Questions and Answers on Regression Models with Lagged ...

[Pages:8]Questions and Answers on Regression Models with Lagged Dependent Variables and ARMA models

L. Magee

Winter, 2013

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1. Consider an AR(1) process: t = t-1 + ut, where E(ut) = 0, E(u2t ) = u2, and E(utus) = 0 for all t = s. Assume that t is stationary. Derive a formula for Cov( t, t-s), the covariance of t and t-s, that holds for s = 0, 1, 2, 3, . . ..

2. Let ut be white noise. That is,

E(ut) = 0 for all t E(u2t ) = 2 for all t E(utut-s) = 0 for all t and s where s = 0

For each of the following time series processes, determine the variance of yt as a function of u2 and of parameters appearing in the equations below. Also derive the first- and second-order autocovariances and autocorrelations. Assume that the time series processes are stationary.

(a) yt = yt-1 + ut (yt is an AR(1) process) (b) yt = + t, where t = t-1 + ut ( t is an AR(1) process) (c) yt = ut + ut-1 (yt is an MA(1) process) (d) yt = ut + 0.6ut-1 + 0.2ut-2 + 0.1ut-3 (yt is an MA(3) process)

3. Consider a stationary AR(2) process

yt = ? + 1yt-1 + 2yt-2 + ut

where 2 = 0. Are there values of 1 and 2 for which this process could be re-written in moving average form as an MA(2) process? If so, what are the values of 1 and 2? If no such values exist, briefly explain why not.

4. An autoregressive distributed lag model is estimated as:

yt = 31.2 + 0.61yt-1 + 0.19yt-2 + 1.40xt + 0.58xt-1 + ut Consider the effect on y of a one-unit increase in x at time t in the following two cases:

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(a) x remains one unit higher permanently after time t. (b) x immediately returns to its former level at time t + 1. Obtain the estimated effect on y in each of these cases at the four time periods: t, t + 1, t + 2, and the long run effect, t + . 5. Consider a regression model with a constant term and three explanatory variables, which include the lagged dependent variable yt-1 and two other variables, x1t and x2t. The estimated model is

yt = 21.0 + 0.6yt-1 + 1.5x1t + 0.75x2t + et

(a) Obtain the estimated effect on y of a permanent one-unit increase in x1 at time t (that is, x1 remains one unit higher permanently after time t) at the four time periods: t; t + 1; t + 2; and the long run effect, t + .

(b) Compare the size of the estimated effect on y of a permanent one-unit increase in x1 to the size of the estimated effect on y of a permanent one-unit increase in x2. Mention their initial (time t) effects and their long run effects. No algebra or calculations are required.

6. For each of the following time series processes (a) yt = ? + yt-1 + ut (b) yt = ? + ut + 0.6ut-1 + 0.2ut-2 derive (i) the unconditional mean, E(yt) (ii) the unconditional variance, Var(yt) (iii) the first-order autocovariance, Cov(yt, yt-1) = E(yt - E(yt))(yt-1 - E(yt-1)) Assume: E(ut) = 0 for all t; E(u2t ) = 2 for all t; E(utut-s) = 0 for all t and s where s = 0; and that the time series processes are stationary.

7. An autoregressive distributed lag model is estimated as

yt = 11 + 0.7yt-1 - 0.4yt-2 + 9xt + 2xt-1 + ut

Consider the effect on y of a one-unit increase in x at time t where x remains one unit higher permanently after time t. Obtain the estimated effect on y at time t, t + 1, t + 2, and the long run effect.

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8. Consider a regression model with a constant term and three explanatory variables, which include the lagged dependent variable yt-1 and two other variables, x1t and x2t. The estimated model is

yt = 2.1 + 0.8yt-1 - 2.0x1t + 0.5x2t + et

(a) Obtain the estimated effect on y of a permanent one-unit increase in x1 at time t (that is, x1 remains one unit higher permanently after time t) at the four time periods: t; t + 1; t + 2; and the long run effect, t + .

(b) Compare the size of the estimated effect on y of a permanent one-unit increase in x1 with the size of the estimated effect on y of a permanent one-unit increase in x2. Mention their initial (time t) effects and their long run effects. No algebra or calculations are required.

9. Suppose t follows a stationary AR(1) process:

t = t-1 + ut, t = 1, . . . , n

where ut is white noise. Let = 0.6 and Var(ut) = 5.

(a) What is the numerical value of the correlation between t and t-3 (b) What is the numerical value of Var( t) (c) Suppose that E(ut) = 10, instead of the usual zero-mean assumption. What is the numerical

value of E( t)?

10. Let ut be white noise, where

E(ut) = 0 for all t E(u2t ) = 20 for all t E(utut-s) = 0 for all t and s where s = 0

Let yt = ut + 0.7ut-1 + 0.1ut-2. Determine the numerical values of

(a) Var(yt) (b) The correlation between yt and yt-1 (c) The covariance between yt and yt-1

Answers

1.

(1 - L) t = ut t = (1 - L)-1ut

t = ut + ut-1 + 2ut-2 + . . .

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Since E( t) = 0 for all t, then

Cov( t, t-s) = E( t t-s) = E(ut + ut-1 + 2ut-2 + . . . + sut-s + . . .) ? (ut-s + ut-s-1 + 2ut-s-2 + . . .)

Because (Eutus) = 0 for all t = s, the only terms with non-zero expectations in this product are those with equal subscripts on the u's. Then the above expression simplifies to

Cov( t, t-s) = E(su2t-s + s+2u2t-s-1 + s+4u2t-s-2 + . . .)

= s(u2 + 2u2 + 4u2 + . . .)

= su2(1 + 2 + 4 + . . .)

=

(1

s - 2)

u2

,

for all s = 0, 1, 2, . . .

2. (a) Var(yt) = Var(yt-1 + ut) = 2Var(yt-1) + u2 = 2Var(yt) + u2

so Var(yt) = u2/(1 - 2)

Cov(yt, yt-1) = E(yt ? yt-1) (since Eyt = 0) = E(yt-1 + ut)yt-1 = E(yt2-1) = Var(yt)

For Cov(yt, yt-2), use: yt = yt-1 + ut = (yt-2 + ut-1) + ut = 2yt-2 + ut-1 + ut Then

Cov(yt, yt-2) = E(ytyt-2) = E(2yt-2 + ut-1 + ut)yt-2 = 2E(yt2-2) = 2Var(yt)

Substitutions then give:

Corr(yt, yt-1) = Cov(yt, yt-1) = Var(yt)Var(yt-1)

and similarly Corr(yt, yt-2) = 2

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(b) (yt - ) = t = t-1 + ut

= (yt-1 - ) + ut

This is like (a) except now E(yt) = instead of = 0 and we now have replacing part (a)'s . Then

Var(yt) = u2/(1 - 2) Cov(yt, yt-1) = E(yt - )(yt-1 - ) = Var(yt) Cov(yt, yt-2) = 2Var(yt) Corr(yt, yt-1) = Corr(yt, yt-2) = 2

(c) Var(yt) = Var(ut + ut-1) = Var(ut) + 2Var(ut-1) = u2 + 2u2 = (1 + 2)u2

Cov(yt, yt-1) = E(ut + ut-1)(ut-1 + ut-2) = E(u2t-1) = u2

Cov(yt, yt-2) = 0 (yt and yt-2 have no ut's in common and the ut's are not correlated)

Corr(yt, yt-1)

=

u2 (1 + 2)u2

=

1 + 2

Corr(yt, yt-2) = 0

(d) Var(yt) = u2 + (0.6)2u2 + (0.2)2u2 + (0.1)2u2

= (1 + 0.36 + 0.04 + 0.01)u2 = 1.41u2

Cov(yt, yt-1) = E(ut + 0.6ut-1 + 0.2ut-2 + 0.1ut-3)(ut-1 + 0.6ut-2 + 0.2ut-3 + 0.1ut-4) = 0.6u2 + 0.12u2 + 0.02u2 = 0.74u2

Cov(yt, yt-2) = E(ut + 0.6ut-1 + 0.2ut-2 + 0.1ut-3)(ut-2 + 0.6ut-3 + 0.2ut-4 + 0.1ut-5) = 0.2u2 + 0.06u2 = 0.26u2 0.74

Corr(yt, yt-1) = 1.41 = 0.52 0.26

Corr(yt, yt-2) = 1.41 = 0.18

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3. Write this process as

(1 - 1L - 2L2)yt = ? + ut

Invert the lag polynomial to get it in MA form

yt = (1 - 1L - 2L2)-1(? + ut) = ?/(1 - 1 - 2) + (1 - 1L - 2L2)-1ut

The inverse lag polynomial (1 - 1L - 2L2)-1 is an infinite series of the form 1 + 1L + 2L2 + 3L3 + . . ., which is an infinite-order MA, not an MA(2). One way to see this is to factor the original quadratic lag polynomial as (1 - 1L)(1 - 2L) for some 1 and 2 values. 1 and 2 both are non-zero since 2 = 0. The inverse of this factorized lag polynomial is the product of two infinite-term geometric series

(1 - 1L)-1(1 - 2L)-1 = (1 + 1L + 21L2 + 31L3 + . . .)(1 + 2L + 22L2 + 32L3 + . . .)

which itself is an infinite series.

4. (Note that represents the change in y due to a change in x. It does not represent the firstdifference operator here.)

(a)

yt = 1.40xt = 1.40(1) = 1.40

yt+1 = 0.61yt + 1.40xt+1 + 0.58xt = 0.61(1.40) + 1.40(1) + 0.58(1) = 2.834

yt+2 = 0.61yt+1 + 0.19yt + 1.40xt+2 + 0.58xt+1 = 0.61(2.834) + 0.19(1.40) + 1.40(1) + 0.58(1) = 3.975

The permanent effect can be obtained from y = 0.61y + 0.19y + 1.40x + 0.58x, where x is the permanent change in x. Then solve for y:

(1 - 0.61 - 0.19)y = 1.98x 1.98

y = x = 9.9x = 9.90 0.2

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(b)

yt = 1.40xt = 1.40

yt+1 = 0.61yt + 0.58xt (Now xt+1 = 0) = 0.61(1.40) + 0.58(1) = 1.434

yt+2 = 0.61yt+1 + 0.19yt = 0.61(1.434) + 0.19(1.40) = 1.141

The permanent effect is y = 0 since the permanent change in x is x = 0.

5. (a) Effect at time t : 1.5 at time t +1 : 1.5 + 0.6 ? 1.5 = 2.4 at time t +2 : 2.4 + (0.6)2 ? 1.5 = 2.94 at time t + : 1.5/(1 - .6) = 3.75

(b) At every time period, the effects of x2 on y are half as big as the effects of x1 on y. Reason: The coefficient on x2 is half the size of the coefficient on x1, and the dynamic pattern of the effects is the same for both, because that depends only on the coefficient on yt-1.

6. (a) (i) Eyt = ? + Eyt-1 + Eut = Eyt = ? + Eyt = (Eyt)(1 - ) = ? = Eyt = ?/(1 - )

(ii) Var(yt) = 2Var(yt-1) + Var(ut) = Var(yt) = 2Var(y) + 2 = Var(y) = 2/(1 - 2)

(iii) yt - Eyt = ? + yt-1 + ut - E(? + yt-1 + ut) = ? + yt-1 + ut - (? + Eyt-1) = (yt-1 - Eyt-1) + ut

So E(yt - Eyt)(yt-1 - E(yt-1)) = E((yt-1 - Eyt-1) + ut)(yt-1 - Eyt-1) = E(yt-1 - Eyt-1)2 = Var(yt)

(b) (i) Eyt = ? (ii) Var(y) = E(yt - ?)2 = (1 + .62 + .22)2 = 1.42

(iii) E(yt - Eyt)(yt-1 - E(yt-1)) = E(ut + .6ut-1 + .2ut-2)(ut-1 + .6ut-2 + .2ut-3) = .6Eu2t-1 + .12Eu2t-2 = .722

7. at t y = 9 ? 1 = 9

at t + 1, y = 0.7 ? 9 + 9 ? 1 + 2 ? 1 = 17.3

at t + 2, y = 0.7 ? 17.3 - 0.4 ? 9 + 9 ? 1 + 2 ? 1 = 19.51

long

run

effect

is

y

=

9+2 1-.7+.4

=

11 0.7

=

15.71

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8. (a) Effect at time t : -2.0 at time t +1 : -2.0 + 0.8 ? (-2.0) = -3.6 at time t +2 : -2.0 + 0.8 ? (-3.6) = -4.88 at time t + : -2.0/(1 - .8) = -10.0

(b) At every time period, the effect of a change in x2 on y is -0.25 times the effect of a change in x1 on y. This is because the coefficient on x2 is -0.25 times the coefficient on x1. This ratio does not change over time, because the way that the effect changes over time in this model depends only on the coefficient on yt-1, in the same way for both the x1 and x2 effects.

9. (a) When t follows a stationary AR(1) process with first-order autocorrelation coefficient , then Corr( t, t-s) = s. Therefore Corr( t, t-3) = 3 = (0.6)3 = 0.216

(b) Var( t) = 2Var( t-1) + Var(ut)

Var( t) = 0.36Var( t) + 5

Var(

t)

=

5 1-0.36

=

7.81

(c) E( t) = E( t-1) + E(ut)

E( t) = 0.6E( t) + 10

E(

t)

=

10 1-0.6

=

25

10. (a)

Var(yt) = Var(ut) + (.7)2Var(ut-1) + (.1)2Var(ut-2) = 20 + .49(20) + (.01)20

= 20(1 + .5) = 30

(b) Since E(yt) = 0, then

Cov(yt, yt-1) = E(ytyt-1) = E(ut + .7ut-1 + .1ut-2)(ut-1 + .7ut-2 + .1ut-3) = .7E(u2t-1) + (.7)(.1)E(u2t-2) = (.7 + .07)20 = 15.4

(c)

Corr(yt, yt-1) =

Cov(yt, yt-1)

Var(yt)Var(yt-1)

15.4

=

= .513

30 ? 30

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