Econometrics I: Solutions of Homework #13

Econometrics I: Solutions of Homework #13

Hiroki Kato ^* July 15, 2020

Contents

1 Solutions: Question 1 1

1.1 (1) . . . . 1 1.2 (2) . . . . 2 1.3 (3) . . . . 3

2 Solutions: Question 2 4

2.1 (1) . . . . 4 2.2 (2) . . . . 5 2.3 (3) . . . . 6

1 Solutions: Question 1

1.1 (1)

OLS estimator of β is given by

β ˆ = (X

X)

X

y = β + (X

X)

X

u = β + (

1 T

∑

t=1

X

X

)

(

1 T

∑

t=1

X

u

)

, (1)

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where X

is k × 1 vector whose elements are explanatory variables for observetion t. Taking the probability limit to both sides yields

plim

T→∞

β ˆ = β + lim

T→∞

( 1 T

∑

t=1

X

X

)

plim

T→∞

( 1 T

∑

t=1

X

u

)

(2)

First, we assume the following stationarity condition:

lim

T→∞

1 T

∑

t=1

X

X

= M

. (3)

Then, by g(X

) →

g(X) if X

→

X, we have

plim

T→∞

( 1 T

∑

t=1

X

X

)

= M

. (4)

We further assume zero covariance between X and u, that is,

plim

T→∞

( 1 T

∑

t=1

X

u

)

= 0. (5)

By these two assumptions, we finally obtain plim

T→∞

β ˆ = β + M

· 0 = β. (6)

We can show the consistency of OLSE.

1.2 (2)

Greenberg and Webster (1983) states the central limit theorem as follows:

Z

, . . . , Z

are mutually independent. Z

is distributed with mean µ and variance Σ

for

i = 1, . . . , n. Then, we have the follwing result:

Using this theorem, we derive the asymptotic distribution of

X

u. Define Z

= X

u

. By the assumption of u

, µ = E(Z

) = 0 and the variance of Z

is given by

Σ

= V (X

u

) = E[(X

u

)(X

u

)

] = σ

X

X

. (7) Then, we obtain Σ as follows:

Σ = lim

T→∞

( 1 n

∑

t=1

σ

X

X

)

= σ

lim

T→∞

( 1 n

∑

t=1

X

X

)

= σ

M

. (8)

By the central limit theorm, we have

√ 1

n X

u = 1

√ n

∑

t=1

(X

u

− 0) →

N (0, σ

M

). (9)

1.3 (3)

We rewrite the equation (1) as follows:

√ T ( ˆ β − β) = (

1 T

∑

t=1

X

X

)

(

√ 1 T

∑

t=1

X

u

)

. (10)

We use the following proerty: z

= H

y

→

N (Hµ, HΩH

) where H

is an r × k matrix with plim

H

= H and y

is a × 1 vector with y

→

N (µ, Ω). Because the equation (10) is seen as the form z

, √

T ( ˆ β − β) has a limiting normal distribution with zero mean and variance given by σ

M

M

M

= σ

M

. (11) Thus, we can write

√ T ( ˆ β − β) →

N (0, σ

M

). (12)

2 Solutions: Question 2

2.1 (1)

The log­likelihood function is

log L

(θ) = log

∏

i=1

f (X

; θ) =

∑

i=1

log f (X

; θ). (13)

The maximum likelihood estimator is defined by θ ˆ ∈ arg max

θ

1 T

∑

i=1

log f (X

; θ). (14)

Assuming the domain of X

does not depend on the parameter θ, we define the following expec­

tation of log f (X

; θ) as

log L(θ) = E [log f (X

; θ)] =

∫

(log f(X

; θ))f (X

| θ

)dx, (15) where θ

is a unknown true parameter. The function log L(θ) is maximized at θ = θ

, that is, log L(θ) ≤ log L(θ

) for any θ bacuase

E[log f (X

; θ) − log f(X

; θ

)]

=E [

log f (X

; θ) f(X

; θ

)

]

≤ E

[ f (X

; θ) f(X

; θ

) − 1

]

=

∫ ( f(X

; θ) f(X

θ

) − 1

)

f(X

; θ

)dx

=

∫

f(X

; θ)dx −

∫

f(X

; θ

)dx = 0.

Note that log x ≤ x − 1.

By the weak law of large numbers, for any θ, T

log L

(θ) → log L(θ). By defenition, θ ˆ is the

2.2 (2)

Applying the central limit theorem with unequal variance yields

√ T ( 1

T

∂ log L(θ)

∂θ − µ )

= 1

√ T

∑

i=1

( ∂ log f(X

; θ)

∂θ − µ )

→

N (0, Σ), (16)

where

log f(X

; θ) is distributed with mean µ and variance Σ

, and Σ = lim

(T

∑

Σ

).

From (16), we have

T

lim

E [ 1

T

∂ log L(θ)

∂θ ]

= µ, (17)

T

lim

T · Var [ 1

T

∂ log L(θ)

∂θ ]

= Σ. (18)

We need the expectation and variance of

log L(θ). Because L(θ) is a joint distribution, ∫

L(θ)dx = 1. Taking the first­order derivative with respect to θ on both sides yields

∫ ∂L(θ)

∂θ dx = 0

∫ ∂ log L(θ)

∂θ L(θ)dx = 0 E

[ ∂ log L(θ)

∂θ ]

= 0.

Thus, we obtain

T

lim

E [ 1

T

∂ log L(θ)

∂θ ]

= lim

T→∞

1 T E

[ ∂ log L(θ)

∂θ ]

= 0 = µ. (19)

To obtain the variance, taking the second­order derivative of ∫

L(θ)dx = 1 with respect to θ,

∫ ∂

log L(θ)

∂θ∂θ

L(θ)dx +

∫ ∂ log L(θ)

∂θ

∂ log L(θ)

∂θ

L(θ)dx = 0

− E

[ ∂

log L(θ)

∂θ∂θ

]

= E

[ ∂ log L(θ)

∂θ

∂ log L(θ)

∂θ

]

Thus, we obtain the variance as follows:

Var

[ ∂ log L(θ)

∂θ ]

=E

[ ∂ log L(θ)

∂θ

∂ log L(θ)

∂θ

]

− E

[ ∂ log L(θ)

∂θ ]

=E

[ ∂ log L(θ)

∂θ

∂ log L(θ)

∂θ

]

= − E

[ ∂

log L(θ)

∂θ∂θ

]

=I(θ), (20)

where I(θ) is the information matrix. This leads to

T

lim

T · Var [ 1

T

∂ log L(θ)

∂θ ]

= lim

T→∞

1 T Var

[ 1 T

∂ log L(θ)

∂θ ]

= lim

T→∞

1

T I (θ) = Σ. (21) Hence, the asymptotic distribution is

√ 1 T

∑

i=1

∂ log f (X

; θ)

∂θ

→

N (0, Σ) (22)

where Σ = lim

T

I(θ).

2.3 (3)

Taking the first­order approximation of

log L(ˆ θ) = 0 around θ ˆ = θ yields

∂ log L(θ)

∂θ + ∂

log L(θ)

∂θ∂θ

(ˆ θ − θ) = 0.

We rewrite it as follows:

θ ˆ − θ = −

( ∂

log L(θ)

∂θ∂θ

)

∂ log L(θ)

∂θ

√ ( )

( )

Since Σ is symmetric, using the Slutsky’s theorem, we have the asymptotic distribution of √

T (ˆ θ − θ) as follows:

√ T (ˆ θ − θ) →

N (0, Σ

ΣΣ

) (24)

→

N (0, Σ

) (25)