Economics 2010c: Lecture 6Quasi-hyperbolic discounting

David Laibson

September 18, 2014

Outline:

1. Hyperbolic Discounting

2. Hyperbolic Euler Equation

3. Lifecycle simulations

4. Method of Simulated Moments (MSM)

1 Hyperbolic Discounting

Check the item below that you would most prefer.

1. 15 minute free massage now

2. 20 minute free massage in an hour

Check the item below that you would most prefer.

1. 15 minute free massage in 168 hours

2. 20 minute free massage in 169 hours

Read and van Leeuwen (1998): Food experiment.

² Choose for Next Week: Fruit (74%) or Chocolate (26%)

² Choose for Today: Fruit (30%) or Chocolate (70%).

Read, Loewenstein & Kalyanaraman (1999): Video experiment

² Choose for Next Week: Low-brow (37%) or High-brow (63%)

² Choose for Today: Low-brow (66%) or High-brow (34%).

Evidence from gyms (Della Vigna and Malmendier 2004).

² Average cost of gym membership: $75 per month.

² Average number of visits per month: 4.

² Average cost per visit: $19.

² Cost of “pay-per-visit:” $10.

² Quasi-hyperbolic discounting (Phelps and Pollak 1968, Laibson 1997):1 2 3

= () + (+1) + 2(+2) +

3(+3) +

² For exponentials: = 1 = () + (+1) +

2(+2) + 3(+3) +

² For “quasi-hyperbolics”: 1 = () +

h(+1) +

2(+2) + 3(+3) +

i

² To build intution, assume that ' 12 and ' 1

f1 2 3 g = f1 121

21

2 g

= () +1

2[(+1) + (+2) + (+3) + ]

² Relative to the current period, all future periods are worth less (weight 12)

² Most (for this example, all) of the discounting takes place between thecurrent period and the immediate future.

² There is little (for this example, no) additional discounting between futureperiods.

= () +1

2[(+1) + (+2) + (+3) + ]

² Preferences are dynamically inconsistent.

² At date we prefer to be patient between + 1 and + 2

² At date + 1 we want immediate grati…cation at + 1

+1 = (+1) +1

2[(+2) + (+3) + (+4) + ]

Akerlof (1992) and O’Donoghue and Rabin (1999) on procrastination:

² Assume = 12 and = 1.

² Suppose exercise (cost 6) generates delayed bene…ts (value 8).

² Exercise Today? ¡6 + 12[8] = ¡2 0

² Exercise Tomorrow? 0 + 12[¡6 + 8] = 1 0

² Agent would like to exercise tomorrow.

Predictions:

² Procrastination when costs precede bene…ts (Della Vigna and Malmendier2004, 2006; Ariely and Wertenbroch 2002; Augenblick, Niederle, andSprenger 2013)

² Downward sloping consumption paths within pay-cycle (e.g., Shapiro 2005,Mastrobuoni and Weinberg 2009)

² Willingness to use commitment: savings (Ashraf, Karlan, and Yin, 2006;Beshears et al 2013), student productivity (Ariely and Wertenbroch, 2002;Houser et al., 2010, Chow 2011; Augenblick, Niederle, and Sprenger,2013), cigarette smoking (Gine, Karlan, and Zinman, 2010), workplaceproductivity (Kaur, Kremer, and Mullainathan, 2010), and exercise (Milk-man, Minson, and Volpp, 2012; Royer, Stehr, and Sydnor, 2012).

Goal account usage(Beshears et al 2013)

FreedomAccount

FreedomAccount

FreedomAccount

Goal Account10% penalty

Goal account20% penalty

Goal accountNo withdrawal

35% 65%

43% 57%

56% 44%

2 Hyperbolic Euler Equation

² Let represent consumption.

² Let represent cash-on-hand.

² Let ~ represent iid stochastic income.

² Let represent gross interest rate.

² So +1 = ( ¡ ) + ~+1

² A (Markov) strategy is a map from state to control .

² Let be the continuation-value function, be the current-value functionand be the consumption function. Then:

() = ( ()) + E[ ((¡ ()) + )]

() = ( ()) + E[ ((¡ ()) + )]

() = argmax() + E[ ((¡ ) + )]

² Note that accumulates utils exponentially.

² Note that accumulates utils quasi-hyperbolically.

² Envelope Theorem. 0() = 0( ())

² First-order-condition. 0( ()) = E

h 0((¡ ()) + )

i

² Identity linking and () = ()¡ (1¡ )( ())

2.1 Problem is recursive

² Start with .

² Find : () = argmax

() + E[ ((¡ ) + )]

² Find : () = ( ()) + E[ ((¡ ()) + )]

² In this way, generate an operator : 7!

2.2 Can also derive an Euler Equation

We have

0() = Eh 0(+1)

i= E

" 0(+1)¡ (1¡ )0(+1)

+1+1

#

= E

"0(+1)¡ (1¡ )0(+1)

+1+1

#

So,

0() = E"

Ã+1+1

!+

Ã1¡ +1+1

!#0(+1)

See Harris and Laibson (2003).

3 Lifecycle simulations (Angeletos et al 2001)

3.1 Demographic Assumptions

² Mortality

² Retirement (PSID)

² Dependents (PSID)

² Three educational groups: NHS, HS, COLL

² Stochastic labor income (PSID)

3.2 Dynamic Budget Constraints

² Credit limit: (.30)( ¹) (Calibrate from SCF.)

² Real after-tax rate of return: 375%

² Real rate of return on illiquid investment: 500%

² Real credit card interest rate: 1175%

² State variables: liquid wealth, illiquid wealth, autocorrelated income.

² Choice variables: liquid wealth investment, illiquid wealth investment.

3.3 Preferences

² Instantaneous utility function: CRRA=2.

² Quasi-hyperbolic discounting: 1 2 3

² For exponentials: = 1 For hyperbolics: = 07

² For exponentials: = Exponential For hyperbolics: = Hyperbolic

² Calibration: Pick value of Exponential (094) that matches empirical re-tirement wealth: median ‘wealth to income ratio’ ages 50-59.

² Pick Hyperbolic the same way (096)

Figure1: Discount functions

0

0.2

0.4

0.6

0.8

1

1.2

0 5 10 15 20 25 30 35 40 45 50Year

Dis

coun

t fun

ctio

n

Quasi-hyperbolic

Exponential

Hyperbolic

Source: Authors’ calculations. Exponential: δt, with δ=0.939; hyperbolic: (1+αt)-γ/α, with α=4 and γ=1; and quasi-hyperbolic: {1,βδ,βδ2,βδ3,...}, with β=0.7 and δ=0.957.

Source: Authors' simulations.The figure plots the simulated average values of consumption and income for households with high school graduate heads.

Figure 2: Simulated Mean Income and Consumption of Exponential Households

0

5000

10000

15000

20000

25000

30000

35000

40000

45000

20 30 40 50 60 70 80 90

Age

Inco

me,

Con

sum

ptio

n

Income

Consumption

Source: Authors' simulations.The figure plost the simulated life-cycle profiles of consumption and income for a typical household with a high school graduate head.

Figure 3: Simulated Income and Consumption of a Typical Exponential Household

0

10000

20000

30000

40000

50000

60000

70000

20 30 40 50 60 70 80 90

Age

Inco

me,

Con

sum

ptio

n

Income

Consumption

Source: Author's simulations.The figure plots average consumption over the life-cycle for simulated exponential and hyperbolic households with high-school graduate heads.

Figure 4: Mean Consumption of Exponential and Hyperbolic Households

0

5000

10000

15000

20000

25000

30000

35000

40000

45000

20 30 40 50 60 70 80 90

Age

Con

sum

ptio

n

Hyperbolic

Exponential

Source: Authors' simulations.The figure plots the simulated mean level of liquid assets (excluding credit card debt), illiquid assets. total assets, and liquid liabilities for households with high school graduate heads.

Figure 5: Simulated Total Assets, Illiquid Assets, Liquid Assets, and Liquid Liabilities for Exponential Consumers

0

25000

50000

75000

100000

125000

150000

175000

200000

Ass

ets

and

Liab

ilitie

s

Total Assets

Illiquid Assets

Liquid Assets

-2000

-1800

-1600

-1400

-1200

-1000

-800

-600

-400

-200

0

20 30 40 50 60 70 80 90Age

Liquid Liabilities

Source: Author's simulations.The figure plots mean total assets, excluding credit card debt, over the life-cycle for simulated exponential and hyperbolic households with high school graduate heads.

Figure 6: Mean Total Assets of Exponential and Hyperbolic Households

0

20000

40000

60000

80000

100000

120000

140000

160000

180000

200000

20 30 40 50 60 70 80 90

Age

Tot

al A

sset

s

Hyperbolic Total Assets

Exponential Total Assets

Source: Authors' simulations.The figure plots average illiquid wealth over the life-cycle for simulated exponential and hyperbolic households with high school graduate heads.

Figure 7: Mean Illiquid Wealth of Exponential and Hyperbolic Households

0

20000

40000

60000

80000

100000

120000

140000

160000

180000

200000

20 30 40 50 60 70 80 90

Age

Illqu

id A

sset

s

Hyperbolic

Exponential

Source: Authors' simulations.The figure plots average liquid assets (liquid wealth excluding credit card debt) and liabilities (credit card debt) over the life-cycle for simulated exponential and hyperbolic households with high school graduate heads.

Figure 8: Mean Liquid Assets and Liabilities of Exponential and Hyperbolic Households

0

20000

40000

60000

80000

100000

120000

Ass

ets

and

Liab

ilitie

s

Exponential Assests

Hyperbolic Assests

-6000

-5000

-4000

-3000

-2000

-1000

0

20 30 40 50 60 70 80 90

Age

Exponential Liabilities

Hyperbolic liabilities

Exponential Hyperbolic Empirical Data

% with liquid 112 73% 40% 42%

mean liquid assetsliquid + illiquid assets 050 039 008

% borrowing on “Visa” 19% 51% 70%

mean borrowing $900 $3408 $5000+

C-Y comovement 003 017 023

¢ ln() = ¡1¢ln() + +

4 MSM Estimation(Laibson, Repetto, Tobacman 2008):

² Use the Method of Simulated Moments (Pakes and Pollard 1989).

² Pick parameter values to minimize the gap between simulated data andempirical data.

– Substantial retirement wealth accumulation (SCF)

– Extensive credit card borrowing (SCF, Fed, Gross and Souleles 2000,Laibson, Repetto, and Tobacman 2000)

– Consumption-income comovement (Hall and Mishkin 1982, many oth-ers)

4.1 Data

Statistic % borrowing on ‘Visa’? 0.68 0.015

(% Visa)

borrowing / mean income 0.13 0.01( Visa)

C-Y comovement 0.23 0.11(CY )

median 50-59 3.88 0.25weighted mean 50-59 2.60 0.13

(wealth)

4.2 Estimator

Estimate parameter vector and evaluate models wrt data.

² = N empirical moments

² () = analogous simulated moments

² () ´ ( ()¡)¡1 ( ()¡)0, a (scalar) loss function

² Minimize loss function: = argmin()

² is the MSM estimator.

² Pakes and Pollard (1989) prove asymptotic consistency and normality.

² Speci…cation tests: () » 2( ¡#)

4.3 Results

² Exponential ( = 1) case: = 0846 (0.025);

³ 1´= 217

² Hyperbolic case:( = 07031 (0.109) = 0958 (0.007)

³ ´= 301

(Benchmark case:h

i= [10375 005 11152])

Punchlines:

² estimated signi…cantly below 1.

² Reject = 1 null hypothesis with a t-stat of 3.

² Speci…cation tests reject only the exponential model.

BenchmarkModel

Exponential Hyperbolic Data Std err

Statistic:(1 )

= 846

( )

= 703

= 958

% 0.67 0.634 0.68 0.015

0.15 0.167 0.13 0.01

0.293 0.314 0.23 0.11

-0.05 2.69 2.60 0.13

() 217 3