Doing It Now or Later
American Economic Review, 1999 · paper · Economics
Models self-control problems as present-biased preferences when an activity must be done exactly once. Naive people procrastinate on immediate-cost tasks and act too soon on immediate-reward ones; sophistication reduces procrastination but worsens acting too soon.
Checked: PDF from pdf.stafforini.com, fetched 5 Oct 2026. Page numbers are those of that PDF.
3 findings
Substance Footnote 20 says "doing it in period 3 yields the stream of instantaneous utilities (−1, 0, 0, 0)"; under the footnote 12 normalization (u_τ = −c_τ, u_{T+1} = v_τ) the stream for τ=3 with c3=1, v3=0 is (0, 0, −1, 0).
- p. 13 Footnote 20
doing it in period 3 yields the stream of instantaneous utilities (−1, 0, 0, 0)
- p. 6 Footnote 12 (normalization)
the instantaneous utilities are u_τ = −c_τ, u_T+1 = v_τ, and u_t = 0 for all t ∉ {τ, T + 1}
- p. 13 footnote 20
instantaneous utilities (−1, 0, 0, 0)
- p. 13 footnote 20
(-1, 0, 0, 0)
Judge's note: The p.13 image shows (−1, 0, 0, 0). Footnote 12 (p.6) fixes u_τ = −c_τ, u_{T+1} = v_τ, and zero elsewhere. With v=(1,x,0), c=(0,y,1) and T=3: doing it in period 1 gives (0,0,0,1), which the paper prints correctly; doing it in period 3 gives u3=−1 and u4=0, i.e. (0,0,−1,0). The dominance conclusion still holds, but the printed vector is wrong.
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Reference or label In Example 2 the sophisticates' strategy is labelled with the naifs' symbol: "s^n = (Y, Y, Y, Y), so sophisticates see the movie in period τs = 1"; it should read s^s.
- p. 8 Example 2, third strategy line
s^n = (Y, Y, Y, Y), so sophisticates see the
Judge's note: The p.8 image shows superscript n on both the second line (naifs, (N,N,Y,Y)) and the third line (sophisticates, (Y,Y,Y,Y)). Example 1 and Definition 4 use s^s for sophisticates. I checked the backward induction with β=1/2 and v=(3,5,8,13): it gives Y in every period, so the strategy is the sophisticates' one.
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Reference or label Section VII says "A naive person may repeatedly delay quitting smoking believing he will quit tomorrow; and Proposition 4 suggests that this problem could lead to significant welfare losses"; the result for naive procrastination under immediate costs is Proposition 3.
- p. 18 Section VII, addiction paragraph
and Proposition 4 suggests that this problem could lead to significant welfare losses
- p. 18 Section VII, col. 1, addiction paragraph
he will quit tomorrow; and Proposition 4 sug- gests that this problem could lead to significant welfare losses
- p. 12 Proposition 3 and Proposition 4
PROPOSITION 3: Suppose costs are imme- diate
- p. 18 Section VII, addiction paragraph
Proposition 4 suggests that this problem could lead to significant welfare losses
Judge's note: Checked p.18 and p.12. Proposition 3 (immediate costs) has sup[U0(τtc) − U0(τn)] = 2X̄ for naifs, and sophisticates' losses vanish as β→1, which matches "Sophistication should prevent this problem." Proposition 4 (immediate rewards) shows naifs' losses vanish as β→1 while sophisticates' losses can be severe. So the pointer should be to Proposition 3.
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7 minor (typos and the like; most checked by the judge only)
Wording The introduction says "When costs are immediate, a person is always better off with sophisticated beliefs than with naive beliefs", but Section IV states only the weak result "sophisticates always do at least as well as naifs [i.e., U0(τs) ≥ U0(τn)]".
- p. 3 Introduction, col. 1, Section IV preview
When costs are immediate, a person is always better off with sophisticated beliefs than with naive beliefs.
- p. 11 Section IV, col. 1
sophisticates always do at least as well as naifs
Judge's note: p.3 intro vs p.11 Section IV. The body result is a weak inequality; for example the two types are equally well off whenever τs = τn. "Always better off" in the intro reads as strict. This is loose usage that either form could carry, so it is a minor wording point.
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Wording The proof of Prop. 3(1) says "It is straightforward to show we can get arbitrarily close to this bound, so sup(v,c)[U0(τtc) − U0(τs)] = ((1 − β)/β)X̄"; this equality cannot hold for β < 1/2.
- p. 19 Appendix, Proof of Proposition 3, part (1)
It is straightforward to show we can get arbitrarily close to this bound, so sup(v,c)[U0(τtc) − U0(τs)] = ((1 − β)/β)X̄.
Only the authors can say what is right here.
Judge's note: Own working: v_t, c_t ∈ [0, X̄] (Section II; Prop. 3), so U0 = v − c ∈ [−X̄, X̄]. The paper itself says on p.113 (main-p011) that the loss cannot exceed 2X̄. Loss < ((1−β)/β)c_τtc and loss ≤ (X̄ − c_τtc) + X̄ = 2X̄ − c. The minimum of these two is largest at c = 2βX̄ (feasible for β ≤ 1/2), where it equals 2(1−β)X̄ < ((1−β)/β)X̄ when β < 1/2. Counterexample: at β=1/4 the claimed sup is 3X̄ > 2X̄. The stated result is a β→1 limit and needs only the valid upper bound, so it stands. Only this intermediate equality is false for small β, hence S3.
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Wording The proof of Prop. 4(1) says "Since the bound is easily achieved, sup(v,c)[U0(τtc) − U0(τn)] = ((1 − β)/β)X̄"; this equality cannot hold for β < 1/2.
- p. 19 Appendix, Proof of Proposition 4, part (1)
Since the bound is easily achieved, sup(v,c)[U0(τtc) − U0(τn)] = ((1 − β)/β)X̄
Only the authors can say what is right here.
Judge's note: Own working: loss ≤ ((1−β)/β)v_τn, and loss ≤ X̄ − (v_τn − c_τn) ≤ 2X̄ − v_τn. The minimum of the two is at most 2(1−β)X̄ < ((1−β)/β)X̄ for β < 1/2. At β=1/4 the claimed sup is 3X̄, but no loss can exceed 2X̄ (the paper says so on p.113). Proposition 4(1) is a β→1 limit that uses only the valid upper bound, so the stated result stands.
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Typo Example 3 writes the ordered vectors with set braces: "Let v = {12, 18, 18} and c = {3, 8, 13}."
- p. 10 Example 3
Let v = {12, 18, 18} and c = {3, 8, 13}.
Judge's note: The p.10 image confirms the braces. Every other example writes v and c with parentheses, as ordered vectors. This is cosmetic; the results τs=1, τtc=2, τn=3 assume ordered vectors.
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Typo The proof of Proposition 1 defines "t' ≡ max_{τ>t}(v_τ − c_τ)" and then uses t' as a period index (v_t', c_t'); it should be argmax.
- p. 19 Appendix, Proof of Proposition 1(1) and (2)
let t' ≡ max_{τ>t}(v_τ − c_τ)
Judge's note: The p.19 image shows max in both parts (1) and (2). t' is then used as a period, so argmax is meant. The slip is notational and the intended meaning is clear.
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Typo The proof of Prop. 3(2) ends "and τn = T so U0(τn) = −c̄"; the constructed schedule has v_T = 0 and c_T = X̄, so this should be −X̄.
- p. 19 Appendix, Proof of Proposition 3(2), last lines
and τ_n = T so U^0(τ_n) = −c̄
Judge's note: The p.19 image shows −c̄, a symbol defined nowhere. With v_T=0 and c_T=X̄, U0(τn) = −X̄, which is what the next line needs to reach (X̄−ε) − (−X̄) = 2X̄ − ε. This is a symbol slip and the result is unaffected.
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Typo The proof of Proposition 6(2) says "(ii) for salient rewards if TCs do it in period t then naifs do it in period t", where the paper's term, used later in the same proof, is "immediate rewards".
- p. 20 Appendix, Proof of Proposition 6(2)
(ii) for salient rewards if TCs do it in period t then naifs do it in period t
- p. 20 Proof of Proposition 6(2)
for salient rewards if TCs do it in period t then naifs do it in period t
- p. 20 Proof of Proposition 6, part (2), first paragraph
(ii) for salient rewards if TCs
- p. 20 Proof of Proposition 6 (2)
for salient rewards if TCs
Judge's note: The p.20 image confirms "salient rewards". Part (i) says "immediate costs", and the justification a few lines later says "(ii) follows because for immediate rewards". "Salient" appears only in footnote 18, describing Akerlof. This is an inconsistent term.
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