}\left(h(e)+b+\frac{\left(1+g\right)h(e)}{R}-\frac{c_1}{\alpha_1}\right)\frac{\theta \left({c}_1\right)}{\beta {\delta}^2\theta (n)} $$
(13)

Rewriting the third FOC to the previous period and using the fact that it’s in steady state, we have

$$ {V}_1\left(e,b\right)=\frac{u^{\prime}\left({c}_1\right)}{\beta {\delta}^2\varPhi^{\prime }(n)}\left(\frac{\left(1+g\right)h(e)x+e}{R}+\frac{b}{R^2}\right)=\frac{R^2u^{\prime}\left({c}_1\right)}{\theta (n)}\left(\frac{\left(1+g\right)h(e)x+e}{R}+\frac{b}{R^2}\right) $$
(14)

Rewriting the equation above, we have

$$ {V}_1\left(e,b\right)=\frac{\theta \left({c}_1\right)}{\theta (n)}\left[\left(1+g\right)h(e)x+ eR+b\right]\frac{u\left({c}_1\right)}{c_1} $$
(15)

Substituting (14) and (15) into (11), we can get the expression for c1 in the steady state,

$$ {c}_1^c={\alpha}_1\frac{\theta \left({c}_1^c\right)}{\theta \left({n}^c\right)-\theta \left({c}_1^c\right)}\left[\left(1+g\right)h\left({e}^c\right) xR+{e}^cR-h\left({e}^c\right)-\frac{\left(1+g\right)h\left({e}^c\right)}{R}\right] $$
(16)

Combining (12) and (16), we have the steady-state bequest expression

$$ {b}^c=\frac{R^2}{R^2-{n}^c}\left\{\left[\left(1+g\right)h\left({e}^c\right) xR+{e}^cR\right]\left(\frac{\theta \left({c}_1^c\right)}{\theta \left({n}^c\right)-\theta \left({c}_1^c\right)}+\frac{1}{R^2}\right)-\frac{\theta \left({n}^c\right)}{\theta \left({n}^c\right)-\theta \left({c}_1^c\right)}\left[h\left({e}^c\right)+\frac{\left(1+g\right)h\left({e}^c\right)}{R}\right]\right\} $$
(17)

We can see that it is the elasticities of the utility and altruism functions that matter. If utility and altruism functions show constant elasticities of substitution, their specific function forms will not influence the consumption and bequests in the steady state.

Plugging the specific functions into (16) and (17), we can get the specific form of Ic and an equation in which bequests are determined.

Appendix 7. The proof of Proposition 7

From Eq. (21) and the budget constraint,

$$ {c}_1^t={\mu}_1{I}^t,{c}_2^t={\mu}_2{I}^t,{c}_3^t={\mu}_3{I}^t $$
(18)

where μ1, μ2, μ3 are defined as in Proposition 7. \( {I}^t={c}_1^t+\frac{c_2^t}{R}+\frac{c_3^t}{R^2} \). The rest of the proof is similar to that of Proposition 6.

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Cao, J., Li, M. Hyperbolic discounting in an intergenerational model with altruistic parents. J Popul Econ 35, 989–1005 (2022). https://doi.org/10.1007/s00148-021-00824-7

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Keywords

  • Hyperbolic discounting
  • Fertility
  • Human capital
  • Intergenerational model

JEL Classification

  • J11
  • I21
  • D14
  • C61