HW 3 answers!!.pdf

Economics 120C Name: _________________________ Professor Yongil Jeon Winter 2009 Student ID#: _________________________

Answer Key to Homework #3 – Winter 2009, Econ 120C

(A part of Final Exam, Summer 2008-Session 2, Econ 120C) Answer all questions on separate paper. This problem set should be handed in to Professor Jeon at the beginning of your class on Wednesday, March 11th, 2009. Problem sets may not be handed in once solutions have been distributed. Please write down your name and PID clearly. Good luck!

Multiple Choices 1. (3 points) Departures from stationarity

a. jeopardize forecasts and inference based on time series regression. b. occur often in cross-sectional data. c. can be made to have less severe consequences by using log-log

specifications. d. cannot be fixed

Answer: a

2. (3 points). An autoregression is a regression

a. of a dependent variable on lags of regressors. b. that allows for the errors to be correlated. c. model that relates a time series variable to its past values. d. to predict sales in a certain industry.

Answer: c

3. (3 points). The random walk model is an example of a

a. deterministic trend model. b. binomial model. c. stochastic trend model. d. stationary model.

Answer: c

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2 Answer to HW#3, ECO 120C, Winter 2009

4. (10 points) Consider the model 10.2 0.9t t tY Y u−= + + , where )1,0(~ iidut . Find the mean and variance of tY . Answer: Rewrite the AR(1) model as follows

1 2 1

22 1

0.2 0.9 0.2 0.9(0.2 0.9 )

0.2(1 0.9) 0.9 0.9t t t t t t

t t t

Y Y u Y u u

Y u u− − −

− −

= + + = + + + +

= + + + +

Continuing the substitution indefinitely then results in 2 3

00.2(1 0.9 0.9 0.9 ) 0.9 .i

t t ii

Y u∞

−=

= + + + + +∑L

Given the result for the sum of a geometric series, the final expression is

0 0

0.2 0.9 2 0.9 .1 0.9

i it t i t i

i iY u u

∞ ∞

− −= =

= + = +− ∑ ∑

To find the mean and the variance, take first expectations on both sides

( ) ( )0 0

2 0.9 2 0.9 2i it t i t i

i iE Y E u E u

∞ ∞

− −= =

⎛ ⎞= + = + =⎜ ⎟⎝ ⎠∑ ∑ , since ( ) 0tE u = for all t.

To derive the variance, note that ( )0

0.9it t t i

iY E Y u

∞

−=

− =∑ . Hence the variance is

( ) ( ) ( ) ( )22 22 2 2

20 0

10.9 0.9 5.26321 0.9 0.19

i i ut t t i u

i i

E Y E Y E u σσ∞ ∞

−= =

− = = = = =⎡ ⎤⎣ ⎦ −∑ ∑ .

5. (10 points) The moving average model of order q has the form

0 1 1 2 2t t t t q t qY e b e b e b eβ − − −= + + + + +L

where te is a serially uncorrelated random variable with mean 0 and variance 2eσ .

a. (5 points) Show that ( ) 0tE Y β= . b. (5 points) Show the expressions in terms of Y only when applying the difference

operator to the following expressions, giving the variance of tY

( ) ( )2 2 2 21 2var 1t e qY b b bσ= + + + +L .

Answer


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(a) E(Yt) = β0 + E(et) + b1E(et–1) + ⋅⋅⋅ + bqE(et–q) = β0 [because E(et) = 0 for all values of t]. (b)

− − − − − + −= + + + + + +

= + + +

L L

L

2 21 1 1 1 1 1

2 2 21

var( ) var( ) var( ) var( ) 2 cov( , ) 2 cov( , )

(1 )t t t q t q t t q q t q t q

e q

Y e b e b e b e e b b e e

b bσ

because var(et) = 2eσ for all t and cov(et, ei) = 0 for i≠ t.

6 (10 points). A distributed lag model is used with only current and past values of Xt–1 coupled with an AR(1) error model to derive a quasi-difference model, where the error term was uncorrelated. Instead use a static model 0 1t t tY X uβ β= + + here, where the error term follows an AR(1). Since 1φ (the autocorrelation parameter for ut) is unknown, describe the Cochrane-Orcutt estimation procedure.

Answer: In this case, nonlinear least squares has to be used to estimate the three

parameters. One possible feasible GLS estimator in this case is the Cochrane-Orcutt estimator. In the first step, 1φ is set to zero, in which case 0β and 1β can be estimated by OLS. The resulting residuals are then used to calculate the OLS estimator for 1φ . This, in return, can then generate the quasi-differenced variables and OLS is then employed to get the estimate of 0β and 1β .

7 (10 points). Define the difference operator (1 )LΔ = − , where L is the lag operator,

such that jt t jL Y Y −= . In general, (1 )i j i

j LΔ = − , where i and j are typically omitted when they take the value of 1. Show the expressions in Y only when applying the difference operator to the following expressions, and give the resulting expression an economic interpretation, assuming that you are working with quarterly data:

(a) (5 points) 1 4 tYΔ Δ

Answer: 4 4 5

1 4 1 4 5

4 1 5

(1 )(1 ) (1 )( ) ( )

t t t t t t t

t t t t

Y L L Y L L L Y Y Y Y YY Y Y Y

− − −

− − −

Δ Δ = − − = − − + = − − += − − −

This is the quarterly change in the annual change. If Y is in logarithms, then this is the quarterly acceleration or change in the annual growth rate.

(b) (5 points) 2

4 tYΔ


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Answer: 2 4 2 4 84 4 8

4 4 8 4 4 4

(1 ) (1 2 ) 2( ) ( )

t t t t t t

t t t t t t

Y L Y L L Y Y Y YY Y Y Y Y YΔ Δ

− −

− − − −

Δ = − = − + = − += − − − = −

This represents the change in the annual change. If Y is in logarithms, then this is the change in the annual growth rate.


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HW 3 answers!!.pdf