l Sample Inference

Tests for General SEstimation of S

Tests under conditional homoskedasticityThe Delta Method and Non-Linear Hypothesis

Large-Sample Based Inference in the Linear Model

Walter Sosa-Escudero

Econ 507. Econometric Analysis. Spring 2009

February 26, 2009

Walter Sosa-Escudero Large-Sample Based Inference in the Linear Model



Motivation

In the classical linear model with finite n, we had to introducedistributional assumptions to perform basic hypothesis testsand confidence intervals.

A really important advantage of the large sample theory is toprovide a framework for inference, without distributionalassumptions.

We will derive simple test statistics to evaluate linearhypothesis in the linear model.

We will discuss a simple strategy to handle the non-linear case.




Our starting point will be the asympototic normality result:

n(n 0

)p N(0,1x S1x )

where x = E(xixi) and S = V (xiui).

Note that n1X iXi is a consistent estimator for x.Recall that our assumptions for asymptotic analysis allow forconditional heteroskedasticity. Alternative consistentestimators for S depend on what we are willing to assume onthis.

We will use the notation AV (n) 1x S1x andAV (n) = 1x S1x .




Road Map

Road Map

1 Derive general test statistics assuming there is available aconsistent estimator S

p S2 Derive consistent estimators for S with and without

conditional heteroskedasticity.

3 Obtain particular cases of the previous tests for theconditional homoskedastic case.




Test for Linear Hypothesis under general S

Hypothesis about single coefficients:

H0 : j = j0 vs. HA : j 6= j0

From the asymptotic normality, using Slutzkys theorem, it easy tocheck that when H0 : j = j0 holds:

tj =

n(j j0

)AV (j)

d N(0, 1)




Multiple Linear Hypothesis

H0 : R r = 0, R is a q K matrix with (R) = q, and r



Proof: by asympototic normality, under H0

n(Rn r) d N

(0, R AV(n) R

)or [

R AV(n) R]1/2

n(Rn r) d N(0, Ik)By Slutzkys theorem and linearity:[

R AV(n) R]1/2

n(Rn r) d N(0, Ik)

The desired result follows by forming the quadratic form.




Estimation of S

A general estimator

S = E(xiuiuixi) = E(u2ixix

i)

We will need an additional assumption

Assumption 6 (fourth moments): E[(xikxij)2

]exists and is finite

for all k, j = 1, 2, . . . ,K.

Result

Sw 1n

ni=1

e2ixix2i

p S

where ei are the OLS residuals.




Proof:

ei = yi xi = yi xi xi( ) = ui xi( )e2i = u

2i 2uixi( ) + ( )xixi( )

Replacing

1n

ni=1

e2ixixi =

1n

ni=1

u2ixixi

2n

ni=1

uixi( )xixi +

+1n

ni=1

( )xixi( )xixi

First note that1n

ni=1

u2ixixip E(u2ixixi)

by Kolmogorovs LLN, since we assumed finite second moments(expectaction exists) and iid.




We will show the other two terms converge to zero

I) A = 2nn

i=1 ui xi( )xixi

A =2n

ni=1

ui

[Kk=1

xik(k k)]xixi

= 2Kk=1

(k k)[n

i=1 uixik xixi

n

]

Note k k p 0, by consistency. So if we can show[ ]

p



II) B = 1nn

i=1( )xi xi( ) xixiUsing the same trick as before:

B =1n

ni=1

[Kk=1

xik(k k)][

Kk=1

xik(k k)]xixi

Now we have a sum of K2 matrices. The (h, j) element of thek, k summand will be

(k k)(k k) 1n

ni=1

xikxikxihxij

Using again the Cauchy Schwartz inequality and the finite fourthmoments assumption: E|xikxikxihxij |



Then, using Sw as an estimator for S and noting

Sw =1n

ni=1

e2ixix2i =

1n

(X BX)

with B diag(e21, . . . , e2n),

AVw(n) = 1x Sw1x

= n(X X)1n1(X BX)n(X X)1

= n(X X)1(X BX)(X X)1

This is Whites heteroskedasticity consistent estimator for theasymptotic variance of n. Remember that in the derivation of allresult we never ruled out the possibility of conditionalheteroskedasticity, then its consistency does not depend on it.




Returns-to-scale revisited:

. reg ltc lq lpl lpf lpk

------------------------------------------------------------------------------

ltc | Coef. Std. Err. t P>|t| [95% Conf. Interval]

-------------+----------------------------------------------------------------

lq | .7209135 .0174337 41.35 0.000 .6864462 .7553808

lpl | .4559645 .299802 1.52 0.131 -.1367602 1.048689

lpf | .4258137 .1003218 4.24 0.000 .2274721 .6241554

lpk | -.2151476 .3398295 -0.63 0.528 -.8870089 .4567136

_cons | -3.566513 1.779383 -2.00 0.047 -7.084448 -.0485779

------------------------------------------------------------------------------

. reg ltc lq lpl lpf lpk, robust

------------------------------------------------------------------------------

| Robust

ltc | Coef. Std. Err. t P>|t| [95% Conf. Interval]

-------------+----------------------------------------------------------------

lq | .7209135 .0325376 22.16 0.000 .656585 .785242

lpl | .4559645 .260326 1.75 0.082 -.0587139 .9706429

lpf | .4258137 .0740741 5.75 0.000 .2793653 .5722622

lpk | -.2151476 .3233711 -0.67 0.507 -.8544698 .4241745

_cons | -3.566513 1.718304 -2.08 0.040 -6.963693 -.1693331

------------------------------------------------------------------------------




Variance estimation under conditional homoskedasticity

If we further assume E(u2i |xi) = 20, then using LIE:

S = E(u2ixixi) =

20E(xix

i) =

20x

So the asymptotic variance of n reduces to

1x S1x =

201x x

1x =

21x

So a consistent estimator for AV(n) can be

AVh = n s2(X X)1

which is n times the finite sample estimator for the variance of nin the classical linear model.




Tests under conditional homoskedasticity

Under conditional homoskedasticity AV(n) = 21x andAVh = n s2(X X)1 provides a consistent estimator:

a) Single linear hypothesis: Our test for H0 : j = j0

tj =

n(j j0

)AV (j)

reduces to:

tj =

(j j0

)s2ajj

where ajj is the jth element of the main diagonal of X X, so weget our old t test. Note that we do not get the t distribution infinite samples, but instead we get the N(0, 1) distributionasymptotically.




Multiple linear hypothesisOur W statistic

W = n (Rn r)[R AV (n) R

]1(Rn r)

simplifies to:

W = n (Rn r)[R AV (n) R

]1(Rn r)

= (Rn r)[R s2(X X)1 R

]1 (Rn r)= q F

This provides an approximation of our old F statistic for theasymptotic case.




Summary of results

We derived general test for single coefficient (t) and multiple(W ) hypothesis, in our asympotic framework that does notassume normality and allows for conditional heteroskedasticity.

Whites AVw estimator provides a general consistent strategyfor estimating the AVAR of n, robust to the presence ofconditional heteroskedasticity.

Under the conditional homoskedasticity assumption, our thegeneral t test simplifies to the old one, but this is anasympotic result.

W simplifies to nF .




The Delta Method

Suppose we want to perform inference about a non-linear functionof , say a().

Example 1yi = 0 + 1south + zi2 + ui

yi is log-wages, south is a dummy indicating if the person lives in thesouthern region zi is a vector of control variables.The percent difference between south/not south is given by

e1 1

and for small values of 1 is very similar to 1. Suppose we are interested

exactly. A consistent estimator is e1 1. A natural problem is how toconstruct a confidence interval for .




Example 2: consider now

yi = 1 exper + 2 exper2 + zi + ui

where exper is work experience in years. The level of experiencethat maximizes expected wages is:

122

and a consistent estimate is provided by

= 122

How can we construct an estimate for the standard deviation of aconfidence interval for ?




Result (Delta Method): suppose xn is a sequence of randomvector of dimension K such that

xnp and n(xn ) d Z

and a(x) :



Proof: Take a first-order mean value expansion of a(xn) around :

a(xn) = a() +A(yn) (xn )

where the mean value yn is a vector between xn and .From this, get

n[a(xn) a()

]= A(yn) (xn )

Now ynp (why?) so A(yn) p A() by continuous

differentiability.

Then, by the hypothesis of the theorem and Slutzkys Theorem

n[a(xn) a()

] d A()ZWalter Sosa-Escudero Large-Sample Based Inference in the Linear Model



As a simple corollary note that if

n(n ) d N(0,AV(n))

then n[a(n) a()

] d N(0, A()AV(n)A())




Example 1 (Blackburn and Neumark, 1992, also in Wooldridge,2002)

yi = 0 + 1 south + zi2 + ui

a(1) = e1 1with

A(1) = e1

So, according to the delta-method

AV(e1 1

)=[e1]2

AV(1)




. reg lwage exper tenure married black south urban educ

------------------------------------------------------------------------------

lwage | Coef. Std. Err. t P>|t| [95% Conf. Interval]

-------------+----------------------------------------------------------------

exper | .014043 .0031852 4.41 0.000 .007792 .020294

tenure | .0117473 .002453 4.79 0.000 .0069333 .0165613

married | .1994171 .0390502 5.11 0.000 .1227801 .276054

black | -.1883499 .0376666 -5.00 0.000 -.2622717 -.1144281

south | -.0909036 .0262485 -3.46 0.001 -.142417 -.0393903

urban | .1839121 .0269583 6.82 0.000 .1310056 .2368185

educ | .0654307 .0062504 10.47 0.000 .0531642 .0776973

_cons | 5.395497 .113225 47.65 0.000 5.17329 5.617704

------------------------------------------------------------------------------

. nlcom exp(_b[south])-1


-------------+----------------------------------------------------------------

_nl_1 | -.0868943 .0239677 -3.63 0.000 -.1339315 -.0398571




Example 2:

yi = 1 exper + 2 exper2 + zi + ui, a(1, 2) = 1/(22). regress lwage edup edusi edus eduui eduu exper exper2 if muest==1

------------------------------------------------------------------------------


-------------+----------------------------------------------------------------

edup | .2104513 .0629835 3.34 0.001 .0869123 .3339903

edusi | .4148728 .0678469 6.11 0.000 .2817946 .547951

edus | .7587112 .0695764 10.90 0.000 .6222406 .8951817

eduui | 1.018209 .077569 13.13 0.000 .866061 1.170356

eduu | 1.560496 .0769774 20.27 0.000 1.409509 1.711483

exper | .0283668 .0071065 3.99 0.000 .0144279 .0423058

exper2 | -.0002502 .0001509 -1.66 0.098 -.0005462 .0000458

_cons | .2130178 .0934142 2.28 0.023 .0297906 .3962449

------------------------------------------------------------------------------

. nlcom -_b[exper]/(2*_b[exper2])

_nl_1: -_b[exper]/(2*_b[exper2])

------------------------------------------------------------------------------


-------------+----------------------------------------------------------------

_nl_1 | 56.6962 20.7191 2.74 0.006 16.05673 97.33567

------------------------------------------------------------------------------




Summary of variance results

Starting point:n(n 0

)p N(0, AV (n))

AV (n) = 1x S

1x , x = E(xix

i), S = V (xiui) = E(u

2ixix

i).

x =1nX X. S is any consistent estimator of S.

AV (n) = 1x S

1x

Tests

H0 : j = j0, use tj

H0 : R r = 0, use W = n (Rn r)[R AV (n) R

]1

(Rn r)Allowing for conditional heteroskedasticity

Sw =1n

ni=1 e

2ixix

2i =

1n

(X BX)

AVw(n) = 1x Sw

1x

Under homoskedasticity

S = E(u2ixixi) =

20E(xix

i) =

20x, so AV (n) =

1x S

1x =

201x

AV h(n) = n s2(X X)1


Tests for General Estimation of STests under conditional homoskedasticityThe Delta Method and Non-Linear Hypothesis

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l Sample Inference