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Problem Set 1: Unbiased Estimator of the Error Variance This
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All the questions are attached in the attachment named Assignment_1.pdf, and the due date would be this Sunday night, so it still has a plenty of time

Problem Set 1: Unbiased Estimator of the Error

Variance

This assignment will guide you through the derivations needed to determine what is

a unbiased estimator of the error variance in the context of a univariate linear regression.

This assignment may be quite challenging. Good Luck!

Consider the univariate linear regression model

yt = ? + ?xt + ut ,

t = 1, . . . , T.

(1)

where the regressors are non-stochastic (fixed) and the disturbances have zero mean

and are uncorrelated and homoscedastic with variance equal to ? 2 , i.e., E [ut ] = 0,

C [ut , us ] = 0, t 6= s, and V [ut ] = E [u2t ] = ? 2 . The aim of this problem set will be to

derive that

T

1 X 2

2

u?

s =

T ? 2 t=1 t

? t,

is an unbiased estimator of the disturbance variance ? 2 , where u?t = yt ? y?t , y?t = ?

? + ?x

and ?

? and ?? are the OLS estimators of ? and ?, respectively. The OLS estimates ?

?

and ?? are given by

?? =

PT

(yt ? y¯) (xt ? x¯)

,

PT

¯ )2

t=1 (xt ? x

?

? = y¯ ? ??x¯,

t=1

which can also be expressed as

?? =

T

X

wt yt ,

?

?=

t=1

where

wt = PT

qt y t ,

t=1

xt ? x¯

t=1

T

X

2

(xt ? x¯)

1

,

qt =

1

? x¯ · wt .

T

Question 1 - Verify simple properties.

Verify the following 7 properties:

PT

1.

t=1 wt = 0

2.

PT

w t xt = 1

3.

PT

=1

4.

PT

5.

PT

6.

PT

7.

PT

t=1

t=1 qt

t=1 qt xt

t=1

=0

1

wt2 =

2

t=1 qt =

t=1 qt wt

PT

x)2

t=1 (xt ?¯

1

PT

t=1

PTT

x2t

x)2

t=1 (xt ?¯

=

x

PT

x)2

t=1 (xt ?¯

As you work through the problem set, you should always look back at these 8 properties

For example, with properties 1 and 2 in hand, it is easy to show (as we did in class)

that:

T

T

T

X

X

X

?

?=

wt yt =

wt (? + ?xt + ut ) = ? +

wt ut .

t=1

t=1

t=1

Likewise, using properties 3 and 4 it is easy to show (as we did in class) that:

?

?=

T

X

qt yt =

t=1

T

X

qt (? + ?xt + ut ) = ? +

t=1

T

X

qt ut .

t=1

Question 2 - Computing the expression for the sum of

the residuals.

Show that the residuals from the regression described by Eq. 1 can be expressed as:

h



 i

u?t = ? (?

? ? ?) + ?? ? ? xt + ut .

And, therefore,

u?2t

h



 i

 i2

2

?

?

= (?

? ? ?) + ? ? ? xt + ut ? 2 (?

? ? ?) + ? ? ? xt ut

h



2

The sum of the residuals therefore can be expressed as:

T

X

u?2t

=

T h

X

t=1

T

T

 i2 X

h



 i

X

2

?

?

(?

? ? ?) + ? ? ? xt +

ut ?

2 (?

? ? ?) + ? ? ? xt ut



t=1

t=1

t=1

The expected sum of the residuals therefore can be expressed as:

T

X

h

T

T

T

h h



 i i



 i2  X

  X

 2 X

E 2 (?

? ? ?) + ?? ? ? xt ut

E (?

? ? ?) + ?? ? ? xt

+

E u2t ?

E u?t =

t=1

t=1

t=1

t=1

The expression above can be broken down into 3 components:

h h



 i i

P

? ? ?) + ?? ? ? xt ut

? Component 1: Tt=1 E 2 (?

h



 i2 

?

(?

? ? ?) + ? ? ? xt

t=1 E

? Component 2:

PT

? Component 3:

PT

t=1

E [u2t ]

The next 3 questions will make you work on each of these components.

Question 3 - Working with Component 1.

Note that:





(?

? ? ?) + ?? ? ? xt =

T

X

s=1

!

qs u s

+

T

X

!

ws us

s=1

Using the expression above, show that:

hh

E



 i i

(?

? ? ?) + ?? ? ? xt ut = (qt + wt xt ) ? 2 ,

and, therefore, that

T

X

h h



 i i

E 2 (?

? ? ?) + ?? ? ? xt ut = 4? 2 .

t=1

3

xt .

Question 4 - Working with Component 2

Show that

&quot;

#

PT

h

1



 i2 

2

2

x

x

x

¯

·

x

t

t

E (?

? ? ?) + ?? ? ? xt

= ? 2 PTT t=1 t 2 + PT

? 2 PT

,

2

2

(x

?

x

¯

)

(x

?

x

¯

)

(x

?

x

¯

)

t

t

t

t=1

t=1

t=1

and, therefore, that

T

X

h

E

 i2 

?

(?

? ? ?) + ? ? ? xt

= 2? 2 .



t=1

(This can be quite challenging)

Question 5 - Working with Component 3

Show that

T

X

 

E u2t = T ? 2

t=1

(This should be very easy)

Final Step

This final step is easy and I will solve it for you. Once we have obtained the expressions

for the three components in questions 3, 4, and 5 we simply have to put all the pieces

together to obtain an unbiased estimator of the error variance. To be precise:

T

X

t=1

u?2t

T

T

T

h



 i2 X

h h



 i i

X

 2 X

?

?

=

E (?

? ? ?) + ? ? ? xt +

E ut ?

E 2 (?

? ? ?) + ? ? ? xt ut

t=1

t=1

t=1

simplifies to

T

X

 

E u?2t = 2 · ? 2 + T · ? 2 ? 4 · ? 2 = (T ? 2) · ? 2 .

t=1

4

Divide by (T ? 2) on both sides the expression above to obtain the unbiased estimator

of the error variance:

#

&quot;

T

1 X 2

u? = ? 2 .

E

T ? 2 t=1 t

5

STATUS
QUALITY
Approved

This question was answered on: Feb 21, 2020

Solution~00096837.zip (18.37 KB)