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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

 

and see which one can help you in each step.

 

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

 

"

 

#

 

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:

 

#

 

"

 

T

 

1 X 2

 

u? = ? 2 .

 

E

 

T ? 2 t=1 t

 


 

5

 


 

 







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