STAT 420 Fall 2010

Homework #12 (due Friday, December 10, by 3:00 p.m.)

1. Consider the AR ( 2 ) processes

Y& t – 0.3 Y& t – 1 – 0.1 Y& t – 2 = e t

where { e t } is zero-mean white noise ( i.i.d. N ( 0, 2eσ ) ), Y& t = Y t – µ .

a) Based on a series of length N = 100, we observe …, y 98 = 152, y 99 = 156,

y 100 = 147, y = 150. Forecast y 101 and y 102 .

Y N + 1 = µ + φ 1 ( Y N – µ ) + φ 2 ( Y N – 1 – µ ) + e N + 1

1ˆ +Ny = E N ( Y N + 1 ) = µ + φ 1 ( y N – µ ) + φ 2 ( y N – 1 – µ )

Y N + 2 = µ + φ 1 ( Y N + 1 – µ ) + φ 2 ( Y N – µ ) + e N + 2

2ˆ +Ny = E N ( Y N + 2 ) = µ + φ 1 ( 1ˆ +Ny – µ ) + φ 2 ( y N – µ )

101 y = µ + 1 φ ( y 100 – µ ) + 2 φ ( y 99 – µ )

= 150 + 0.3 ( 147 – 150 ) + 0.1 ( 156 – 150 ) = 149.7.

102 y = µ + 1 φ ( 101

y – µ ) + 2 φ ( y 100 – µ )

= 150 + 0.3 ( 149.7 – 150 ) + 0.1 ( 147 – 150 ) = 149.61.

b) Use Yule-Walker equations to find ρ 1 and ρ 2 .

The Yule-Walker equations for an AR(2) process:

ρ 1 = φ 1 + φ 2 ρ 1

ρ 2 = φ 1 ρ 1 + φ 2

ρ 1 = φ 1 + φ 2 ρ 1 ⇒ ρ 1 = 2

1

1 φφ−

= 1.01

3.0

− =

31

.

⇒ ρ 2 = φ 1 ρ 1 + φ 2 = 0.3 ⋅ 1/3 + 0.1 = 0.20.

c) Is this process stationary?

Φ ( B ) = 1 – 0.3 B – 0.1 B 2 = ( 1 – 0.5 B ) ( 1 + 0.2 B )

The roots of Φ ( z ) = 0 are z 1 = 2 and z 2 = – 5.

Both are outside the unit circle. That is, | z 1 | > 1, | z 2 | > 1. ⇒ The process is stationary.

OR

An AR(2) model is stationary if

– 1 < φ 2 < 1, φ 2 + φ 1 < 1, φ 2 – φ 1 < 1.

– 1 < 0.1 < 1, 0.1 + 0.3 < 1, 0.1 – 0.3 < 1. ⇒ This process is stationary.

2. Consider the AR ( 2 ) process

Y t = µ + φ 1 ( Y t – 1 – µ ) + φ 2 ( Y t – 2 – µ ) + e t

Based on a series of length N = 60, we observe …, y 59 = 190, y 60 = 215, y = 200.

a) Suppose r 1 = 0.40, r 2 = – 0.26. Use Yule-Walker equations to estimate φ 1 and φ 2.

The Yule-Walker equations for an AR ( 2 ) process are given by:

ρ 1 = φ 1 + φ 2 ρ 1

ρ 2 = φ 1 ρ 1 + φ 2

0.40 = φ 1 + 0.40 φ 2 × 5 2 = 5 φ 1 + 2 φ 2

– 0.26 = 0.40 φ 1 + φ 2 × 2 – 0.52 = 0.80 φ 1 + 2 φ 2

⇒ 2.52 = 4.2 φ 1 ⇒ 1 φ = 2.52/4.2 = 0.60

⇒ 0.40 = 0.60 + 0.40 φ 2 – 0.20 = 0.40 φ 2

⇒ 2 φ = – 0.50.

b) If φ 1 and φ 2 are equal to your answers to part (a), is this process stationary?

t Y& – 0.60 1 Y −t& + 0.50 2

Y −t& = e t

Φ ( z ) = 1 – 0.60 z + 0.50 z 2 = 0.

Roots 64.1 60.064.1 60.050.02

50.01460.0 60.0

2

2,1z i±=−±=−±

=⋅

⋅⋅

are outside of the unit circle: ( )2

222,1

64.160.0 z += = 2 > 1.

⇒ This process is stationary.

OR

An AR(2) model is stationary if

– 1 < φ 2 < 1, φ 2 + φ 1 < 1, φ 2 – φ 1 < 1.

– 1 < – 0.50 < 1, – 0.50 + 0.60 < 1, – 0.50 – 0.60 < 1.

⇒ This process is stationary.

c) Use your answers to part (a) to forecast y 61 , y 62 , and y 63 .

y 59 = 190, y 60 = 215, y = 200.

61 y = µ + 1 φ ( y 60 – µ ) + 2 φ ( y 59 – µ )

= 200 + 0.60 ( 215 – 200 ) – 0.50 ( 190 – 200 ) = 214.

62 y = µ + 1 φ ( 61

y – µ ) + 2 φ ( y 60 – µ )

= 200 + 0.60 ( 214 – 200 ) – 0.50 ( 215 – 200 ) = 200.9.

63 y = µ + 1 φ ( 62

y – µ ) + 2 φ ( 61 y – µ )

= 200 + 0.60 ( 200.9 – 200 ) – 0.50 ( 214 – 200 ) = 193.54.

3. 10.24 (a) Consider the ARMA ( 1, 1 ) model

( Y t – 60 ) + 0.3 ( Y t – 1 – 60 ) = e t – 0.4 e t – 1 which was fitted to a time series where the last 10 values are

60, 57, 52, 59, 62, 59, 63, 67, 61, 58

and the last residual is e N = – 2.

Calculate the forecasts of the next two observations, and indicate how forecasts can be

calculated for lead times greater than two. Show what happens to the forecasts as the

lead time becomes arbitrarily large.

( Y N + 1 – 60 ) = – 0.3 ( Y N – 60 ) + e N + 1 – 0.4 e N

1ˆ +Ny = E N ( Y N + 1 ) = 60 – 0.3 ( y N – 60 ) – 0.4 e N

= 60 – 0.3 ( 58 – 60 ) – 0.4 ( – 2 ) = 61.4.

( Y N + 2 – 60 ) = – 0.3 ( Y N + 1 – 60 ) + e N + 2 – 0.4 e N + 1

2ˆ +Ny = E N ( Y N + 2 ) = 60 – 0.3 ( 1ˆ +Ny – 60 )

= 60 – 0.3 ( 61.4 – 60 ) = 59.58.

( Y N + l – 60 ) = – 0.3 ( Y N + l – 1 – 60 ) + e N + l – 0.4 e N + l – 1

l > 2 lNy +ˆ – 60 = – 0.3 ( 1ˆ

−+lNy – 60 )

⇒ lNy +ˆ – 60 = ( – 0.3 ) l – 1 ( 1ˆ +Ny – 60 )

⇒ lNy +ˆ = 60 + 1.4 × ( – 0.3 ) l – 1

⇒ lNy +ˆ → 60 as l → ∞

For fun:

b)* Given 2 ˆeσ = 4, calculate 90-percent probability limits for the next two observations.

Interpret these limits.

Y t – 60 = ( ) 4.013.01

1B

B−

+⋅ e t

= ( ) ( )

4.01...00243.00081.0027.009.03.01 5432 BBBBBB −+−+−+− ⋅ e t

= ( 1 – 0.70 B + 0.21 B 2 – 0.063 B

3 + … ) e t

= e t – 0.70 e t – 1 + 0.21 e t – 2 – 0.063 e t – 3 + …

Var ( Y N + 1 – 1ˆ +Ny ) = 2 eσ . 2 ˆeσ = 4.

61.4 ± 1.645 × 4 61.4 ±±±± 3.29

Var ( Y N + 2 – 2ˆ +Ny ) = ( 1 + ( – 0.70 ) 2

) 2 eσ .

( 1 + ( – 0.70 ) 2

) 2 ˆeσ = 5.96.

59.58 ± 1.645 × 96.5 59.58 ±±±± 4.016

4. https://netfiles.uiuc.edu/stepanov/www/ur.dat contains the U.S.

unemployment rate series. These are seasonally adjusted quarterly rates from 1948-1978.

ur = scan(" ... /ur.dat")

ur.ts = ts(ur, start=1948, frequency=4) #make a time series object

par(mfrow=c(3,1)) #3 plots per page

plot(ur.ts) #plots time series

title("U.S. Unemployment Rate")

a) Model Identification:

acf(ur.ts)

pacf(ur.ts)

Based on the sample ACF and PACF, what is an appropriate model?

b) Estimation:

Fit an AR ( 2 ) model to the data using MLE.

AR ( 2 ): ( Y t – µ ) – φ 1 ( Y t – 1 – µ ) – φ 2 ( Y t – 2 – µ ) = e t

E ( e t ) = 0, Var ( e t ) = 2 σe for all t

E ( e t e t' ) = 0, for t ≠ t'

E ( e t Y t' ) = 0, for t' < t

fit = arima(ur.ts,order=c(2,0,0)) #fits an AR(2) with mean included

fit #prints results

c) Diagnostic Checking:

tsdiag(fit) #diagnostic plots

Based on the first two diagnostics plots, is AR(2) an appropriate model?

d) Forecasting:

Now, we will forecast the next 2 quarters ahead using predict function.

predfit2 = predict(fit,n.ahead=2) #forecasts 2 steps ahead

predfit2 #prints results

> ur <- scan("http://www.stat.uiuc.edu/~stepanov/ur.dat")

Read 121 items

> ur.ts <- ts(ur, start=1948, frequency=4)

> par(mfrow=c(3,1))

> plot(ur.ts)

> title("U.S. Unemployment Rate")

> acf(ur.ts)

> pacf(ur.ts)

ACF smoothly “dies out”.

PACF “cuts off” after lag 2.

This is consistent with an AR(2) model.

> fit <- arima(ur.ts,order=c(2,0,0))

> fit Call:

arima(x = ur.ts, order = c(2, 0, 0)) Coefficients:

ar1 ar2 intercept

1.5499 -0.6472 5.0815

s.e. 0.0681 0.0686 0.3269 sigma^2 estimated as 0.1276: log likelihood = -48.76, aic = 105.53

> tsdiag(fit)

Standardized Residuals plot does not appear to have noticible patterns. Standardized

Residuals seem “random” and bell-shared (a lot of them are close to zero, with just a few

away from zero). Most of the Standardized Residuals are between – 2 and 2, only one is

outside ( – 3, 3 ).

The ACF of the Residuals looks like the ACF of white noise – the only significant

autocorrelation coefficient is ρ 0 = 1.

This suggests that AR(2) is indeed an appropriate model.

> predfit2 <- predict(fit,n.ahead=2)

> predfit2

$pred

Qtr2 Qtr3

1978 5.812919 5.491268

$se

Qtr2 Qtr3

1978 0.3572328 0.6589087