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3.2.2 Random E ff ect Model (

ランダム効果モデル

) Model:

y

it

= X

it

β + v

i

+ u

it

, i = 1 , 2 , · · · , n , t = 1 , 2 , · · · , T where i indicates individual and t denotes time.

The assumptions on the error terms v

i

and u

it

are:

E(v

i

| X) = E(u

it

| X) = 0 for all i ,

V(v

i

| X) = σ

2v

for all i , V(u

it

| X) = σ

2u

for all i and t ,

Cov(v

i

, v

j

| X) = 0 for i , j , Cov(u

it

, u

js

| X) = 0 for i , j and t , s , Cov(v

i

, u

jt

| X) = 0 for all i, j and t .

Note that X includes X

it

for i = 1 , 2 , · · · , n and t = 1 , 2 , · · · , T .

(2)

In a matrix form with respect to t = 1 , 2 , · · · , T , we have the following:

y

i

= X

i

β + v

i

1

T

+ u

i

, i = 1 , 2 , · · · , n ,

where y

i

=

 







y

i1

y

i2

...

y

iT

 





 , X

i

=

 







X

i1

X

i2

...

X

iT

 





 and u

i

=

 







u

i1

u

i2

...

u

iT

 





 are T × 1, T × k and T × 1, respectively.

u

i

N(0 , σ

2u

I

T

) and v

i

1

T

N(0 , σ

2v

) = ⇒ v

i

1

T

+ u

i

N(0 , σ

2v

1

T

1

0T

+ σ

2u

I

T

) . Again, in a matrix form with respect to i, we have the following:

y = X β + v + u ,

where y =

 







y

1

y

2

...

 





 , X =

 







X

1

X

2

...

 





 , v =

 







v

1

1

T

v

2

1

T

...

 





 and u =

 







u

1

u

2

...

 





 are nT × 1, nT × k, nT × 1 and

(3)

nT × 1, respectively.

The distribution of u + v is given by:

v + uN (

0 , I

n

⊗ ( σ

2v

1

T

1

0T

+ σ

2u

I

T

) ) The likelihood function is given by:

L( β, σ

2v

, σ

2u

) = (2 π )

−nT/2

I

n

⊗ ( σ

2v

1

T

1

0T

+ σ

2u

I

T

)

1/2

× exp (

− 1

2 (y − X β )

0

(

I

n

⊗ ( σ

2v

1

T

1

0T

+ σ

2u

I

T

) )

−1

(y − X β ) ) . Remember that f (x) = (2 π )

k/2

|Σ|

1/2

exp (

12

(x − µ )

0

Σ

1

(x − µ ) )

when XN( µ, Σ ),

where X denotes a k-variate random variable.

(4)

The estimators of β , σ

2v

and σ

2u

are given by maximizing the following log-likelihood function:

log L( β, σ

2v

, σ

2u

) = − nT

2 log(2 π ) − 1

2 log I

n

⊗ ( σ

2v

1

T

1

0T

+ σ

2u

I

T

)

− 1

2 (y − X β )

0

(

I

n

⊗ ( σ

2v

1

T

1

0T

+ σ

2u

I

T

) )

−1

(y − X β ) . MLE of β , denoted by ˜ β , is given by:

β ˜ = ( X

0

(

I

n

⊗ ( σ

2v

1

T

1

0T

+ σ

2u

I

T

) )

−1

X )

−1

X

0

(

I

n

⊗ ( σ

2v

1

T

1

0T

+ σ

2u

I

T

) )

−1

y

= ( ∑

n

i=1

X

i0

( σ

2v

1

T

1

0T

+ σ

2u

I

T

)

−1

X

i

)

1

( ∑

n

i=1

X

i0

( σ

2v

1

T

1

0T

+ σ

2u

I

T

)

−1

y

i

) , which is equivalent to GLS.

Note that ˜ β is not operational, because ˆ β depends on σ

2v

and σ

2u

.

(5)

3.3 Hausman’s Specification Error (

特定化誤差

) Test

Regression model:

y = X β + u , y : n × 1 , X : n × k , β : k × 1 , u : n × 1 . Suppose that X is stochastic.

If E(u | X) = 0, OLSE ˆ β is unbiased because of ˆ β = (X

0

X)

1

X

0

y = β + (X

0

X)

1

X

0

u and E((X

0

X)

1

X

0

u) = 0.

However, If E(u | X) , 0, OLSE ˆ β is biased and inconsistent.

Therefore, we need to check if X is correlated with u or not.

= ⇒ Hausman’s Specification Error Test

(6)

The null and alternative hypotheses are:

H

0

: X and u are independent, i.e., Cov(X , u) = 0,

H

1

: X and u are not independent.

Suppose that we have two estimators ˆ β

0

and ˆ β

1

, which have the following properties:

• β ˆ

0

is consistent and e ffi cient under H

0

, but is not consistent under H

1

,

• β ˆ

1

is consistent under both H

0

and H

1

, but is not e ffi cient under H

0

. Under the conditions above, we have the following test statistic:

( ˆ β

1

− β ˆ

0

)

0

(

V( ˆ β

1

) − V( ˆ β

0

) )

1

( ˆ β

1

− β ˆ

0

) −→ χ

2

(k) . Example: β ˆ

0

is OLS, while ˆ β

1

is IV such as 2SLS.

Hausman, J.A. (1978) “Specification Tests in Econometrics,” Econometrica, Vol.46,

(7)

3.4 Choice of Fixed E ff ect Model or Random E ff ect Model

3.4.1 The Case where X is Correlated with u — Review

The standard regression model is given by:

y = X β + u , uN(0 , σ

2

I

n

) OLS is:

β ˆ = (X

0

X)

1

X

0

y = β + (X

0

X)

1

X

0

u .

If X is not correlated with u, i.e., E(X

0

u) = 0, we have the result: E( ˆ β ) = β . However, if X is correlated with u, i.e., E(X

0

u) , 0, we have the result: E( ˆ β ) , β .

= ⇒ β ˆ is biased.

(8)

Assume that in the limit we have the followings:

( 1

n X

0

X)

1

−→ M

xx1

, 1

n X

0

u −→ M

xu

, 0 when X is correlated with u.

Therefore, even in the limit,

plim ˆ β = β + M

xx1

M

xu

, β, which implies that ˆ β is not a consistent estimator of β .

Thus, in the case where X is correlated with u, OLSE ˆ β is neither unbiased nor con-

sistent.

(9)

3.4.2 Fixed E ff ect Model or Random E ff ect Model

Usually, in the random e ff ect model, we can consider that v

i

is correlated with X

it

. [Reason:]

v

i

includes the unobserved variables in the ith individual, i.e., ability, intelligence, and so on.

X

it

represents the observed variables in the ith individual, i.e., income, assets, and so on.

The unobserved variables v

i

are related to the observed variables X

it

. Therefore, we consider that v

i

is correlated with X

it

.

Thus, in the case of the random e ff ect model, usually we cannot use OLS or GLS.

In order to use the random e ff ect model, we need to test whether v

i

is uncorrelated

with X

it

.

(10)

Apply Hausman’s test.

H

0

: X

it

and e

it

are independent ( −→ Use the random e ff ect model),

H

1

: X

it

and e

it

are not independent ( −→ Use the fixed e ff ect model), where e

it

= v

i

+ u

it

.

Note that:

• We can use the random e ff ect model under H

0

, but not under H

1

.

• We can use the fixed e ff ect model under both H

0

and H

1

.

• The random e ff ect model is more e ffi cient than the fixed e ff ect model under H

0

. Therefore, under H

0

we should use the random e ff ect model, rather than the fixed

ff

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