ETC3400-BEX3400-ETC5340: PRINCIPLES OF ECONOMETRICS

Solutions to Assignment 3, Semester 2, 2020

1. Suppose that the discrete random variable Y has a one parameter probability density

function which is given by

f (yj) = (1 )y ; y = 0; 1; 2; :::; 0 < < 1:

The mean and variance 2 of this distribution are equal to

1 and

(1)2 respectively.

Suppose that we have a sample of n i.i.d observations on this variable, fy1; y2; :::; yng.

(a) Show that the maximum likelihood estimator of is

^MLE =

y

y + 1

;

where y is the sample mean of fy1; y2; :::; yng :

L(; y1; y2; :::; yn) =

i=nY

i=1

(1 )yi = (1 )nyi ; =)

l(; y1; y2; :::; yn) = n log(1 ) +

i=nX

i=1

yi log ; )

@l

@

=

n

(1 ) +

1

i=nX

i=1

yi; so setting

@l

@

= 0 =)

n

(1 ) =

1

i=nX

i=1

yi; =) = (1 )y ) (1 + y) = y =)

^MLE =

y

y + 1

as required. To check that we have a maximum,

@2l

@2

=

n

(1 )2

1

2

i=nX

i=1

yi < 0, (since yi 0), so we have a maximum.

(b) Use the Weak Law of Large Numbers for sequences of i.i.d. random variables

together with Slutskys theorem to show that ^MLE is a consistent estimator of

0; carefully explaining each of the steps.

The WLLN tells us that plim(y) = E(Y ) = 0

10 :

1

Hence, since ^MLE =

y

y+1

; from Slutskys theorem (or the CMT) we obtain

p lim(^MLE) = p lim(

y

y+1

) = p lim(y)

p lim(y+1)

=

0

10

p lim(y)+1

=

0

10

0

10+1

=

0

10

1

10

= 0:

Therefore, since p lim(^MLE) = 0, we can conclude that ^MLE is a consistent

estimator of 0:

(c) Find the information matrix I () (a 1 1 matrix in this case), and hence write

down the limiting distribution for

p

n

^MLE 0

:

I () j=0 = E(H())j=0 = E( @

2l

@2

)j=0 = E( n(1)2 12

i=nP

i=1

yi)j=0

= n

(10)2 +

nE(Y )

20

= n

(10)2 +

n

20

0

10 =

n

(10)

1

(10) +

1

0

= n

(10)20 :

Thus (I () j=0)1 = (10)

20

n

=) (i () j=0)1 = (1 0)20; and

p

n

^MLE 0

d! N(0; (1 0)20):

(d) De
ne b2N = (^MLE) = bMLE(1bMLE)2 ; and suppose that n = N , where N is a large

but
nite number. Find the "asymptotic" distribution of b2N :

We apply the delta method here, and note that

@ ()

@

=0

=

@

@

(1 )2

=0

=

(1 )2 + 2(1 )

(1 )4

=0

=

(1 2)

(1 )4

=0

=

(1 + )

(1 )3

=0

Hence the limit distribution is

p

n(b2MLE 0(10)2 )) d! N

0;

(1+0)

(10)3

2

(i(0))

1

i.e.

p

n(b2MLE 0(10)2 )) d! N

0;

(1+0)

(10)3

2

(1 0)20

using the result from

(c), so

p

n(b2MLE 0(10)2 )) d! N

0; 0

(1+0)

(10)2

2

and the requested "asymp-

totic" distribution is b2N N 0(10)2 ; 0N (1+0)(10)22

:

2

(e) Consider maximum likelihood estimation of () =

2

, and provide the limit

distribution for

p

n

bMLEb2MLE

2

in terms of the observations and the

parameter :

We set ()| {z }

21

=

1()

2()

=

"

1

(1)2

#

, and given regularity, we have

p

n( (MLE)

(0))

d! N(0; ()| {z }

21

i(0)

1| {z }

11

(0)

0| {z })

12

; where (0) =

@

@0

=0

=

@ 1

@

@ 2

@

=0

:

Here, the details for (0) are (0) ="

@

@

(

1)

@

@

(

(1)2 )

#

=0

=

"

1+

(1)2

(1)2+2(1)

(1)4

#

=0

=

"

1

(1)2

12

(1)4

#

=0

=

"

1

(10)2

120

(10)4

#

;

so using results from (c) again, the variance of the limiting distribution becomes"

1

(10)2

120

(10)4

#

(1 0)20

h

1

(10)2

120

(10)4

i

=

"

1

(10)2

120

(10)4

#

h

0

(120)0

(10)2

i

=

"

0

(10)2

(1+0)0

(10)3

(1+0)0

(10)3

(1+0)20

(10)4

#

:

Hence

p

n

bMLEb2MLE

"

0

10

0

(10)2

#!

d! N

0

0

;

"

0

(10)2

(1+0)0

(10)3

(10)0

(10)3

(1+0)

20

(10)4

#!

:

2. Consider the CNLM,

y

(n1)

= X

(nk)

(k1)

+ u

(n1)

(1)

u

(n1)

N(0; 2In)

with all notation as de
ned in lectures, and X being treated as
xed. Perform the

following tasks:

(a) Derive the MLE of = (0; )0 (noting that we are interested in estimating

rather than 2). You can assume that the second order conditions for a maximum

are satis
ed.

3

The de
nition of u implies that ui

i:i:d: N(0; 2) for i = 1; 2; :::; n; and since we

are treating X as
xed and yi = x

0

i+ui, we can transform the pdf of ui to a pdf

of yi via

f(yi) = f(ui)

@[email protected]

= f(ui = yi x0i) 1

where f(ui) = (22)1=2 expf 1

22

ui

2g:

so f (yij; ) =

22

1=2

exp

1

22

yi x0i

2

Hence, the log likelihood function can be expressed as

l = logL (jyi; y2; :::; yn)

= log

nQ

i=1

f (yij; )

= log

nQ

i=1

22

1=2

exp

1

22

yi x0i

2

= n

2

log (2) n

2

log

2

1

22

nP

i=1

yi x0i

2

= n

2

log (2) n

2

log

2

1

22

(y X)0 (y X) :

The MLE of = (0; )0, b = (b0; b)0; is obtained by solving the set of
rst order

conditions @l

@

= 0 and @l

@

= 0; where

@l

@

= 1

22

@

@

(y X)0 (y X)

= 1

22

@

@

(y0 0X0) (y X)

= 1

22

@

@

(y0y y0X 0X0y + 0X0X)

= 1

22

@

@

(y0y 20X0y + 0X0X)

= 1

22

(2X0y + 2X0X)

=

1

2

(X0y X0X) (2)

and

@l

@

= n

+

1

3

(y X)0 (y X) (3)

4

We set (2) to be 0 and
nd that

1b2

X0y X0Xb = 0

so b = (X0X)1X0y

(assuming that X has full rank so that (X0X)1 will exist).

Next, we set (3) to be 0

nb + 1b3

y Xb0 y Xb = 0

b =

vuuty0 b0X0y Xb

n

(b) Consider the following vector of parameter functions:

() =

1()

2()

=

=4

[2]

1

Specify all components of the limiting distribution of the MLE of () :

p

n( (b) (0)) d! N(0; (0)i(0)1 (0)0): (4)

with all notation as de
ned in lectures.

We use the delta method for vector MLE and need to
nd

(0) =

@

@

=4 @

@

=4

@l

@

2 @

@

2

=0

=

Ik=

4 4=5

0 23

=0

=

"

Ik

04

40

50

0 230

#

The estimated and true values of the parameters of interest are

(bMLE) = " bMLEb4MLEb2MLE

#

and (0) =

"

0

04

20

#

The (per observation) information matrix is

i(0) =

1

n

(

E

@2l

@@0

=0

)

=

1

n

8><>:E

24 @[email protected]@0 @[email protected]@

@2l

@[email protected]

@2l

@()2

35

=0

9>=>;

5

where

@2l

@@0

=

@

@

@l

@

0

=

@

@

1

2

(X0y X0X)

0

= X

0X

2

@2l

@@

=

@2l

@@

=

@

@

@l

@

=

@

@

1

2

X0u

= 2

3

X0u

@2l

@ ()2

=

@

@

@l

@

=

@

@

n

+

1

3

(y X)0 (y X)

=

n

2

3

4

(y X)0 (y X)

=

n

2

3

4

u0u

so that i(0) is therefore

i(0) =

1

n

(

E

X0X

2

2

3

X0u

2

3

X0u n

2

3

4

u0u

=0

)

=

1

n

(

E

X0X

2

2

3

X0u

2

3

X0u n

2

+ 3

4

u0u

=0

)

=

1

n

(

X0X

2

0

0 n

2

+ 3

4

E [u0u]

=0

)

=

1

n

(

X0X

2

0

0 n

2

+ 3

4

n2

=0

)

=

1

n

(

X0X

2

0

0 n

2

+ 3n

2

=0

)

=

1

n

(

X0X

2

0

0 2n

2

=0

)

=

"

X0X

n20

0

0 2

20

#

The variance of the limit distribution is (0)i(0)1 (0)0

=

"

Ik

40

40

50

0 230

#

"

X0X

n20

0

0 2

20

#1

"

Ik

40

40

50

0 230

#0

6

="

Ik

40

40

50

0 230

#

"

n20(X

0X)1 0

0

20

2

#

" Ik

40

0

4

0

0

50

230

#

=

"

Ik

40

40

50

0 230

#

24 n(X0X)120 0

2

0

0

30

10

35 =

24 n(X0X)160 + 800080 4060

4

0

0

60

240

35

Hence

p

n

" bMLEb4MLEb2MLE

#

"

0

04

20

#!

d! N

[email protected] 0

0

;

24 n(X0X)160 + 800080 4060

4

0

0

60

240

35 1A :

3. Assume n i:i:d: draws from Y Exponential() with pdf:

f(yj) = 1

exp(1

y) for y > 0

E(Y j) = ; var(Y j) = 2

and consider the use of s2 =

nP

i=1

Yi Y

2

= (n 1) as an estimator of var(Y j) = 2:

(a) It is well known that the sample variance is an unbiased estimator for the pop-

ulation variance, and for the exponential distribution it is also the case that

var(s2) =

4

n

h

9 (n3)

(n1)

i

: Use these two pieces of information to explain why the

su¢ cient conditions for s2 to be a consistent estimator of 2 hold.

The su¢ cient conditions for s2 to be a consistent estimator for 2 are that

(i) s2 is an unbiased estimator for 2; (i.e. E(s2) = 2), and since our
rst piece

of information says that s2 is an unbiased estimator for var(Y j); we know

that s2 is an unbiased estimator for 2 so that E(s2) = 2 holds true; and

(ii) var(s2)! 0 as n!1; which also holds since

lim

n!1

var(s2) = lim

n!1

1

n

94 4 (n 3)

(n 1)

(5)

= lim

n!1

1

n

94 4 (n 1 2)

(n 1)

(6)

= lim

n!1

4

n

9

1 2

(n 1)

(7)

= 0 (8)

Hence, s2 is a consistent estimator of var(Y ) = 2:

7

The above two conditions ensure that the distribution of s2 will become degenerate

as n!1, with all probability mass of s2 concentrated on 2:

b. UseM = 5000 independent samples of size n = 100 to estimate E(s2) and var(s2)

from an exponential distribution with = 5: (You will need to generate samples

of s2 to do this). Report your estimates, compare them with their theoretical

values, and also provide kernal estimates of the distribution of your simulated s2

values. Repeat this exercise with n = 1000, and then again with n = 10000:

Summary statistics from the three simulations are provided on the last page of

these answers, and the requested sample and theoretical results are tabulated

below:

n bs2 2 \V ar(s2) 4

n

h

9 (n3)

(n1)

i

100 24.8362 25.0 48.4487 50.1262

1000 24.9521 25.0 4.8407 5.0012

10000 24.9948 25.0 0.4866 0.5000

The kernal densities of simulated s2 values for the three sample sizes are:

8

(c) Outline how the results found in part (b) provide numerical demonstration of the

consistency of s2:

Comparison of the simulated s2 and 2 shows that there is very little apparent

bias, and further, as the sample size grows, this evidence of bias is disappearing.

This provides numerical evidence that E(s2) = 2.

Looking at the simulated \V ar(s2); we see that they are following their theoretical

values, and in particular, they are falling as the sample size grows. Thus it appears

that var(s2)! 0 as n!1:

Taking these two numerical observations together they provide evidence that the

two su¢ cent conditions for s2 to be consistent for 2 are satis
ed.

9

Further evidence that s2 is a consistent estimator for 2 is that as the sample size

grows the kernal densities become more concentrated around the theoretical value

for 2, which is 25 in this case.

10

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