Panel Data Analysis Part III Modern Moment Estimation Arellano - - PowerPoint PPT Presentation

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Panel Data Analysis Part III Modern Moment Estimation Arellano - - PowerPoint PPT Presentation

Panel Data Analysis Part III Modern Moment Estimation Arellano and Honor (2000) James J. Heckman University of Chicago Econ 312 This draft, May 8, 2006 = + + = 1


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SLIDE 1

Panel Data Analysis

Part III — Modern Moment Estimation Arellano and Honoré (2000)

James J. Heckman University of Chicago Econ 312 This draft, May 8, 2006

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= + + = 1 I = + = 1 is strictly exogenous if ( |

) = 0

  • = (1 )
  • OLS identifies

1

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SLIDE 3

Panel Data Setting: Strictly exogenous given if ( |

) = 0

= 1 for all

but not necessarily for

2

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SLIDE 4

Consequence: First dierence eliminates eects: ( 1 |

) = 0.

Multivariate regression with cross equation restrictions. (This is essentially all the information). Partial Adjustment Model With Strictly Exogenous Variable = (1) + 0 + 11 + + . Assume ( |

) = 0

= 2 Does not restrict the serial correlation in . 3

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SLIDE 5

Restriction: ( |

) = 0

Model identified for 3 = 3 case we acquire orthogonality restrictions ((3 2 03 12) = 0 = ((3)) = 0 = 1 2 3 Use these in GMM to identify model. 3 equations in 3 un- knowns and we acquire exact identification. Note: Strict exogeneity enables us to identify dynamic eect

  • f on with arbitrary serial correlation in the errors;

Price: Assumes not influenced by past values of and . 4

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SLIDE 6

Definition: is predetermined if (*) ( |

1

  • ) = 0

= 2

  • = (1

) 1

  • = (1

1

  • ).

Current shocks are uncorrelated with past values of and cur- rent and past values of . Feedback from lagged dependent variables to future not ruled out.

  • Eg. Euler equations. (Information set of agents uncorrelated

with current and future idiosyncratic shocks but not past shocks). 5

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SLIDE 7

Example: Euler Equation:

  • ()

1

  • (1)

| I1

¸ = 1 Power Utility: = ()1 1 1 = ()

1 ¶

  • | I1

# = 0 6

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SLIDE 8

[

() | I1] = 0

  • £
  • | I1

¤ = 0 Instruments that are ecient:

  • £
  • 1

¤ = 0 1 is in the information set. Crucial that instruments don’t include variables that cause the innovation. 7

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SLIDE 9

Implication: ( 1 | 1

  • 2
  • ) = 0, = 3

For = 3 we acquire 0 =

  • 1

1 2 (3 2 03 12)

  • This condition is not the same as that in strictly exogenous

models: We acquire 3 moments only 2 in common with last (across strictly exogenous and these models). Standard errors are consistent with arbitrary serial correlation. 8

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SLIDE 10

If we ruled out arbitrary serial correlation, by (

| 1 1

  • ) = 0

= 2 we acquire superset of all conditions. (*) and previous ones. (*) = ( ) = 0 1 because we have that the covariances are zero ( 1

  • ) = 0
  • ( 1) = 0 generically.

9

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SLIDE 11

Observe that in the predetermined case we can have special cases of serial correlation. e.g. = 4 ( ) = 0 2 first order MA. Valid orthogonality conditions: 3 2 03 12 = 3 2 4 3 04 13 = 4 3 10

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SLIDE 12

Orthogonality conditions: (14) = 0 (14) = 0 (24) = 0 Other orthogonality conditions. 4 = 3 + 04 + 13 + + 4 3 = 2 + 02 + 11 + + 3 4 = (3 2) + 0(4 3) +1(32 2) + (4 3) 11

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SLIDE 13

Suppose Uncorrelated Eects (Some uncorrelated with ) [

(2 1 02 11)] = 0

  • rthogonality conditions for each regressor. Predetermined

variables could be uncorrelated with fixed eects = (1) + (1) + + if = 0 would be uncorrelated with 12

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SLIDE 14

Adds more orthogonality restrictions: (1(2 1 02 11)) = 0 (( 1 0 11)) = 0 = 2 Only identified when 3. 13

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SLIDE 15

Statistical Definitions: Strict Exogeneity: ( |

) = ( | )

  • (+1 |

)

= ((+1) | (y does not Granger cause ) 14

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SLIDE 16

Let (+1)

  • = (+1 )

if ( |

) = 0

  • + 0

(+1)

  • +

and ((+1) |

) = 0 + 0

  • +

= 0 = 0. 15

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SLIDE 17

AR-1 Models Balestra - Nerlove Problem // Nickell = 1 + + = 1 I ; = 2 (A-1) ( | 1

  • ) = 0

= 2 () = , (2

) = 2

  • () = 2
  • and freely correlated

(2

| 1

  • ) need not coincide with 2

.

16

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SLIDE 18

We get ( 1)( 2)2 moment restrictions: (2

  • ( 1)) = 0 etc., etc.

Using minimum discrepancy (CMD) methods take = 1 + + . For we obtain: = 1 + + () = (1) + () + ()

=0

() = [(1) = 1]() = ] 17

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We take × μ + 1 2 ¶ distinct elements of = (0

).

For = 3, we obtain 31 = 21 + 1 21 = 11 + 1 = 31 21 21 11 = (21 11) (21 11) 1 = 31 21 2 = 32 22

  • model just identified.

Fit discrepancies between the population moments and fitted moments. 18

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SLIDE 20

Other Restrictions Lack of correlation between eects and errors ( | 1

  • ) = 0

= 2 0 = [( 1) [1 2]] quadratic (in ) restrictions: because (1) = 0. 19

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SLIDE 21

When (Ahn-Schmidt). Multiply = 1 + + by = 1 + 2

+

= 1 + 2

.

For = 3, imposes no further restrictions. 2

= (32 21) (22 11)

20

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SLIDE 22

Other Restrictions: Homoscedasticity: (2

) = 2

= 2 (2

2 0) = 0, etc.

Time Series Homoscedasticity: (2

) = 2

= 2(1)(1) + 2

+ 2 + 21 q

21

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SLIDE 23

Change in uncorrelated with fixed eect ( 1 | ) = 0 = 2 adds (to A-1) the following moment conditions: [( 1)(1)] = 0 = 3 will satisfy = (22 21)1(32 31) Full stationarity: 11 = 2

  • (1 )2 +

2 1 2 22

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SLIDE 24

Unit Roots Case = 1 + + for | | 1 we can write =

+

= 1 + where

=

  • 1 Substitute:

= (1 ) +

= (1 + ) + . Now when = 1, we get a distinction: 23

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SLIDE 25

Model I: =

+

= 1 + . A model with an initial random intercept; (now) we ignore the restriction that | | 1). Model II: = 1 + + (heterogenous linear growth). 24

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SLIDE 26

Empirical Features of II: = ( 1) () = + ( 1) () 1 and ( 1) 0 (But rarely found) 25

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SLIDE 27

Observe that we fail the rank condition for applicate of (A-1) in Model I using: ([ 1]) = 0 1 Why? ( 1) = ( 1 2) = ( 1) = 0 we divide by zero in forming this function. 26

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SLIDE 28

Model II: Passes the rank condition: (( 1)) = ( 1 2) for = 2 satisfy rank condition. Observe that with mean stationarity rank condition is satisfied: [(1)( 1)] = 0 [[1](1)] = [[1 2] [ + 1]] 6= 0 (1 2) = 1 2 + 1 = 1

  • If maintained, can test null hypothesis of random walk with-
  • ut drift against mean stationarity.

27

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SLIDE 29

GMM Estimation (See Part IV for details on GMM) Consider = 0 + = + iid across ( |

) = 0

are the instruments: ( × 1) = 0 + = (1 )0 = (0

1 0 )

= (1 )0. We get I.V. for models in dierences. 28

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SLIDE 30

Let be an upper triangular ( 1)× transformation matrix

  • f rank 1 Such that = 0 is × 1 vector of 1’s. is

first dierence operator or forward dierence operator. Orthogonality restrictions: [0

] = 0

is block diagonal

  • 1
  • 29
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SLIDE 31

Optimal GMM: ˆ = (0

)1

=

  • X

=1

  • =
  • X

=1 .

30

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SLIDE 32

is estimate of inverse of (0

0)

(Just an application of our GMM notes). Robust version use ˜ in place of (as in Eicker-White) ˜ = ˜ . Optimal variance: Var(ˆ ) = £ (0

0) [(0 0]1 (0 )

¤ can be shown to be invariant to . 31

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SLIDE 33

Orthogonal Deviations (Forward Dierencing)

  • 1 =
  • 1

(+1 + + ) ¸ 2

=

+ 1 (a) Preserves the orthogonality of transformed errors, gets rid

  • f fixed eect.

0 = μ( 1)

  • 1

2 ¶¸12

  • 32
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SLIDE 34

+ =

  • 1

( 1)1 ( 1)1 ( 1)1 1 ( 2)1 ( 2)1 1 12 12 1 1

  • 00

0 = 1 00 = 0

= within operator. = 1 + +

  • =

+

33

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SLIDE 35

= 1 +

=

(

) = (1 ) +

= 1 + (1 )

+ : correct

34

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SLIDE 36

Consider = 1 I. =

+

= 1 + Random Initial Condition Model?

  • r

II. = 1 + + 35

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SLIDE 37

Random Walk with Heterogenous Drift. Model I is more relevant: Why? Model II has autocorrelation 1 = ( 1) () = + ( 1) () 1 when 1. = + Correl ( 1) 0. But usually found to be negative. Correl ( 1) = Correl( 1) = ( 1). 36

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SLIDE 38

Using Stationarity Restrictions: Consider a model = 0 + + if ( |

) = 0

  • = ( 1 1).

Moments are [1

  • ( 0)] = 0.

If ( |

) = 0

37

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SLIDE 39

Add to this Ahn-Schmidt ( 0)(1 1)] = 0 if ( | ) = 0 we get (()( 0)) = 0 38