金融计量经济学双语版全套

5-1课件Chapter

5Univariate

time

seriesmodelling

and

forecasting5-2课件1

introduction单变量时间序列模型只利用变量的过去信息和可能的误差项的当前和过去值来建模和预测的一类模型(设定)。与结构模型不同；通常不依赖于经济和金融理论用于描述被观测数据的经验性相关特征

ARIMA（AutoRegressive

Integrated

Moving

Average）是一类重要的时间序列模型Box-Jenkins

1976当结构模型不适用时，时间序列模型却很有用如引起因变量变化的因素中包含不可观测因素，解释变量等观测频率较低。结构模型常常不适用于进行预测本章主要解决两个问题一个给定参数的时间序列模型，其变动特征是什么？给定一组具有确定性特征的数据，描述它们的合适模型是什么？5-3A

Strictly

Stationary

ProcessA

strictly

stationary

process

is

one

whereFor

any

t1

,t2

,…,

tn

∈

Z,

any

m

∈

Z,n=1,2,…A

Weakly

Stationary

ProcessIf

a

series

satisfies

the

next

three

equations,

it

is

said

toor

covariance

stationary1.

E(yt)

=2.3.,

t

=

1,2,...,t1

,

t22

Some

Notation

and

Concepts课件5-4

So

if

the

process

is

covariance

stationary,

all

the

varianthe

same

and

all

the

covariances

depend

on

the

differencebetween

t1

and

t2.

The

moments,

s

=

0,1,2,

...are

known

as

the

covariance

function.The

covariances,

s,

are

known

as

autocovariances.

However,

the

value

of

the

autocovariances

depend

on

the

unof

measurement

of

yt.

It

is

thus

more

convenient

to

use

the

autocorrelations

whiare

the

autocovariances

normalised

by

dividing

by

thevariance:,

s

=

0,1,2,

...If

we

plots

against

s=0,1,2,...

then

we

obtain

theautocorrelation

function

(acf)

or

correlogram.Some

Notation

and

Concepts课件5-5A

white

noise

process

is

one

with

no

discernible

structure.

Thus

the

autocorrelation

function

will

be

zero

apart

from

asingle

peak

of

1

at

s

=

0.如果假设yt服从标准正态分布,

则 approximately

N(0,1/T

We

can

use

this

to

do

significance

tests

for

the

autocorrelacoefficients

by

constructing

a

confidence

interval.a

95%

confidence

interval

would

be

given

by

.

If

the

sample

autocorrelation

coefficient, ,

falls

outsiregion

for

any

value

of

s,

then

we

reject

the

null

hypothesisthe

true

value

of

the

coefficient

at

lag

s

is

zero.A

White

Noise

Process课件5-6We

can

also

test

the

joint

hypothesis

that

all

m

of

thekcorrelation

coefficients

are

simultaneously

equal

to

zerthe

Q-statistic

developed

by

Box

and

Pierce:where

T

=

sample

size,

m

=

maximum

lag

lengthThe

Q-statistic

is

asymptotically

distributed

as

a

.

However,

the

Box

Pierce

test

has

poor

small

sample

propertso

a

variant

has

been

developed,

called

the

Ljung-Boxstatistic:

This

statistic

is

very

useful

as

a

portmanteau

(general)

tlinear

dependence

in

time

series.Joint

Hypothesis

Tests课件5-7课件Question:Suppose

that

we

had

estimated

the

first

5

autocorrelationcoefficients

using

a

series

of

length

100

observations,

andthem

to

be

(from

1

to

5):

0.207,

-0.013,

0.086,

0.005,

-0.022Test

each

of

the

individual

coefficient

for

significance,

aboth

the

Box-Pierce

and

Ljung-Box

tests

to

establish

whetheare

jointly

significant.Solution:A

coefficient

would

be

significant

if

it

lies

outside

(-0.1

at

the

5%

level,

so

only

the

first

autocorrelation

coeffici

significant.Q=5.09

and

Q*=5.26Compared

with

a

tabulated

2(5)=11.1

at

the

5%

level,

so

th5

coefficients

are

jointly

insignificant.An

ACF

Example

(p234)5-8

Let

ut

(t=1,2,3,...)

be

a

sequence

of

independently

andidentically

distributed

(iid)

random

variables

with

E(ut)and

Var(ut)=2,

thenyt

=

+

ut

+

2ut-2

+

...

+1ut-1

+

qut-qis

a

qth

order

moving

average

model

MA(q).Or

using

the

lag

operator

notation:Lyt

=

yt-1

Liyt

=

yt-i通常，可以将常数项从方程中去掉，而并不失一般性。3

Moving

Average

Processes课件5-9移动平均过程的性质

Its

properties

areE(

yt

)=0

=

(1+Var(

yt

)

=Covariances)2自相关函数课件5-10课件Consider

the

following

MA(2)

process:where

ut

is

a

zero

mean

white

noise

process

with

variance

.Calculate

the

mean

and

variance

of

XtDerive

the

autocorrelation

function

for

this

process

(iexpress

the

autocorrelations,

1,

2,

...

as

functions

of

thparameters

1

and

2).If

1

=

-0.5

and

2

=

0.25,

sketch

the

acf

of

Xt.Example

of

an

MA

Process5-11(i)

If

E(ut

)=0,

then

E(ut-i

)=0

i.

So1ut-1+E(Xt

)

=

E(ut

+

2ut-2

)=

E(ut

)+1E(ut-1

)+2E(ut-2

)Var(Xt

)Var(Xt)=

E[Xt

-E(Xt

)][Xt

-E(Xt

)]=

E[(Xt

)(Xt

)]1ut-1+

1ut-1+=

E[(ut

+

2ut-2)(ut

+=

E[

+cross-products]2ut-2)]But

E[cross-products]=0

,since

Cov(ut,ut-s)=0

for

s0.

SoVar(Xt

)===

0=

E

[

]Solution课件5-12(ii)

The

acf

of

Xt1=

E[Xt-E(Xt

)][Xt-1-E(Xt-1

)]=

E[Xt

][Xt-1

]1ut-1+2ut-2

)(ut-1

+)]=

E[(ut

+=

E[(==1ut-2+2ut-3

)]2=

E[Xt

-E(Xt

)][Xt-2

-E(Xt-2

)]=

E[Xt

][Xt-2

]=

E[(ut

+=

E[(=1ut-1+

2ut-2

)(ut-2

+

1ut-3+

2ut-4

)])]Solution

(cont’d)课件5-133=

E[Xt

][Xt-3

]=

E[(ut

+=

01ut-1+2ut-2

)(ut-3

+1ut-4+2ut-5

)]So

s

=

0

for

s

>

2.now

calculate

the

autocorrelations:Solution

(cont’d)课件5-14(iii)

For

1

=

-0.5

and

2

=

0.25,

substituting

these

into

tformulae

above

gives

1

=

-

0.476,

2

=

0.190.Thus

the

acf

plot

will

appear

as

follows:ACF

Plot课件5-15

An

autoregressive

model

of

order

p,

AR(p)

can

be

expresseasOr

using

the

lag

operator

notation:Lyt

=

yt-1

Liyt

=

yt-iororwhere4

Autoregressive

Processes课件5-16课件

平稳性使AR模型具有一些很好的性质。如前期误差项对当前值的影响随时间递减。

The

condition

for

stationarity

of

a

general

AR(p)model

isthe

roots

of特征方程all

lie

outside

the

unit

circle.Example

1:

Is

yt

=

yt-1

+

ut

stationary?The

characteristic

root

is

1,

so

it

is

a

unit

root

process

(stationary)Example

2:

p241

A

stationary

AR(p)

model

is

required

for

it

to

have

an

MA(representation.The

Stationary

Conditionfor

an

AR

Model5-17

States

that

any

stationary

series

can

be

decomposed

into

sum

of

two

unrelated

processes,

a

purely

deterministic

pand

a

purely

stochastic

part,

which

will

be

an

MA(

).,

ignoring

the

intercept,For

the

AR(p)

model,the

Wold

decomposition

iswhere,可以证明，算子多项式R(L)的集合与代数多项式R(z)的集

合是同结构的，因此可以对算子L做加、减、乘和比率运算。Wold’s

Decomposition

Theorem课件5-18

If

the

AR

model

is

stationary,

the

autocorrelation

functiondecay

exponentially

to

zero.The

Moments

of

an

Autoregressive

Process

The

moments

of

an

autoregressive

process

are

as

follows.

Thmean

is

given

by*

The

autocovariances

and

autocorrelation

functions

can

beobtained

by

solving

what

are

known

as

the

Yule-Walkerequations:

*课件5-19Consider

the

following

simple

AR(1)

modelCalculate

the

(unconditional)

mean

of

yt.For

the

remainder

of

the

question,

set =0

forsimplicity.Calculate

the

(unconditional)

variance

of

yt.Derive

the

autocorrelation

function

for

yt.Sample

AR

Problem课件5-20(i) E(yt)=

E(

+1yt-1)=

+

1E(yt-1)But

alsoE(yt)=

+1

(

+

1E(yt-2))=

+1+

2

E(y

))1

t-2=

++2

(1+

1E(yt-3))1=

+

+121+

3

E(y

)1

t-3An

infinite

number

of

such

substitutions

would

giveE(yt)=

(1+

1+2

+...)

+

y1

1

0So

long

as

the

model

is

stationary,

i.e. ,

then1=

0So

E(yt)=

(1+

1+2

+...)

=1Solution课件5-21(ii)

Calculating

the

variance

of

yt

:*From

Wold’s

decomposition

theorem:So

long

as,

this

will

converge.Solution

(cont’d)课件5-22Var(yt)

=

E[yt-E(yt)][yt-E(yt)]but

E(yt)

=

0,

since

we

are

setting =

0.*有简便方法]Var(yt)

=

E[(yt)(yt)]=

E[=

E[=

E[===Solution

(cont’d)课件5-23(iii)Turning

now

to

calculating

the

acf,first

calculate

theautocovariances:（*用简便方法）1

=

Cov(yt,

yt-1)

=

E[yt-E(yt)][yt-1-E(yt-1)]1

=

E[ytyt-1]1

=

E[=

E[==]Solution

(cont’d)课件5-24Solution

(cont’d)For

the

second

autocorrelation

coefficient,2

=

Cov(yt,

yt-2)

=

E[yt-E(yt)][yt-2-E(yt-2)]Using

the

same

rules

as

applied

above

for

the

lag

1

covariance2

=

E[yt

yt-2]=

E[=

E[==]=课件5-25Solution

(cont’d)

If

these

steps

were

repeated

for

3,

the

following

expreswould

be

obtained3

=and

for

any

lag

s,

the

autocovariance

would

be

given

bys

=The

acf

can

now

be

obtained

by

dividing

the

covariances

bythe

variance:课件5-26Solution

(cont’d)0

=1

=2

=3

=…s

=课件5

The

Partial

Autocorrelatio5-n27课件

Measures

the

correlation

between

an

observation

k

periodago

and

the

current

observation,

after

controlling

forobservations

at

intermediate

lags

(i.e.

all

lags

<

k).yt-k与yt之间的偏自相关函数kk

是在给定yt-k+1

,yt-k+2

,…,yt的条件下，yt-k与yt之间的部分相关。So

kk

measures

the

correlation

between

yt

and

yt-k

afterremoving

the

effects

of

yt-k+1

,

yt-k+2

,

…,

yt-1

.2)

/

(1-2-

12)1或者说，偏自相关函数

kk

是对yt-k与yt之间未被yt-k+1

,

ytk+2

,

…,

yt-1所解释的相关的度量。At

lag

1,

the

acf

=

pacf

alwaysAt

lag

2,

22

=

(For

lags

3+,

the

formulae

are

more

complex.Function

(denotedkk)5-28课件

The

pacf

is

useful

for

telling

the

difference

between

an

Aprocess

and

an

MA

process.In

the

case

of

an

AR(p),

there

are

direct

connections

betwyt

and

yt-s

only

for

s

p.So

for

an

AR(p),

the

theoretical

pacf

will

be

zero

after

la

In

the

case

of

an

MA(q),

this

can

be

written

as

an

AR(

),

sthere

are

direct

connections

between

yt

and

all

its

previovalues.

For

an

MA(q),

the

theoretical

pacf

will

be

geometricallydeclining.The

Partial

Autocorrelation

Function5-29The

invertibility

condition

If

MA(q)

process

can

be

expressed

as

an

AR(∞),

then

MA(q)

iinvertible.根的绝对值大于1。MA(q)的可逆性条件：特征方程从而有课件5-30

By

combining

the

AR(p)

and

MA(q)

models,

we

can

obtain

anARMA(p,q)

model:whereandorwith6

ARMA

Processes课件5-31课件ARMA过程的特征是AR和MA的组合。

可逆性条件：Similar

to

the

stationarity

condition,wetypically

require

the

MA(q)part

of

the

model

to

have

root(z)=0

greater

than

one

in

absolute

value.The

mean

of

an

ARMA

series

is

given

by

The

autocorrelation

function

for

an

ARMA

process

willdisplay

combinations

of

behaviour

derived

from

the

AR

andMA

parts,

but

for

lags

beyond

q,

the

acf

will

simply

beidentical

to

the

individual

AR(p)

model.ARMA过程的特征5-32课件Summary

ofthe

Behaviour

of

theacfand

pacf

for

AR

and

MA

ProcessesAn

autoregressive

process

hasa

geometrically

decaying

acf:拖尾number

of

spikes尖峰信号of

pacf=AR

order：截尾A

moving

average

process

hasNumber

of

spikes

of

acf=MA

order：截尾a

geometrically

decaying

pacf：拖尾A

ARMA

process

hasa

geometrically

decaying

acf:拖尾a

geometrically

decaying

pacf：拖尾5-33The

acf

and

pacf

are

estimated

using

100,000

simulated

observations

withdisturbances

drawn

from

a

normal

distribution.ACF

and

PACF

for

an

MA(1)

Model:

yt

=

–

0.5ut-1

+

utSome

sample

acf

and

pacf

plotsfor

standard

processes课件ACF

and

PACF

for

an

MA(2)

Model:5-34yt

=

0.5ut-1

-

0.25ut-2

+

ut课件5-35课件ACF

and

PACF

for

a

slowly

decaying

AR(1)Model:

yt

=

0.9

yt-1

+

ut5-36ACF

and

PACF

for

a

more

rapidly

decayingAR(1)

Model:

yt

=

0.5

yt-1

+

ut课件5-37ACF

and

PACF

for

a

AR(1)

Model

withNegative

Coefficient:yt

=

-0.5

yt-1

+

ut课件5-38课件ACF

and

PACF

for

a

Non-stationaryModel

(

a

unit

coefficient):

yt

=

yt-1

+

utACF

and

PACF

for

an

ARMA(1,1):5-39yt

=

0.5yt-1

+

0.5ut-1

+

ut课件6-40©

Chris

Brooks

2002Chapter

6Multivariate

models6-411

MotivationsAll

the

models

we

have

looked

at

thus

far

have

been

singleequations

models

of

the

formy

=

X

+

u

All

of

the

variables

contained

in

the

X

matrix

are

assumebe

EXOGENOUS.由系统外因素决定的变量

y

is

an

ENDOGENOUS

variable.既影响系统同时又被该系统及其外部因素所影响的变量.An

example

-

the

demand

and

supply

of

a

good:(1)(2)(3)、 =

quantity

of

the

good

demanded

/

suppliedPt

=

price，

St

=

price

of

a

substitute

goodTt

=

some

variable

embodying

the

state

of

technology©

Chris

Brooks

2002Simultaneous

Equations

Models6:-42The

Structural

Form

Assuming

that

the

market

always

clears,

and

dropping

thetime

subscripts

for

simplicity(4)(5)This

is

a

simultaneous

STRUCTURAL

FORM

of

the

model.

The

point

is

that

price

and

quantity

are

determinedsimultaneously

(price

affects

quantity

and

quantity

affeprice).

P

and

Q

are

endogenous

variables,

while

S

and

T

areexogenous.We

can

obtain

REDUCED

FORM

equations

corresponding

to(4)

and

(5)

by

solving

equations

(4)

and

(5)

for

P

and

for

Q©

Chris

Brooks

20026-43Solving

for

Q,(6)Solving

for

P,(7)Rearranging

(6),(8)Obtaining

the

Reduced

Form©

Chris

Brooks

20026-44Multiplying

(7)

through

by

,(9)(8)

and

(9)

are

the

reduced

form

equations

for

P

and

Q.Obtaining

the

Reduced

Form©

Chris

Brooks

20026-45©

Chris

Brooks

2002

But

what

would

happen

if

we

had

estimated

equations

(4)

an(5),

i.e.

the

structural

form

equations,

separately

using

Both

equations

depend

on

P.

One

of

the

CLRM

assumptionswas

that

E(X

u)

=

0,

where

X

is

a

matrix

containing

all

thevariables

on

the

RHS

of

the

equation.It

is

clear

from

(8)

that

P

is

related

to

the

errors

in

(4)-

i.e.

it

is

stochastic.

What

would

be

the

consequences

for

the

OLS

estimator,

,we

ignore

the

simultaneity?2

Simultaneous

Equations

Bias6-46Recall

that

andSo

thatTaking

expectations,If

the

X’s

are

non-stochastic,

E(Xu)

=

0,

which

wouldthe

case

in

a

single

equation

system,

so

that

,which

is

the

condition

for

unbiasedness.Simultaneous

Equations

Bias©

Chris

Brooks

20026-47©

Chris

Brooks

2002

But

....

if

the

equation

is

part

of

a

system,

then

E(X

u)general.

Conclusion:

Application

of

OLS

to

structural

equationswhich

are

part

of

a

simultaneous

system

will

lead

to

biasecoefficient

estimates.Is

the

OLS

estimator

still

consistent,

even

though

it

is

bNo

-

In

fact

the

estimator

is

inconsistent

as

well.Hence

it

would

not

be

possible

to

estimate

equations

(4)

a(5)

validly

using

OLS.Simultaneous

Equations

Bias6-483

Avoiding

Simultaneous

Equations

BiasSo

What

Can

We

Do?Taking

equations

(8)

and

(9),

we

can

rewrite

them

as(10)(11)

We

CAN

estimate

equations

(10)

&

(11)

using

OLS

since

allthe

RHS

variables

are

exogenous.

But

...

we

probably

don’t

care

what

the

values

of

thecoefficients

are;

what

we

wanted

were

the

originalparameters

in

the

structural

equations

-

,

,

,

,©

Chris

Brooks

20026-49Supply

equationDemand

equationWe

cannot

tell

which

is

which!(12)(13)Both

equations

are

UNIDENTIFIED

or

UNDERIDENTIFIED.

The

problem

is

that

we

do

not

have

enough

information

fromthe

equations

to

estimate

4

parameters.

Notice

that

we

woulnot

have

had

this

problem

with

equations

(4)

and

(5)

since

thhave

different

exogenous

variables.4

Identification

of

Simultaneous

EquationCan

We

Retrieve

the

Original

Coefficients

from

the

’s?

Short

answer:

sometimes.we

sometimes

encounter

another

problem:

identification.*Consider

the

following

demand

and

supply

equations©

Chris

Brooks

20026-50©

Chris

Brooks

2002What

Determines

whether

anEquation

is

Identified

or

not?We

could

have

three

possible

situations:1.

An

equation

is

unidentified· like

(12)

or

(13)· we

cannot

get

the

structural

coefficients

from

the

redform

estimates2.

An

equation

is

exactly

identified· e.g.

(4)

or

(5)· can

get

unique

structural

form

coefficient

estimates3.

An

equation

is

over-identifiedExample

given

later

More

than

one

set

of

structural

coefficients

could

beobtained

from

the

reduced

form.6-51©

Chris

Brooks

2002What

Determines

whether

anEquation

is

Identified

or

not?How

do

we

tell

if

an

equation

is

identified

or

not?There

are

two

conditions

we

could

look

at:The

order阶condition-is

a

necessary

but

not

sufficient condition

for

an

equation

to

be

identified.The

rank秩condition-is

a

necessary

and

sufficient

condi for

identification.在G个内生变量、G个方程的联立方程组模型中，某一方程是可识别的，当且仅当该方程没有包含的变量在其他方程中对应系数组成的矩阵的秩为G-1。对于相对简单的方程系统，这两个规则将得到同样的结论。事实上，大多数经济和金融方程系统都是过度识别的。6-52Statement

of

the

Order

ConditionLet

G

denote

the

number

of

structural

equations.

An

equation

is

just

identified

if

the

number

of

variablesexcluded

from

an

equation

is

G-1.If

more

than

G-1

are

absent,

it

is

over-identified.If

less

than

G-1

are

absent,

it

is

not

identified.Example

the

Y’s

are

endogenous,

while

the

X’s

are

exogenous.Determine

whether

each

equation

is

over-,

under-,

or

just-identified.(14)-(16)Statement

of

the

order

condition©

Chris

Brooks

20026-53SolutionG

=

3;If

#

excluded

variables

=

2,

the

eqn

is

just

identified

If

#

excluded

variables

>

2,

the

eqn

is

over-identified

If

#

excluded

variables

<

2,

the

eqn

is

not

identifiedEquation

14:

Not

identifiedEquation

15:

Just

identifiedEquation

16:

Over-identified如果模型中每个结构方程都是可识别的，则称结构型联立方程组模型是可识别的。Example

ofthe

order

condition©

Chris

Brooks

20026-54©

Chris

Brooks

20025

外生性的定义Leamer（1985）：p310变量X对变量Y是外生的，如果变量Y关于X的条件分布不随产生X的过程的变化而改变。外生性的两种形式：前定变量：与方程中的当前和未来误差项独立。严格外生变量：与方程中任何时期的误差项独立。前定变量的通常定义：包括外生变量和滞后的内生变量6-55

How

do

we

tell

whether

variables

really

need

to

be

treatedendogenous

or

not?

Consider

again

equations

(14)-(16).

Equation

(14)

contaiand

Y3

-

but

do

we

really

need

equations

for

them?We

can

formally

test

this

using

a

Hausman

test

as

follows:1.

Obtain

the

reduced

form

equations

corresponding

to

(14(16).

The

reduced

forms

turn

out

to

be:(17)-(19)Estimate

the

reduced

form

equations

(17)-(19)

using

OLS,obtain

the

fitted

values,5

Tests

for

Exogeneity©

Chris

Brooks

20026-56Run

the

regression

corresponding

to

equation

(14)Run

the

regression

(14)

again,

but

now

also

includingthe

fitted

values as

additional

regressors:(20)Use

an

F-test

to

test

the

joint

restriction

that

2

=

03

=

0.

If

the

null

hypothesis

is

rejected,

Y2

and

Y3

shoube

treated

as

endogenous.Tests

for

Exogeneity©

Chris

Brooks

20026-57Consider

the

following

system

of

equations:(21-23)

Assume

that

the

error

terms

are

not

correlated

with

each

othCan

we

estimate

the

equations

individually

using

OLS?Equation

21:

Contains

no

endogenous

variables,

so

X1

and

X2are

not

correlated

with

u1.

So

we

can

use

OLS

on

(21).Equation

22:

Contains

endogenous

Y1

together

with

exogenouX1

and

X2.

We

can

use

OLS

on

(22)

if

all

the

RHS

variables

in(22)

are

uncorrelated

with

that

equation’s

error

term.

InY1

is

not

correlated

with

u2

because

there

is

no

Y2

term

inequation

(21).

So

we

can

use

OLS

on

(22).6

Recursive

Systems©

Chris

Brooks

20026-58©

Chris

Brooks

2002

Equation

23:

Contains

both

Y1

and

Y2;

we

require

these

tobe

uncorrelated

with

u3.

By

similar

arguments

to

the

aboveequations

(21)

and

(22)

do

not

contain

Y3,

so

we

can

use

OLon

(23).

This

is

known

as

a

RECURSIVE

or

TRIANGULAR

system.We

do

not

have

a

simultaneity

problem

here.

But

in

practice

not

many

systems

of

equations

will

berecursive...Recursive

Systems6-597

Estimation

procedures

for

Systems©

Chris

Brooks

2002Indirect

Least

Squares

(ILS)

Cannot

use

OLS

on

structural

equations,

but

we

can

validlapply

it

to

the

reduced

form

equations.

If

the

system

is

just

identified,

ILS

involves

estimatingreduced

form

equations

using

OLS,

and

then

using

them

tosubstitute

back

to

obtain

the

structural

parameters.However,

ILS

is

not

used

much

becauseSolving

back

to

get

the

structural

parameters

can

betedious.Most

simultaneous

equations

systems

are

over-identifi6-60©

Chris

Brooks

2002

In

fact,

we

can

use

this

technique

for

just-identified

anover-identified

systems.Two

stage

least

squares

(2SLS

or

TSLS)

is

done

in

two

stagStage

1:

Obtain

and

estimate

the

reduced

form

equations

using

OLS.Save

the

fitted

values

for

the

dependent

variables.Stage

2:

Estimate

the

structural

equations,

but

replace

any

RHSendogenous

variables

with

their

stage

1

fitted

values.Estimation

of

SystemsUsing

Two-Stage

Least

Squares6-61Estimation

of

SystemsUsing

Two-Stage

Least

SquaresExample:

Say

equations

(14)-(16)

are

required.Stage

1:

Estimate

the

reduced

form

equations

(17)-(19)

individualOLS

and

obtain

the

fitted

values,

.Stage

2:

Replace

the

RHS

endogenous

variables

with

their

stage

1estimated

values:(24)-(26)

Now

and will

not

be

correlated

with

u1, will

not

becorrelated

with

u2

,

and will

not

be

correlated

with

u3

.©

Chris

Brooks

20026-62©

Chris

Brooks

2002Estimation

of

SystemsUsing

Two-Stage

Least

SquaresTSLS是比较经济、易用的方法。如果在第一阶段估计时所得到的R2非常高，那么古典OLS估计量与TSLS估计量将非常接近；如果在第一阶段估计时所得到的R2非常低，TSLS估计量将没有太大的实际意义。TSLS估计量是有偏估计量，但却是一致估计量。

It

is

still

of

concern

in

the

context

of

simultaneous

systwhether

the

CLRM

assumptions

are

supported

by

the

data.

If

the

disturbances

in

the

structural

equations

areautocorrelated,

the

2SLS

estimator

is

not

even

consisten

The

standard

error

estimates

also

need

to

be

modifiedcompared

with

their

OLS

counterparts,

but

once

this

hasbeen

done,

we

can

use

the

usual

t-

and

F-tests

to

testhypotheses

about

the

structural

form

coefficients.6-63

Recall

that

the

reason

we

cannot

use

OLS

directly

on

thestructural

equations

is

that

the

endogenous

variables

arecorrelated

with

the

errors.

One

solution

to

this

would

be

not

to

use

Y2

or

Y3

,

but

ratheto

use

some

other

variables

instead.

We

want

these

other

variables

to

be

(highly)

correlated

wiY2

and

Y3,

but

not

correlated

with

the

errors

-

they

are

calINSTRUMENTS.

Say

we

found

suitable

instruments

for

Y2

and

Y3,

z2

and

z3respectively.

We

do

not

use

the

instruments

directly,

butregressions

of

the

form(27)

&(28)Instrumental

Variables©

Chris

Brooks

20026-64©

Chris

Brooks

2002

Obtain

the

fitted

values

from

(27)

&

(28),

and ,

andreplace

Y2

and

Y3

with

these

in

the

structural

equation.

It

is

typical

to

use

more

than

one

instrument

per

endogenovariable.

If

the

instruments

are

the

variables

in

the

reduced