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14.387 Spring 2014

Problem Set 2 Solutions

1.

(a) Fits and residuals:

x i = zi π

ξ i = x i −x i

cov x i , ξ i = 0

Plug these in to the IV estimand:

β IV = cov (yi , x i )

cov (x i , x i )

= cov (yi , x i )

cov x i + ξ i , x i

= cov (yi , x i )

cov (x i , x i ) + cov ξ, x i

= cov (yi , x i )

var (x i )= β 2 SLS

(b) Fits:

x i = zi π

x ∗

i = zi π

Preliminaries:

π p

→ π1N

Z Z p

→ QZ Z

1N

Z ξ p

→ 0

1√ N

Z η d

→ N (0, ΩZ η )

Using these we get:

β IV = X X − 1

Xy

= ( π Z X )− 1

π Z y

= ( π Z (Zπ + ξ ))− 1

π Z (Xβ + η)

= β + ( π Z Zπ + π Z ξ )− 1

π Z η

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√ N β IV −β = √ N (π Z Zπ + π Z ξ )− 1

π Z η

= π 1N

Z Zπ + π 1N

Z ξ − 1

π 1√ N

Z η

d

→ (π Q Z Z π )− 1

π N (0, ΩZ η )

= N 0, (π Q Z Z π )− 1

π ΩZ η π (π Q Z Z π )− 1

β ∗IV = X ∗ X − 1

X ∗ y

= ( π Z X )− 1

π Z y

= ( π Z (Zπ + ξ ))− 1

π Z (Xβ + η)

= β + ( π Z Zπ + π Z ξ )− 1

π Z η

√ N β ∗IV −β = √ N (π Z Zπ + π Z ξ )− 1

π Z η

= π 1N

Z Zπ + π 1N

Z ξ − 1

π 1√ N

Z η

d

→ (π Q Z Z π )− 1

π N (0, ΩZ η )

= N 0, (π Q Z Z π )− 1

π ΩZ η π (π Q Z Z π )− 1

Thus, the limiting distributions of β IV and β ∗IV are the same. We know that β IV and β 2 SLS are the same, andthus also the limiting distributions of β ∗IV and β 2 SLS are the same.

(c) Reduced form:

yi

= ziπβ + ν

iν i = ξ i β + ηi

Preliminaries:

1N

Z Z p

→ QZ Z

1√ N

Z ξ d

→ N (0, ΩZ ξ )

1√ N

Z ν d

→ N (0, ΩZ ν )

Using these we get:

β ∗2 SLS = X ∗ X ∗− 1

X ∗ y

= ( π Z Zπ )− 1

π Z y

= ( π Z Zπ )− 1

π Z (Zπβ + ν )

= β + ( π Z Zπ )− 1

π Z ν

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√ N β ∗2 SLS −β = √ N (π Z Zπ )− 1

π Z ν

= π 1N

Z Zπ− 1

π 1√ N

Z ν

d

→ (π Q Z Z π )− 1

π N (0, ΩZ ν )

= N (π Q Z Z π )− 1

π ΩZ ν π (π Q Z Z π )− 1

Notice that

Σ ν = E (ξβ + η) (ξβ + η)

= E β 2 ξξ + ηη + 2 βξη

= β 2 E [ξξ ] + E [ηη ] + 2βE [ξη ]

= Σ η + β 2 Σ ξ + 2 β Σ ηξ

and

ΩZ ν = E 1N

Z Σ ν Z

= E 1N

Z Σ η + β 2 Σ ξ + 2 β Σ ηξ Z

= E 1N

Z Σ η Z + β 2 E 1N

Z Σ ξ Z + 2 βE 1N

Z Σ ηξ Z

= Ω Z η + β 2 ΩZ ξ + 2 β ΩZ ηξ .

Thus, the limiting distributions of β 2 SLS and β ∗2 SLS are different unless Σν = Σ η which is equivalent to β 2 Σ ξ +2β Σ ηξ = 0 .

Both of these estimators are consistent, but only β ∗2 SLS is unbiased in nite sample (assuming Q > 1).Therelationship between the asymptotic (as well as nite sample) variances of the estimators is ambiguous. For in-stace, if we assume homoscedasticity, the sign of the difference between the asymptotic variances of β 2 SLS andβ ∗2 SLS depends on the sign of β 2 σ 2

ξ + 2βσ ηξ which can be either postitive or negative.

2.

(a) Model:

D i = D 0 i + ( D 1 i −D 0 i ) Z i

Y 1 i = Y 0 i + ρ

Y i = Y 0 i + ( Y 1 i −Y 0 i ) D i

= Y 0 i + ρD i

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Wald estimand:

E [Y i |Z i = 1] −E [Y i |Z i = 0]E [D i |Z i = 1] −E [D i |Z i = 0]

= E [Y 0 i + ρD i |Z i = 1] −E [Y 0 i + ρD i |Z i = 0]

E [D 0 i + ( D 1 i −D 0 i ) Z i |Z i = 1] −E [D 0 i + ( D 1 i −D 0 i ) Z i |Z i = 0]

= E [ρD 1 i ]−E [ρD 0 i ]

E [D 1 i

−D 0 i ]

= ρ E [D 1 i −D 0 i ]E [D 1 i −D 0 i ]

= ρ

Notice that monotonicity played no role above.

Bias formula in the general case:

E [Y i |Z i = 1] −E [Y i |Z i = 0]E [D i |Z i = 1] −E [D i |Z i = 0]

= E [Y 0 i + ( Y 1 i −Y 0 i ) D i |Z i = 1] −E [Y 0 i + ( Y 1 i −Y 0 i ) D i |Z i = 0]E [D 0 i + ( D 1 i −D 0 i ) Z i |Z i = 1] −E [D 0 i + ( D 1 i −D 0 i ) Z i |Z i = 0]

= E [(Y 1 i −Y 0 i ) (D 1 i −D 0 i )]

E [D 1 i −D 0 i ]= E [Y 1 i −Y 0 i |D 1 i > D 0 i ]

P [D 1 i > D 0 i ]P [D 1 i > D 0 i ]−P [D 1 i < D 0 i ]

−E [Y 1 i −Y 0 i |D 1 i < D 0 i ] P [D 1 i < D 0 i ]

P [D 1 i > D 0 i ]−P [D 1 i < D 0 i ]= E [Y 1 i −Y 0 i |D 1 i > D 0 i ]

+ P [D 1 i < D 0 i ]

P [D 1 i > D 0 i ]−P [D 1 i < D 0 i ]

×E [Y 1 i −Y 0 i |D 1 i > D 0 i ]−E [Y 1 i −Y 0 i |D 1 i < D 0 i ]

WALD −LATE =

P [D 1 i < D 0 i ]P [D 1 i > D 0 i ]−P [D 1 i < D 0 i ]

= ×E [Y 1 i −Y 0 i |D 1 i > D 0 i ]−E [Y 1 i −Y 0 i |D 1 i < D 0 i ]Thus, the failure of monotonicity is not a huge concern if either E [Y 1 i −Y 0 i |D 1 i > D 0 i ] ≈E [Y 1 i −Y 0 i |D 1 i < D 0 i

or P [D 1 i > D 0 i ] is large relative to P [D 1 i < D 0 i ].

(b) Notice that in general

E [Y i |Z i = 1] −E [Y i |Z i = 0] = E [Y 1 i −Y 0 i |D 1 i > D 0 i ]P [D 1 i > D 0 i ]

−E [Y 1 i −Y 0 i |D 1 i < D 0 i ]P [D 1 i < D 0 i ] .

Suppose Y 1 i −Y 0 i > 0 for all i . Then,

E [Y i |Z i = 1] −E [Y i |Z i = 0] ≤ 0

⇔ E [Y 1 i −Y 0 i |D 1 i > D 0 i ]E [Y 1 i −Y 0 i |D 1 i < D 0 i ] ≤

P [D 1 i < D 0 i ]P [D 1 i > D 0 i ]

.

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(c) No always-takers: D 0 i = 0 for all i which implies that D i = 1⇔D 1 i = 1 , Z i = 1 . Thus,

E [Y 1 i −Y 0 i |D 1 i > D 0 i ] = E [Y 1 i −Y 0 i |D 1 i = 1]

= E [Y 1 i −Y 0 i |D 1 i = 1 , Z i = 1]

= E [Y 1 i

−Y 0 i

|D i = 1] .

No never-takers: D 1 i = 1 for all i which implies that D i = 0⇔D 0 i = 1 , Z i = 0 . Thus,

E [Y 1 i −Y 0 i |D 1 i > D 0 i ] = E [Y 1 i −Y 0 i |D 0 i = 0]

= E [Y 1 i −Y 0 i |D 0 i = 1 , Z i = 0]

= E [Y 1 i −Y 0 i |D i = 0] .

3.

(a) Difficulties in determining the causal effect of female employment on divorce:

• Reverse causality (divorce causing employment)

• Confounding factors

Experiment to capture the causal effect of interest:

• Ideally would like to randomize the employment status of married women

• More realistically need an instrument that affects employment probability (rst stage) but does not affectdivorce probability directly or through any other channel (exclusion restriction)

(b) Notation:

• D i : divorce dummy

• E i : employment dummy

• X i : covariate vector

• Z i : instrument dummy

• i , ν i : unobserved random errors

Observed divorce and employment status:

D i = 1 [X i β 0 + β 1 E i > i ]

E i = 1 [X i γ 0 + γ 1 Z i > ν i ]

Potential divorce and employment status:

D 1 i = 1 [X i β 0 + β 1 > i ]D 0 i = 1 [X i β 0 > i ]

E 1 i = 1 [X i γ 0 + γ 1 > ν i ]

E 0 i = 1 [X i γ 0 > ν i ]

(c)

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• The population for the study should be working-aged married women (panel data) or working-aged ever-married women (cross-sectional data).

• In the structural model the endogeneity of employment means that i and ν i are correlated.

• In the causal model the endogeneity of employment means that potential outcomes E 0 i and E 1 i are not(mean-)independent of the treatment status D i .

(d) Distributional assumption:

i

ν i

iid∼ N

00

, σ2 ρσ ν

ρσ ν σ2ν

Notation:

• Φ (·)= CDF of standard univariate normal distribution

• Φb (·, ·)= CDF of standard bivariate normal distributions

AT E = E [D 1 i −D 0 i ]

= E 1 [X i β 0 + β 1 > i ]−1 [X i β 0 > i ]= E E 1

X i β 0 + β 1σ

> i

σ | X i −E 1X i β 0

σ>

i

σ | X i

= E ΦX i β 0 + β 1

σ −ΦX i β 0

σ

ET T = E [D 1 i −D 0 i |E i = 1]

= E 1 [X i β 0 + β 1 > i ]−1 [X i β 0 > i ] | X i γ 0 + γ 1 Z i > ν i= E E 1

X i β 0 + β 1σ

> i

σ | X i , Z i , X i γ 0 + γ 1 Z i

σν >

ν iσν

−E E 1X i β 0

σ>

i

σ | X i , Z i , X i γ 0 + γ 1 Z i

σν >

ν iσν

= E Φb

X i β 0 + β 1

σ , X i γ 0 + γ 1 Z iσ ν −Φb

X i β 0

σ , X i γ 0 + γ 1 Z iσ ν

Φ X i γ 0 + γ 1 Z iσ ν

LATE = E [D 1 i −D 0 i |E 1 i > E 0 i ]

= E 1 [X i β 0 + β 1 > i ]−1 [X i β 0 > i ] | X i γ 0 + γ 1 > ν i > X i γ 0 = E E 1

X i β 0 + β 1σ

> i

σ | X i , X i γ 0 + γ 1

σν >

ν iσν

> X i γ 0

σν

−E E 1X i β 0

σ>

i

σ | X i , X i γ 0 + γ 1

σν >

ν iσν

> X i γ 0

σν

= E Φb

X i β 0 + β 1

σ , X i βγ 0 + γ 1σ ν −Φb

X i β 0 + β 1

σ , X i β + γ 0σ ν − Φb

X i β 0

σ , X i βγ 0 + γ 1σ ν −Φb

X i β 0

σ , X i

Φ X i βγ 0 + γ 1σ ν −Φ X i β + γ 0

σ ν

ATE is a weighted average of ETT and the average effect on the untreated. ETT is a weighted average of LATE and the average effect on the always-takers. In general we cannot say anything about the relative

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magnitudes of ATE, ETT and LATE.

(e) See attached code.

4.

(a) Model:

yit = α i + x it β + it

where i = 1 , . . . , N and t = 1 , 2.

Notation:

yit = yit −yi

x it = x it −x i

∆ yi = yi 2 −yi 1

∆ x i = x i 2 −x i 1

Notice that

x i 1 = x i 1 − 12

(x i 1 + x i 2 )

= 1

2 (x i 1 −x i 2 )

= −12

∆ x i

x i 2 = x it 2 − 12 (x i 1 + x i 2 )

= 1

2 (x i 2 −x i 1 )

= 1

2∆ x i

yi 1 = yi 1 − 12

(yi 1 + yi 2 )

= 1

2 (yi 1 −yi 2 )

= −12

∆ yi

˜y

i 2 = y

it 2 − 1

2 (y

i 1 +y

i 2 )=

12

(yi 2 −yi 1 )

= 1

2∆ yi .

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Using this we get

β F E = N

i =1

x i 1 x i 1 + x i 2 x i 2

− 1 N

i =1

x i 1 yi 1 + x i 2 yi 2

=

N

i =1

14∆ x i ∆ x i +

14∆ x i ∆ x i

− 1 N

i =1

14∆ x i ∆ yi +

14∆ x i ∆ yi

= N

i =1

∆ x i ∆ x i

− 1 N

i =1

∆ x i ∆ yi

= β F D .

(b) Notation:

• Y ist : employment dummy for individual i in state s in year t

• F A s : fraction of teenagers in the affected wage range in state s

• D t : dummy for year 1990

Micro-level model:

Y ist = γ s + λ t + δ (F A s ·D t ) + ist

Since all the variables live on the state-year level, we can average over individuals (teenagers) within state-yearcells without any loss of information (need to weight by cell sizes to match exactly):

Y st = γ s + λ t + δ (F A s ·D t ) + ¯ st

We can get rid of the state xed effect by rst-differencing the data:

Y s, 1990 − Y s, 1989 = ( λ 1990 −λ 1989 ) + δF A s + (¯ s, 1990 − s, 1989 )

∆ Y s = λ∗ + δF A s + ∆¯ s

Card (1992) weights the differenced equation according to the average sample size in the two periods. Runningthe FE model and weighting according to the sample sizes in each state-year cell might work better (althoughunlikely to matter much in practice).

5. See attached code.

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cd " H: \ Teachi ng\ 14. 387_Spri ng_2014\ Pr obl em_Set_ 2"

/ ***************// * EXCERCI SE 3 *// ***************/

/ * P a r t ( e ) * /

use "m_d_806. dt a", cl eardesccountdest ri ng, repl ace

r epl ace qt r mar=qtr mar - 1gen yom = yobm+ agemar i f qt r bthm <= qtr marrepl ace yom = yobm + agemar + 1 i f qt rbthm> qt r margen dom_q = yom + ( qt r mar/ 4)gen do1b_q = yobk + ( qt r bki d/ 4)gen i l l egi t = 1 i f ( dom_q- do1b_q)>0

gen boy1st = ( sexk == 0)gen boy2nd = ( SEX2ND == 0)gen boys2 = ( ( sexk == 0) & ( SEX2ND == 0) )gen gi r l s2 = ( ( sexk == 1) & ( SEX2ND == 1) )

gen samesex = ( ( boys2 == 1) | ( gi r l s2 == 1) )gen yobd = 80 - aged i f qt r bt hd == 0r epl ace yobd = 79 - aged i f qt r bt hd ! = 0

gen ageqm = 4*( 80- yobm) - qt r bt hm- 1gen ageqd = 4*( 80- yobd) - qtr bthdgen agefst m = i nt ( ( ageqm- ageqk)/ 4)gen agefst d = i nt ( ( ageqd- ageqk)/ 4)

gen mul t i 2nd = ( AGEQ2ND == AGEQ3RD)gen moreki ds = ki dcount > 2gen workedm = ( weeksm > 0)

gen educm = gradem- 3repl ace educm= gr adem - 2 i f f i ngradm == 2 | f i ngradm == 1repl ace educm= 0 i f educm< 0

gen hsgrad = ( educm == 12)gen moret hs = ( educm > 12)gen col l egegr ad = ( educm >= 16)

gen bl ackm = ( r acem == 2)gen hi spm = ( racem == 12)

keep i f aged ! = .keep i f t i mesmar == 1keep i f mari t al == 0keep i f i l l egi t ! = 1keep i f agef st m >= 15keep i f agef st d >= 15

keep i f agem >= 21 & agem <= 35keep i f ki dcount >= 2 & ki dcount ! = .keep i f AGEQ2ND > 4keep i f asex == 0 & aage == 0 & ASEX2ND == 0 & AAGE2ND == 0 & aqt r br t h == 0

count

gl obal covs = " ageqm agef st m boy1st boy2nd"

est cl earr eg wor kedm mor eki deststoprobi t workedm moreki dsmf xeststo

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r eg wor kedm mor eki ds $covseststoprobi t workedm moreki ds $covsmf xeststoest t ab, b( 3) s e( 3) compress margi n keep( moreki ds)

est cl ear

i vr eg2 workedm ( moreki ds = samesex)eststobi pr obi t ( workedm = moreki ds) ( moreki ds = samesex)mf x comput e, predi ct ( pcond1)eststoi vr eg2 workedm ( moreki ds = samesex) $covseststobi pr obi t ( workedm = moreki ds $covs) ( moreki ds = samesex $covs)mf x comput e, predi ct ( pcond1)eststoest t ab, b( 3) s e( 3) compress margi n keep( moreki ds)

/ ***************// * EXCERCI SE 5 *// ***************/

use " i ndi vi d_f i nal . dta" , c l ear

gen above = ( di f shar e >= 0)gen above_di f shar e = above*di f shar ef orval i = 2/ 4

gen di f share`i ' = di f share^ i 'gen above_di f shar e`i ' = above*di f share^ i '

l ogi t myout comenext above di f shar e* above_di f shar e*mf x, at ( zero)

r d myout comenext di f shar e, c( 0)

l ocal h_opt = . 1514835257991828gen l ambda = ( 1 - abs( di f shar e/ `h_opt ' ) ) *( abs( di f shar e/ `h_opt ' ) <= 1)r eg myout comenext above di f shar e above_di f shar e [ aw = l ambda] , r obust