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Two-step Estimators Hypothesis Testing Small Sample Properties Conditional Moments

Estimadores ExtremosGMM (Parte 2)

Cristine Campos de Xavier Pinto

CEDEPLAR/UFMG

Maio/2010

Cristine Campos de Xavier Pinto Institute

Estimadores Extremos
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The two step-estimators can be seen as a GMM estimatorwhen we stack the moments functions from the rst andsecond steps in a vector of moment function.

A general type of estimators

b is the one that with probability

approaching one solves the equation:

1

N

N

i=1

gZi,b,b = 0

where g(Zi,

,

)is a vector of function with the same

dimension of and b is a rst step estimator.


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The estimator can be a part of the joint GMM estimator if bsolves the following moment condition with probabilityapproaching one:

1

N

N

i=1

m (Zi,b) = 0where m (Zi,) is a vector with the same dimension as .


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Lets stack this two vector of functions,

eg(Zi, ,) =

m (Zi,)

0, g(Zi, ,)

0

0

and1

N

N

i=1

gZi,b,b = 0

In this case, the two-step estimator is a GMM estimator.


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To get the asymptotic distribution, we need to formulate theregularity conditions by applying the asymptotic normalityresults for GMM to the vector of stacked moments.

Notation:

G = E [rg(Zi, 0,0)]G = E [rg(Zi, 0,0)]g(Z) = g(Z, 0,0)M= E [rm (Z,0)] (z) =

M1m (Z,0)



S S C
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With probability one b ,b is a GMM estimator withmoment function eg(Zi, ,) = m (Zi,)0 , g(Zi, ,)00with cW equal to an identity matrix.From last class, we know the asymptotic distribution of the

GMM estimator is:pNbGMM 0!d N

0,

G00W0G0

1G00W0W0G0

G00W0G01

where E

g(Zi, 0) g(Zi, 0)

0 .



T E i H h i T i S ll S l P i C di i l M


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In this particular case,G00W0G0

1G00W0 =

G00G0

1G00 = G10

and the asymptotic variance of the estimator isG00W0G0

1G00W0W0G0

G00W0G0

1= G10 E

eg(Zi, 0,0)

eg(Zi, 0,0)

0

G100



T t E ti t H th i T ti S ll S l P ti C diti l M t


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whereG0 = E

"eg(Zi, 0,0)0,0

0 # = G G0 MG10 =

G1 G1 GM10 M

1 =

E m (Zi,0)m (Zi,0)0

E m (Zi,0) g(Zi, 0,0)0

E g(Zi, 0,0)m (Zi,0)0 E g(Zi, 0,0) g(Zi, 0,0)0



Two step Estimators Hypothesis Testing Small Sample Properties Conditional Moments
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The asymptotic variance ofb is the upper left block of thejoint variance matrix, and it is equal to

V0 = G1 E

m (Zi,0)m (Zi,0)

0

G10

G1

GM1

E g(Zi, 0,0)m (Zi,0)0G10 G1 E

m (Zi,0) g(Zi, 0,0)

0G10

+ G1 GM1Eg(Zi, 0,0) g(Zi, 0,0)

0M10G0G

10

= G1 E fg(Z) + G (z)g fg(Z) + G (z)g0

G10





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Suppose we want to test a set of restrictions:

H0 : c(0)rx1 = 0

against the alternatives

H1 : c(0)rx1 6

= 0

or asymptotically local alternatives of the form

H1N :pNrx1

6= 0

The null hypothesis may be linear or nonlinear.

Assume that Crxk rc(0) has rank r.



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Dene the constrained extremum estimator:

N = arg max2

QN () subject to c(0) = 0

Under the conditions necessary to consistent of the extremum

estimator, and assuming that c(0) = pN, including the nullwhen = 0, with c continuously dierentiable and C rank r,

N !p 0

We look at the trinity of tests.



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In the GMM context....

Wald statistics: deviations of the unconstrained estimatorsfrom values consistent with the null.

Lagrange Multiplier or score statistics: deviations of theconstrained estimates from values solving the unconstrained

problems.Distance metric statistics: dierence in the GMM criterionbetween the unconstrained and the constrained estimates.

In the case of MLE, the distance metric statistic is

asymptotically equivalent to the likelihood ratio statistics.In the GMM set-up, they are all asymptotically equivalent.





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p yp g p p



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Recall that represents the asymptotic variance of

pNrbQN () . Lets assume that the generalize informationinequality holds: = HLet b be an estimator of based on b and be an estimatorbased on .

In this case, the 3 statistics will be:

W = N cb0 hbCb1bC0i1 cb

LM= Nr

bQN 1rbQN DM= 2N

hbQN b bQN i





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One drawback of the Wald test is that is not invariant howthe nonlinear restrictions are imposed. We can change theoutcome of a hypothesis test by redening the constraintfunction, a (.) .

Score statistics that use the outer product of the score to

estimate are invariant to reparametrizations. Scorestatistics based on the estimated Hessian are not generalinvariant to reparametrization, since we need to get thesecond derivatives of the objective function.

We can use the same test as IV to test if all the momentshold in the overidentifying case.



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Example: Tests for overidentifying restrictions

Null hypothesis:

H0 : E [m (Z, 0)] = 0

Suppose that we partition the moments

m (Z,

) = m1 (Z, )m2 (Z, )where m1 (Z, ) is px1 and m2 (Z, ) is (r p)x1.Augmented Model

em (Z, ) = m1 (Z, )m2 (Z, ) +

where is an (r p) vector of additional parameters.

Cristine Campos de Xavier Pinto InstituteEstimadores Extremos

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In this augmented model, we want to test: H0 : = 0

In this case

DM = 2NheSN bN, 0 eSN N,Ni 2NSN

bN d! X2rp


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Notation:

=G

0G

1, F=

G

0G

1G

0

P = 1 1G0G1

and

a =1

2tr

"E

2g(0)

0

0#, FW = G

0W

G

0WG

1and ej is the jth unit vector.




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Under the assumptions that we use to derive the asymptoticproperties of GMM, under other smooth assumptions andassumptions about cWN, Newey and Smith (2004) show thatwe can write two-step GMM estimator (optimal GMM):

pNb = Ni=1 (Zi, 0)pN

+ Q1 e, 0, ap

N+ Q

2 e, 0N

+RN

where E [ (Zi, 0)] = 0, and Q1 is a quadratic function, Q2

is a cubic function and RN

OpN 32


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Using this expansion, they show that

BiasbGMM = BI + BG + B + BWwhere

BI =F (a+E [GiFgi])

N

BG = E [GiPgi]N

B =FE

hgig

0iPgi

iN

BW = HP

j=1

rj (FW F)0ej

N


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BI: asymptotic bias for a GMM estimator with the optimalvariance.

BG: bias due to the estimation ofG

B: bias dues to estimate the second moment matrix .

BW : bias due to the choice of preliminary estimation.

Note that if you are in the just identied case, P= 0, andBG = 0 and B = 0.


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Suppose that 0 2 Rp satisfy strong conditions0 = E [u(Zi, 0)jXi]

u(Zi,

0) is a qx1 of functions0 is a px1 vector of parameters

Since E [u(Zi, 0)jXi] is a random variable, we need tointerpret this inequality as holding with probability one.




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Assuming that the function u(.) is bounded above by some

square-integral function,

sup

ku(Zi, 0)k b(Zi) , Ehb(Zi)

2i<

Using the law of iterated expectation,

0 = E [h (Xi) u(Zi, 0)] E [m (Zi, 0)]where h (.) is rxq matrix of functions ofXi that satises

E hkh (Xi)k2i E trh (Xi) h (Xi)0 < In this case, q can be lower than p, but r p


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For a given choice ofh (X), we can apply the GMM theory above

b (h) = arg min [mN ()]0 bV1 mN ()mN () =

1N

Ni=1 h (Xi) u(Zi, 0)

plim bV = V0V0 = Var[h (Xi) u(Zi, 0)] = E [h (Xi) (Xi) h (Xi)] (Xi) = E

u(Zi, 0) u(Zi, 0)0

XiThe asymptotic distribution is

pNb 0 d! N0, D00V10 D01where D0 = E

hh (Xi) u(Zi,0) 0

i.


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How do we choose the optimal h (Xi)?

We can choose the one that minimizes the asymptoticvariance

h (Xi) = arg minHD00V

10 D0

1where

D00V10 D0

1=

"E

h (Xi) u(Zi, 0)

0

0E [h (Xi) (Xi) h (Xi)]1

E h (Xi) u(Zi, 0) 0

1


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The best choice ofh (Xi) will make E hh (Xi) u(Zi,0)

0 i0

equals to E [h (Xi) (Xi) h (Xi)]1.The optimal h (Xi) is

h (Xi) = E u(Zi, 0)

0 Xi (Xi)1

Using the optimal matrix h (Xi), the asymptotic variance of

b

reduces to

E E u(Zi, 0) 0 Xi (Xi)1 E u(Zi, 0) 0 Xi

1


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Example: Linear model with endogeneity

Yi = X0i+ i

whereE [ ijWi] = 0

In this case,u(Zi, 0) = Yi X0i




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This conditional moment implies the unconditional moment:

E [h (Xi)u(Zi, 0)] = 0

where

h (Xi) = Eu(Zi, 0)

0

Wi

(Xi)1

= E X0i Wi E hYi X0i02Wii= E X0i Wi 2 (Wi)

In the case that Xi = Wi,

E X0i (Yi X0i)2 (Xi)

= 0which is the moment solved by the weighted least squaresestimator.




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References

Amemya: 4

Wooldridge: 14

Rudd: 14 and15

Newey, W. and D. McFadden (1994). "Large SampleEstimation and Hypothesis Testing", Handbook ofEconometrics, Volume IV, chapter 36.

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