How the Future Shapes Consumption with Time-Inconsistent Preferences

James Feigenbaum; Sepideh Raei

doi:10.1515/bejte-2022-0115

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How the Future Shapes Consumption with Time-Inconsistent Preferences

James Feigenbaum and Sepideh Raei

Published/Copyright: December 29, 2023

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From the journal The B.E. Journal of Theoretical Economics Volume 24 Issue 1

Abstract

Time-inconsistent preferences, which are modeled by relative discount functions, are a common explanation for the empirical finding that lifecycle profiles of household consumption are typically hump-shaped rather than monotonic. More precisely, time-inconsistent preferences that are present-biased often generate a hump-shaped consumption profile over the lifecycle. We develop a general framework for understanding present bias in consumption through a future weighting factor that perturbs the discount factor of utility at future periods away from exponential discounting. Using our framework we derive necessary and sufficient conditions on the future weighting factors for the log consumption profile to be locally concave. We find that these conditions, which are necessary for the consumption profile to be hump-shaped, are stronger than just assuming a present bias. Furthermore, we explore the conditions under which the consumption profile determined in the first period of life Pareto dominates the realized consumption profile. Lastly, we explore the interconnections between these two sets of conditions, elucidating the linkages between the determinants of hump-shaped consumption profiles and the conditions necessary for the initial consumption path to achieve Pareto dominance.

Keywords: present bias; time-inconsistent preferences; consumption hump; commitment mechanisms; welfare comparison

JEL Classification: D60; D90

Corresponding author: Sepideh Raei, Huntsman School of Business, Utah State University, Logan, UT, USA, E-mail: sepideh.raei@usu.edu

We would like to thank Frank Caliendo and Scott Findley for their input. We are grateful to our editor, Dr. Ronald Peeters, and the two anonymous reviewers for their excellent comments. Further thanks goes to participants at the SABE 2022 conference for their helpful discussion.

Appendix A: Simplifying the Concavity Condition

The log consumption profile is concave at t + 1 iff we have

∑ z ′ = 1 T − t − 1 D 1 z ′ 1 + ε z ′ ∑ s ′ = 1 T − t D 1 s ′ 1 + ε s ′ ∑ s = 1 T − t D 1 s ( 1 + ε s − 1 ) ∑ z = 1 T − t − 1 D 1 z ( 1 + ε z − 1 ) ≤ 1 .

We can rearrange this inequality as follows.

∑ z ′ = 1 T − t − 1 D 1 z ′ 1 + ε z ′ ∑ s ′ = 1 T − t D 1 s ′ 1 + ε s ′ ≤ ∑ z = 1 T − t − 1 D 1 z ( 1 + ε z − 1 ) ∑ s = 1 T − t D 1 s ( 1 + ε s − 1 )

1 − D 1 T − t ( 1 + ε T − t ) ∑ s ′ = 1 T − t D 1 s ′ 1 + ε s ′ ≤ 1 − D 1 T − t ( 1 + ε T − t − 1 ) ∑ s = 1 T − t D 1 s ( 1 + ε s − 1 )

1 + ε T − t − 1 ∑ s = 1 T − t D 1 s ( 1 + ε s − 1 ) ≤ 1 + ε T − t ∑ s ′ = 1 T − t D 1 s ′ 1 + ε s ′

We wish to isolate ɛ _T−t, which appears in both the numerator and the denominator of the right-hand side.

1 + ε T − t − 1 ∑ s = 0 T − t − 1 D 1 s + 1 ( 1 + ε s ) ≤ 1 + ε T − t ∑ s ′ = 1 T − t − 1 D 1 s ′ 1 + ε s ′ + D 1 T − t ( 1 + ε T − t )

∑ s ′ = 1 T − t − 1 D 1 s ′ 1 + ε s ′ + D 1 T − t ( 1 + ε T − t ) ∑ s = 0 T − t − 1 D 1 s + 1 ( 1 + ε s ) ( 1 + ε T − t − 1 ) ≤ 1 + ε T − t

∑ s ′ = 1 T − t − 1 D 1 s ′ 1 + ε s ′ ∑ s = 0 T − t − 1 D 1 s + 1 ( 1 + ε s ) ( 1 + ε T − t − 1 ) ≤ 1 + ε T − t 1 − D 1 T − t ( 1 + ε T − t − 1 ) ∑ s = 0 T − t − 1 D 1 s + 1 ( 1 + ε s )

∑ s ′ = 1 T − t − 1 D 1 s ′ 1 + ε s ′ ∑ s = 0 T − t − 1 D 1 s + 1 ( 1 + ε s ) ( 1 + ε T − t − 1 ) ≤ 1 + ε T − t ∑ z ′ = 0 T − t − 2 D 1 z ′ + 1 1 + ε z ′ ∑ z = 0 T − t − 1 D 1 z + 1 ( 1 + ε z )

Thus we obtain the condition

(47) ∑ z = 1 T − t − 1 D 1 z ( 1 + ε z ) ∑ z ′ = 0 T − t − 2 D 1 z ′ + 1 1 + ε z ′ ( 1 + ε T − t − 1 ) ≤ 1 + ε T − t

for concavity at t + 1.

Appendix B: Derivation of Eq. (37)

ε ̲ T τ = ∑ i = 0 T − 2 D 1 i ( 1 + ε i + 1 ) ∑ t = τ T − 1 D 1 t − 1 ( 1 + ε t − τ ) ( 1 + ε T − τ ) − 1 = ∑ i = 0 T − 2 D 1 i ( 1 + ε i + 1 ) ∑ t = 0 T − τ − 1 D 1 t + τ − 1 ( 1 + ε t ) ( 1 + ε T − τ ) − 1 = D 1 1 − τ ∑ i = 0 T − 2 D 1 i ( 1 + ε i + 1 ) ∑ t = 0 T − τ − 1 D 1 t ( 1 + ε t ) ( 1 + ε T − τ ) − 1 = D 1 1 − τ ∑ i = 0 T − 2 D i ϕ i ∑ t = 0 T − τ − 1 D t ( 1 + ε T − τ ) − 1 = D 1 T − τ ( 1 + ε T − τ ) D 1 T − 1 ( 1 + ε T − 1 ) ∑ t = 0 T − 2 D t ∑ s = 0 T − τ − 1 D s ∑ i = 0 T − 2 D i ϕ i ∑ j = 0 T − 2 D j ( 1 + ε T − 1 ) − 1 = D T − τ D T − 1 ∑ t = 0 T − 2 D t ∑ s = 0 T − τ − 1 D s ϕ ̄ T − 2 ( 1 + ε T − 1 ) − 1 .

Appendix C: Derivation of Limits of ΔU _τ

lim ε T → ∞ Δ U τ = lim ε T → ∞ ∑ t = τ T ∑ i = 0 t − 1 D t − τ ⁡ ln ϕ ̄ T − t + i ϕ i = ∑ t = τ T ∑ i = 0 t − 1 D t − τ lim ε T → ∞ ln ϕ ̄ T − t + i − ∑ t = τ T ∑ i = 0 t − 1 D t − τ lim ε T → ∞ ln ϕ i = ∑ t = τ T ∑ i = 0 t − 2 D t − τ lim ε T → ∞ ln ϕ ̄ T − t + i + ∑ t = τ T D t − τ lim ε T → ∞ ln ϕ ̄ T − t + ( t − 1 ) − ∑ t = τ T − 1 ∑ i = 0 t − 1 D t − τ lim ε T → ∞ ln ϕ i − ∑ i = 0 T − 1 D T − τ lim ε T → ∞ ln ϕ i = ∑ t = τ T ∑ i = 0 t − 2 D t − τ ⁡ ln ϕ ̄ T − t + i + ∑ t = τ T D t − τ lim ε T → ∞ ln ϕ ̄ T − 1 − ∑ t = τ T − 1 ∑ i = 0 t − 1 D t − τ ⁡ ln ϕ i − ∑ i = 0 T − 2 D T − τ ⁡ ln ϕ i − D T − τ lim ε T → ∞ ln ϕ T − 1

lim ε T → ∞ ln ϕ T − 1 = lim ε T → ∞ ln ε T − ln ( 1 + ε T − 1 ) = lim ε T → ∞ ln ε T

lim ε T → ∞ ln ϕ ̄ T − 1 = lim ε T → ∞ ln ∑ s = 0 T − 1 D s ϕ s ∑ s ′ = 0 T − 1 D s ′ = lim ε T → ∞ ln D 1 T − 1 ( 1 + ε T ) − ln ∑ s ′ = 0 T − 1 D s ′ = ln ⁡ D 1 T − 1 − ln ∑ s ′ = 0 T − 1 D s ′ + lim ε T → ∞ ln ε T = lim ε T → ∞ ln ε T

For τ < T,

lim ε T → ∞ Δ U τ = ∑ t = τ T ∑ i = 0 t − 2 D t − τ ⁡ ln ϕ ̄ T − t + i − ∑ t = τ T − 1 ∑ i = 0 t − 1 D t − τ ⁡ ln ϕ i − ∑ i = 0 T − 2 D T − τ ⁡ ln ϕ i + ∑ t = τ T − 1 D t − τ lim ε T → ∞ ln ε T = ∑ t = τ T − 1 D t − τ lim ε T → ∞ ln ε T > 0 .

Appendix D: Gradient of ΔU ₁ at the Origin

Δ U τ = ∑ t = τ T ∑ i = 0 t − 1 D t − τ ⁡ ln ϕ ̄ T − t + i ϕ i .

Δ U 1 = ∑ s = 1 T ∑ i = 0 s − 1 D s − 1 ⁡ ln ϕ ̄ T − s + i ϕ i

∂ Δ U 1 ∂ ε t = ∑ s = 1 T ∑ i = 0 s − 1 ∂ D s − 1 ∂ ε t ln ϕ ̄ T − s + i ϕ i + D s − 1 × ∂ ⁡ ln ϕ ̄ T − s + i ∂ ε t − 1 1 + ε i + 1 ∂ ε i + 1 ∂ ε t + 1 1 + ε i ∂ ε i ∂ ε t

ϕ ̄ s = ∑ i = 0 s D i ϕ i ∑ j = 0 s D j

∂ ⁡ ln ϕ ̄ s ∂ ε t = ∑ i = 0 s ∂ D i ∂ ε t ϕ i + D i ∂ ϕ i ∂ ε t ∑ i ′ = 0 s D i ′ ϕ i ′ − ∑ j = 0 s ∂ D j ∂ ε t ∑ j ′ = 0 s D j ′

∂ ϕ i ∂ ε t = ∂ ∂ ε t 1 + ε i + 1 1 + ε i = δ i + 1 , t 1 + ε i − 1 + ε i + 1 ( 1 + ε i ) 2 δ i , t

∂ ϕ i ∂ ε t ε = 0 = δ i + 1 , t − δ i , t

∂ ⁡ ln ϕ ̄ s ∂ ε t ε = 0 = ∑ i = 0 s ∂ D i ∂ ε t + D 1 i ( δ i + 1 , t − δ i , t ) ∑ i ′ = 0 s D 1 i ′ − ∑ j = 0 s ∂ D j ∂ ε t ∑ j ′ = 0 s D 1 j ′ = ∑ i = 0 s D 1 i ( δ i + 1 , t − δ i , t ) ∑ i ′ = 0 s D 1 i ′

∂ Δ U 1 ∂ ε t ε = 0 = ∑ s = 1 T ∑ i = 0 s − 1 D 1 s − 1 ∂ ⁡ ln ϕ ̄ T − s + i ∂ ε t − δ i + 1 , t + δ i , t = ∑ s = 1 T ∑ i = 0 s − 1 D 1 s − 1 ∑ j = 0 T − s + i D 1 j ( δ j + 1 , t − δ j , t ) ∑ s ′ = 0 T − s + i D 1 s ′ − δ i + 1 , t + δ i , t

Suppose T = 2.

∂ Δ U 1 ∂ ε 2 ε = 0 = ∑ s = 1 2 ∑ i = 0 s − 1 D 1 s − 1 ∑ j = 0 2 − s + i D 1 j ( δ j + 1,2 − δ j , 2 ) ∑ s ′ = 0 2 − s + i D 1 s ′ − δ i + 1,2 + δ i , 2

∂ Δ U 1 ∂ ε 2 ε = 0 = ∑ i = 0 0 D 1 0 ∑ j = 0 1 + i D 1 j ( δ j + 1,2 − δ j , 2 ) ∑ s ′ = 0 1 + i D 1 s ′ − δ i + 1,2 + δ i , 2 + ∑ i = 0 1 D 1 1 ∑ j = 0 i D 1 j ( δ j + 1,2 − δ j , 2 ) ∑ s ′ = 0 i D 1 s ′ − δ i + 1,2 + δ i , 2 = ∑ j = 0 1 D 1 j ( δ j + 1,2 − δ j , 2 ) ∑ s ′ = 0 1 D 1 s ′ + D 1 ∑ j = 0 0 D 1 j ( δ j + 1,2 − δ j , 2 ) ∑ s ′ = 0 0 D 1 s ′ + D 1 ∑ j = 0 1 D 1 j ( δ j + 1,2 − δ j , 2 ) ∑ s ′ = 0 1 D 1 s ′ − 1 = D 1 1 + D 1 + D 1 D 1 1 + D 1 − 1 = D 1 + D 1 2 1 + D 1 − D 1 = 0

∂ Δ U 1 ∂ ε t ε = 0 = ∑ s = 1 T ∑ i = 0 s − 1 D 1 s − 1 ∑ j = 0 T − s + i D 1 j ( δ j + 1 , t − δ j , t ) ∑ s ′ = 0 T − s + i D 1 s ′ − δ i + 1 , t + δ i , t

S = { ( s , i ) ∈ Z 2 : 1 ≤ s ≤ T ∧ 0 ≤ i ≤ s − 1 } S ′ = { ( s , i ) ∈ Z 2 : 0 ≤ i ≤ T − 1 ∧ i + 1 ≤ s ≤ T }

If (s, i) ∈ S, 1 ≤ s ≤ T ∧ 0 ≤ i ≤ s − 1. Thus 0 ≤ i ≤ s − 1 ≤ T − 1, and i + 1 ≤ s ≤ T, so (s, i) ∈ S′.

If (s, i) ∈ S′, 0 ≤ i ≤ T − 1 ∧ i + 1 ≤ s ≤ T, 1 ≤ i + 1 ≤ s ≤ T and 0 ≤ i ≤ s − 1.

∂ Δ U 1 ∂ ε t ε = 0 = ∑ i = 0 T − 1 ∑ s = i + 1 T D 1 s − 1 ∑ j = 0 T − s + i D 1 j ( δ j + 1 , t − δ j , t ) ∑ s ′ = 0 T − s + i D 1 s ′ − δ i + 1 , t + δ i , t

∑ i = 0 T − 1 ∑ s = i + 1 T D 1 s − 1 ( δ i + 1 , t − δ i , t ) = ∑ s = t T D 1 s − 1 − ( 1 − δ t T ) ∑ s = t + 1 T D 1 s − 1 = D 1 t − 1 + δ t T ∑ s = t + 1 T D 1 s − 1 = D 1 t − 1

V 1 = ∑ s = 1 T ∑ i = 0 s − 1 ∑ j = 0 T − s + i D 1 s + j − 1 ( δ j + 1 , t − δ j , t ) ∑ s ′ = 0 T − s + i D 1 s ′

Let z = s − i, so i = s − z

V 1 = ∑ s = 1 T ∑ z = 1 s ∑ j = 0 T − z D 1 s + j − 1 ( δ j + 1 , t − δ j , t ) ∑ s ′ = 0 T − z D 1 s ′

S = { ( z , j ) ∈ Z 2 : 1 ≤ z ≤ s ∧ 0 ≤ j ≤ T − z } S ′ = { ( z , j ) ∈ Z 2 : 0 ≤ j ≤ T − 1 ∧ 1 ≤ z ≤ min { s , T − j } }

If (z, j) ∈ S, 1 ≤ z ≤ s ∧ 0 ≤ j ∧ j ≤ T − z. Thus 0 ≤ j ≤ T − z ≤ T − 1. 1 ≤ z, z ≤ s, and z ≤ T − j. Thus 1 ≤ z ≤ min{s, T − j}. So (z, j) ∈ S′.

If (z, j) ∈ S′, 0 ≤ j ≤ T − 1 ∧ 1 ≤ z ≤ min{s, T − j}. Thus 1 ≤ z ≤ s. Since z ≤ T − j, we have j ≤ T − z. Thus 0 ≤ j ≤ T − z. Thus (z, j) ∈ S.

V 1 = ∑ s = 1 T ∑ j = 0 T − 1 ∑ z = 1 min { s , T − j } D 1 s + j − 1 ( δ j + 1 , t − δ j , t ) ∑ s ′ = 0 T − z D 1 s ′ = ∑ j = 0 T − 1 ∑ s = 1 T ∑ z = 1 min { s , T − j } D 1 s + j − 1 ( δ j + 1 , t − δ j , t ) ∑ s ′ = 0 T − z D 1 s ′ = ∑ s = 1 T ∑ z = 1 min { s , T − t + 1 } D 1 s + t − 2 ∑ s ′ = 0 T − z D 1 s ′ − ( 1 − δ t , T ) ∑ s = 1 T ∑ z = 1 min { s , T − t } D 1 s + t − 1 ∑ s ′ = 0 T − z D 1 s ′ = ∑ s = 1 T ∑ z = 1 min { s , T − t + 1 } D 1 s + t − 2 ∑ s ′ = 0 T − z D 1 s ′ − ∑ z = 1 min { s , T − t } D 1 s + t − 1 ∑ s ′ = 0 T − z D 1 s ′ + δ t , T ∑ z = 1 min { s , T − t } D 1 s + t − 1 ∑ s ′ = 0 T − z D 1 s ′ = ∑ s = 1 T ∑ z = 1 min { s , T − t + 1 } D 1 s + t − 2 ∑ s ′ = 0 T − z D 1 s ′ − ∑ z = 1 min { s , T − t } D 1 s + t − 1 ∑ s ′ = 0 T − z D 1 s ′

If T = t = 2,

V 1 = ∑ s = 1 2 ∑ z = 1 min { s , 1 } D 1 s + 2 − 2 ∑ s ′ = 0 2 − z D 1 s ′ − ∑ z = 1 min { s , 0 } D 1 s + 2 − 1 ∑ s ′ = 0 2 − z D 1 s ′ = ∑ s = 1 2 ∑ z = 1 min { s , 1 } D 1 s ∑ s ′ = 0 2 − z D 1 s ′ = ∑ z = 1 min { 1,1 } D 1 1 ∑ s ′ = 0 2 − z D 1 s ′ + ∑ z = 1 min { 2,1 } D 1 2 ∑ s ′ = 0 2 − z D 1 s ′ = D 1 1 + D 1 + D 1 2 1 + D 1 = D 1

V 1 = D 1 t − 1 ∑ s = 1 T ∑ z = 1 min { s , T − t + 1 } D 1 s − 1 ∑ s ′ = 0 T − z D 1 s ′ − ∑ z = 1 min { s , T − t } D 1 s ∑ s ′ = 0 T − z D 1 s ′ = D 1 t − 1 ∑ s = 1 T ∑ z = 1 min { s , T − t + 1 } D 1 s − 1 ∑ s ′ = 0 T − z D 1 s ′ − ∑ s = 1 T ∑ z = 1 min { s , T − t } D 1 s ∑ s ′ = 0 T − z D 1 s ′ = D 1 t − 1 ∑ s = 0 T − 1 ∑ z = 1 min { s + 1 , T − t + 1 } D 1 s ∑ s ′ = 0 T − z D 1 s ′ − ∑ s = 1 T ∑ z = 1 min { s , T − t } D 1 s ∑ s ′ = 0 T − z D 1 s ′ = D 1 t − 1 ∑ z = 1 min { 1 , T − t + 1 } D 1 0 ∑ s ′ = 0 T − z D 1 s ′ + ∑ s = 1 T − 1 D 1 s ∑ s ′ = 0 T − min { s , T − t } − 1 D 1 s ′ − ∑ z = 1 min { T , T − t } D 1 T ∑ s ′ = 0 T − z D 1 s ′ = D 1 t − 1 1 ∑ s ′ = 0 T − 1 D 1 s ′ + ∑ s = 1 T − 1 D 1 s ∑ s ′ = 0 T − min { s , T − t } − 1 D 1 s ′ − ∑ z = 1 T − t D 1 T ∑ s ′ = 0 T − z D 1 s ′

Suppose T = 3 and t = 2.

V 1 = ∑ s = 1 3 ∑ z = 1 min { s , 2 } D 1 s ∑ s ′ = 0 3 − z D 1 s ′ − ∑ z = 1 min { s , 1 } D 1 s + 1 ∑ s ′ = 0 3 − z D 1 s ′

V 1 = ∑ z = 1 min { 1,2 } D 1 1 ∑ s ′ = 0 3 − z D 1 s ′ − ∑ z = 1 min { 1,1 } D 1 2 ∑ s ′ = 0 3 − z D 1 s ′ + ∑ z = 1 min { 2,2 } D 1 2 ∑ s ′ = 0 3 − z D 1 s ′ − ∑ z = 1 min { 2,1 } D 1 3 ∑ s ′ = 0 3 − z D 1 s ′ + ∑ z = 1 min { 3,2 } D 1 3 ∑ s ′ = 0 3 − z D 1 s ′ − ∑ z = 1 min { 3,1 } D 1 4 ∑ s ′ = 0 3 − z D 1 s ′ = D 1 1 + D 1 + D 1 2 − D 1 2 1 + D 1 + D 1 2 + D 1 2 1 + D 1 + D 1 2 + D 1 2 1 + D 1 − D 1 3 1 + D 1 + D 1 2 + D 1 3 1 + D 1 + D 1 2 + ̇ D 1 3 1 + D 1 − D 1 4 1 + D 1 + D 1 2 = D 1 − D 1 4 1 + D 1 + D 1 2 + D 1 2 = D 1 − D 1 4 + D 1 2 + D 1 3 + D 1 4 1 + D 1 + D 1 2 = D 1 + D 1 2 + D 1 3 1 + D 1 + D 1 2 = D 1

If T = t = 3,

V 1 = ∑ s = 1 3 ∑ z = 1 min { s , 1 } D 1 s + 1 ∑ s ′ = 0 3 − z D 1 s ′ − ∑ z = 1 min { s , 0 } D 1 s + 2 ∑ s ′ = 0 3 − z D 1 s ′ = D 1 2 1 + D 1 + D 1 2 + D 1 3 1 + D 1 + D 1 2 + D 1 4 1 + D 1 + D 1 2 = D 1 2

V 1 = D 1 t − 1 1 ∑ s ′ = 0 T − 1 D 1 s ′ + ∑ s = 1 T − 1 D 1 s ∑ s ′ = 0 T − min { s , T − t } − 1 D 1 s ′ − ∑ z = 1 T − t D 1 T ∑ s ′ = 0 T − z D 1 s ′

If T = 3 and t = 2,

V 1 = D 1 1 ∑ s ′ = 0 2 D 1 s ′ + ∑ s = 1 2 D 1 s ∑ s ′ = 0 3 − min { s , 1 } − 1 D 1 s ′ − ∑ z = 1 1 D 1 3 ∑ s ′ = 0 3 − z D 1 s ′ = D 1 1 1 + D 1 + D 1 2 + D 1 1 + D 1 + D 1 2 1 + D 1 − D 1 3 1 + D 1 + D 1 2 = D 1 1 − D 1 3 1 + D 1 + D 1 2 + D 1 = D 1 1 − D 1 3 + D 1 + D 1 2 + D 1 3 1 + D 1 + D 1 2 = D 1

V 1 = ∑ s = 1 T ∑ z = 1 s ∑ j = 0 T − z D 1 s + j − 1 ( δ j + 1 , t − δ j , t ) ∑ s ′ = 0 T − z D 1 s ′

Let S = {(s, z):1 ≤ s ≤ T ∧ 1 ≤ z ≤ s} and S′ = {(s, z):1 ≤ z ≤ T ∧ z ≤ s ≤ T}. Let (s, z) ∈ S. Then 1 ≤ s ≤ T ∧ 1 ≤ z ≤ s, so 1 ≤ z ≤ s ≤ T and z ≤ s ≤ T, so (s, z) ∈ S′.

Let (s, z) ∈ S′. Then 1 ≤ z ≤ T ∧ z ≤ s ≤ T, so 1 ≤ z ≤ s ≤ T and 1 ≤ z ≤ s.

V 1 = ∑ z = 1 T ∑ s = z T ∑ j = 0 T − z D 1 s + j − 1 ( δ j + 1 , t − δ j , t ) ∑ s ′ = 0 T − z D 1 s ′ = ∑ z = 1 T ∑ s = 0 T − z ∑ j = 0 T − z D 1 s + z + j − 1 ( δ j + 1 , t − δ j , t ) ∑ s ′ = 0 T − z D 1 s ′ = ∑ z = 1 T ∑ j = 0 T − z ∑ s = 0 T − z D 1 s + z + j − 1 ( δ j + 1 , t − δ j , t ) ∑ s ′ = 0 T − z D 1 s ′ = ∑ z = 1 T ∑ j = 0 T − z D 1 z + j − 1 ( δ j + 1 , t − δ j , t )

Let T = t = 2.

V 1 = ∑ z = 1 2 ∑ j = 0 2 − z D 1 z + j − 1 ( δ j + 1,2 − δ j , 2 ) = ∑ j = 0 1 D 1 j ( δ j + 1,2 − δ j , 2 ) + ∑ j = 0 0 D 1 j + 1 ( δ j + 1,2 − δ j , 2 ) = D 1

Let T = t = 3

V 1 = ∑ z = 1 3 ∑ j = 0 3 − z D 1 z + j − 1 ( δ j + 1,3 − δ j , 3 ) = ∑ j = 0 2 D 1 j ( δ j + 1,3 − δ j , 3 ) + ∑ j = 0 1 D 1 j + 1 ( δ j + 1,3 − δ j , 3 ) + ∑ j = 0 0 D 1 j + 2 ( δ j + 1,3 − δ j , 3 ) = D 1 2

Let T = t = 2

V 1 = ∑ z = 1 3 ∑ j = 0 3 − z D 1 z + j − 1 ( δ j + 1,2 − δ j , 2 ) = ∑ j = 0 2 D 1 j ( δ j + 1,2 − δ j , 2 ) + ∑ j = 0 1 D 1 j + 1 ( δ j + 1,2 − δ j , 2 ) + ∑ j = 0 0 D 1 j + 2 ( δ j + 1,2 − δ j , 2 ) = D 1 − D 1 2 + D 1 2 = D 1

V 1 = ∑ z = 1 T ∑ j = 0 T − z D 1 z + j − 1 ( δ j + 1 , t − δ j , t )

S = { ( z , j ) : 1 ≤ z ≤ T ∧ 0 ≤ j ≤ T − z } S ′ = { ( z , j ) : 0 ≤ j ≤ T − 1 ∧ 1 ≤ z ≤ T − j }

Let (z, j) ∈ S. Then 1 ≤ z ≤ T ∧ 0 ≤ j ≤ T − z. So z ≤ T − j, and 1 ≤ z ≤ T − j while 0 ≤ j ≤ T − z ≤ T − 1. Thus (z, j) ∈ S′.

Let (z, j) ∈ S′. Then 0 ≤ j ≤ T − 1 ∧ 1 ≤ z ≤ T − j. So j ≤ T − z, so 0 ≤ j ≤ T − z. 1 ≤ z ≤ T − j ≤ T. Thus (z, j) ∈ S.

V 1 = ∑ j = 0 T − 1 ∑ z = 1 T − j D 1 z + j − 1 ( δ j + 1 , t − δ j , t ) = ∑ z = 1 T − ( t − 1 ) D 1 z + t − 2 − ( 1 − δ T t ) ∑ z = 1 T − t D 1 z + t − 1

If T = t,

V 1 = ∑ z = 1 1 D 1 z + t − 2 = D 1 T − 1

If t < T,

V 1 = ∑ z = 1 T − t + 1 D 1 z + t − 2 − ∑ z = 1 T − t D 1 z + t − 1

Let s = z − 1, so z = s + 1.

V 1 = ∑ s = 0 T − t D 1 s + 1 + t − 2 − ∑ z = 1 T − t D 1 z + t − 1 = ∑ s = 0 T − t D 1 s + t − 1 − ∑ z = 1 T − t D 1 z + t − 1 = D 1 t − 1

Thus

∂ Δ U 1 ∂ ε t ε = 0 = ∑ i = 0 T − 1 ∑ s = i + 1 T D 1 s − 1 ∑ j = 0 T − s + i D 1 j ( δ j + 1 , t − δ j , t ) ∑ s ′ = 0 T − s + i D 1 s ′ − δ i + 1 , t + δ i , t = D 1 t − 1 − D 1 t − 1 = 0

Appendix E: Hessian of ΔU ₁ at the Origin for T = 3

Δ U 1 = ∑ s = 1 T − 1 ln ϕ ̄ s ∑ t = max { T − s − 1,0 } T − 1 D t − ln ϕ s ∑ t = max { 0 , s } T − 1 D t = ∑ s = 1 T − 1 ln ϕ ̄ s ∑ t = T − 1 − s T − 1 D t − ln ϕ s ∑ t = s T − 1 D t

Only ϕ _T−1 and ϕ ̄ T − 1 will depend on ɛ ₃, so

Δ U 1 = ln ϕ ̄ T − 1 ∑ t = 0 T − 1 D t − D T − 1 ⁡ ln 1 + ε T 1 + ε T − 1

ϕ ̄ T − 1 = ∑ z = 0 T − 1 D z ϕ z ∑ z = 0 T − 1 D z

∂ ⁡ ln ϕ ̄ T − 1 ∂ ε T = D T − 1 1 1 + ε T − 1 ∑ z = 0 T − 1 D z ϕ z = D 1 T − 1 ∑ z = 0 T − 1 D z ϕ z

∂ Δ U 1 ∂ ε T = D 1 T − 1 ∑ z = 0 T − 1 D z ϕ z ∑ t = 0 T − 1 D t − D T − 1 1 + ε T

If ɛ ₂ = … = ɛ _T−1 = 0,

∂ Δ U 1 ∂ ε T = D 1 T − 1 ∑ t = 0 T − 1 D 1 t ∑ z = 0 T − 2 D 1 z + D 1 T − 1 ( 1 + ε T ) − D 1 T − 1 1 + ε T = D 1 T − 1 ( 1 + ε T ) ∑ t = 0 T − 1 D 1 t − ∑ z = 0 T − 2 D 1 z + D 1 T − 1 ( 1 + ε T ) ∑ z ′ = 0 T − 1 D 1 z ′ + D 1 T − 1 ε T ( 1 + ε T ) = D 1 T − 1 ( 1 + ε T ) ∑ t = 0 T − 2 D 1 t − ∑ z = 0 T − 2 D 1 z ∑ z ′ = 0 T − 1 D 1 z ′ + D 1 T − 1 ε T ( 1 + ε T ) = D 1 T − 1 ∑ t = 0 T − 2 D 1 t ∑ z ′ = 0 T − 1 D 1 z ′ + D 1 T − 1 ε T ( 1 + ε T ) ε T

This is positive except when ɛ _T = 0.

∂ 2 Δ U 1 ∂ ε T 2 = D 1 T − 1 ∑ t = 0 T − 2 D 1 t ∑ z = 0 T − 1 D 1 z + D 1 T − 1 ε T ( 1 + ε T ) − ε T ∑ z = 0 T − 1 D 1 z + D 1 T − 1 ε T + D 1 T − 1 ( 1 + ε T ) ∑ z ′ = 0 T − 1 D 1 z ′ + D 1 T − 1 ε T ( 1 + ε T ) 2

∂ 2 Δ U 1 ∂ ε T 2 ε = 0 = D 1 T − 1 ∑ t = 0 T − 2 D 1 t ∑ z = 0 T − 1 D 1 z ∑ z ′ = 0 T − 1 D 1 z ′ 2 > 0

Thus if ɛ ₂ = … = ɛ _T−1 = 0, ΔU ₁ ≥ 0 with equality only if ɛ _T = 0.

Δ U 1 = ( D 1 + D 1 2 ( 1 + ε 2 ) ) ln 1 + D 1 1 + D 1 ε 2 − ln ( 1 + ε 2 ) + ( 1 + D 1 + D 1 2 ( 1 + ε 2 ) ) ln 1 + D 1 + D 1 2 + D 1 ε 2 + D 1 2 ε 3 1 + D 1 + D 1 2 + D 1 2 ε 2 − D 1 2 ( 1 + ε 2 ) ln 1 + ε 3 1 + ε 2

∂ Δ U 1 ∂ ε 2 = D 1 2 ln 1 + D 1 1 + D 1 ε 2 − ln ( 1 + ε 2 ) + ( D 1 + D 1 2 ( 1 + ε 2 ) ) × D 1 1 + D 1 1 + D 1 1 + D 1 ε 2 − 1 1 + ε 2 + D 1 2 ⁡ ln 1 + D 1 + D 1 2 + D 1 ε 2 + D 1 2 ε 3 1 + D 1 + D 1 2 + D 1 2 ε 2 + ( 1 + D 1 + D 1 2 ( 1 + ε 2 ) ) × D 1 1 + D 1 + D 1 2 + D 1 ε 2 + D 1 2 ε 3 − D 1 2 1 + D 1 + D 1 2 + D 1 2 ε 2 − D 1 2 ⁡ ln 1 + ε 3 1 + ε 2 + D 1 2 1 + ε 2 1 + ε 2

∂ Δ U 1 ∂ ε 2 = D 1 2 ln 1 + D 1 1 + D 1 ε 2 − ln ( 1 + ε 2 ) + ( D 1 + D 1 2 ( 1 + ε 2 ) ) × D 1 1 + D 1 + D 1 ε 2 − 1 1 + ε 2 + D 1 2 ⁡ ln 1 + D 1 + D 1 2 + D 1 ε 2 + D 1 2 ε 3 1 + D 1 + D 1 2 + D 1 2 ε 2 + ( 1 + D 1 + D 1 2 ( 1 + ε 2 ) ) × D 1 1 + D 1 + D 1 2 + D 1 ε 2 + D 1 2 ε 3 − D 1 2 1 + D 1 + D 1 2 + D 1 2 ε 2 − D 1 2 ⁡ ln 1 + ε 3 1 + ε 2 + D 1 2

As a check,

∂ Δ U 1 ∂ ε 2 ε 2 = ε 3 = 0 = D 1 + D 1 2 D 1 1 + D 1 − 1 + 1 + D 1 + D 1 2 D 1 − D 1 2 1 + D 1 + D 1 2 + D 1 2 = − D 1 + D 1 2 1 + D 1 + D 1 − D 1 2 + D 1 2 = 0

∂ Δ U 1 ∂ ε 2 = ( D 1 + D 1 2 ( 1 + ε 2 ) ) D 1 1 + D 1 + D 1 ε 2 − 1 1 + ε 2 + D 1 2 + D 1 2 ⁡ ln 1 + D 1 1 + D 1 ε 2 1 + D 1 + D 1 2 + D 1 ε 2 + D 1 2 ε 3 1 + D 1 + D 1 2 + D 1 2 ε 2 ( 1 + ε 3 ) + ( 1 + D 1 + D 1 2 ( 1 + ε 2 ) ) × D 1 1 + D 1 + D 1 2 + D 1 ε 2 + D 1 2 ε 3 − D 1 2 1 + D 1 + D 1 2 + D 1 2 ε 2

∂ 2 Δ U 1 ∂ ε 2 ∂ ε 3 = D 1 2 D 1 2 1 + D 1 + D 1 2 + D 1 ε 2 + D 1 2 ε 3 − 1 1 + ε 3 − D 1 3 1 + D 1 + D 1 2 ( 1 + ε 2 ) 1 + D 1 + D 1 2 + D 1 ε 2 + D 1 2 ε 3 2

∂ 2 Δ U 1 ∂ ε 2 ∂ ε 3 ε 2 = ε 3 = 0 = D 1 2 D 1 2 1 + D 1 + D 1 2 − 1 − D 1 3 1 + D 1 + D 1 2 1 + D 1 + D 1 2 2 = − D 1 2 1 + D 1 1 + D 1 + D 1 2 − D 1 3 1 + D 1 + D 1 2

∂ 2 Δ U 1 ∂ ε 2 ∂ ε 3 ε 2 = ε 3 = 0 = − D 1 2 1 + D 1 + D 1 2 ( 1 + 2 D 1 )

∂ 2 Δ U 1 ∂ ε 3 2 ε 2 = ε 3 = 0 = D 1 2 1 + D 1 + D 1 2 ( 1 + D 1 )

Meanwhile,

∂ 2 Δ U 1 ∂ ε 2 2 ε 2 = ε 3 = 0 = D 1 1 + D 1 + 4 D 1 2 + 3 D 1 3 ( 1 + D 1 ) ( 1 + D 1 + D 1 2 )

Δ U 1 = D 1 2 + D 1 3 2 1 + D 1 + D 1 2 ε 3 2 − D 1 2 + 2 D 1 3 1 + D 1 + D 1 2 ε 2 ε 3 + D 1 + D 1 2 + 4 D 1 3 + 3 D 1 4 2 ( 1 + D 1 ) ( 1 + D 1 + D 1 2 ) ε 2 2 + O ( ε 3 ) = 1 2 1 1 + D 1 + D 1 2 ε 2 ε 3 D 1 2 ( 1 + D 1 ) − D 1 2 ( 1 + 2 D 1 ) − D 1 2 ( 1 + 2 D 1 ) D 1 + D 1 2 + 4 D 1 3 + 3 D 1 4 ( 1 + D 1 ) ε 2 ε 3 + O ( ε 3 )

D 1 2 ( 1 + D 1 ) − D 1 2 ( 1 + 2 D 1 ) − D 1 2 ( 1 + 2 D 1 ) D 1 + D 1 2 + 4 D 1 3 + 3 D 1 4 ( 1 + D 1 ) = D 1 3 + D 1 4 + 4 D 1 5 + 3 D 1 6 − D 1 4 ( 1 + 2 D 1 ) 2 = D 1 3 + D 1 4 + 4 D 1 5 + 3 D 1 6 − D 1 4 − 4 D 1 5 − 4 D 1 6 = D 1 3 − D 1 6 = D 1 3 1 − D 1 3

If D ₁ > 1, ΔU ₁ < 0 is possible. However, if D ₁ < 1, the determinant is nonnegative. Thus in a deleted neighborhood of (ɛ ₂, ɛ ₃) = (0, 0), ΔU ₁ must be strictly nonnegative.

Appendix F: Sufficient Upper Bound on ɛ _T for Pareto Dominance of the Commitment Path

Δ U τ = ∑ t = τ T ∑ i = 0 t − 1 D t − τ ⁡ ln ϕ ̄ T − t + i ϕ i .

We can rewrite this as

Δ U τ = ∑ t = τ T ∑ j = T − t T − 1 D t − τ ⁡ ln ϕ ̄ j − ∑ t = τ T ∑ i = 0 t − 1 D t − τ ⁡ ln ϕ i ,

where j = T − t + i. The first terms are all positive while the second terms are all negative.

S = {(t, i):τ ≤ t ≤ T ∧ 0 ≤ i ≤ t − 1}. S′ = {(t, i):0 ≤ i ≤ T − 1 ∧ max{τ, i + 1} ≤ t ≤ T}. Let (t, i) ∈ S, so τ ≤ t ≤ T ∧ 0 ≤ i ≤ t − 1. Then 0 ≤ i ≤ t − 1 ≤ T − 1. We have both τ ≤ t and i + 1 ≤ t, so max{τ, i + 1} ≤ t ≤ T. Thus (t, i) ∈ S′.

Now let (t, i) ∈ S′, so 0 ≤ i ≤ T − 1 ∧ max{τ, i + 1} ≤ t ≤ T. Then τ ≤ t ≤ T. 0 ≤ i ≤ t − 1. Thus (t, i) ∈ S.

Let S = {(t, j):τ ≤ t ≤ T ∧ T − t ≤ j ≤ T − 1}. Let S′ = {(t, j):0 ≤ j ≤ T − 1 ∧ max{τ, T − j} ≤ t ≤ T}. Let (t, j) ∈ S. Then τ ≤ t ≤ T ∧ T − t ≤ j ≤ T − 1. So 0 ≤ T − t ≤ j ≤ T − 1, and we have both τ ≤ t and T − j ≤ t, so max{τ, T − j} ≤ t ≤ T. Thus (t, j) ∈ S′. Let (t, j) ∈ S′. Then 0 ≤ j ≤ T − 1 ∧ max{τ, T − j} ≤ t ≤ T. τ ≤ t ≤ T, and T − t ≤ j ≤ T − 1. Thus (t, j) ∈ S.

Δ U τ = ∑ j = 0 T − 1 ∑ t = max { τ , T − j } T D t − τ ⁡ ln ϕ ̄ j − ∑ i = 0 T − 1 ∑ t = max { τ , i + 1 } T D t − τ ⁡ ln ϕ i .

The first terms are all positive and the second terms are all negative. Let us define

(48) P i τ = ∑ t = max { τ , T − i } T D t − τ

and

(49) Q i τ = ∑ t = max { τ , i + 1 } T D t − τ

Thus we have

Δ U τ = ∑ i = 0 T − 1 P i τ ⁡ ln ϕ ̄ i − Q i τ ⁡ ln ϕ i = ∑ i = 0 T − 1 P i τ ln ϕ ̄ i ϕ i + ln ϕ i − Q i τ ⁡ ln ϕ i = ∑ i = 0 T − 1 P i τ ln ϕ ̄ i ϕ i + ∑ j = i T − 1 ln ϕ j − ∑ j = i + 1 T − 1 ln ϕ j − Q i τ ⁡ ln ϕ i = ∑ i = 0 T − 1 P i τ ln ϕ ̄ i ϕ i + ln 1 + ε T 1 + ε i − ∑ j = i + 1 T − 1 ln ϕ j − Q i τ ⁡ ln ϕ i

Δ U τ = ∑ i = 0 T − 1 P i τ ⁡ ln ( 1 + ε T ) + ∑ i = 0 T − 1 P i τ ln ϕ ̄ i ϕ i − ln ( 1 + ε i ) − ∑ j = i + 1 T − 1 ln ϕ j − Q i τ ⁡ ln ϕ i

Suppose that s < T − 1. Suppose that

ε T ≤ B s τ = exp ∑ i = 0 s P i τ ln ϕ i ϕ ̄ i + ln ( 1 + ε i ) + ∑ j = i + 1 s ln ϕ j + Q i τ ⁡ ln ϕ i ∑ i ′ = 0 T − 1 P i ′ τ − 1 .

Then we will have

∑ i = 0 T − 1 P i τ ⁡ ln ( 1 + ε T ) + ∑ i = 0 s P i τ ln ϕ ̄ i ϕ i − ln ( 1 + ε i ) − ∑ j = i + 1 s ln ϕ j − Q i τ ⁡ ln ϕ i ≤ 0

since

0 ≥ ∑ i = 0 s P i τ ln ϕ ̄ i ϕ i − ln ( 1 + ε i ) − ∑ j = s + 1 T − 1 ln ϕ j − Q i τ ⁡ ln ϕ i + ∑ i = s + 1 T − 1 P i τ ln ϕ ̄ i ϕ i − ln ( 1 + ε i ) − ∑ j = i + 1 T − 1 ln ϕ j − Q i τ ⁡ ln ϕ i

we have

0 ≥ ∑ i = 0 T − 1 P i τ ⁡ ln ( 1 + ε T ) + ∑ i = 0 s P i τ ln ϕ ̄ i ϕ i − ln ( 1 + ε i ) − ∑ j = i + 1 s ln ϕ j − Q i τ ⁡ ln ϕ i + ∑ i = 0 s P i τ ln ϕ ̄ i ϕ i − ln ( 1 + ε i ) − ∑ j = s + 1 T − 1 ln ϕ j − Q i τ ⁡ ln ϕ i + ∑ i = s + 1 T − 1 P i τ ln ϕ ̄ i ϕ i − ln ( 1 + ε i ) − ∑ j = i + 1 T − 1 ln ϕ j − Q i τ ⁡ ln ϕ i = ∑ i = 0 T − 1 P i τ ⁡ ln ( 1 + ε T ) + ∑ i = 0 s P i τ ln ϕ ̄ i ϕ i − ln ( 1 + ε i ) − ∑ j = i + 1 T − 1 ln ϕ j − Q i τ ⁡ ln ϕ i + ∑ i = s + 1 T − 1 P i τ ln ϕ ̄ i ϕ i − ln ( 1 + ε i ) − ∑ j = i + 1 T − 1 ln ϕ j − Q i τ ⁡ ln ϕ i = ∑ i = 0 T − 1 P i τ ⁡ ln ( 1 + ε T ) + ∑ i = 0 T − 1 P i τ ln ϕ ̄ i ϕ i − ln ( 1 + ε i ) − ∑ j = i + 1 T − 1 ln ϕ j − Q i τ ⁡ ln ϕ i = Δ U τ .

References

Attanasio, Orazio, and Martin Browning. 1995. “Consumption Over the Life Cycle and Over the Business Cycle.” The American Economic Review 85 (5): 1118–37.Search in Google Scholar

Attanasio, Orazio, and Guglielmo Weber. 1995. “Is Consumption Growth Consistent with Intertemporal Optimization? Evidence from the Consumer Expenditure Survey.” Journal of Political Economy 103 (6): 1121–57. https://doi.org/10.1086/601443.Search in Google Scholar

Attanasio, Orazio P., James Banks, Costas Meghir, and Guglielmo Weber. 1999. “Humps and Bumps in Lifetime Consumption.” Journal of Business & Economic Statistics 17 (1): 22–35. https://doi.org/10.1080/07350015.1999.10524794.Search in Google Scholar

Bernheim, B. Douglas, and Debraj Ray. 1987. “Economic Growth with Intergenerational Altruism.” The Review of Economic Studies 54 (2): 227–43. https://doi.org/10.2307/2297513.Search in Google Scholar

Browning, Martin, and Thomas Crossley. 2001. “Unemployment Insurance Benefit Levels and Consumption Changes.” Journal of Public Economics 80 (1): 1–23. https://doi.org/10.1016/s0047-2727(00)00084-0.Search in Google Scholar

Browning, Martin, Angus Deaton, and Margaret Irish. 1985. “A Profitable Approach to Labor Supply and Commodity Demands over the Life-Cycle.” Econometrica: Journal of the Econometric Society 53 (3): 503–43, https://doi.org/10.2307/1911653.Search in Google Scholar

Bullard, James, and James Feigenbaum. 2007. “A Leisurely Reading of the Life-Cycle Consumption Data.” Journal of Monetary Economics 54 (8): 2305–20. https://doi.org/10.1016/j.jmoneco.2007.06.025.Search in Google Scholar

Caliendo, Frank, and David Aadland. 2007. “Short-Term Planning and the Life-Cycle Consumption Puzzle.” Journal of Economic Dynamics and Control 31 (4): 1392–415. https://doi.org/10.1016/j.jedc.2006.05.002.Search in Google Scholar

Caliendo, Frank, and T. Scott Findley. 2019. “Commitment and Welfare.” Journal of Economic Behavior & Organization 159: 210–34, https://doi.org/10.1016/j.jebo.2019.01.008.Search in Google Scholar

Campbell, John Y., and N. Gregory Mankiw. 1989. “Consumption, Income, and Interest Rates: Reinterpreting the Time Series Evidence.” NBER Macroeconomics Annual 4: 185–216. https://doi.org/10.1086/654107.Search in Google Scholar

Cao, Dan, and Iván Werning. 2018. “Saving and Dissaving with Hyperbolic Discounting.” Econometrica 86 (3): 805–57. https://doi.org/10.3982/ECTA15112.Search in Google Scholar

Carroll, Christopher D. 1994. “How Does Future Income Affect Current Consumption?” Quarterly Journal of Economics 109 (1): 111–47. https://doi.org/10.2307/2118430.Search in Google Scholar

Carroll, Christopher. 1997. “Buffer-Stock Saving and the Life Cycle/Permanent Income Hypothesis.” Quarterly Journal of Economics 112 (1): 1–55. https://doi.org/10.1162/003355397555109.Search in Google Scholar

Carroll, Christopher, and Lawrence Summers. 1991. “Consumption Growth Parallels Income Growth: Some New Evidence.” In National Saving and Economic Performance, 3090. Cambridge: National Bureau of Economic Research, Inc.Search in Google Scholar

Deaton, Angus. 1992. Understanding Consumption. Oxford University Press. https://EconPapers.repec.org/RePEc:oxp:obooks:9780198288244.10.1093/0198288247.001.0001Search in Google Scholar

Dewatripont, Mathias, Isabelle Brocas, and Juan Carrillo. 2004. “Commitment Devices Under Self-Control Problems: An Overview.” Ulb institutional repository. ULB – Universite Libre de Bruxelles.Search in Google Scholar

Drouhin, Nicolas. 2020. “Non-Stationary Additive Utility and Time Consistency.” Journal of Mathematical Economics 86: 1–14. https://doi.org/10.1016/j.jmateco.2019.10.005.Search in Google Scholar

Feigenbaum, James. 2008. “Can Mortality Risk Explain the Consumption Hump?” Journal of Macroeconomics 30 (3): 844–72. https://doi.org/10.1016/j.jmacro.2007.08.010.Search in Google Scholar

Feigenbaum, James. 2016. “Equivalent Representations of Non-Exponential Discounting Models.” Journal of Mathematical Economics 66: 58–71, https://doi.org/10.1016/j.jmateco.2016.08.001.Search in Google Scholar

Feigenbaum, James, and Sepideh Raei. 2023. “Lifecycle Consumption and Welfare with Nonexponential Discounting in Continuous Time.” Journal of Mathematical Economics 107: 102869, https://doi.org/10.1016/j.jmateco.2023.102869.Search in Google Scholar

Feldstein, Martin. 1985. “The Optimal Level of Social Security Benefits.” Quarterly Journal of Economics 100 (2): 303–20. https://doi.org/10.2307/1885383.Search in Google Scholar

Fernandez-Villaverde, Jesus, and Dirk Krueger. 2011. “Consumption and Saving over the Life Cycle: How Important are Consumer Durables?” Macroeconomic Dynamics 15 (5): 725–70. https://doi.org/10.1017/s1365100510000180.Search in Google Scholar

Friedman, Milton. 2018. Theory of the Consumption Function. Princeton: Princeton university press.Search in Google Scholar

Gourinchas, Pierre-Olivier, and Jonathan A. Parker. 2002. “Consumption Over the Life Cycle.” Econometrica 70 (1): 47–89. https://doi.org/10.1111/1468-0262.00269.Search in Google Scholar

Grenadier, Steven R., and Neng Wang. 2007. “Investment Under Uncertainty and Time-Inconsistent Preferences.” Journal of Financial Economics 84 (1): 2–39. https://doi.org/10.1016/j.jfineco.2006.01.002.Search in Google Scholar

Griffin, Barbara, Beryl Hesketh, and Vanessa Loh. 2012. “The Influence of Subjective Life Expectancy on Retirement Transition and Planning: A Longitudinal Study.” Journal of Vocational Behavior 81 (2): 129–37. https://doi.org/10.1016/j.jvb.2012.05.005.Search in Google Scholar

Gul, Faruk, and Wolfgang Pesendorfer. 2004. “Self-Control, Revealed Preference and Consumption Choice.” Review of Economic Dynamics 7 (2): 243–64. https://doi.org/10.1016/j.red.2003.11.002.Search in Google Scholar

Guo, Jang-Ting, and Alan Krause. 2015. “Dynamic Nonlinear Income Taxation with Quasi-Hyperbolic Discounting and No Commitment.” Journal of Economic Behavior & Organization 109: 101–19. https://doi.org/10.1016/j.jebo.2014.11.002.Search in Google Scholar

Hansen, Gary, and Selahattin Imrohoroglu. 2008. “Consumption Over the Life Cycle: The Role of Annuities.” Review of Economic Dynamics 11 (3): 566–83. https://doi.org/10.1016/j.red.2007.12.004.Search in Google Scholar

Harris, Christopher, and David Laibson. 2013. “Instantaneous Gratification.” Quarterly Journal of Economics 128 (1): 205–48. https://doi.org/10.1093/qje/qjs051.Search in Google Scholar

Heckman, James. 1974. “Life Cycle Consumption and Labor Supply: An Explanation of the Relationship Between Income and Consumption Over the Life Cycle.” The American Economic Review 64 (1): 188–94.Search in Google Scholar

Hong, Eunice, and Sherman D. Hanna. 2014. “Financial Planning Horizon: A Measure of Time Preference or a Situational Factor?” Journal of Financial Counseling and Planning 25 (2): 184–96.Search in Google Scholar

Hubbard, R. Glenn, Jonathan Skinner, and Stephen P. Zeldes. 1994. “The Importance of Precautionary Motives in Explaining Individual and Aggregate Saving.” In Carnegie-Rochester Conference Series on Public Policy, Vol. 40, 59–125. Elsevier.10.1016/0167-2231(94)90004-3Search in Google Scholar

Laibson, David. 1994. “Hyperbolic Discounting and Consumption.” PhD diss., Massachusetts Institute of Technology.Search in Google Scholar

Laibson, David I. 1996. “Hyperbolic Discount Functions, Undersaving, and Savings Policy.” Working Paper 5635. National Bureau of Economic Research.10.3386/w5635Search in Google Scholar

Laibson, David. 1997. “Golden Eggs and Hyperbolic Discounting.” Quarterly Journal of Economics 112 (2): 443–78. https://doi.org/10.1162/003355397555253.Search in Google Scholar

Laibson, David. 1998. “Life-Cycle Consumption and Hyperbolic Discount Functions.” European Economic Review 42 (3): 861–71. https://doi.org/10.1016/s0014-2921(97)00132-3.Search in Google Scholar

Laibson, David I., Andrea Repetto, Jeremy Tobacman, Robert E. Hall, William G. Gale, and George A. Akerlof. 1998. “Self-Control and Saving for Retirement.” Brookings Papers on Economic Activity 1998 (1): 91–196. https://doi.org/10.2307/2534671.Search in Google Scholar

Lane, John, and Tapan Mitra. 1981. “On Nash Equilibrium Programs of Capital Accumulation Under Altruistic Preferences.” International Economic Review 22 (2): 309–31. https://doi.org/10.2307/2526279.Search in Google Scholar

Leininger, Wolfgang. 1986. “The Existence of Perfect Equilibria in a Model of Growth with Altruism Between Generations.” The Review of Economic Studies 53 (3): 349–67. https://doi.org/10.2307/2297633.Search in Google Scholar

Marin-Solano, Jesus, and Jorge Navas. 2009. “Non-Constant Discounting in Finite Horizon: The Free Terminal Time Case.” Journal of Economic Dynamics and Control 33 (3): 666–75. https://doi.org/10.1016/j.jedc.2008.08.008.Search in Google Scholar

Modigliani, Franco, and Richard Brumberg. 1954. “Utility Analysis and the Consumption Function: An Interpretation of Cross-Section Data.” Franco Modigliani 1 (1): 388–436.Search in Google Scholar

Mu, Congming, Jinqiang Yang, and Jingiang Yang. 2016. “Optimal Contract Theory with Time-Inconsistent Preferences.” Economic Modelling 52: 519–30. https://doi.org/10.1016/j.econmod.2015.09.032.Search in Google Scholar

Nagatani, Keizo. 1972. “Life Cycle Saving: Theory and Fact.” The American Economic Review 62 (3): 344–53.Search in Google Scholar

O’Donoghue, Ted, and Matthew Rabin. 1999. “Doing it Now or Later.” The American Economic Review 89 (1): 103–24. https://doi.org/10.1257/aer.89.1.103.Search in Google Scholar

O’Donoghue, Ted, and Matthew Rabin. 2000.“The Economics of Immediate Gratification.” Journal of Behavioral Decision Making 13 (2): 233–50. https://doi.org/10.1002/(sici)1099-0771(200004/06)13:2<233::aid-bdm325>3.0.co;2-u.10.1002/(SICI)1099-0771(200004/06)13:2<233::AID-BDM325>3.0.CO;2-USearch in Google Scholar

O’Donoghue, Ted, and Matthew Rabin. 2001. “Choice and Procrastination.” Quarterly Journal of Economics 116 (1): 121–60. https://doi.org/10.1162/003355301556365.Search in Google Scholar

O’Donoghue, Ted, and Matthew Rabin. 2015. “Present Bias: Lessons Learned and to be Learned.” The American Economic Review 105 (5): 273–9. https://doi.org/10.1257/aer.p20151085.Search in Google Scholar

Phelps, Edmund S., and Robert A. Pollak. 1968. “On Second-Best National Saving and Game-Equilibrium Growth.” The Review of Economic Studies 35 (2): 185–99. https://doi.org/10.2307/2296547.Search in Google Scholar

Prelec, Drazen. 2004. “Decreasing Impatience: A Criterion for Non-Stationary Time Preference and “Hyperbolic” Discounting.” The Scandinavian Journal of Economics 106 (3): 511–32. https://doi.org/10.1111/j.0347-0520.2004.00375.x.Search in Google Scholar

Richter, Michael. 2020. “A Time-Inconsistent First Welfare Theorem: Efficiency and the Convexity of Patience.” working paper.Search in Google Scholar

Samuelson, Paul A. 1937. “A Note on Measurement of Utility.” The Review of Economic Studies 4 (2): 155–61. https://doi.org/10.2307/2967612.Search in Google Scholar

Strotz, Robert Henry. 1955. “Myopia and Inconsistency in Dynamic Utility Maximization.” The Review of Economic Studies 23 (3): 165–80. https://doi.org/10.2307/2295722.Search in Google Scholar

Thurow, Lester C. 1969. “The Optimum Lifetime Distribution of Consumption Expenditures.” The American Economic Review 59 (3): 324–30.Search in Google Scholar

Received: 2022-10-10

Accepted: 2023-10-22

Published Online: 2023-12-29

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Articles in the same Issue

https://doi.org/10.1515/bejte-2022-0115

Keywords for this article

present bias; time-inconsistent preferences; consumption hump; commitment mechanisms; welfare comparison

How the Future Shapes Consumption with Time-Inconsistent Preferences

Article

Abstract

Appendix A: Simplifying the Concavity Condition

Appendix B: Derivation of Eq. (37)

Appendix C: Derivation of Limits of ΔU τ

Appendix D: Gradient of ΔU 1 at the Origin

Appendix E: Hessian of ΔU 1 at the Origin for T = 3

Appendix F: Sufficient Upper Bound on ɛ T for Pareto Dominance of the Commitment Path

References

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Appendix C: Derivation of Limits of ΔU _τ

Appendix D: Gradient of ΔU ₁ at the Origin

Appendix E: Hessian of ΔU ₁ at the Origin for T = 3

Appendix F: Sufficient Upper Bound on ɛ _T for Pareto Dominance of the Commitment Path