Econometrics I: Solutions of Homework 7

Econometrics I: Solutions of Homework 7

Hiroki Kato ^* May 29, 2020

Contents

1 Solutions 1

1.1 (1) . . . . 1

1.2 (2) . . . . 2

1.3 (3) . . . . 3

1.4 (4) . . . . 3

1.5 (5) . . . . 4

1.6 (6) . . . . 4

1.7 (7) . . . . 4

1.8 (8) . . . . 5

1.9 (9) . . . . 6

1 Solutions

1.1 (1)

A transformation matrix M is defined by

M = I

− 1 T ii

,

*

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where I

is T × T identity matrix

. Then, M i = I

i − 1

T i(i

i) = i − 1

T iT = 0, and

M e = I

e − 1

T i(i

e) = e − 1 T i

∑

e

= e.

Note that third equality comes from the property of the OLS estimator, that is, ∑

t

e

/T = ¯ e = 0.

1.2 (2)

First, I will show the first equality. By M e = e, premultiplying e on M yields M e = M y − M X β ˆ

e = M y − M X β. ˆ Then, we obtain

e

e = y

M

M y − β ˆ

X

M

M X β ˆ = y

M y − βX ˆ

M X β ˆ

The second equality comes from the fact that M is symmetric and idempotent. The M is idempotent because

M

= (

I

− 1 T ii

)(

I

− 1 T ii

)

= I

− 1

T ii

− 1

T ii

+ 1

T

i(i

i)i

= I

− 1

T ii

− 1

T ii

+ 1 T

i(T )i

= I

− 1

T ii

= M.

Second, I will show the second equality. By M e = e and M i = 0, premultiplying e = y − i β ˆ

− X

β ˆ

on M gives

M e = M y − M i β ˆ

− M X

β ˆ

e = M y − M X

β ˆ

1

This matrix is different from the matrix

that I used in solutions to HW5.

Then, we obtain

e

e = (M y − M X

β ˆ

)

(M y − M X

β ˆ

)

= y

M y − β ˆ

X

M y − y

M X

β ˆ

+ ˆ β

X

M X

β ˆ

= y

M y − β ˆ

X

M X

β ˆ

− β ˆ

X

M X

β ˆ

+ ˆ β

X

M X

β ˆ

= y

M y − β ˆ

X

M X

β ˆ

.

The third equality comes from X

M y = X

M X

β ˆ

. This holds because

X

e = X

M y − X

M X

β ˆ

0 = X

M y − X

M X

β ˆ

Note that X

e = 0 holds since X

e = X

y − X

X β ˆ = (X

X) ˆ β − X

X β ˆ = 0 by β ˆ = (X

X)

X

y.

1.3 (3)

Since y

M y is a scalar,

R

= 1 − e

e

y

M y = y

M y

y

M y − e

e

y

M y = y

M y − e

e y

M y =

β ˆ

X

M X

β ˆ

y

M y

1.4 (4)

R β ˆ = R(X

X)

X

y = R(X

X)

X

(Xβ + u) = Rβ + R(X

X)

X

u

Since u is normally distributed, Rb is also normally distributed. Expectation and variance of Rb are as follows:

E(R β) = ˆ Rβ

V (R β) = ˆ E[(Rb − Rβ)(Rb − Rβ)

] = E[R(X

X)

X

uu

X

(X

X)

R

] = σ

R(X

X)

R

Thus, the distribution of Rb is

R β ˆ ∼ N (Rβ, σ

R(X

X)

R

).

1.5 (5)

By the question (4), we can replace Rβ by r if the null hypothesis is correct. Thus,

R β ˆ ∼ N (r, σ

R(X

X)

R

),

or

(R β ˆ − r) ∼ N (0, σ

R(X

X)

R

),

1.6 (6)

R =



 

 

 

 

 



0 1 0 · · · 0 0 0 1 · · · 0 .. . .. . .. . .. . .. . 0 0 0 · · · 1



 

 

 

 

 



= (0, I

)

where R is (k − 1) × k matrix. Thus, G = k − 1 and r = 0.

1.7 (7)

We will show that, given R and r,

(R β ˆ − r)

(R(X

X)

R

)

(R β ˆ − r) = ˆ β

X

M X

β ˆ

.

By the solution to (6), define R = (0, I

) and r = 0. Then, R β ˆ − r = ˆ β

.

Next, given R and r as defined above, we will show (R(X

X)

R

)

= X

M X

. First, by X = (i, X

),

(X

X)

=



 





 

 i

X



 

 (

i X

) 

 



−1

=



 



i

i i

X

X

i X

X



 



−1

=



 



B

B

B

B



 



where B

is unknown matrices. Then, we have

R(X

X)

R

= (

0 I

) 

 



B

B

B

B



 





 

 0 I



 

 = (

B

B

) 

 

 0 I



 

 = B

Thus, we only need to calculate B

. By the property of the inverse of a partitioned matrix,

B

= (X

X

− X

i(i

i)

i

X

)

= (X

I

X

− X

( 1

T ii

)X

)

= (X

(I

− 1

T ii

)X

)

= (X

M X

)

Hence, (R(X

X)

R

)

= ((X

M X

)

)

= X

M X

. Finally, given R = (0, I

) and r = 0, we obtain

(R β ˆ − r)

(R(X

X)

R

)

(R β ˆ − r) = ˆ β

X

M X

β ˆ

.

1.8 (8)

By solutions to (6) and (7), test statistic for H

: β

= 0 is β ˆ

X

M X

β ˆ

/(k − 1)

e

e/(T − k) ∼ F (k − 1, T − k) We will show that

R

/(k − 1) (1 − R

)/(T − k) =

β ˆ

X

M X

β ˆ

/(k − 1) e

e/(T − k) ,

where R

is the coefficient of determination.

By solutions to (3), we obtain

(y

M y )R

= ˆ β

X

M X

β ˆ

,

and,

(y

M y)(1 − R

) = e

e.

Since y

M y is scalar, we can obtain

β ˆ

X

M X

β ˆ

/(k − 1)

e

e/(T − k) = R

/(k − 1) (1 − R

)/(T − k)

1.9 (9)

First, we test β = 0, using t­statistic. Recall that t­statistic is given by

t =

β ˆ − β s/ √∑

t

(X

− X) ¯

∼ t(T − k)

where s

is unbiased and consistent estimator of σ

. Since V ( ˆ β) = σ

/ ∑

t

(X

− X) ¯

, we can calculate t­statistic as follows:

t =

β ˆ − β SE( ˆ β) .

Thus,

t = 0.65 − 0

0.240 = 2.70833

Under the degree of freedom is 4 − 2 = 2, the test statistic at 1%, 5%, and 10% significance level is 9.9248, 4.3072, and 2.9200, respectively. Thus, we cannot reject the null hypothesis β = 0.

Second, we test β = 0, using F ­statistic with R

. By solutions to (8), a test statistic is given by R

/(k − 1)

(1 − R

)/(T − k) = 0.786/(2 − 1)

(1 − 0.786)/(4 − 2) = 7.345794

Under F ∼ F (1, 2), the test statistic at 1%, 5%, and 10% significance level is 98.50251, 18.51282, and 8.526316, respectively. Thus, we cannot reject the null hypothesis β = 0.

Overall, F ­test obtains the same result as t­test. Note that the square of t­statistic is approximate

to F ­statistic, that is, t

= (2.70833)

= 7.335069 ≈ 7.345794.