Econometrics Assignment Sample Online

1. Consider a general time series model of the form

Yt = mt + st + et; (1)

where fmtg is the trend component of the form mt = _0+_1 (t?12)+_2 (t?12)2

with _i 6= 0 (i = 0; 1; 2) being unknown parameters, fstg is the seasonal

component satisfying st+12 = st and

P12

j=1 sj = 0, and fetg is a sequence of

stationary residuals with E[e1] = 0 and E[e21

] = 1.

(a) Is fYtg stationary ? Give your reasoning.

(b) Is the _rst{order di_erenced version of Yt, Zt = 512Yt = Yt ? Yt?12,

stationary? Give your detailed reasoning.

(c) Is the second{order di_erenced version of Yt, Wt = 52

12Yt = (1?B12)2Yt,

stationary? Give your detailed reasoning.

1

2. The stationary process fZtg is said to be white noise with mean 0 and variance

_2, written

Zt _ WN(0; _2);

if and only if fZtg has zero mean and covariance function given by

(h) =

8><

>:

_2 if h = 0

0 otherwise.

(a) Consider an ARMA(1,1) model of the form

Xt = _Xt?1 + Zt + _Zt?1;

where j_j < 1 and j_j < 1.

Find the auto{correlation coe_cient function (ACF) of fXtg, _(k), for

k = 1; 2; _ _ _.

(b) Consider an MA(1) model with drift of the form Xt = _ + Zt + _Zt?1.

Find the ACF. Does it depend on _ ?

(c) Consider a time series model of the form: (1 ? B)(1 ? 0:2B)Xt = (1 ?

0:5B)Zt. Classify the model as an ARIMA(p,d,q) model (i.e., give your

reasoning for the speci_cation of (p; d; q)).

3. Consider an auto{regressive model of order one (AR(1)) of the form

Xt = _Xt?1 + Zt; (2)

where fZtg is a sequence of white noises with E[Zt] = 0 and 0 < _2 = E[Z2

t ],

and j_j < 1 is an unknown parameter.

(a) Derive the autocorrelation function _(k) for all k _ 1.

(b) State the necessary and su_cient condition such that fXtg is stationary.

(c) Give some detailed description for each of the possible estimation meth-

ods you have learned.

(d) Write down the corresponding code functions from R for the possible

estimation methods to be implemented in R.

(e) Using at least one of the estimation methods, write down some detailed

formulae for the estimators of the unknown parameters _ and _2.

4. (a) Let fZtg be a sequence of random errors satisfying

E[ZtjFt?1] = 0: (3)

In addition, we allow for a heteroscedastic structure of the form

Z2

t = _0 + _1Z2

t?1 + ut; (4)

where futg is a sequence of white noises, and _0 > 0 and _1 _ 0.

The process fZtg satisfying (3){(4) is called an auto{regressive conditional

heteroscedastic model of order one, simply, ARCH(1).

_ Rewrite model (4) as an auto{regressive model of order one (AR(1)).

_ Give some detailed description for each of the possible estimation

methods you have learned.

_ Write down the corresponding code functions from R for the possible

estimation methods to be implemented in R.

_ Under the conditions: 0 < _1 < 1 and 3_2

1 < 1, _nd the second and

fourth moments1:

E[Z2

t ] and E[Z4

t ]:

(b) Let fZtg be a sequence of random errors satisfying

E[ZtjFt?1] = 0: (5)

In addition, we allow for a heteroscedastic structure of the form

The process fZtg satisfying (5){(6) is called a generalized auto{regressive

conditional heteroscedastic model of order (r1; r2), simply, GARCH(r1; r2).

Consider a GARCH(1,1) model of the form

Z2

t = _0 + (_1 + _1)Z2

t?1 + ut ? _1ut?1; (7)

where ut _ WN(0; _2).

_ Find the conditions such that Z2

t is stationary and 0 < E[Z2

t ] < 1.

1The derivation for the fourth moment is optional.

5. (a) Consider a seasonal ARIMA (SARIMA) model of the form

_2(B)_3(B12)Yt = _1(B)_2(B12)Zt; (8)

where B denotes the backward shift operator, _2, _3, _1 and _2 are

polynomials of order 2, 3, 1 and 2, respectively, fZtg _ WN(0; _2), and

Yt = 5252

12Xt = (I ?B)2(I ?B12)2Xt. This model is called a SARIMA

model of order (2; 2; 1) _ (3; 2; 2)12 for fXtg.

_ Does fYtg follow an ARIMA model of ARIMA(b1; b2; b3) ? If so, can

you specify the values of bi for i = 1; 2; 3 ?

_ Does Wt = 52

12Xt follow an ARIMA model of ARIMA(c1; c2; c3) ? If

so, can you specify the values of ci for i = 1; 2; 3 ?

_ Based on your own understanding and experience, write down the

main steps for you to identify and then estimate a seasonal ARIMA

model of the form Xt _ SARIMA(2; 2; 1) _ (3; 2; 2)12.

(b) The real data set USAccDeaths was _tted by a seasonal ARIMA model

with the following summary:

> USAccDeaths

> usa.arima1<-arima(USAccDeaths, order=c(0,1,1),

seasonal = list(order=c(0,1,1), period =12))

> usa.arima1

Call:

arima(x = USAccDeaths, order = c(0, 1, 1),

seasonal = list(order = c(0, 1, 1), period = 12))

Coefficients:

ma1 sma1

-0.4303 -0.5528

s.e. 0.1228 0.1784

sigma^2 estimated as 99347:

log likelihood = -425.44, aic = 856.88

> usa.fore<-predict(arima(USAccDeaths, order =

c(0,1,1), seasonal = list(order=c(0,1,1),

period =12)), n.ahead = 12)

$pred

Jan Feb Mar Apr May Jun

8336.061 7531.829 8314.644 8616.868 9488.912 9859.757

Jul Aug Sep Oct Nov Dec

10907.470 10086.508 9164.958 9384.259 8884.973 9376.573

$se

Jan Feb Mar Apr May Jun

315.4481 363.0054 405.0164 443.0618 478.0891 510.7197

Jul Aug Sep Oct Nov Dec

541.3871 570.4081 598.0224 624.4167 649.7397 674.1121

> ts.plot(window(USAccDeaths,1973-1978), usa.fore$pred,

usa.fore$pred + 2*usa.fore$se,

usa.fore$pred - 2*usa.fore$se)

Using the summarized information given above, answer the following

questions:

_ Which seasonal ARIMA model was used? Give your identication

of (p; d; q) _ (P;D;Q)s.

_ Write down an explicit expression for the fitted model.

1.

1. a) Mt = + (t-12) + 2 (t -12)²

St =0 and St+ 12 = 0

St+12 = St

= 0

While et is normally distributed with mean 0 and variance 1

Yt = Mt + St + et becomes Yt = + (t-12) + 2 (t -12)² + et

The equation is Not strictly stationary since the joint distribution of Yt1, Yt2 cannot be the same as the joint distribution of Yt1+h Yt2+h , …This cannot be transformed into an AR(2) equation and is therefore not 2^nd order stationary.

b)

The non-stationary second order equation can only be removed by getting the second order difference and therefore the first order difference does not make the equation stationary

1. c) the second order difference makes the time series stationary the trend is quadratic and is removed by getting the second order difference.

2.

1. a) Mean is given by ,EXt = EWt(1 ? Wt?1)Zt = (EWt)(1 ? EWt?1)(EZt) = 0.

Auto Cov(s, t) = E Ws(1 ? Ws?1)ZsWt(1 ? Wt?1)Zt = E Ws(1 ? Ws?1)Wt(1 ? Wt?1) · EZsZt.

EZsZt = EZs · EZt = 0.

When s = t then EZsZt = EZ 2 t = 1

cov(t, t) = EW2 t (1 ? Wt?1) 2 = 1 4 .

{Xt} is not independently and identically distributed

Xt?1 = 1 shows that Wt?1 = 1,

This means that Xt = 0.

b)

The autocovariance function does not depend on the mean

Finding the mean, we have that aoutocovariance = 1 + ² when ?=0

When ?= o otherwise.

C)

1. The auto covariance = Cov (zt, zt?k) = E(zt ? µ)(zt?k ? µ).

The auto correlation = Cov(zt, zt?k)/ ( [Var(zt) · Var(zt?k)]^1/2].

This is an ARIMA p model

3.

1. a) Mean is given by ,EXt = EWt(1 ? Wt?1)Zt = (EWt)(1 ? EWt?1)(EZt) = 0.

Auto Cov(s, t) = E Ws(1 ? Ws?1)ZsWt(1 ? Wt?1)Zt = E Ws(1 ? Ws?1)Wt(1 ? Wt?1) · EZsZt.

E(ZsZt)= E(Zs) · E(Zt) = 0.

When s = t then E(ZsZt) = EZ 2 t = 1

Cov(t, t) = EW2 t (1 ? Wt?1) 2 = 1 4 .

{Xt} is not independently and identically distributed

Xt?1 = 1 shows that Wt?1 = 1,

This means that Xt = 0.

1. b) The condition for stationary is that, the mean must be constant while the auto covariance function must not depend on t and should instead depend on the time difference

c)yule walker estimation and maximum likelihood estimation

1. d) An R code for estimation of AR(1)

x=scan("quakes.dat")

x=ts(x)

plot(x, type="b") install.packages("astsa")

library(astsa)

lag1.plot(x,1)

acf(x, xlim=c(1,19)) xlag1=lag(x,-1) y=cbind(x,xlag1) ar1fit=lm(y[,1]~y[,2])summary(ar1fit) plot(ar1fit$fit,ar1fit$residuals)

acf(ar1fit$residuals, xlim=c()

e)

r1 =

4.

1. a) Z2 = ?0 + ?1 Z.
2. b) The method of likelihood estimation involves the use of a combined function of the time series created by multiplying the functions of the individual functions.
3. c) R code for the estimation of AR(2)

mort=scan("cmort.dat")

plot(mort, type="o")

mort=ts(mort)

mortdiff=diff(mort,1)

plot(mortdiff,type="o")

acf(mortdiff,xlim=c(1,24))

mortdifflag1=lag(mortdiff,-1)

y=cbind(mortdiff,mortdifflag1)

mortdiffar1=lm(y[,1]~y[,2])

summary(mortdiffar1)

acf(mortdiffar1$residuals, xlim = c(1,24))