Econometrics I: Solutions of Homework #11

Econometrics I: Solutions of Homework #11

Hiroki Kato ^* July 3, 2020

Contents

1 Solutions 1

1.1 (1) . . . . 1

1.2 (2) . . . . 2

1.3 (3) . . . . 3

1.4 (4) . . . . 4

1 Solutions

1.1 (1)

Since we assume u

is normally distributed with E(u

) = 0 and V (u

) = σ

, the density function of u

is given by

f (u

) = 1

(2πσ

)

exp (

− 1 2σ

u

) .

By the mutual independent assumption, the joint density function of u

, . . . , u

is given by

f(u

, . . . , u

) = 1

(2πσ

)

exp (

− 1 2σ

∑

u

)

*

e­mail: [email protected]. Room 503. If you find any errors in handouts and materials,

please contact me via e­mail.

Transforming u

, we have the following likelihood function;

L(α, β, σ

| y

, . . . , y

) ≡ 1

(2πσ

)

exp (

− 1 2σ

∑

(y

− α − βx

)

)

. (1)

1.2 (2)

To obtain the log­likelihood function, we take a natural logarithm of (1) as follows:

log L(α, β, σ

| y

, . . . , y

) = − T

2 log(2π) − T

2 log(σ

) − 1 2σ

∑

(y

− α − βx

)

Taking a first­order derivative with respect to unknown parameters (α, β, σ

) yields

∂L(α, β, σ

| y

, . . . , y

)

∂α = 1

σ

∑

(y

− α − βx

) = 0 (2)

∂L(α, β, σ

| y

, . . . , y

)

∂β = 1

σ

∑

(y

− α − βx

)x

= 0 (3)

∂L(α, β, σ

| y

, . . . , y

)

∂σ

= − T

2 1 σ

+ 1

2σ

∑

(y

− α − βx

)

= 0 (4)

Rewriting the equation (2) and the equation (3) yields

T y ¯ − αT ˜ − βT ˜ x ¯ = 0,

∑

t

y

x

− αT ˜ x ¯ − β ˜ ∑

t

x

= 0,

where y ¯ = ∑

t

y

/T and x ¯ = ∑

t

x

/T .

First, using the first equation, we can obtain the ML estimator of α, denoted by α, as follows; ˜

˜

α = ¯ y − β ˜ x. ¯

Substituting this estimator into the second equation, we obtain the ML estimator of β, denoted by

β, as follows; ˜

− β[ ˜ ∑

t

x

− T x ¯

] + ∑

t

y

x

− T x¯ ¯ y = 0 β ˜ =

∑

t

y

x

− T x ¯ y ¯

∑

t

x

− T x ¯

β ˜ =

∑

t

(y

− y)(x ¯

− x) ¯

∑

t

(x

− x) ¯

Solving the equation (4) gives the ML estimator of σ

, denoted by σ ˜

, as follows;

˜ σ

=

∑

t=1

(y

− α ˜ − βx ˜

)

T .

In summary, the ML estimator of θ is

θ ˜ = ( ˜ α, β, ˜ σ ˜

)

= (

¯ y − β ˜ x, ¯

∑

t

(y

− y)(x ¯

− x) ¯

∑

t

(x

− x) ¯

,

∑

t=1

(y

− α ˜ − βx ˜

)

T

)

.

1.3 (3)

By the equation (2), (3), and (4),

∂

L(α, β, σ

| y

, . . . , y

)

∂α

= − T

σ

∂

L(α, β, σ

| y

, . . . , y

)

∂β

= −

∑

t

x

σ

∂

L(α, β, σ

| y

, . . . , y

)

∂(σ

)

= T 2σ

− 1

σ

∑

(y

− α − βx

)

= T 2σ

− 1

σ

∑

u

∂

L(α, β, σ

| y

, . . . , y

)

∂α∂β = −

∑

t

x

σ

∂

L(α, β, σ

| y

, . . . , y

)

∂α∂σ

= − 1

σ

∑

(y

− α − βx

) = − 1 σ

∑

u

∂

L(α, β, σ

| y

, . . . , y

)

∂β∂σ

= − 1

σ

∑

(y

− α − βx

)x

= − 1 σ

∑

u

x

By the distributional assumption, we have ∑

t

E(u

) = 0, ∑

t

E(u

)x

= 0, and ∑

t

E(u

) =

∑

t

[E(u

) − E(u

)] = ∑

t

V (u

) = T σ

. Thus, the information matrix I(θ) is given by

I(θ) = − E

( ∂

L(α, β, σ

| y

, . . . , y

)

∂θ∂θ

)

=



 

 

 



T σ²

∑

txt

σ²

0

∑

txt

σ²

∑

tx²_t

σ²

0

0 0



 

 

 



1.4 (4)

Step 1: Variance­Covariance Matrix of MLE

By the Cramer­Rao lower bound theorem, the inverse of information matrix I (θ)

provides a lower bound of the variance­covariance matrix for unbiased estimators of θ.

Then, the variance­

covariance matrix of θ ˜ is



 

 

 



σ² T

∑

tx²_t

∑

t(xt−x)¯²

− σ

t(xt−x)¯ ²

0

− σ

t(xt−x)¯ ²

σ²

∑

t(xt−¯x)²

0

0 0



 

 

 



Step 2: Derive the expectation and variance of OLSE

The OLS estimator of β, denoted by β, is equivalent to the ML estimator of ˆ β, denoted by β. That ˜ is,

β ˆ =

∑

t

(y

− y)(x ¯

− x) ¯

∑

t

(x

− x) ¯

. Thus, we have E ( ˆ β) = β and V ( ˆ β) = σ

/( ∑

t

(x

− x) ¯

). This implies that there is no difference between the MLE and the OLSE of β.

The OLS estimator of σ

is given by

ˆ

σ

= 1 T − 2

∑

(y

− α ˜ − βx ˜

)

.

This estimator is unbiased: E(ˆ σ

) = σ

. Since the OLSE is different from the ML estimator of β, β. the ML estimator is biased estimator of ˜ σ

, that is,

E(˜ σ

) = T − 2

T E( ˆ σ

) = T − 2

T σ

.