How the Future Shapes Consumption with Time-Inconsistent Preferences

James Feigenbaum; Sepideh Raei

doi:10.1515/bejte-2022-0115

40% Rabatt

auf Fachbücher bei De Gruyter Brill *

Artikel

How the Future Shapes Consumption with Time-Inconsistent Preferences

James Feigenbaum und Sepideh Raei

Veröffentlicht/Copyright: 29. Dezember 2023

Veröffentlicht von

Veröffentlichen auch Sie bei De Gruyter Brill

Manuskript einreichen Informationen für Autor*innen Erkunden Sie dieses Fachgebiet

Aus der Zeitschrift The B.E. Journal of Theoretical Economics Band 24 Heft 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 .

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Received: 2022-10-10

Accepted: 2023-10-22

Published Online: 2023-12-29

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

Schlagwörter für diesen Artikel

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

How the Future Shapes Consumption with Time-Inconsistent Preferences

Artikel

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